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Define thermodynamics.
The branch of science which deals with the study of different forms of energy and the quantitative relationships between them is known as thermodynamics.
What do you understand by the terms: the system and surroundings?
A system in thermodynamics refers to that part of the universe in which observations are made and remaining universe constitutes the surrounding. The surroundings include everything other than the system. System and the surroundings together constitute the universe.
Classify the following into different types of systems:
(i) Tea placed in a cup
(ii) Tea placed in a tea-pot
(iii) Tea placed in a thermosflask.
(i) Open system.
(ii) Closed system.
(iii) Isolated system.
Define a closed system.
Define an isolated system.
A sytem which can neither exchange energy nor matter with its surroundings is an isolated system
Define intensive properties.
Define extensive properties.
Why all living systems need to be 'open systems'?
Living systems transact both energy and mass with the surroundings for their survival. Thus, a system needs to open.
Define state variables of a system.
Variables like pressure (P), volume (V) and temperature (T) are called state variables of a system because their values depend only on the state of the system and not on how the state has been reached.
From thermodynamic point of view, to which system the animals and plants belong?
Open system.
What is a state function?
A state function is one which depends only on the initial and final state of the system and is independent of the path followed.
B.
whose value is independent of pathWhat is reversible process in thermodynamics?
Define irreversible change.
Give one difference between an isothermal and an adiabatic process.
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Define internal energy of a system
Define enthalpy.
Why is enthalpy considered more useful than internal energy in chemical reactions?
In which ways the internal energy of a system can be changed?
By heating or cooling the system and keeping volume constant, the internal energy of the system increases or decreases respectively.
What happens to the internal energy of a system if:
(i) Work is done on the system
(ii) Work is done by the sytem?
Why internal energy is a state function but is not work?
What is the relationship between internal energy change and enthalpy change?
No, because the value of internal energy depends on.
The chemical nature of substance, amount of the substance and the condition of temperature pressure.
What is heat(q)?
The change in internal energy of the system by transfer of heat from the surroundings to the system or vice-versa without the expenditure of work. This exchange of energy, which is a result of temperature difference is called heat, q.
For the same increase in volume, why work done is more if the gas is allowed to expand reversibly at higher temperature?
Define the law of conservation of energy.
What is the main limitation of the first law of thermodynamics?
Write mathematical statement of first law of thermodynamics.
In a process, 100J of heat energy is supplied and 50J of work is done on the system. What is the change in internal energy of the system ?
What is the difference between heat and work?
Define extensive property ?
An extensive property is a property whose value depends on the quantity or size of matter present in the system. For example, heat capacity etc. is extensive properties.
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Why heat changes reported are usually enthalpy changes and not internal energy?
Give a chemical reaction in which ∆ H and ∆E are equal.
Under what conditions is ∆H = ∆U in a chemical reaction?
What is given of w if:
(i) work is done by the system.
(ii) work is done on the system.
(i) w is negative if work is done by the system.
(ii) w is positive if work is done on the system.
What is the mode of transference of energy, when petrol is subjected to combustion in an internal combustion engine ?
It is a partly as work and partly as heat.
Define heat capacity of a system.
It is the amount of heat required to raise the temperature by one degree celcius or kelvin. It is expressed in units of thermal energy per degree temperature.
Define specific heat of a substance.
It is the amount of heat required to raise the temperature of 1 gram of a substance through l°C.
Define molar heat capacity of a substance.
Differentiate between molar heat capacity and molar solution.
Molar heat capacity: Molar heat capacity means the amount of heat absorbed by one of the substance to raise its temperature through 1°.
Molar solution. The molar solution means a solution containing one mole of the solute per litre of the solution.
What is the realtionship between Cp and Cv?
What are exothermic and endothermic reactions?
Exothermic reactions are those reactions which are accompanied by the evolution of heat.
Endothermic reactions are those reactions which are accompanied by the absorption of heat.
What is the standard state of the substance?
Exothermic reactions are accompanied by the evolution of heat. It means that enthalpy of the reactants is more than the enthalpy of the products i.e.
Hr > Hp
But 
State with reason whether the work is done by the system on the surroundings or on the system by the surroundings if the following change occurs at constant pressure:
Work is done on the system by surroundings since there is 2 mol of gaseous reactants while there is no gaseous product. The result is decreased in volume i.e. w is positive.
What is the enthalpy of formation of the most stable form of an element in its standard state?
The enthalpy of formation of the most stable form of an element is Zero.
is -285. 9 kJ
Comment on the thermodynamic stability of NO(g), given
Out of the following different forms of oxygen, which will have the standard enthalpy of the formation to be 0.0 kJ; O, O2(g), O3(g), O2(l)?
Out of graphite and diamond, which one refers to the standard rate?
Graphite is referred to the standard rate.
Why the heat of neutralisation is less than 57.1 kJ if either the acid or the base or both are weak?
If acid or base is weak its ionisation is not complete in aqueous solutions. Therefore,a part of the energy liberated during the combination of H+ ions and OH- ions is utilised for the ionisation of weak acid or base. Hence, the heat of neutralisation is less than 57.1KJ.
Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example temperature, density, pressure etc.
What is the importance of ∆fH°?
Will the heat evolved be same in the following two cases?
If not, in which case it will be greater and why?
Which of the following is/are exothermic and which is/are endothermic/
(i) Endothermic (ionisation enthalpy is required).
(ii) Exothermic (first electron gain enthalpy is energy released).
(iii) Endothermic (higher electron gain enthalpies are energy required)
What is the significance of enthalpy of fusion?
What is the significance of enthalpy of combustion?
Under what conditions, the heat absorbed by the system is equal to the work done by the system?
State Hess's law of constant heat summation.
What is calorific value?
It is defined as the quantity of heat produced by the combustion of one gram of fuel or food. It is expressed in joules per kilogram.
The heat of combustion of glucose (C6H12,O6) is 2840 kJ mol–1. What is its calorific value?
Molar mass of glucose (C6H12,O6) = 180
Define bond enthalpy.
Is the bond enthalpy of all the four C – H bonds in CH4 molecule equal ? If not, then why ? How is the C – H bond energy then reported ?
What are renewable sources of energy?
Renewable energy is generally defined as energy that is collected from resources which are naturally replenished on a human timescale, such as sunlight, the wind, rain, tides, waves, and geothermal heat.
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When does bond enthalpy become equal to bond dissociation enthalpy?
Bond enthalpy is equal to the bond dissociation energy for diatomic molecules e.g.
Explain spontaneous process.
Explain non-spontaneous process.
Define entropy.
Define entropy.
Entropy is a measure of randomness or disorder of the system. It can also be regarded as the measure of ‘unavailable energy’.
Entropy change, 
Why does entropy of a solid increase on fusion?
On fusion, a more ordered solid form of a substance is changed to less ordered (or more disordered) liquid form. So entropy increases.
Why entropy is a state function?
An endothermic reaction 
proceeds to completion. What is the sign of 
?
Under what condition ∆Ssystem is equal and opposite in sign to ∆Ssurroundings?
What is the effect of temperature on spontaneity?
What is the relation between enthalpy change and entropy change for a process at equilibrium?
A reaction 
is found to have a positive entropy change. The reaction will be
possible at high temperature
D.
possible at any temperature
Why should entropy increase on expansion of a gas?
Define standard enthalpy of vaporization ?
It is defined as the amount of heat required to vaporize one mole of a liquid at constant temperature and under standard enthalpy of vaporization or molar enthalpy of vaporization,
Although dissolution of ammonium chloride in water is endothermic yet it dissolves. Why?
What is the standard state of the substance?
What are exothermic and endothermic compounds?
What is the physical significance of entropy?
What is the sign of ∆S for a spontaneous process in an isolated system?
Give a statement which includes the main ideas of the first and second laws of thermodyanimcs.
Define standard entropy (S0).
State the effect of increased temperature on the entropy of the substance.
Entropy of diamond is less than entropy of graphite. What conclusion do you draw from this?
State the second law of thermodynamics.
What is Gibb's free energy?
Gibb's free energy (G) is a state function related to enthalpy, entropy and temperature as
G = H - TS
What is Gibb's free energy change?
Write an expression that connects free energy, entropy and enthalpy.
The relationship of free energy, entropy and enthalpy,
G = H - TS and 
What is the driving force of a reaction?
The overall tendency for a reaction to occur is known as driving force of the reaction. It is the resultant of enthalpy change and entropy change.
All the naturally occurring processes proceed spontaneously in a direction which leads to
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State the thermodynamic conditions for spontaneous process to occur.
(i) For spontaneous process to occur ∆H < 0 (–ve)
(ii) ∆S of the process must be > 0 i.e. ∆S is +ve.
(iii) ∆G of the process must be < 0 i.e. ∆G is –ve.
How can a chemical reaction with positive enthalpy and entropy changes be made entropy drives?
For a reaction both ∆H and T∆S are positive. Under what conditions will the reaction be spontaneous?
What is the criteria for spontaneity in terms of free energy change?
Criteria for spontaneity in terms of free energy change:
(i) If ∆G is negative, the process is spontaneous.
(ii) If ∆G is positive, the direct process is non-spontaneous.
(iii) If ∆G is zero, the process is in equilibrium.
What is the significance of free energy change?
Define standard free energy change.
It is the free energy change for a process in which reactants in their standard state are converted into the products in their standard states.
At what temperature entropy of a perfectly crystalline substance is zero?
Which thermodynamic quantity is required for the calculation of absolute entropies ?
Heat capacity at constant pressure (Cp).
For which molecule bond energy equal to bond dissociation energy?
Bond energy becomes equal to bond dissociation energy for the diatomic molecules. For example, H2, O2. Cl2, N2 etc.
What will be the sign of ∆G for the melting of ice at 267K and at 276K?
Melting point temperature of ice = 273 K.
Thus, at 267 K, the process is non-spontaneous (∆G = +ve) while at 276K, it becomes spontaneous (∆G = –ve).
How can a non-spontaneous reaction be made spontaneous?
