Q13 / 90 · KVPY 2017 (19 Nov SB/SX) · Chemistry
The maximum work (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) that can be derived from complete combustion of $1 \mathrm{~mol}$ of $\mathrm{CO}$ at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ is
[Standard enthalpy of combustion of $\mathrm{CO}=-283.0 \mathrm{~kJ} \mathrm{~mol}^{-1} ;$ standard molar entropies at $298 \mathrm{~K} ; \mathrm{S}_{\mathrm{O} 2}=205.1 \mathrm{~J} \mathrm{~mol}^{-1}$ $\left.\mathrm{S}_{\mathrm{CO}}=197.7 \mathrm{~J} \mathrm{~mol}^{-1}, \mathrm{~S}_{\mathrm{CO} 2}=213.7 \mathrm{~J} \mathrm{~mol}^{-1}\right]$
- Correct answerA257
- B227
- C57
- D127
Solution
$\mathrm{Co}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) \Delta \mathrm{N}=-283.5$
$\Delta \mathrm{S}=213.7-197.7-\frac{20501}{2}=-86.5$
$\Delta \mathrm{G}=-283-\frac{298 \times(-86.5)}{1000}=-2570 \mathrm{~kJ}$
$\mathrm{~W}_{\max }=-\Delta \mathrm{G}=-(-257 \mathrm{~kJ})=257 \mathrm{~kJ}$
Hence, the correct answer is option: 'A'.