Question

The maximum work ( in $kJ\, mol ^{-1}$ ) that can be derived from complete combustion of 1 mole of $CO$ at $298\, K$ and 1 atm is. [ Standard enthalpy of combustion of $CO =-283.0\, kJ\, mol ^{-1}$; standard molar entropies at $298\, K : S _{ O _{2}}=205.1\, J\, mol ^{-1},\, S _{ CO }=197.7\, J\, mol ^{-1} S _{ CO _{2}}=213.7\, J\, mol^{-1}$]

• A. 257
• B. 227
• C. 57
• D. 127

Answer 1

Answer: (A)

For the reaction,
$CO (g)+\frac{1}{2} O _{2}(g) \longrightarrow CO _{2}(g)\Delta H_{C}=-283.5$
$\Delta S =\Sigma S_{P}-\Sigma S_{R}$
$=213.7-\left(197.7+\frac{1}{2} \times 205.1\right)$
$=-86.5\, J\, mol ^{-1}$
$\Delta G =\Delta H-T \Delta S$
$=-283-298 \times-86.5 \times 10^{-3}$
$=-257.0\, kJ\, mol ^{-1}$
$w_{\max }=-\Delta G=-(-2570.0)=257\, kJ\, mol ^{-1}$