Question

For the reaction,
$CO(g)+\frac{1}{2}{O}_2(g)⟶C{O}_2(g)$
If the standard entropy change at $298K$ is $-0.082kJmo{l}^{-1}{K}^{-1}$ and standard Gibb's energies of for formation of $C{O}_2$ and $CO$ are $-298.4$ and $-114.6kJ/mol$ respectively, then the enthalpy change for the given reaction will be

• A. $314.5kJ/mol$
• B. $159.8kJ/mol$
• C. $208.24kJ/mol$
• D. $418.3kJ/mol$

Answer 1

Answer: (C)

As we know that,
$\Delta G_{\text{(reaction) }}^{\circ }={\Sigma }_f{G}^{\circ }{}_{\text{(products) }}-{\Sigma }_f{G}^{\circ }{}_{(\text{reactants) }}$
$\therefore$ For the reaction,
$CO(g)+\frac{1}{2}{O}_2(g)⟶C{O}_2(g)$
$\Delta G_{(\text{reaction) }}^{\circ }=\Sigma G_f^{\circ }\left(C{O}_2\right)-\left[G_f^{\circ }(CO)+\frac{1}{2}{G}_f^{\circ }\left(O_2\right)\right]$
$\Rightarrow -298.4-[-114.6+0]$
$\left[\because G_f^{\circ }\left(C{O}_2\right)=0\right]$
$\Rightarrow -183.8kJ/mol$
Also, $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$
$\Rightarrow -183.8=\Delta H^{\circ }-[298\times (-0.082)]$
$\Rightarrow \Delta H^{\circ }=208.24kJ/mol$