Question

Calculate the entropy change for $ C{{H}_{4}}(g)+{{H}_{2}}O(g)\xrightarrow{{}}3{{H}_{2}}(g)+CO(g) $ ,using the following data:

Substance	$CH_4(g)$	$H_2O(g)$	$H_2(g)$	$CO(g)$
$S^{\circ}/JK^{-1} mol^{-1}$	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)

$\Delta S^{\circ} _{\text{reaction}} = \Sigma\Delta S^{\circ} _{\text{Products}} - \Delta S^{\circ}_{\text{reactants}}$
Thus, entropy change for,
$CH _{4}(g)+ H _{2} O (g) \longrightarrow 3 H _{2}(g)+ CO (g)$
$\Delta S_{\text {reaction }}^{\circ}=\left[3 \times S_{ H _{2}(g)}^{\circ}+S_{ CO (g)}^{\circ}\right]$
$-\left[S_{ CH _{4(g)}}^{\circ}+S_{ H _{2} O (g)}^{\circ}\right]$
$=[3 \times 130.6+197.6]-[186.2+188.7]$
$=589.4-374.9$
$=214.5\, JK ^{-1} \,mol ^{-1}$