Thermodynamics

Sponsor Area

Question
CBSEENCH11008171

(∆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

Solution

A.

–1238.78 J mol–1

∆H −∆U =∆ngRT
= (-1 x 8.314 x 298)/2
= - 1238.78

Sponsor Area

Question
CBSEENCH11008066

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

Solution

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]

Question
CBSEENCH11008174

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

Solution

C.

∆H < ∆U

∆H = ∆U + ∆nRT
∆n = -2
∆H = ∆U - 2RT
∆H < ∆U

Question
CBSEENCH11008107

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

Solution

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

Question
CBSEENCH11008061

For the complete combustion of ethanol, C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2(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

Solution

A.

-1366.95 kJ mol-1

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2(l),
∆U = - 1364.47 kJ/mol
∆H = ∆U +∆ngRT
∆ng = -1
∆H = - 1364.47 +fraction numerator negative 1 space straight x space 8.314 space straight x space 298 over denominator 1000 end fraction
[Here, value of R in unit of J must be converted into kJ]
 = - 1364.47-2.4776
 = -1366.94 kJ/mol

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