Given that: S∘H2=131JK−1mol−1S∘Cl2=223JK−1mol−1andS∘HCl=187JK−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)
A
20JK−1
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B
10JK−1
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C
187JK−1
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D
374JK−1
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Solution
The correct option is B10JK−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]=10JK−1