InterviewSolution
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InterviewSolution
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Calculate `Delta_(r)G^(@)` for `(NH_(4)Cl,s)` at 310K. Given :`Delta_(r)H^(@)(NH_(4)Cl,s)` =-314 kj/mol,`Delta_(r)C_(p)=0` `S_(N_(2)(g))^(@)=192 JK^(-1mol^(-1)),S_(H_(2)(g))^(@)=130.5JK^(-1)mol^(-1),` `S_(Cl_(2)(g))^(@)=233JK mol^(-1), S_(NH_(4)Cl(s))^(@)=99.5JK^(-1)mol^(-1)` All given data at 300KA. `-198.56 KJ//"mol"`B. `-426.7 KJ//"mol"`C. `-202.3 KJ//"mol"`D. None of these |
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Answer» Correct Answer - A `Delta_(f)S^(@) (NH_(4)Cl,s) "at" 300k` `= S_(NH_(4)Cl(s))^(@) - [(1)/(2) S_(N_(2))^(@) + 2S_(H_(2))^(@) + (1)/(2)S_(Cl_(2))^(@)] =- 374 Jk^(-1) mol^(-1)` `Delta_(f)C_(P) =0` `therefore Delta_(f)S_(310)^(@)=Delta_(f)S_(300)^(@)` `=-374JK^(-1) mol^(-1)` `Delta_(f)H_(300)^(@) =Delta_(f)H_(300)^(@) =- 314.5` `Delta_(f)G_(310)^(@) = Delta_(f)H^(@) - 310 DeltaS^(@) =- (310(-374))/(1000) =- 198.56 KJ//"mol"` . |
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