What will be the value of free energy change when the reaction is in equilibrium?
What are the types of systems?
Types of systems. The exchange of energy between the system and its surroundings usually takes place in the form of heat or work or both. Based upon this, the systems have been classified into three types:
(i) Open system (ii) Closed system (iii) Isolated system.
(i) Open system: A system is said to be an open system if it exchanges the matter (mass) as well as the energy with its surroundings. All chemical reactions carried in open containers constitute the open system. For example,
(a) Combustion of carbon in an open tube.
(b) Tea placed in a cup.
(ii) Closed system: A system is said to be a closed system if the only exchange of energy is possible between system and surroundings and exchange of matter (mass) is not possible. All chemical reactions carried in closed containers constitute closed system e.g.
(a) Decomposition of calcium carbonate in a closed tube.
(b) Tea placed in a tea-pot.
(iii) Isolated system: A system which can neither exchange energy nor matter (mass) with its surroundings is called an isolated system. All chemical reactions carried in a closed container insulated from all sides represent the isolated system. For example.
(a) Neutralisation reaction between NaOH and HCl carried in a thermos flask.
(b) Tea placed in a thermos flask.
(i) State variables : The measurable properties required to describe the state of the system are called state variables. For example temperature, pressure, volume, composition etc. are state variables.
(ii) State functions: A state function is a property of the system whose value depends on only upon the state of the system and is independent of the path or manner by which the state is reached. For example, pressure, volume, temperature, internal energy (E), enthalpy (H), entropy (S) etc. are state functions.
What are homogeneous and heterogeneous systems?
Homogeneous system: A system is said to be homogeneous if all the constituents present are in the same phase and the composition of the system is uniform throughout.
Heterogeneous system: A system is said to be heterogeneous if it consists of two or more phases and its composition is also not uniform.
What do you understand by:
(i) Isothermal process
(ii) Adiabatic process
(iii) Isobaric process
(iv) Isochoric process
(v) Cyclic process?
(i) Isothermal process: A process is said to be isothermal if the temperature of the system remains constant i.e. operation is carried at a constant temperature. For an isothermal process, dT = 0 where dT is the change in temperature.
(ii) Adiabatic process: In an adiabatic process, no heat can flow from the system to the surrounding or vice versa i.e. the system is completely insulated from the surroundings.
(iii) Isobaric process: It is the process during which the pressure of the system is kept constant i.e. dP = 0. The volume may or may not change.
(iv) Isochoric process: It is a process during which the volume of the system is kept constant i.e dV = 0.
(v) Cyclic process. When a system undergoes a series of changes and finally returns to its initial state, it is an example of the cyclic process. In cyclic process dU = 0.
Give the points of difference between reversible process and irreversible process.
|
Reversible process |
Irreversible process |
|
1. Such processes take place infinitesimally slowly and their direction at any point can be reversed. |
1. The process cannot be reversed. |
|
2. It is a slow process. |
2. It is a fast process. |
|
3. This process is in equilibrium at all stages of operation. |
3. There is no equilibrium. |
|
4. This is an ideal and imaginary process. |
4. It is a real process. |
Classify the following processes as reversible or irreversible:
(i) mixing of two gases by diffusion.
(ii) dissolution of sodium chloride in water
(iii) evaporation of water at 373 K at 1 atm pressure
(iv) expansion of a gas in vacuum.
(i) Irreversible
(ii) Irreversible.
(ii) Reversible
(iv) Irreversible.
How will you distinguish between intensive and extensive properties?
(i) Intensive properties: These are the properties which depend on only upon the nature of the substance and are independent of the amount of the substance present is the system. The common examples of these properties are temperature, pressure, refractive index, viscosity, density, surface tension, specific heat, freezing point, boiling point etc.
(ii) Extensive properties: These are the properties which depend on upon the quantity of the matter contained in the system. The common examples of these properties are mass, volume, heat capacity, internal energy, enthalpy, entropy, Gibb’s free energy etc.
What do you mean by internal energy and internal energy change?
the extra energy possessed by the system in the initial state (or the reactants) would be given out and 
will be negative. 
energy will be absorbed in the process and 
will be positive.
Derive an expression for work done in isothermal reversible expansion of ideal gas.
Total work done when the gas expands from initial volume V1 to final volume V2, will be
Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversibly at 27°C from a volume of 15L to 25L.
We have given,
Volume (V2) = 25L
Volume (V1) =15L
Number of moles =0.75
using the equation,
Substituting the values, we have,
State first law of thermodynamics in different ways.
(i) Energy can neither be created nor destroyed, though it may be converted from one form into another.
(ii) The total energy of an isolated system remains constant, although there may be internal changes in the state of the system.
(iii) It is impossible to construct a perpetual motion machine i.e. a machine that can produce work without any expenditure of energy.
(iv) Total energy of a system and its surrounding such as universe remains constant in any physical or chemical process.
What informations are provided by the First law of Thermodynamics?
From the first law, we learn that:
(i) Different forms of energy are interconvertible i.e. one form of energy can be converted into another.
(ii) Whenever one form of energy disappears, an equivalent amount of energy in another form appears.
(iii) Total energy of the universe i.e. energy of the system and surroundings taken together remains constant.
What are the limitations of First law of Thermodynamics?
(i) The first law of thermodynamics does not tell us anything about the extent and direction of the convertibility of one form of energy into another. For example when two bodies which are capable of exchanging heat energy are brought into contact. First law tells only that heat gained by one must be equal to that lost by the other, but it does not tell us which of the two would lose or gain heat energy. It also does not tell us how much heat energy would be transferred from one to the other.
(ii) It provides no information regarding the feasibility of the process.
Derive mathematical form of First law of Thermodynamics.
Or
Derive the relationship between heat, internal energy and work.
Mathematical form:
The internal energy of the system can be changed in two ways:
(i) by allowing heat to flow into the system or out of the system,
(ii) by work is done on the system or work done by the system.
Let us consider a system which undergoes a change from one state to another. Let the initial internal energy of the system be U1. Now if a system is supplied q amount of heat, then internal energy of the system increases and becomes U1 + q.
If work (w) is done on the system, then its internal energy further increases and becomes U2. The energy U2 is the energy in the final state.
The above expression is the mathematical form of the first law of thermodynamics. w includes all kinds of work such as pressure-volume work, electrical work etc.
If the work is done in the above process is only pressure-volume work,
Compute internal energy change of a system if it
(i) absorbs 500 kJ heat and does 300 kJ work.
(ii) loses 200 kJ heat and has 450 kJ of work done on it.
(i) Here, q = 500 kJ, w = -300 kJ
According to the first law of thermodynamics,
(ii) Here, q = -200 kJ, w = 450 kJ
Here,
q = +701 J, w = -394 J
According to first law of thermodynamics,
i.e. internal energy of the system increases by 307J
A sample of gas is compressed by an average pressure of 0.50 bar so as to decrease its volume from 450 cm3 to 250 cm3. During the process 6.00 J of heat flows out to surroundings. Calculate the change in internal energy of the system.
State whether each of the following will increase or decrease the total energy content of the system:
(i) heat transferred to the surroundings
(ii) work done on the system
(iii) work done by the system.
(i) When heat is transferred to surroundings the total energy content of the system decreases.
(ii) When work is done on the system, the energy content of the system increases.
(iii) When work is done by the system, the energy content of the system decreases.
(i) ∆U = wad, wall is adiabatic.
(ii) ∆U = –q, thermally conducting walls.
(iii) ∆U = q – w, closed system.
Enthalpy: Enthalpy is the total energy associated with the system and is defined as the sum of internal energy and product of its pressure and volume. It is denoted by H.
Mathematically,
H = U + PV
where U is the internal energy change, P and V are respectively the pressure and volume of the system. H is also called heat content of the system. Enthalpy is a state function and every substance has a definite value of enthalpy in a particular state. The absolute value of enthalpy of a substance cannot be determined, but the change in enthalpy (∆ H) accompanying a process can be determined.
Change in enthalpy (∆H): It may be defined as the difference in the enthalpies of the product and reactant species taking part in a chemical reaction at a constant pressure. For example, if HR represents the enthalpies of the reactants and HP represents the enthalpies of the products, then ∆H = Hp – Hr where ∆H gives the change in enthalpy.
The enthalpy change (∆H) is equal to the heat evolved or absorbed in a chemical reaction at constant pressure and constant temperature,
Under what conditions
(i)∆H < ∆U
(ii) ∆H > ∆U
(iii) ∆H = ∆U?
We know that ∆H = ∆U + ∆ng RT where ∆ng represents the difference between the number of moles of the gaseous products and of gaseous reactants.
(i) ∆H < ∆U
When ∆ng = –ve i.e. when there is a decrease in the number of moles of the gaseous components.
when ∆ng = + ve i.e. when the reaction proceeds by an increase in the number of moles of the gaseous components.
(iii) ∆ng = 0 i.e. when the number of moles or the gaseous reactants is equal to the number of moles of the gaseous products.
The enthalpy change (∆H) for the reaction
is -92.38 kJ at 298 K. what is the 
at 298 K?
Substituting the values, we have,
of combustion of methane is –X kJ mol–1. The value of 
is :
What do you understand by:
(i) Heat capacity of a substance
(ii) Heat capacity at constant volume
(iii) Heat capacity at constant pressure?
(i) Heat capacity: The heat capacity (C) of a sample of a substance is the quantity of heat needed to raise its temperature by 1°C (or one kelvin).
(ii) Heat capacity at constant volume: The heat supplied to a system to raise its temperature through 1°C keeping the volume of the system constant is called heat capacity at constant volume.
(iii) Heat capacity at constant pressure: The heat supplied to a system to raise its temperature through 1 °C keeping the external pressure constant is called heat capacity at constant pressure.
If q is the amount of heat supplied to a sample and as a result, if the temperature of the sample changes from initial temperature to a final temperature tf, then the heat capacity is given by
What do you understand by: (i) Specific heat capacity (ii) Molar heat capacity?
