Question

Calculate the entropy change for $C{H}_4(g)+H_{2}O(g)\to 3H_2(g)+CO(g)$ ,using the following data:

Substance	$C{H}_4(g)$	$H_{2}O(g)$	$H_2(g)$	$CO(g)$
$S^{\circ }/J{K}^{-1}mo{l}^{-1}$	186.2	188.7	130.6	197.6

The entropy change is :

• A. $-46J{K}^{-1}mo{l}^{-1}$
• B. $+46J{K}^{-1}mo{l}^{-1}$
• C. $-214.5J{K}^{-1}mo{l}^{-1}$
• D. $+214.5J{K}^{-1}mo{l}^{-1}$

Answer 1

Answer: (D)

$\Delta S_{\text{reaction}}^{\circ }=\Sigma \Delta S_{\text{Products}}^{\circ }-\Delta S_{\text{reactants}}^{\circ }$
Thus, entropy change for,
$C{H}_4(g)+H_{2}O(g)⟶3H_2(g)+CO(g)$
$\Delta S_{\text{reaction }}^{\circ }=\left[3\times S_{H_2(g)}^{\circ }+S_{CO(g)}^{\circ }\right]$
$-\left[S_{C{H}_{4(g)}}^{\circ }+S_{H_{2}O(g)}^{\circ }\right]$
$=[3\times 130.6+197.6]-[186.2+188.7]$
$=589.4-374.9$
$=214.5J{K}^{-1}mo{l}^{-1}$