Question

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

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

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,
$CO (g)+\frac{1}{2} O _{2}(g) \longrightarrow CO _{2}(g)$
$\Delta G_{(\text {reaction) }}^{\circ}=\Sigma G_{f}^{\circ}\left( CO _{2}\right)-\left[G_{f}^{\circ}( CO )+\frac{1}{2} G_{f}^{\circ}\left( O _{2}\right)\right]$
$\Rightarrow-298.4-[-114.6+0]$
$\left[\because G_{f}^{\circ}\left( CO _{2}\right)=0\right]$
$\Rightarrow-183.8\, kJ / mol$
Also, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$
$\Rightarrow -183.8=\Delta H^{\circ}-[298 \times(-0.082)]$
$\Rightarrow \Delta H^{\circ}=208.24\, kJ / mol$