Question

The maximum work (in $kJmo{l}^{-1})$ that can be derived from complete combustion of $1mol$ of $CO$ at $298K$ and $1atm$ is :
[Standard enthalpy of combustion of $CO=-283.0kJmo{l}^{-1}$; standard molar entropies at $298K:S_{O_2}=205.1Jmo{l}^{-1},S_{CO}=197.7Jmo{l}^{-1},S_{C{O}_2}=213.7Jmo{l}^{-1}]$

• A. $257$
• B. $227$
• C. $57$
• D. $127$

Answer 1

The correct option is A $257$

$CO+\frac{1}{2}{O}_2\to C{O}_2$

Now,
$\Delta S^0=S_{CO2}-[S_{CO}+\frac{1}{2}{S}_{O2}]$

$\Delta S^0=213.7-[197.7+\frac{1}{2}205.1]$

$\Delta S^0=-86.55J/mol/K$

$\Delta H^0=-283.0kJ/mol=-283.0kJ/mol\times 1000J/kJ=-283,000J/mol$

Again,

$\Delta G^0=\Delta H^0-T\Delta S^0$
$\Delta G^0=-283,000J/mol-[298K\times -86.55J/mol/K]$

$\Delta G^0=-257208J/mol$

$\Delta G^0=-257208J/mol\times \frac{1kJ}{1000J}$

$\Delta G^0=-257J/mol$

The maximum work $=\Delta G^0=-257J/mol$