Thermodynamics

Sponsor Area

Question
CBSEENCH11008456

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

Solution

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

Sponsor Area

Question
CBSEENCH11008262

A reaction having equal energies of activation for forward and reverse reactions has

  • ΔS =0
  • ΔG =0
  • ΔH = 0
  • ΔH = ΔG=ΔS = 0

Solution

C.

ΔH = 0

Energy 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

Question
CBSEENCH11008437

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
CBSEENCH11008339

Consider the following processes Δ H (kJ/mol)

1/2 A →                               +150
3B   → 2 C + D                     -125
E + A  → 2D                        +350

For  B + D   → E + 2C, ΔH will be 

  • 525 kJ/mol

  • -175 kJ/mol

  • -325 kJ /mol

  • 325 kJ/mol

Solution

B.

-175 kJ/mol

1 half space straight A space space rightwards arrow space straight B semicolon space space space space space space space space space space space space space space space space space space space space space space increment space equals space 150 space space kJ divided by space mol space space.. space left parenthesis straight i right parenthesis

3 straight B space rightwards arrow space 2 space straight C space plus space straight D space semicolon space space space space space space space space space space space increment space straight H space equals space minus space 125 space kJ divided by mol space... space left parenthesis ii right parenthesis

straight E space plus straight A space rightwards arrow space 2 straight D semicolon space space space space space space space space space space space space space space space space increment space straight H space equals space plus 350 space kJ divided by mol space.... space left parenthesis iii right parenthesis
________________________________________________
By space left square bracket space 2 space straight x space left parenthesis straight i right parenthesis space plus space left parenthesis ii right parenthesis space minus space left parenthesis iii right parenthesis comma space we space have


space straight B space plus space straight D space space rightwards arrow space straight E space plus space thin space 2 straight C
therefore space increment straight H space equals space 150 space straight x space 2 space space plus space left parenthesis negative 125 right parenthesis space minus space 350
equals space minus 175 space kJ divided by mol

Question
CBSEENCH11008430

Consider the following reactions:

straight i right parenthesis space straight H to the power of plus space left parenthesis straight g right parenthesis thin space space plus space OH to the power of minus space left parenthesis aq right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus space straight x subscript 1 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
ii right parenthesis space straight H subscript 2 space left parenthesis straight g right parenthesis thin space space plus space straight O subscript 2 space left parenthesis straight q right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus space straight x subscript 2 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
iii right parenthesis space CO subscript 2 space left parenthesis straight g right parenthesis thin space space plus space straight H subscript 2 space left parenthesis straight g right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space plus space CO subscript 2 space left parenthesis straight g right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus space straight x subscript 3 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
iv right parenthesis straight C subscript 2 straight H subscript 2 space left parenthesis straight g right parenthesis thin space space plus 5 over 2 space straight O subscript 2 space left parenthesis straight g right parenthesis space equals space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space plus 2 CO subscript 2 space left parenthesis straight g right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space plus space straight x subscript 4 space kJ space mol to the power of blank to the power of negative 1 end exponent end exponent
Enthalpy of formation of H2O (l) is:

  • - x2 kJ mol-1

  • + x3 kJ mol-1

  • - x4 kJ mol-1

  • + x1 kJ mol-1

Solution

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:
straight H subscript 2 space left parenthesis straight g right parenthesis space space plus space 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis thin space rightwards arrow space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space semicolon increment straight H equals negative straight x subscript 2 space kJ space mol to the power of negative 1 end exponent