60.8 J of energy is required to change the temperature of 25.0 g of ethylene glycol (a compound used as an antifreeze in automobile engines) by 1.0 K. Calculate heat capacity of ethylene glycol.
We know that
Here, q = 68.8J; m = 25.0 g,
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Substituting the values in eq. (1), we have,
We know,
...(1)
Substituting the values in eq. (1), we get
= 1066.7 J = 1.07 kJ
We know that
Here 
m = 10.0 g
Substituting the values in (1), we have,
We know that
Substituting the values in (1), we have,
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C.
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Heat capacity at constant pressure is greater than heat capacity at constant volume. Why?
Derive a relationship between Cp and Cv for an ideal gas.
Substituting the values of ∆H and ∆U in eq. (1), we get,
Cp∆T – Cv ∆T = R∆T or Cp – Cv = R (for one mole of an ideal gas)
Thus Cp is greater than Cv by the gas constant R i.e. approximately 2 calories or 8.314 Joules.
Define the terms:
(i) Chemical energetics
(ii) Thermochemistry.
Whenever a chemical reaction occurs, some bonds present between atoms of reacting molecules are rearranged to form products. Energy is required to break bonds while it is released during bond formation. Thus during the rearrangement of bonds in reactants and products, sometimes energy is released and sometimes it is absorbed. Therefore, there is an energy change in chemical reactions.
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Comment on:
(i) ∆U is –ve for exothermic reactions and
(ii) ∆U is +ve for endothermic reactions.
Give the points of difference between exothermic reactions and enodthermic reactions.
| Exothermic reactions | Endothermic reactions |
| 1. In such reactions, heat is evolved | 1. In such reactions, heat is absorbed. |
| 2. ∆H has –ve value. | 2. ∆H has a +ve value. |
| 3. ∆U has a –ve value. | 3. ∆U has a +ve value. |
Explain the origin of enthalpy change in a direction. interest
During the reformation of chemical bonds, when the energy required to break bonds is less than the energy released in the formation of new bonds, the chemical reaction would be exothermic i.e. heat would be evolved. e.g.
Similarly, when the energy required to break bonds is more than the energy released in the formation of new bonds for products, a chemical reaction would be endothermic and ∆H would be positive. For example,
In gaseous phase, the enthalpy change (∆H) of a chemical reaction can be written as
What is standard enthalpy change (∆rH0) of a reaction?
It should be noted that the magnitude of ∆H for a reaction varies with the temperature and therefore for comparison, the values of ∆H for various reactions are expressed at the standard state. A substance is said to be in the standard state when it is present in its most stable state generally at 298K and under I bar pressure.
The standard enthalpy of a reaction is the enthalpy change for a reaction when all the participating substances (elements and compounds) are in their standard states (i.e. at 298K and 1 bar pressure). It is denoted by ∆r H0. For example
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What do you mean by the standard enthalpy of combustion?
What will be the amount of heat evolved when 39 gm of C6H6(l) are burnt? Given that:
would be = 
How much heat is evolved when 204g of ammonia are produced according to the equation,
2 mol i.e. 34 g of NH3 produced are accompanied by the evolution of 92.6 kJ of heat.
So, 204 g of NH3 produced would be accompanied by evolution of
Explain:
(i) Enthalpy (heat) of the formation.
(ii) Standard enthalpy of formation. How is it helpful in calculating the enthalpy of a reaction?
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Given:
What is the standard enthalpy of formation of NH3 gas?
The enthalpy of formation of ammonia is the enthalpy change for the formation of 1 mole of ammonia from its elements.
Given that enthalpy of formation for 2 moles of NH3 = –92.4 kJ
Therefore, standard enthalpy of formation for
Hence standard enthalpy of formation for NH3
The combustion of one mole of benzene takes place at 298K and 1 atm. After combustion, CO2(g) and H2O(l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation of ∆fH° benzene, given that standard enthalpy of formation of CO2(g) and H2O(l) are –393.5 kJ mol–1 and –285.83 kJ mol–1respectively.
Define enthalpy of neutralization.
and NaOH is -55.2 kJ
Explain why enthalpy of neutralization of a strong acid and strong base remains the same and the value changes if one of them is weak.
Enthalpy of neutralisation for a strong acid and a strong base is always constant: This is because in dilute solution all strong acids and strong bases are completely ionised. The neutralisation of a strong acid and strong base simply involves the combination of H+ions (from acid) and OH– ions (from base) to form unionised water molecules with the evolution of 57.1 kJ heat.
Since the same reaction takes place during neutralisation of all strong acids and strong bases, the value of enthalpy of neutralisation is constant.
The neutralisation of HCl and NaOH can be represented as:
Cancelling the common ions,
Enthalpy of neutralisation if either acid or base is weak : If one of the acid or bases is weak, then its ionisation is not complete in solution. Therefore, a part of the energy liberated is utilised for the ionisation of a weak acid (or base). Consequently, the value of enthalpy of neutralisation of the weak acid-strong base or strong acid-weak base is numerically less than 57.1 kJ. For example, neutralisation of acetic acid and sodium hydroxide can be represented as:
Thus, enthalpy of neutralisation of acetic acid and sodium hydroxide is –55.2 kJ.
Similarly, in the neutralisation of NH4OH and HCl, 5.6 kJ of heat is used up for the dissociation of weak base i.e. NH4OH. Hence enthalpy of neutralisation, in this case, is –57.1 + 5.6 = –51.5 kJ.
Explain the following:
(i) Enthalpy of fusion
(ii) Enthalpy of vaporization
(iii) Enthalpy of sublimation.
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Differentiate between change of state and phase change.
A swimmer coming out from a pool is covered with a film of water weighing about 80 g. How much heat must be supplied to evaporate this water?
When 1 gm of liquid naphthalene (C10H8) solidifies, 49 joules of heat is evolved. Calculate the enthalpy of fusion of naphthalene.
How many grams of methane and volume of oxygen at N.T.P. would be required to produce 445.15 kJ of heat in the following reaction?
The thermochemical reaction is,
Step I. Amount of methane required:
890.3 kJ of heat is produced from methane = 16 g
445.15 kJ of heat is produced from methane
Step II. Volume of oxygen required:
890.3 kJ heat is evolved from oxygen = 44.8 litres
445.15 kJ of heat is evolved from oxygen
Standard vaporisation enthalpy of benzene at its boiling point is 30.8 kJ mol-1 ; for how long would a 100W electric heater has to operate in order to vaporise a 100g sample of benzene at its boiling temperature?
State and explain Hess's law of constant heat summation.
This law states that if a reaction is the sum of two or more constituent reactions, then ∆ H for the overall process must be the sum of all ∆H of the constituent reactions. In brief, the enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps.
Consider a reaction in which reactant S1 is converted to product S4. There are two reactions for converting S1 to S4:
(i) Direct reaction: Reactant S1 is directly converted to product S4 in one step.
Let ∆rH° be the enthalpy change during this conversion.
(ii) Indirect reactions: Let reactant S1 be directly converted to S4 in a number of steps.
be the enthalpy changes in the first (a), second (b) and third reaction (c) respectively.
According to Hess's law
Here a, b, c .....(as stated above) refer to the balanced thermochemical equation that can be summed to give the equation for the desired reaction.
Illustration of Hess’ law: Carbon when burnt in oxygen forms CO2 in two ways: First way:
Second way: Carbon may be first converted into CO and then changed into CO2.
Total amount of enthalpy changes in a second way
= -110.5 - 283.0
= -393.5 kJ
which is the same as in the first case.
The thermochemical equations for the formation of water are:
How much energy is needed to convert 2 mol of liquid water to water vapour?
We aim at equation
Given that 
Subtracting equation (ii) from equation (i), we have,
Calculate the heat change accompanying the transformation of C(graphite) to C(diamond). You are given:
The equation,
Given that
Subtracting equation (ii) from equation (i), we have, 
i.e. 1.9 kJ of heat is absorbed in the process.
We aim at the equation
Here we want one mole of C(graphite) as a reactant, so write equation (i) as such. We want two moles of H2(g) as a reactant ; so multiply equation (ii) by 2, we want one mole of CH4(g) as a product, so write equation (iii) as such.
Adding equations (i), (ii) and (iii), we have,
We have the equation,
Given that:
Multiplying equation (ii) by 2, we have,
Adding equations (i) and (iv), we have,
Subtracting equation (iii) from equation (v), we have,
Enthalpy of combustion of methane is -217.3 kJ mol-1.
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We aim at the equation
Multiplying equation (i) by 6 and also equation (ii) by 6,
Subtracting equation (iii) from equation (vi),
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We aim at the equation
Given that:
Multiplying equation (iii) by 2, we have
Adding equations (ii) and (iv),
Subtracting equation (i) from equation (v), we have
We have equation,
Multiplying equation (iii) by 2,
Adding equations (ii) and (iii),
Subtracting equation (i) from equation (ii),
The combustion of one mole of benzne takes place at 298K and 1 atm. After combustion, CO2(g) and H2O (l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ∆fH° of benzene. Standard enthalpies of formation of CO2(g) and H2O(l) are – 393.5 kJ mol–1 and –285.83 kJ mor–1respectively.
We just aim at the equation
Here,
What do you mean by bond enthalpy? When is bond enthalpy equal to bond dissociation energy?
Chemical reactions involve the breaking and making of chemical bonds. Energy is required to break a bond and energy is released when a bond is formed. The bond enthalpy is the amount of energy necessary to break bonds in one mole of a gaseous covalent substance to form products in the gaseous state. The bond enthalpy for H2 is 435.0 kJ mol–1.
This endothermic reaction (∆H0 is positive) can be written as,
Bond energy of a diatomic molecule (e.g. H2, Cl2, O2 etc.) is equal to its dissociation energy. The dissociation energy of a bond is defined as the enthalpy change involved in breaking the bond between atoms of a gaseous homonuclear molecule.
Calculation of enthalpy of reaction from bond energies: We know that when a reaction takes place, some bonds are broken while some new bonds are formed. Energy is required to the dissociation of bonds while energy is released when the bonds are formed.
