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The entropy values (in JK^-1 mol^-1) of H_2(g)=130.6, Cl_2(g) = 223.0 and HCl _(g) = 186.7 at 298 K and 1atm pressure . Then entropy change for the reaction $H_{2 (g)} + {Cl}_{2 (g)} \to 2 {HCl}_{g}$

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answer is D.

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Detailed Solution

$\begin{array}{l} △ S^{0} = 2 S^{0} HCl - (S^{0} H_{2} + S^{0} {Cl}_{2}) \\ = 2 \times 186 .7 - (130 .6 + 223 .0) = 19 .8 {JK}^{- 1} {mol}^{- 1} \end{array}$

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