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Question
Given that:
S∘H2=131 JK−1mol−1S∘Cl2=223 JK−1mol−1and S∘HCl=187 JK−1mol−1
The standard entropy change in formation of 1 mole of HCl(g) from H2(g) and Cl2(g) will be:
(Given, reaction for formation of HCl : H2+Cl2→2HCl)
Given that:
S∘H2=131 JK−1mol−1S∘Cl2=223 JK−1mol−1and S∘HCl=187 JK−1mol−1
The standard entropy change in formation of 1 mole of HCl(g) from H2(g) and Cl2(g) will be:
(Given, reaction for formation of HCl : H2+Cl2→2HCl)
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Solution
The correct option is B 10 JK−1
Balanced chemical equation for formation of one mole of HCl is :
12H2(g)+12Cl2(g)→HCl(g)ΔS∘=ΣS∘(Product)−ΣS∘(reactants) =S∘HCl−[12S∘H2+12S∘Cl2] =187−12[131+223]=10 JK−1
Balanced chemical equation for formation of one mole of HCl is :
12H2(g)+12Cl2(g)→HCl(g)ΔS∘=ΣS∘(Product)−ΣS∘(reactants) =S∘HCl−[12S∘H2+12S∘Cl2] =187−12[131+223]=10 JK−1
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