How does bond enthalpy vary with:
(a) size of atoms
(b) electronegativity of atoms?
(a) It decreases with the increase in the size of atoms.
(b) It increases with the increase in the electronegativity of atoms.
Compute the enthalpy of hydrogenation of ethylene if the bond enthalpies of H – H, C – H, C – C and C = C bonds are 436, 485, 347 and 619 kJ mol–1 respectively.
The required thermochemical equation is
Thus enthalpy of hydrogenation of ethylene is -262 kJ.
Calculate 
for the reaction 
The bond enthalpies of C - H, O = O, C = O, O - H and C = C bonds are 414, 499, 460 and 619 kJ mol-1 respectively.
In the thermochemical equation,
four C-H bonds, one C = C bond and three O = O bonds are broken while four C = O bonds and four O - H bonds are formed. Thus
is enthalpy of atomisation and 
We aim at the equation
The given data implies as under
Multiplying equaton (iv) by 2,
Subtracting equation (i) and (ii) from equation (vi),
We aim at the equation,
Multiplying equation (ii) by 2 and equation (iii) by 3, we have,
Subtracting (i) from (vi), we have,
Ethane molecule contains one C - C bond and six C - H bonds
Bond dissociation energy of one C - H bond = 416 kJ
Bond dissociation energy of six C - H bonds = 6 x 416 = 2496 kJ
What do you mean by calorific value of foods and fuels?
Coal, petroleum, natural gas etc. are regarded as fuels for many machines and industrial purposes. The oxidation or combustion of all these fuels releases energy. This energy is measured in terms of calorific value. It is defined as the amount of heat produced by the complete combustion of one gram of the substance (food or fuel) in oxygen. It is generally expressed in kilocalories per gram (k cal g–1) or kJ g–1. The calorific value of fuel or food is measured by bomb calorimeter. Greater the calorific value, better is the fuel.
Food is regarded as fuel for the animal or human body. It has been found that fats and carbohydrates are the main sources of energy which have calorific values of about 9 k cal g–1 and 4 k cal g–1 respectively. Proteins are also the main constituents of our food but have a low calorific value of 4 k cal g–1.
The enthalpies of combustion of methane and ethane are –890.3 and –1559.7 kJ mol–1 respectively. Which of the two has a greater efficiency of the fuel per gram?
The fuel efficiency can be predicted from the amount of heat evolved for every gram of fuel consumed.
(i) Thermochemical equation for the combustion of methane is,
Gram molecular mass of methane 
= 12 + 4 X 1 = 16 g
(ii) Thermochemical equation for combustion of ethane is
Gram molecular mass of ethane 
= 2 x 12 + 6 x 1 = 30 g
Thus, methane has greater fuel efficiency than ethane.
What are important consequences of lattice enthalpies?
Important consequences of lattice enthalpies:
(i) The greater the lattice enthalpy, more stable is the ionic compound.
(ii) The lattice enthalpy is greater, for ions of higher charge, and smaller radii.
(iii) The lattice enthalpies effect the solubilities of ionic compounds.
In 1919, Born and Haber proposed a method in which lattice enthalpy of an ionic crystal is related to certain thermodynamic parameters. The formation of ionic crystal from its elements by applying different thermochemical quantities is called Born-Haber cycle.
Consider the enthalpy change during the formation of ionic solid MX from its elements M and X.
Formation of ionic solid MX may be done by two different methods:
(A) Indirect method
(B) Direct method
(A) Indirect method:
The various terms are
Since energy is required for the process (endothermic), therefore, 
is taken as a positive quantity.
Since the process is endothermic, therefore ionisation enthalpy is taken as a positive quantity.
Since the process is exothermic, therefore electron gain enthalpy is taken as negative quantity.
Energy is released in this process i.e. process is exothermic, therefore 
is always taken as negative quantity.
Direct method: Let 
be the enthalpy of formation of 1 mol of MX(s) from its constituent species.
where 
is the enthalpy change for lattice formation from,
The reverse of the above equation is
Thus by using the Born-Haber cycle, one can determine the lattice enthalpy of an ionic compound.
According to Hess’s law, the enthalpy of formation of one mole of ionic solid MX should be the same irrespective of the fact whether it takes place directly in one step (Direct method) or through a number of steps (Indirect method).
How can the lattice enthalpy of an ionic NaCl be determined by using Born-Haber cycle?
Formation of NaCl may be done by two different methods:
(a) Indirect method
(b) Direct method.
(a) Indirect method: Various terms involved are:
Since enthalpy is required for the process (endothermic), therefore 
is taken as a positive quantity.
(ii) 
The process is endothermic, therefore 
is taken as a positive quantity. 
Since the process is endothermic, therefore, ionisation enthalpy is taken as a positive quantity.
Since the process is exothermic, therefore electron gain enthalpy is taken as a negative quantity. 
Enthalpy is released in this process i.e. process is exothermic, therefore 
is always taken as a negative quantity.
(b) Direct method: Let 
be the enthalpy of formation of 1 mol of NaCl(s) from its constituent species.
According to Hess's law, the enthalpy of formation of one mole of sodium chloride should be the same irrespective of the fact whether it takes place directly in one step.
(Direct method) or through a number of steps (Indirect method).
where 
is the enthalpy change for lattice formation from
The reverse of the above equation i.e.
defines the lattice enthalpy of NaCl.
Thus by using the Born- Haber cycle, one can determine the lattice enthalpy of an ionic compound.
Calculate the lattice enthalpy of KCl crystal from the following data:
Sublimation enthalpy of pottasium (K) = +89 kJ mol-1
Dissociation enthalpy of 
= +122 kJ mol-1
Ionisation enthalpy of K(g) +425 kJ mol-1
Electron gain enthalpy of Cl(g) = -355 kJ mol-1
Enthalpy of formation of 
Using Born-Haber cycle for KCl,
Substituting the values we have,
The reverse of the above equation i.e.
Hence the lattice enthalpy of KCl = +719 kJ mol-1
Calculate the lattice enthalpy of LiF, given that the enthalpy of
(i) sublimation of lithium is 155.2 kJ mol–1
(ii) dissociation of 1/2 mol of F2is 75.3 kJ
(iii) ionization of lithium is 520 kJ mol–1
(iv) electron gain of 1 mol of F(g) is –333 kJ
(v) ∆f H0 ∆fH0 overall is –795 kJ mol–1
Using Born-Haber cycle to LiF
Substituting the values, we have
The reverse of the above equation i.e.
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defines the lattice enthalpy of LiF
Hence lattice enthalpy of LiF = +1212.5 kJ mol-1
Calculate the lattice enthalpy of MgBr2 from the given data. The enthalpy of formation of MgBr2 according to the reaction
Using Born-Haber cycle of MgBr2,
Substituting the values, we have,
= -524 - 148 - 2187 - 31 - 193 - 2 (-331)
= -2421 kJ mol-1
Hence lattice enthalpy = 
Spontaneous process. A process which has a natural tendency to take place either by itself or after initiation under a given set of conditions is known as a spontaneous process. For example,
(i) Sugar dissolve in water and forms a solution.
Sugar + water 
Solution of sugar in water
(ii) In the domestic oven, coal keeps on burning once ignited.
Non-spontaneous process: A process that has no natural tendency or urge to take place is called non-spontaneous process. However, many non-spontaneous processes can also be made to take place by supplying energy continuously from some external sources. For example:
(i) Water can be made to flow uphill by the use of an electric motor pump.
(ii) Electrolysis of water to separate it into pure hydrogen and pure oxygen.
The tendency of a system to acquire a state of maximum randomness is the sole criterion for the spontaneity of a process. Comment.
No, there is a natural tendency of a system to attain a state of the greater randomness i.e. more disordered state. For example,
(i) There is more randomness on mixing of two gases (which do not react chemically).
(ii) Evaporation of water. The evaporation of water results in increase of randomness because the molecules in the vapour state have more randomness than in the liquid state
(iii) Dissolution of ammonium chloride in water. Solid ammonium chloride has less randomness while in solution ammonium chloride particles move freely as 
and hence randomness increases.
If the randomness factor were the only criterion, then the process like liquefaction of gas or solidification of a liquid would not have been feasible since these were accompanied by a decrease in randomness. Hence the tendency of a system to acquire a state of maximum randomness is not the sole criterion for determining the spontaneity of the process.
The overall tendency of a process to occur can be expressed on the resultant of two tendencies namely:
(i) the tendency to acquire a state of minimum energy, and
(ii) the tendency to acquire a state of maximum randomness or disorder.
The overall tendency of a process to take place by itself is called the driving force.
It should be noted that:
(i) the two tendencies act independent of each other,
(ii) the two tendencies may work in the same direction or opposite direction in a process and
(iii) the driving force is the resultant of the magnitude of the two tendencies. When the two tendencies act in the opposite direction, the tendency with the greater magnitude determines whether the process is feasible or not. For example,
(a) Evaporation of water. Evaporation of water is endothermic, therefore, energy factor opposes the process. But it is favoured by randomness factor.
Since the process is known to take place, randomness factor must be greater than energy factor.
(b) The reaction between hydrogen and oxygen to form water. It is an exothermic reaction, therefore, favours the process, but randomness factor opposes the reaction.
As the reaction takes place, the energy factor must be greater than the randomness factor.
Define entropy and entropy change. What are the units of entropy?
The degree of randomness or disorderliness is expressed by a thermodynamic function called entropy. It may be defined as the property of a system which measures the degree of randomness or disorderliness in the system. It is denoted by S. It is a state function. It depends only on the initial and final state of the system and not on the path followed.
If SA and SB are the entropies at state A and B, then entropy change ∆S is given by![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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For a chemical reaction, 
For a reversible process at equilibrium, the change in entropy may be expressed as,
Where qrev represents the total heat absorbed reversibly and isothermally at temperature T.
What is the physical significance of entropy?
