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.