If the standard molar entropies (S^o) for H_2 | vedclass

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Question

(A) The reaction is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$
The formula for the change in standard entropy is: $\Delta S^o = \sum S^o(	ext{pr

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$30$

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(A) The reaction is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$
The formula for the change in standard entropy is: $\Delta S^o = \sum S^o(	ext{products}) - \sum S^o(	ext{reactants})$
$\Delta S^o = [2 	imes S^o(HCl)] - [S^o(H_2) + S^o(Cl_2)]$
Substitute the given values: $\Delta S^o = [2 	imes 0.19] - [0.13 + 0.22]$
$\Delta S^o = 0.38 - 0.35 = 0.03 \, kJ \, K^{-1} \, mol^{-1}$
Convert to $J \, K^{-1} \, mol^{-1}$ by multiplying by $1000$: $0.03 	imes 1000 = 30 \, J \, K^{-1} \, mol^{-1}$.

If the standard molar entropies $(S^o)$ for $H_2$,$Cl_2$,and $HCl$ are $0.13$,$0.22$,and $0.19 \, kJ \, K^{-1} \, mol^{-1}$ respectively,what is the total change in standard entropy $(\Delta S^o)$ for the reaction $H_2 + Cl_2 \rightarrow 2HCl$ in $J \, K^{-1} \, mol^{-1}$?

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Given:
$\Delta S_A^o = 100 \ J \ mol^{-1} \ K^{-1}$
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$\Delta S_C^o = 200 \ J \ mol^{-1} \ K^{-1}$
$\Delta S_D^o = 150 \ J \ mol^{-1} \ K^{-1}$

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