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Question

Given that:
SH2=131 JK1mol1SCl2=223 JK1mol1and SHCl=187 JK1mol1
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+Cl22HCl)

A
20 JK1
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B
10 JK1
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C
187 JK1
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D
374 JK1
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Solution

The correct option is B 10 JK1
Balanced chemical equation for formation of one mole of HCl is :
12H2(g)+12Cl2(g)HCl(g)ΔS=ΣS(Product)ΣS(reactants) =SHCl[12SH2+12SCl2] =18712[131+223]=10 JK1

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