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Question

$$S^O_{H_{2(g)}}=130.6$$J $$K^{-1}$$ $$mol6{-1}$$; $$S^O_{H_2O_{(l)}}=69.9$$J $$K^{-1}mol^{-1}$$ $$S^O_{O_{2(g)}}=205$$ J$$K^{-1}$$$$mol^{-1}$$, then the absolute entropy change of $$H_{2_{(g)}}+\dfrac{1}{2}O_{2(g)}\rightarrow H_2O_{(l)}$$ is?


A
163.2J mol1K1
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B
+163.2J mol1K1
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C
33J mol1K1
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D
+303J mol1K1
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Solution

The correct option is D $$-163.2$$J $$mol^{-1}K^{-1}$$
$$ H_{2} $$ + $$ \dfrac{1}{2} O_{2} $$ -------> $$ H_{2}O $$
Entropy Change of reaction = Entropy of products - Entropy of reactants
$$ \Delta S^{o}  =  \Delta S^{o}_{H_{2}O}  -  \dfrac{1}{2} \times \Delta S^{o}_{O_{2}}  -  \Delta S^{o}_{H_{2}} $$
$$ \Delta S^{o}  =  69.9 JK^{-1}mol^{-1}  -  \dfrac{1}{2} \times 205 JK^{-1}mol^{-1}  - 130.6 JK^{-1}mol^{-1} $$
$$ \Delta S^{o} = -163.2 Jmol^{-1}K^{-1} $$

Chemistry

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