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The entropy values $({in JK}^{-} {mol}^{-}) of H_{2} (g)$ $= 130.6; {Cl}_{2} (g) = 223.0 . HCl (g) = 186.7 at 298 K$ and 1 $atm$ pressure. Then entropy change for the reaction. $H_{2} (g) + {Cl}_{2} (g) ⟶ 2 HCl (g)$ is.

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$+ 540.3$

$- 106.9$

$+ 19.8$

$+ 727.3$

answer is D.

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Detailed Solution

$∆ S^{o} = 2 S_{HCl}^{o} - (S_{H_{2}}^{o} + S_{{Cl}_{2}}^{o}) = 2 \times 186.7 - (130.6 + 223.0) = 19.8 J K^{-} {mol}^{-}$

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