Calculate `Delta G` (in kJ) for the reaction at 300 K, `H_(2)(g)+Cl_(2)(g) rarr 2HCl (g)` Given at 300 K, `BE_(H-H) = 435 kJ mol^(-1), BE_(Cl-Cl) = 240 kJ mol^(-1), BE_(HCl) = 430 kJ mol^(-1)` Entropies of `H_(2), Cl_(2)` and HCl are 131, 223 and `187 JK^(-1) mol^(-1)` respectively.A. 191B. 291C. `-191`D. None of these

Calculate `Delta G` (in kJ) for the reaction at 300 K, `H_(2)(g)+Cl_(2

1.	Calculate `Delta G` (in kJ) for the reaction at 300 K, `H_(2)(g)+Cl_(2)(g) rarr 2HCl (g)` Given at 300 K, `BE_(H-H) = 435 kJ mol^(-1), BE_(Cl-Cl) = 240 kJ mol^(-1), BE_(HCl) = 430 kJ mol^(-1)` Entropies of `H_(2), Cl_(2)` and HCl are 131, 223 and `187 JK^(-1) mol^(-1)` respectively.A. 191B. 291C. `-191`D. None of these
Answer» Correct Answer - C `DeltaH=(435+240-2xx430)=- 185" kJ mol"^(-1)` `DeltaS=(2xx187-(131+223))/(1000)=20/1000"kJ mol"^(-1)` `DeltaG=DeltaH-TDeltaS` `=-185-(300xx20)/(1000)=-185-6=-191" kJ mol"^(-1)`

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