Physical significance: Entropy has been regarded as a measure of disorder or randomness of a system. Thus when a system goes from a more orderly to less orderly state, there is an increase in its randomness and hence entropy of the system increases. Conversely, if the change is one in which there is an increase in orderliness, there is a decrease in entropy. For example, when a solid changes to a liquid, an increase in entropy takes place, because with the breaking of the orderly arrangement of the molecules in the crystal to the less orderly liquid state, the randomness increases. The process of vaporisation produces an increase in randomness in the distribution of molecules, hence an increase in entropy. When two gases are mixed, the molecules of the gases intermix to achieve more randomness.
Thus, this concept of entropy (measure of randomness) has led to the conclusion that all substances in their normal crystalline state at absolute zero temperature would be in the condition of maximum orderly arrangement, because all motion has essentially ceased at ‘0 K.’ In other words, entropy of a substance at 0 K is minimum.
Account for the following:
(i) Why does real crystal have more entropy than an ideal crystal?
(ii) Why the entropy of a pure substance is taken as zero at absolute zero?
(iii) Why the entropy of the universe is continuously increasing?
(i) An ideal crystal has a perfect order of its constituent particles while a real crystal has less order because of some defects. Therefore, a real crystal has more entropy than an ideal crystal.
(ii) Because at 0K, there is a complete order in all the crystals and no randomness, therefore, entropy is taken as zero.
(iii) Because, in the universe, almost all processes are spontaneous and for all spontaneous processes, there is an increase in entropy.
Which of the following reaction will have a greater change in entropy? Explain.![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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Predict the change in entropy for the system in which the following physical changes occur:
(i) Dissolving cream in a cup of coffee.
(ii) The beating of an egg for an omellete.
(iii) The formation of a rain drop in a cloud.
(iv) The crystallization of a metal alloy from the molten state.
Changes in entropy of the given system:
(i) Increases
(ii) Increases
(iii) Decreases
(iv) Decreases
In the following changes, state whether order has increased or decreased and consequently the direction of change of entropy of the system:
(i) The order has increased, entropy decreases in the forward direction.
(ii) The order has decreased, entropy increases in the forward direction.
(iii) The order has increased, entropy decreases in the forward direction.
In the following changes, state whether order has increased or decreased and consequently by the direction of change of entropy of the system:
(i) The order has increased, entropy decreases in the forward direction.
(ii) The order has increased, entropy decreases in the forward direction.
(iii) The order has increased entropy decreases in the forward direction.
(iv) The order has increased, entropy decreases in the forward direction.
[Hint. Boiled egg is solid and less disordered as an ordinary normal egg which is a liquid].
Will entropy increased or decrease during boiling of an egg? Comment on the statement.
The entropy of steam is more than that of water at its boiling point. Explain.
For an isolated system, ∆U = 0, what will be ∆S ?
Consider two bulbs each having gas connected by a stop-cock and isolated from the surrounding as an example of an isolated system. On opening the stop cock, the two gases mix up. As a result, more space is available for each gas to move apart i.e. system becomes more disordered. This shows that ∆S > 0 i.e. ∆S is positive.
For the reaction 
what are the signs of 
?
Since the given reaction![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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represents the formation of bonds, therefore energy is released i.e. ∆H is –ve, further 2 moles of atoms have greater randomness than 1 mole of molecules. Hence randomness decreases, i.e. ∆S is < 0 i.e. negative.
Prove that in a reversible process:
∆(system) + ∆S(surroundings) = 0
Or
Prove that there is no net change in entropy in a reversible process.
Suppose heat is absorbed by the system reversible and the heat is lost by the surroundings also reversibly (process occurs under complete reversible condition).
If qrev is the heat absorbed by the system reversibly, then the heat lost by the surroundings will also be qrev. If the process takes place isothermally at T kelvin, then Entropy change of the system
Entropy change of the surroundings![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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Hence the total entropy change for the combined system and surroundings will be:
Hence in a reversible process, the net entropy change for the combined system and the surroundings is zero i.e. there is no net change in entropy.
Prove that in an irreversible process:
∆S(system) + ∆S(surroundings) > 0
If any part of the process is irreversible, the process as a whole is irreversible. Suppose the total heat lost by the surrounding is qirrev. This heat is absorbed by the system. However, entropy change of the system is always calculated from the heat absorbed reversibly.
Entropy change of the surroundings is given by![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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This is because of the large size of the surroundings due to which heat lost (qirrev) by the surroundings can be considered as the heat lost reversibly and isothermally at temperature T.
The total entropy change for the combined system and surroundings is
We know that the work done in a reversible process is the maximum work.
Also, internal energy (U) is a state function, the value of U is same whether the process is carried out reversibly or irreversibly. Hence
From relation (4) and (5), we conclude that
Combining the result with the result given in equation (3),
Thus, in an irreversible process, the entropy change for the combined system and the surroundings i.e. an isolated system is greater than zero i.e. an irreversible process is accompanied by a net increase of entropy.
What do you understand by:
(i) The entropy of fusion?
(ii) The entropy of vapourisation ?
(i) The entropy of fusion. It may be defined as the entropy change when one mole of the solid substance changes into liquid form at its melting point. For example when ice melts as
where ∆Hfusion is the enthalpy of fusion and Tf is the fusion temperature. Since ∆Hfus is +ve, therefore ∆Sjfus is +ve, hence the process of fusion is accompanied by an increase of entropy.
(ii) The entropy of vapourisation. It may be defined as the entropy change when one mole of the liquid changes into vapour at its boiling point.
Since ∆Hvap is +ve, therefore ∆Svap is +ve, hence the process of vapourisation is accompanied by an increase of entropy.
You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below:
Liquid 
The thermochemical equation is,
From the above equation, it is clear that when 1 mol of H2O(l) is formed, 286 kJ of heat is released. The same amount of heat is absorbed by the surroundings.
What are the two tendencies which determine the feasibility of process? How are the two related to each other?
The following two tendencies are responsible for determining the feasibility of a process:
(i) The tendency of a system to acquire a state of minimum energy i.e. energy factor. It is expressed in terms of enthalpy change (∆H). A negative value of ∆H suggests that system has the tendency to proceed.
(ii) The tendency of a system to acquire a state of maximum randomness i.e. randomness factor. It is expressed by T∆S where T is the absolute temperature and ∆S is the change in entropy. A positive value of T∆S indicates an inherent tendency of a process to occur.
Relation between two tendencies:
The overall tendency or the driving force of a process is expressed in terms of free energy change (∆G). This is expressed as ∆G = ∆H – T∆S
This equation is called Gibb’s Helmholtz equation. For a spontaneous process, ∆G should have a negative value i.e. the system should undergo a decrease in its free energy.
what is free energy? Prove that ∆G = ∆H – T∆S.
Explain Gibb's Helmholtz equation.
Gibbs-Helmholtz equation is
Significance of 
The value of ∆G can be zero, positive or negative.
(i) If ∆G = 0, the reaction is in the state of equilibrium i.e. ∆H = T∆S.
(ii) If ∆G = – ve, the reaction is spontaneous in the forward direction.
(iii) If ∆G = +ve, the reaction is non-spontaneous i.e. reaction is spontaneous in the reverse direction.
(a) If ∆G < 0 i.e. negative, the reaction will be spontaneous. ∆G can be negative if:
(i) Both energy and the entropy factors are favourable and may have any magnitude i.e. ∆H is –ve and T∆S are +ve.
(ii) When energy factor favours (∆H= –ve), but entropy factor opposes (T∆S = –ve), but ∆H ( –ve) > T∆S (–ve).
(iii) When energy factor is not favouring (∆H = +ve) but entropy factor favours (T∆S = +ve). But T∆S (+ve) > ∆H (+ve).
(b) ∆G = 0, the reaction is in an equilibrium state and thus there is no net change in either direction.
This happens when one of the factors is favourable and the other is opposite but they are equal in magnitude.
(c) ∆G > 0 i.e. +ve, the reaction will not be spontaneous, ∆G can be positive if:
(i) both the factors oppose i.e. ∆H is +ve and T∆S is –ve,
(ii) both the factors i.e. ∆H and T∆S have –ve sign and T∆S > ∆H,
(iii) both the factors i.e. ∆H and T∆S have a +ve sign and ∆H > T∆S.
When ∆G is positive, the process is always non-spontaneous. Explain.
∆G can be positive only if;
(i) both the factors oppose i.e. ∆H is +ve and T∆S is –ve,
(ii) both the factors i.e. ∆H and T∆S have a negative sign and T∆S > ∆H,
(iii) both the factors i.e. ∆H and T∆S have a +ve sign and ∆H > T∆S.
Prove that ∆G = –T∆Stotal.
The change in entropy in a process carried out in a non-isolated system is given by
∆Stotal = ∆Ssystem + ∆Ssurroundings ...(1)
Consider an isothermal process carried out at constant pressure in which heat q is transferred by the surroundings to the system.
Substituting this value in (1), we have,
It has been shown that 
must be positive for a process to be spontaneous. Therefore, equation (2) becomes useful in predicting the spontaneity of a process in terms of ∆G.
What is free energy change? Show that the change in free energy is equal to useful work done.
or
prove that –∆G = w(useful work)
(i) The overall criterion of a process to occur spontaneously i.e. driving force is expressed in terms of free energy change (∆G).
Free energy change (∆G) is related to enthalpy change (∆H) and entropy change (∆S) as ∆G = ∆H – T∆S [Gibbs-Helmoltz equation] For a spontaneous process, ∆G should have negative value.
(ii) According to first law of thermodynamics,
where q is heat absorbed by the system. ∆U is the change in internal energy and w is the work done on the system.
If we are to determine the work a system can do, then w should be taken as – w.
Now w includes expansion as well as non-expansion work. 
But work due to expansion at constant temperature is given by P∆V.
wnon-exp is also called useful work because this type of work can be used for useful effect e.g. electrical work.
Substituting this value in (2), we have,
or 
This means that decrease in free energy of a system (–∆G) is a measure of the non-expansion or useful work done by the system in a process or –∆G = wuseful work
Predict the enthalpy change, free energy change and entropy change when ammonium chloride is dissolved in water and the solution becomes colder.
