Given the entropy of H_2 and HBr are 130.6 , | vedclass

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(A) The entropy change for the reaction is given by $\Delta S^{o} = \sum S^{o}_{products} - \sum S^{o}_{reactants}$.For the reaction $H_{2(g)} + Br_{2

Vedclass · Accepted Answer

$246.3$

Vedclass · Answer

(A) The entropy change for the reaction is given by $\Delta S^{o} = \sum S^{o}_{products} - \sum S^{o}_{reactants}$.For the reaction $H_{2(g)} + Br_{2(g)} ightarrow 2HBr_{(g)}$,the equation is $\Delta S^{o} = [2 	imes S^{o}(HBr)] - [S^{o}(H_2) + S^{o}(Br_2)]$.Substituting the given values: $20.1 = [2 	imes 198.5] - [130.6 + S^{o}(Br_2)]$.$20.1 = 397 - 130.6 - S^{o}(Br_2)$.$20.1 = 266.4 - S^{o}(Br_2)$.$S^{o}(Br_2) = 266.4 - 20.1 = 246.3 \, JK^{-1}$.

Given the entropy of $H_2$ and $HBr$ are $130.6 \, JK^{-1}$ and $198.5 \, JK^{-1}$ respectively,calculate the entropy of $Br_{2(g)}$ in the reaction $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ where $\Delta S^{o} = 20.1 \, JK^{-1}$.

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