The entropy values (in J K^{-1} mol^{-1}) of | vedclass

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(D) The entropy change for the reaction is given by the formula: $\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}$For the re

Vedclass · Accepted Answer

$+19.8$

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(D) The entropy change for the reaction is given by the formula: $\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}$For the reaction $H_{2(g)} + Cl_{2(g)} 	o 2HCl_{(g)}$,the expression is: $\Delta S^{\circ} = 2 S^{\circ}_{HCl} - (S^{\circ}_{H_2} + S^{\circ}_{Cl_2})$Substituting the given values: $\Delta S^{\circ} = 2 	imes 186.7 - (130.6 + 223.0)$$\Delta S^{\circ} = 373.4 - 353.6 = 19.8 \ J K^{-1} mol^{-1}$

The entropy values (in $J K^{-1} mol^{-1}$) of $H_{2(g)} = 130.6$,$Cl_{2(g)} = 223.0$ and $HCl_{(g)} = 186.7$ at $298 \ K$ and $1 \ atm$ pressure. Then entropy change for the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$ is :-

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