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NEET chemistry

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Question
CBSEENCH11008436
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Identify the correct statement for the change of Gibbs energy for a system (ΔGsystem) at constant temperature and pressure

  • If ΔGsystem > 0, the process is spontaneous

  • If ΔGsystem =0, the system has attained equilibrium

  • If ΔGsystem < 0, the system is still moving in a particular direction

  • If ΔGsystem < 0, the process is not spontaneous

Solution

B.

If ΔGsystem =0, the system has attained equilibrium

If the Gibbs free energy for a system (ΔGsystem) is equal to zero, then system is present in equilibrium at a constant temperature and pressure.

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Question
CBSEENCH11008437
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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)

Solution

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 

Question
CBSEENCH11008438
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Given: The mass of electron is 9.11 x 10-31 kg
Planck constant is 6.626 x 10-34 Js,
the uncertainty involved in the measurement of velocity within a distance of 0.1 A is:

  • 5.79 x 106 ms-1

  • 5.79 x 107 ms-1

  • 5.79 x 108 ms-1

  • 5.79 x 105 ms-1

Solution

A.

5.79 x 106 ms-1

By Heisenberg's uncertainty principle
increment straight p space. increment straight x space greater or equal than space fraction numerator straight h over denominator 2 straight pi end fraction or space increment straight v. increment straight x space greater or equal than fraction numerator straight h over denominator 4 πm end fraction increment straight p space rightwards arrow space uncertainty space in space momentum increment straight x space rightwards arrow space uncertainty space in space position increment straight v space rightwards arrow space uncertainty space in space velocity straight m rightwards arrow space mass space of space particle Given comma increment straight x space equals space 0.1 space straight A space equals space 0.1 space straight x space 10 to the power of negative 10 end exponent space straight m straight m equals space 9.11 space straight x space 10 to the power of negative 31 end exponent space kg straight h space equals space planck space constant space equals space 6.626 space straight x space 10 to the power of negative 34 end exponent space Js straight pi space equals space 3.14 In space uncertain space position space increment straight v. increment straight x space equals space fraction numerator straight h over denominator 4 πm end fraction increment straight v space straight x space 0.1 space straight x space 10 to the power of negative 10 end exponent space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent over denominator 4 space straight x space 3.14 space straight x space 9.11 space straight x space 10 to the power of negative 31 end exponent end fraction increment straight v space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent over denominator 4 space straight x space 3.14 space straight x space 911 space straight x space 10 to the power of negative 31 end exponent straight x space 0.1 space straight x space 10 to the power of negative 1 end exponent end fraction ms to the power of negative 1 end exponent equals space 5.785 space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent 5.79 space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent

Question
CBSEENCH11008439
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The enthalpy and entropy change for the reaction:

Br2 (l) + Cl2 (g)→ 2BrCl (g)

are 30 kJ mol-1 and 105 JK-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is:

  • 285.7 K 

  • 273 K

  • 450 K

  • 300 K

Solution

A.

285.7 K 

At equilibrium, Gibbs free energy change (ΔGo) is equal to zero. The following thermodynamic relation is used to show the relation of ΔGo with the enthalpy change (ΔHo) and entropy change (ΔSo)
ΔGo  = ΔHo - ΔSo
0 = 30 x 103 (J mol-1) - T x 105 (J K-1) mol-1
Therefore comma straight T space equals space fraction numerator 30 space straight x space 10 cubed over denominator 105 end fraction space straight K thin space equals space space 285.71 space straight K

Question
CBSEENCH11008441
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For the reaction,

CH4  (g) + 2 O2 (g) ⇌ CO2 (g) + 2H2O (l), 

ΔrH = - 170. 8 kJ mol-1

Which of the following statements is not true? 

  • At equilibrium, the concentrations of CO2 (g) and H2O (l) are not equal

  • The equilibrium constant for the reaction is given by Kp = fraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O subscript 2 right square bracket end fraction

  • Addition of CH4 (g) or O2 (g) at equilibrium will cause a shift to the right

  • The reaction is exothermic

Solution

B.

The equilibrium constant for the reaction is given by Kp = fraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O subscript 2 right square bracket end fraction

For the reaction,
CH4  (g) + 2 O2 (g) ⇌ CO2 (g) + 2H2O (l), 
ΔrH = - 170. 8 kJ mol-1
This equilibrium is an example of heterogeneous chemical equilibrium.Hence, for it
straight K subscript straight c space equals space fraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O right square bracket squared end fraction space... space left parenthesis straight i right parenthesis left parenthesis equilibrium space constant space on space the space basis space of space conc. right parenthesis and space straight K subscript straight p space equals space fraction numerator straight P subscript Co subscript 2 end subscript over denominator straight P subscript CH subscript 4 end subscript space straight x space straight P subscript straight O subscript 2 end subscript squared space end fraction space space space space.... space left parenthesis ii right parenthesis left parenthesis Equilibrium space constant space according space to space partial space pressure right parenthesis Thus space in space it space concentration space of space CO subscript 2 space left parenthesis straight g right parenthesis space and space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space are space not space equilibrium The space equilibrium space constant space left parenthesis straight K subscript straight p right parenthesis space equals space fraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O subscript 2 right square bracket space end fraction
not correct expression.
In addition of CH4 (g) or O2 (g) at equilibrium Kc = will be decreased according to expression (i) but Kc remains constant at constant  at constant temperature fro a reaction, so for maintaining the constant value of Kc, the concentration of CO2 will increase in same order. Hence, on the addition of CH4 or O2 equilibrium will cause to the right.
This reaction is an example of an exothermic reaction.

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