Question

The maximum work (in $kJ\,mo{{l}^{-1}}$) that can be derived from complete combustion
of 1 mol of CO at 298 K and 1 atm is [ Standard enthalpy of combustion of CO = $-283.0\,kJ\,mo{{l}^{-1}}$; standard molar entropies at 298 K : ${{S}_{{{O}_{2}}}}=205.1\,J\,mo{{l}^{-1}}$, ${{S}_{CO}}=197.7\,J\,mo{{l}^{-1}}$, ${{S}_{C{{O}_{2}}}}=213.7\,J\,mo{{l}^{-1}}$]
a) 257
b) 227
c) 57
d) 127

Answer 1

Hint : The Gibbs free energy of a system is given as the sum of its enthalpy (H) with the product of the temperature (T in Kelvin) and the entropy (S) of the system:
G = H - TS
Free energy of reaction
$\Delta G=\Delta H-T\Delta S$
Standard-state free energy of reaction
$\Delta {{G}^{\circ }}=\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}$

Complete answer : We have been given in the question that the:
Standard enthalpy of combustion of CO = $-283.0\,kJ\,mo{{l}^{-1}}$
Standard molar entropies at 298 K
${{S}_{{{O}_{2}}}}=205.1\,J\,mo{{l}^{-1}}$
${{S}_{CO}}=197.7\,J\,mo{{l}^{-1}}$
${{S}_{C{{O}_{2}}}}=213.7\,J\,mo{{l}^{-1}}$

Now the reaction for complete combustion of Carbon monoxide can be given by and the Standard enthalpy of the reaction as provided:
$CO(g)+\frac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)$ $\Delta {{H}^{\circ }}_{rxn}=-283.0\,kJ\,mo{{l}^{-1}}$
After that we will calculate the standard entropy change for the reaction for the reaction, which can be given as:
$\Delta {{S}^{\circ }}_{rxn}={{S}_{C{{O}_{2}}}}-\left[ {{S}_{CO}}+\frac{1}{2}({{S}_{{{O}_{2}}}}) \right]$
Putting the values in the equation above,
$\Delta {{S}^{\circ }}_{rxn}=213.7\,J\,mo{{l}^{-1}}-\left[ 197.7\,J\,mo{{l}^{-1}}+\frac{1}{2}(205.1\,J\,mo{{l}^{-1}}) \right]$
$\Delta {{S}^{\circ }}_{rxn}=-86.5\,J\,mo{{l}^{-1}}$

Now to determine the maximum work, $\Delta {{G}^{\circ }}_{rxn}$, for the reaction at 298 K we can write the Gibbs free energy as:
$\Delta {{G}^{\circ }}_{rxn}=\Delta {{H}^{\circ }}_{rxn}-T\Delta {{S}^{\circ }}_{rxn}$
Since, $\Delta {{H}^{\circ }}_{rxn}=-283.0\,kJ\,mo{{l}^{-1}}=-283.0\,kJ\,mo{{l}^{-1}}\times \frac{1000J}{kJ}=283,000J\,mo{{l}^{-1}}$
Temperature T = 298 K
$\Delta {{S}^{\circ }}_{rxn}=-86.5\,J\,mo{{l}^{-1}}$
Putting the values in equation,
$\Delta {{G}^{\circ }}_{rxn}=283,000J\,mo{{l}^{-1}}-(293\,K)(-86.5\,J\,mo{{l}^{-1}})$
$\Delta {{G}^{\circ }}_{rxn}=257,208\,J\,mo{{l}^{-1}}$
$\Delta {{G}^{\circ }}_{rxn}=257,208\,J\,mo{{l}^{-1}}\times \frac{1kJ}{1000J}$
$\Delta {{G}^{\circ }}_{rxn}=257.2\,kJ\,mo{{l}^{-1}}\approx 257\,kJ\,mo{{l}^{-1}}$
Therefore, the maximum work $\Delta {{G}^{\circ }}_{rxn}=257\,kJ\,mo{{l}^{-1}}$.

So, the correct option is (a).

Note : In this case, we can also determine whether the reaction is spontaneous, non-spontaneous or at equilibrium.

If a reaction is favourable for both enthalpy ($\Delta {{H}^{\circ }}$ < 0) and entropy ($\Delta {{S}^{\circ }}$> 0), then the reaction will be spontaneous ($\Delta {{G}^{\circ }}$< 0) at any temperature.
If a reaction is unfavourable for both enthalpy ($\Delta {{H}^{\circ }}$ > 0) and entropy ($\Delta {{S}^{\circ }}$< 0), then the reaction will be non-spontaneous ($\Delta {{G}^{\circ }}$> 0) at any temperature.
If a reaction is favourable for only one of either entropy or enthalpies, the standard-state free energy equation must be used to determine whether the reaction is spontaneous or not.