(i) Enthalpy change is positive since the solution becomes colder due to intake of heat from the solution.
(ii) Free energy change is negative since the dissolution of ammonium chloride takes place spontaneously.
(iii) Entropy change is positive since more ordered solid ammonium chloride is changed to less ordered solution of ammonium chloride containing larger number of ions 
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, ice and water are in equilibrium and 
for the process 
Calculate 
for the conversion of ice to liquid water.
At equilibrium state,
For the reaction
calculate 
at 700K when enthalpy and entropy changes are -113 kJ mol-1 and -145 JK-1 mol-1 respectively.
According to Gibb's Helmoltz equation,
From the following values of ∆H and ∆S, decide whether or not these reactions will be spontaneous at 298 K:
Reaction A:
∆H = – 10.5 X 103 J mol–1
∆S = + 31 JK–1 mol–1
Reaction B:
∆H = – 11.7 X 103 J mol–1 ;
∆S = –105 jK–1 mol–1.
(i) Reaction A:
According to Gibb’s Helmholtz equation,
Since ∆G is –ve, the reaction A will be spontaneous at 298 K.
(ii) Reaction B:
As ∆G is +ve, the reaction B will not be spontaneous at 298 K.
Calculate the free energy change of the reaction 
According to Gibb's Helmoltz equation,
Here 
Substituting the values in (1), we have,
For the reaction:
Calculate the temperature at which it attains equilibrium.
According to Gibb's Helmoltz equation,
which it attains equilibrium, 
At what temperature, reduction of lead oxide to lead by carbon
becomes spontaneous? For this reason 
are 108.4 kJ mol-1 and 190.0 JK-1 mol-1 respectively. 
According to Gibb's Helmoltz equation 
At equilibrium,
Above this temperature, ∆G will be negative and the reaction will become spontaneous.
For the reaction at 298 K
At what temperature will the reaction become spontaneous considering 
to be constant over the temperature range?
According to Gibb's Helmholtz equation,
At equilibrium,
Substituting the values in equation (1), we get
Above this temperature, ∆G will be negative and reaction becomes spontaneous.
For the reaction
Calculate 
for the reaction and predict whether the reaction may occur spontaneously.
We know,
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Substituting the values in equation (1), we get
According to Gibb's Helmholtz equation,
Since ∆G0 is positive, the reaction will not occur spontaneously.
For the reaction![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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, calculate the temperature at which 
is equal to zero. Also predict the direction of the reaction at:
(i) temperature
(ii) below this temperature
Let 
be zero at temperature TK.
(i) At this temperature, the reaction proceeds in either direction at the same rate and it represents the state of equilibrium.
(ii) Below this temperature,
T = 462 K (say)
As ∆G is +ve below this temperature, the reaction will not be spontaneous.
For the melting of ice at 
the enthalpy of fusion is 
and entropy of fusion is 25.4 
Calculate the free energy change and predict whether the melting of ice is spontaneous or not at this temperature.
According to Gibb's Helmoltz equation
Substituting the values in (1), we have,
What do you understand by:
(i) Standard free energy change (∆rG0)
(ii) Standard free energy of formation (∆fG0).
(i) Standard free energy change (∆G0): It is defined as ‘the free energy change for a process at 298K in which the reactants in their standard states are converted into the products in their standard states’. Thus,
The value of ∆G0 for a reaction can be derived from the standard free energies of formation (∆fG0).
(ii) Standard free energy of formation (∆fG0) of a compound is defined as ‘the change in free energy when 1 mole of a compound is formed from its constituent elements in their standard states”. Thus,
= [Sum of the standard free energies of formation of products] - [Sum of the standard free energies of formation of reactants]
It may be pointed here that ‘the standard free energy of formation of an element in its standard state is zero’.
Calculate the standard free energy change for the reaction
Given that the standard free energies of formation 
for 
are -16.8, +86.7 and -237.2 kJ mol-1 respectively. Predict the feasibility of the above reaction at the standard state.
We know, 
Since 
is negative, the process is feasible.
For the equilibrium
,
What is the ∆G0 for the reaction?
Here, 
We know, 
Substituting the values, we get,
We know,
Substituting the values, we get,
Calculate the equilibrium constant for the reaction at 400 K:
We know,
...(1)
Substituting the values in (1), we have,
Find out the value of equilibrium constant for the following reaction at 298 K:
Standard Gibbs energy change, 
at the given temperature, is -13.6 kJ mol-1.
We know,
Substituting the values in equation (1), we get
The equilibrium constant at 
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Calculate the value of 
In which direction is the reaction spontaneous when reactants and products are under standard conditions?
is negative, the reaction is spontaneous in the forward direction.N2O4 (g) 
2NO2(g)
Initial 1 mole 0
At eqn. 1 - 0.5 2 x 0.5
= 0.5 mol = 1 mol
Total number of moles = 0.5 + 1 = 1.5 mol
According to law of chemical equilibrium
We know,
Substituting the values in equation (1), we get
State second law of thermodynamics in different ways.
(i) It is impossible to convert heat completely into an equivalent amount of work without producing some change in some other part of the system.
(ii) All natural and spontaneous processes take place in one direction only and are thermodynamically irreversible.
(iii) It is impossible to convert heat from a reservoir into work by a cyclic process without transferring to a colder reservoir.
(iv) Heat cannot pass itself from a colder to a hotter body.
(v) It is impossible for a self-acting machine, unaided by any external agency, to convey heat from a body at lower temperature to a body at a higher temperature.
State second law of thermodynamics in terms of entropy.
All spontaneous processes are accompanied by a net increase of entropy, i.e. for all spontaneous processes, the total entropy change (sum of the entropy changes of the system and the surroundings) is positive.
Since all naturally occurring processes are spontaneous and are accompanied by a net increase of entropy, therefore the second law of thermodynamics may also be stated as: “The entropy of the inverse is continuously increasing and tends to be maximum.
(i) All spontaneous processes (or naturally occurring processes) are thermodynamically irreversible.
(ii) The entropy of the universe is continuously increasing.
(iii) Total energy absorbed by a system cannot be converted completely into work.
Third law of thermodynamics states: “The entropy of all perfectly crystalline solids may be taken as zero at absolute zero.”
Molecular interpretation: Entropy is a measure of disorder. Thus, at absolute zero a perfectly crystalline solid has a perfect order of its constituent particles i.e. there is no disorder at all. Hence absolute entropy is taken as zero.
Application. It helps in the calculation of the absolute entropies of the substances at room temperature (or at any temperature T). A simplified expression for the absolute entropy of solids at temperature T is given by,
where Cp is the heat capacity of the substance at constant pressure.
Explain the effect of temperature on feasibility for:
(i) endothermic process
(ii) exothermic process in terms of Gibb’s Helmoltz equation.
When does endothermic processs become spontaneous? Explain.
(a) What are Exergonic and Endergonic reactions?
(b) Given that the standard heat of formation of NH3(g) as represented by the equation
is – 46.191 kJ. The standard entropies of N2(g), H2(g) and NH3(g) are 191.62, 130.12 and 193.3 JK–1 mol–1 respectively. Calculate the standard free energy of formation (∆G0) for NH3. Is the reaction feasible?
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(i) Define first and the second law of thermodynamics in the combined form.
(ii)
(a) Write an expression for the entropy change of an ideal gas for isothermal change.
(b) Write an expression for entropy change for an ideal gas for isobaric change.
(iii) Calculate the maximum work obtained when 0.75 ml of an ideal gas expands isothermally and reversibly at 27°C from a volume of 15L to 25L.
(i) The energy of the universe is constant but the entropy of the universe is continuously increasing and tends to a maximum.
(ii)
(a)
Substituting the values, we have, 
(i) What are the limitations of criterion for randomness?
(ii) Calculate the standard free energy of formation of 
The free change for the reaction.
Given:
What is the standard enthalpy of formation of 
gas?
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form its constituent elements in their standard state.
writing the given equation for 1 mole of NH3 (g).
For an isolated system, ![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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what will be 
?
Here, will be positive i.e., greater than zero
since,, will be positive and the reaction will be spontaneous.
For the reaction,
H and 
S are negative.
The given reaction represents the formation of chlorine molecule from chlorine atoms. here, the bond formation is taking place. Therefore, energy is being released. Hence, 
H is negative.
Also, two moles of atoms have more randomness than one mole of molecules. Since spontaneity is decreased, 
S is negative for the given reaction.
The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:
676.5
-676.5
-110.5
110 s
C.
-110.5
C(s) + O2 (g) → CO2 (g); ΔH = -393.5 kJ mol-1 ... (i)
CO + O2/2 → CO2 (g); ΔH = - 283.5 kJ mol-1 ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
C (s) + O2/2 (g) → CO (g)
ΔH = (-393.5 + 283.5) kJ mol-1 = - -110 kJ mol-1
The following reaction is performed at 298 K
2NO(g) + O2 (g) ⇌ 2NO2 (g)
The standard free energy of formation of NO (g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2 (g) at 298 K? (KP = 1.6 x 1012)
R (298) In (1.6 x 1012)-86600
86600 + R (298) In (1.6 x 1012)
86600 - In(1.6 x 1012)/R(298)
0.5[2 x 86600-R(298)In (1.6 x 1012)]
D.
0.5[2 x 86600-R(298)In (1.6 x 1012)]
For the given reaction,
2NO(g) + O2 (g) ⇌ 2NO2 (g)
Given ,
Now, we have,
The standard Gibbs energy change at 300 K for the reaction, 2A ⇌ B +C is 2494.2J at a given time, the composition of the reaction mixture is
and 
, The reaction proceeds in the [R= 8.314 JK/mol, e = 2.718]
forward direction because Q>Kc
reverse direction because Q>Kc
forward direction because Q < Kc
reverse direction because Q < Kc
B.
reverse direction because Q>Kc
We know,
ΔG = ΔGo + RTlnQ .. (i)
Given,
ΔGo = 2494.2J
thus,
putting the value in equation (i)
= 2494.2 +8.314 + 300 In 4
= 28747.27 J
= positive value
Also, we have
If ΔG is positive, Q >K
therefore, reaction shifts in the reverse direction
For the complete combustion of ethanol, C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l), the amount of heat produced as measured in a bomb calorimeter, is 1364.47 kJ mol-1 at 25oC. Assuming ideality the enthalpy of combustion, ∆CH, for the reaction will be (R = 8.314 JK-1 mol-1)
-1366.95 kJ mol-1
-1361.95 kJ mol-1
-1460.50 kJ mol-1
-1350.50 kJ mol-1
A.
