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) = 24

Question

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) = 24

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. 191
B. 291
C. `-191`
D. None of these

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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)`

6 views Answered 2023-02-05 15:29