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 300K

Options

A. `-198.56 KJ//"mol"`
B. `-426.7 KJ//"mol"`
C. `-202.3 KJ//"mol"`
D. None of these

Answer

A

Explanation

`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"` .