-1366.95 kJ mol-1
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l),
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.00C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be:(R = 8.314 J/mol K) ( ln 7.5 = 2.01)
q =+208J, W = - 208 J
q =-208 J, W =-208 J
q=-208J, W = +208 J
q =+208 J, W = +208 J
A.
q =+208J, W = - 208 J
From first law of thermodynamics, ΔE = q+W for an isothermal expansion.
Hence, q =-W
q= +208 J
W =-208 J [expansion work]
The incorrect expression among the following is
In isothermal process
C.
Option C has incorrect expression. The correct expression is,
The standard reduction potentials for Zn2+/ Zn, Ni2+/ Ni, and F2+/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction X + Y2+ → X 2+ + Y will be spontaneous when
X = Ni, Y = Fe
X = Ni, Y = Zn
X =Fe, Y= Zn
X = Zn, Y = Ni
D.
X = Zn, Y = Ni
X = Zn, Y = Ni
Zn + Ni2+ →Zn2+ + Ni
The solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol–1)to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is
1.2 x 10-10 g
1.2 x 10-9 g
6.2 x 10-5 g
5.0 x 10-8 g
A.
1.2 x 10-10 g
Ksp of AgBr = [Ag+][Br-] = 5.0 x 10-13
[Ag+] = 0.05 M
Moles of KBr = 1 x 10-11 x 1 = 1 x 10-11
weight of KBr 1 x 10-11 x 120 = 1.2 x 10-19 g
For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when
Te>T
T >Te
Te is 5 times T
T=Te
A.
Te>T
For a particular reversible reaction at T temperature
ΔG = ΔH-TΔS
When ΔH, S is positive
ΔG = +ΔH- T (+ΔS)
For a spontaneous process, ΔG must be negative, it is possible only at high temperature.
That mean T> Te
Δ U is equal to
Isochoric work
Isobaric work
Adiabatic work
Isothermal work
C.
Adiabatic work
From 1st law thermodynamics:
ΔU = q + w
For adiabatic process :
q = 0
∴ ΔU = w
∴ Work involves in the adiabatic process is at the expense of a change in internal energy of the system.
Given
C(grahite) + O2(g) → CO2(g)
ΔrH° = - 393.5 kJ mol-1
H2(g) + 1/2O2(g) → H2O (l)
ΔrH° = +890.3 kJ mol-1
Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction
C(grahite) + 2H2(g) →CH4 will be
+74.8 kJ mol–1
+144.0 kJ mol–1
–74.8 kJ mol–1
–144.0 kJ mol–1
C.
–74.8 kJ mol–1
CO2(g) + 2H2O (l)→ CH4 (g) + 2O2(g);
ΔrH°= 890.3
ΔfH° –393.5 –285.8 ? 0
ΔrH°= Σ(ΔfH°)products -Σ(ΔfH°)reactant
890.3 = [ 1 x(ΔfH°)CH4 + 2x0]-[1x(-393.5)+2(-285.8)]
(ΔfH°)CH4 = 890.3-965.1 = -74.8 kJ/mol
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)
At 298 K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2(g) are –166.2, –237.2 and –394.4 kJ mol–1 respectively. If
standard enthalpy of combustion of methanol is –726 kJ mol–1, efficiency of the fuel cell will be:
80%
97%
87%
90%
B.
97%
CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)
Also ∆G°f CH3 OH( l) = -166.2 kJ mol-1
∆Gf°H2O (l ) = -237.2 kJ mol-1
∆Gf°CO2 (l ) = -394.4 kJ mol-1
∆G = Σ∆Gf° products −Σ∆Gf° reactants
= -394.4 -2 (237.2) + 166.2
= −702.6 kJ mol-1
now Efficiency of fuel cell = ∆G/∆H×100
= (702.6/726) x100
= 97%
In the reaction,
2Al(s) + 6HCl(aq) → 2Al3+ (aq) + 6Cl¯(aq) + 3H2(g)
6L HCl(aq) is consumed for every 3L H2(g) produced
33.6 L H2(g) is produced regardless of temperature and pressure for every mole Al that reacts
67.2 L H2(g) at STP is produced for every mole Al that reacts
11.2 L H2(g) at STP is produced for every mole HCl (aq) consumed
D.
11.2 L H2(g) at STP is produced for every mole HCl (aq) consumed
2Al(s) + 6HCl(aq) → 2Al3+ (aq) + 6Cl¯(aq) + 3H2(g)
For each mole of HCl reacted, 0.5 mole of H2 gas is formed at STP.
1 mole of an ideal gas occupies 22.4 lit at STP.
Volume of H2 gas formed at STP per mole of HCl reacted is 22.4 × 0.5 litre
Identify the correct statement regarding a spontaneous process –
For a spontaneous process in an isolated system, the change in entropy is positive
Endothermic processes are never spontaneous
Exothermic processes are always spontaneous
Lowering of energy in the reaction process is the only criterion for spontaneity
A.
For a spontaneous process in an isolated system, the change in entropy is positive
In the conversion of limestone to lime, CaCO3(s) → CaO(s) + CO2(g) the values of ∆Hº and ∆Sº are + 179.1 kJ mol–1 and 160.2 J/K respectively at 298K and 1 bar. Assuming that ∆Hº and ∆Sº do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is
1008 K
1200 K
845 K
1118 K
D.
1118 K
We know, ∆G = ∆H-T∆S
So, lets find the equilibrium temperature, i.e. at which
∆G = 0
∆H-T∆S
T= 179.1 x 1000/160.2
= 1118 K
So, at the temperature above this, the reaction becomes will spontaneous.
The standard enthalpy of formation (∆fHo) at 298 K for methane, CH4(g), is –74.8 kJ mol–1. The additional information required to determine the average energy for C – H bond formation would be
the dissociation energy of H2 and enthalpy of sublimation of carbon
latent heat of vapourization of methane
the first four ionization energies of carbon and electron gain enthalpy of hydrogen
the dissociation energy of hydrogen molecule, H2
A.
the dissociation energy of H2 and enthalpy of sublimation of carbon
(∆H −∆U) for the formation of carbon monoxide (CO) from its elements at 298 K is
(R = 8.314 J K–1 mol–1)
–1238.78 J mol–1
1238.78 J mol–1
–2477.57 J mol–1
2477.57 J mol–1
A.
–1238.78 J mol–1
∆H −∆U =∆ngRT
= (-1 x 8.314 x 298)/2
= - 1238.78
Consider the reaction: N2 +3H2 → 2NH3 carried out at constant temperature and pressure. If ∆H and ∆U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?
∆H = 0
∆H = ∆U
∆H < ∆U
∆H >∆U
C.
∆H < ∆U
∆H = ∆U + ∆nRT
∆n = -2
∆H = ∆U - 2RT
∆H < ∆U
Which one of the following statements is NOT true about the effect of an increase in temperature on the distribution of molecular speeds in a gas?
The most probable speed increases
The fraction of the molecules with the most probable speed increases
The distribution becomes broader
The area under the distribution curve remains the same as under the lower temperature
B.
The fraction of the molecules with the most probable speed increases
Most probable velocity increase and a fraction of a molecule possessing most probable velocity decreases.
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ∆f H for the formation of XY is -200 kJ mole-1. The bond dissociation energy of X2 will be
100 kJ mol-1
800 kJ mol-1
300 kJ mol-1
400 kJ mol-1
B.
800 kJ mol-1
XY→ X(g) + Y(g) ; ∆H +a kJ/ mole ............(i)
X2 → 2X; ∆H = +a kJ/mole........(ii)
Y2 → 2Y; ∆H =+0.5a kJ/mole.......(iii)
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide per mole is
110.5 kJ
-110.5 kJ
-676.5 kJ
676.5 kJ
B.
-110.5 kJ
Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?
A and D
A and B
B and C
C and D
B.
A and B
Therefore ln K vs 1/T the graph will be a straight line with slope equal to .Since reaction is
exothermic, therefore itself will be negative resulting in positive slope.
The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 250C; the heat of combustion (in kJ mol–1) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1)
–3267.6
4152.6
–452.46
3260
A.
–3267.6
Δng = 6 - 7.5
= -1.5 (change in gaseous mole)
ΔU or ΔE = - 3263.9 kJ
ΔH = ΔU + ΔngRT
Δng = - 1.5
R = 8.314 JK-1 mol-1
T = 298 K
So ΔH = -3263.9 + (-1.5) 8.314 x 10-3 x 298
= -3267.6 kJ
ΔH = Heat at constant pressure
ΔU/ΔE = Heat at constant volume
R = gas constant
Which of the following statements is correct for a reversible process in a state of equilibrium?
ΔG = 2.303RT log K
ΔG0 = -2.303RT log K
ΔG0 = 2.303RT log K
C.
ΔG0 = -2.303RT log K
ΔG =ΔG0+2.303RT log K log Q
At equilibrium when Δ G = 0 and Q = K
then ΔG = ΔG0 +2.303 RT log K = 0
ΔG0 = -2.303 RT log K
A reaction having equal energies of activation for forward and reverse reactions has
C.
ΔH = 0Energy profile diagram for are reaction is as from the figure it is clear that
(Ea)b = (Ea)f +ΔH
[Here (Ea)b = activation energy of backward reaction and (Ea)f = activation energy of forward reaction].
If (Ea)b = (Ea)b = (Ea)f
then ΔH = 0
In which of the following reactions, standard reaction entropy changes (ΔSo) is positive and standard Gibb's energy change (ΔGo) decreases sharply with increasing temperature?
