Question

Calculate $△G^{\ominus }$ for the following reaction:
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)$; $△H^{\ominus }$ = -282.84 kJ
Given,
$S^{\ominus }C{o}_2=213.8J$ $K^{-1}mo{l}^{-1}$, $S^{\ominus }CO\left(g\right)$ = 197.9 J $K^{-1}mo{l}^{-1}$,
$S^{\ominus }C{o}_2=205.0J$ $K^{-1}mo{l}^{-1}$

• A. -357 KJ
• B. -383 KJ
• C. -257 KJ
• D. + 357 KJ

Answer 1

The correct option is C

-257 KJ

How do we proceed in these kind of questions?
We can calculate $△S^{\ominus }$
$△S$ is a state function.
$\therefore △S$ = $△S$ final-$△S$ initial
$\Rightarrow$$△S$ = $△S$ products - $△S$ reactants and then apply $△G$ = $△H$ - $T△S$
because we know $△H$ = - 282.84 KJ and T = 298 K (room temp).
So, let's do it.
$△S^{\ominus }$ = $\sum △S^{\ominus }$ (products) - $△S^{\ominus }$ (reactants)
$=\left[S_{C{O}_2}^{\ominus }\right]-[S_{CO}^{\ominus }+\frac{1}{2}{S}_{O_2}^{\ominus }]$
$=213.8-[197.9+\frac{1}{2}205]=-86.6J{K}^{-1}$
According to Gibbs-Helmholtz equation,
$△G^{\ominus }=△H^{\ominus }-T△S^{\ominus }$
= $-282.84-298\times (-86.6\times {10}^{-3})$
= -282.84 + 25.87
= -257.033 kJ