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If $S^{0} for H_{2}, {Cl}_{2} and HCl$ are 0.13, 0.22 and 0.19 kJ $K^{- 1} {mol}^{- 1}$ respectively. The total change in standard entropy for the reaction $H_{2} + {Cl}_{2} \overset{}{\to} 2 HCl$ is

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$20 {JK}^{- 1} {mol}^{- 1}$

$30 {JK}^{- 1} {mol}^{- 1}$

$40 {JK}^{- 1} {mol}^{- 1}$

$60 {JK}^{- 1} {mol}^{- 1}$

answer is A.

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Detailed Solution

$∆ S = S_{P} - S_{R} = (2 \times 0.19) - 0.13 - 0.22 = 0.03 kJ K^{- 1} {mol}^{- 1} = 30 J K^{- 1} {mol}^{- 1}$

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