C (graphite) +1/2 O2 (g) → CO (g)
CO (g) +1/2 (g) → CO2 (g)
Mg(s) +1/2O2 (g) → MgO (s)
1/2 C (graphite) +1/2 O2 (g) → 1/2 CO2 (g)
A.
C (graphite) +1/2 O2 (g) → CO (g)
Among the given reactions only in te case of
C (graphite) +1/2 O2 (g) → CO (g)
entropy increases because randomness (disorder) increases. Thus, standard entropy change (ΔSo) is positive.
Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, ie. ΔHo=-ve
We know that
ΔGo = ΔHo-TΔSo
ΔGo = -ve-T(+ve)
Thus, as the temperature increases, the value of ΔGo decreases.
The enthalpy of fusion of water is 1.435 Kcal/mol. The molar entropy change for the melting of ice of at 0o C is
10.52 cal/(mol K)
21.04 cal/(mol K)
5.260 cal/ (mol K)
0.526 cal / (mol K)
C.
5.260 cal/ (mol K)
Molar entropy change for the melting of ice,
Which of the following is the correct option for free expansion of an ideal gas under an adiabatic condition?
B.
For an adiabatic process, q = 0 and for free expansion, W = 0,
Therefore ΔT = 0.
Enthalpy change for the reaction,
4H (g) → 2H2 (g) is - 869.6 kJ
The dissociation energy of H - H bond is
-869.6 kJ
+434.8 kJ
+217.4 kJ
-434.8 kJ
B.
+434.8 kJ
4H (g) → 2H2 (g); ΔH = - 869.6 kJ
2H2 (g) → 4 H (g); ΔH = 869.6 kJ
H2 (g) → 2 H (g); ΔH = 869.6/2 = 434.8 kJ
The formation of the oxide ion O2- (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below,
Thus, the process of formation of O2- in the gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that
electron repulsion outweighs the stability gained by achieving a noble gas configuration
O- ion has comparatively smaller size that oxygen atom
oxygen is more electronegative
the addition of electron in oxygen result in the large size of the ion
A.
electron repulsion outweighs the stability gained by achieving a noble gas configuration
Electron repulsion predominates over the stability gained by achieving noble gas configuration. Hence, the formation of O2- in the gas phase is unfavourable.
Three moles of an ideal gas expanded spontaneously into the vacuum. The work done will be
infinte
3 J
9 J
zero
D.
zero
W = pext. ΔV
For vaccum,
ptext = 0
therefore,
W = 0 ΔV = 0
Match List I (equation) with List II (types of the process) and select the correct option.
|
List I (Equations)
|
List II (Types of process)
|
||
|
A |
KP > Q |
1 |
Non-spontaneous |
|
B |
ΔGo < RT In Q |
2 |
Equilibrium |
|
C |
KP = Q |
3 |
Spontaneous and endothermic |
|
D |
T> ΔH/ ΔS |
4 |
Spontaneous |
|
A
|
B
|
C
|
D
|
|
1
|
2
|
3
|
4
|
| A | B | C | D |
| 3 | 4 | 2 | 1 |
| A | B | C | D |
| 4 | 1 | 2 | 3 |
| A | B | C | D |
| 2 | 1 | 4 | 3 |
C.
| A | B | C | D |
| 4 | 1 | 2 | 3 |
A) When KP > Q, the reaction goes in the forward direction, ie, the reaction is spontaneous.
B) Given ΔGo < RT lnQ, thus, ΔGo = + ve and hence, the reaction is non-spontaneous.
C) At equilibrium, Kp = Q
The values of ΔH and ΔS for the reaction, C(graphite) + CO2(g) → 2 CO (g) are 170 kJ and 170 JK-1 respectively. This reaction will be spontaneous at
710 K
910 K
1110 K
510 K
C.
1110 K
For spontaneous process, ΔG < 0
Which of the following are not state functions?
I) q + W
II) q
III) W
IV) H-TS
(I) and (IV)
(II) (III) and (IV)
(I), (II) and (III)
(II) and (III)
D.
(II) and (III)
A state function is the property of the system whose value depends only on the initial and final state of the system and is independent of the path.
Therefore,
Internal energy (ΔE) = q +W
It is a state function because it is independent of the path. It is an extensive property.
∴ Gibbs energy (G) = H -TS
It is also a state function because it is independent of the path. It is also extensive property.
Heat (q) and work (W) are not state functions being path dependent.
For the gas phase, reaction,
PCl5 (g) ⇌ PCl3 (g) Cl2 (g)
Which of the following conditions are correct?
Δ H = 0 and ΔS > 0
ΔH >0 and ΔS > 0
Δ H < 0 and ΔS < 0
ΔH > 0 and ΔS < 0
B.
ΔH >0 and ΔS > 0
ΔH = ΔE + ΔnRT
Δn = number of moles of product - number of moles of reactants
ΔG = ΔH - TΔS
For a spontaneous process, ΔG must be negative
PCl5 (g) ⇌ PCl3 (g) Cl2 (g)
In this reaction
Δn = 2-1 = 1
Thus, ΔH is positive, ie, >0
If ΔH is positive, then to maintain the value of ΔG negative, ΔS should be positive , ie, ΔS>0.
Consider the following reactions:
Enthalpy of formation of H2O (l) is:
- x2 kJ mol-1
+ x3 kJ mol-1
- x4 kJ mol-1
+ x1 kJ mol-1
A.
- x2 kJ mol-1
Enthalpy of formation: The amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent elements is known as heat of formation. SO, the correct answer is:
Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE?
H2 (g) + Br2 (g) →2HBr (g)
C (s) + 2 H2O (g) → 2 H2 (g) + CO2 (g)
PCl5 (g) →PCl3 (g) + Cl2 (g)
2CO (g) + O2 (g) → 2 CO2 (g)
A.
H2 (g) + Br2 (g) →2HBr (g)
As we know that
ΔH = ΔE + PΔV
ΔH = ΔE +ΔnRT ..(1)
where ΔH → change in enthalpy of the system (standard heat at constant pressure)
Δ E → change in internal energy of system (Standard heat at constant volume)
Δn → no. of gaseous moles of product - no. of gaseous moles of reactant
R → gas constant
T → absolute temperature
If Δ n = 0 for reactions which is carried out in an open container, therefore, Δn = 0 for reactions which are carried out in an open container, therefore, ΔH =ΔE
so for reaction (1) Δn = 2-2 = 0
Hence, for reaction (1) , ΔH =ΔE
The enthalpy of combustion of H2, cyclohexene (C6H10) and cyclohexene (C6H12) are -241, -3800 and -3920 kJ per mol respectively.The heat of hydrogenation of cyclohexane is:
-212 kJ mol
+121 kJ mol
+242 kJ per mol
-242 kJ per mol
A.
-212 kJ mol
ΔH= [ΔH of combustion of cyclohexane -(ΔH of combustion of cyclohexene +ΔH of combustion of H2)]
= -[-3920 -(3800-24)] kJ
= - [3920 + 4041] kJ
=-[121] kJ
=--121 kJ
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be
1136.5 J
-500 J
-505 J
+505 J
C.
-505 J
ΔU = q + w
For adiabatic process, q = 0
∴ ΔU = w
= – P·ΔV
= –2.5 atm × (4.5 – 2.5) L
= –2.5 × 2 L-atm
= –5 × 101.3 J
= –506.5 J
= –505 J
For a given reaction, ΔH = 35.5 kJ mol–1 and ΔS = 83.6 JK–1 mol–1. The reaction is spontaneous at (Assume that ΔH and ΔS do not vary with temperature)
T < 425 K
T > 425 K
All temperatures
T > 298 K
B.
T > 425 K
∵ ΔG = ΔH – TΔS
For a reaction to be spontaneous, ΔG = –ve
i.e., ΔH < TΔS
therefore, T> ΔH/ΔS
= 35.5 x 103J/83.6JK-1
i.e T>425 K
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
Then the species undergoing disproportionation is
HBrO
Br2
C.
HBrO
Calculate corresponding to each compound undergoing disproportionation reaction. The reaction for which comes out +ve is spontaneous.
Which one of the following conditions will favour maximum formation of the product in the reaction,
Low temperature and high pressure
Low temperature and low pressure
High temperature and low pressure
High temperature and high pressure
A.
Low temperature and high pressure
On increasing pressure equilibrium shift in a direction where number of moles decreases i.e. forward direction.
On decreasing temperature, equilibrium shift n exothermic direction i.e., forward direction.
So high pressure and lower temperature favour maximum formation of product.
The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be
200 kJ mol-1
100 kJ mol-1
400 kJ mol-1
800 kJ mol-1
D.
800 kJ mol-1
Let B.E of x2,y2 and xy are x kJ mol-1, 0.5 x kJ mol-1 and x kJ mol-1 respectively
The correction factor ‘a’ to the ideal gas equation corresponds to
Density of the gas molecules
Volume of the gas molecules
Forces of attraction between the gas molecules
Electric field present between the gas molecules
C.
Forces of attraction between the gas molecules
In the real gas equation, van der Waal's constant, 'a' signifies intermolecular forces of attraction.
The energy released when 6 moles of octane is burnt in air will be [Given, ΔHf for CO2 (g). H2O(g) and C8H18 (l), respectively are -490, -240 and +160J/mol]
-37.4 kJ
-20 kJ
-6.2 kJ
-35.5 kJ
D.
-35.5 kJ
On applying, 8 x Eq. (i) + 9x Eq. (ii) -Eq. (iii), we get
Hence, energy exchange when 6 moles of octane is burnt in air = - 5920 x 6 = -35520 J = -35.5 kJ
Sponsor Area
Sponsor Area
at 
,
for:
if the heats of formation of 
respectively.
at 298K, given that the enthalpies of formation at 298 K for 
for the reaction:
N2O(g) + 3CO2(g)
of tge reaction
Calculate its melting point.
for conversion of oxygen to ozone, 
If 
for this conversation is 
in 
