Calcualate `Delta_(f)G^(@) "for" (NH_(4)Cl,s)` at 310 K. Given : `Delta_(f)H^(@)(NH_(4)Cl,s)=-314.5 KJ//"mol," "Delta_(r)C_(p)=0` `S_(N_(2)(g))^(@)=19
Calcualate `Delta_(f)G^(@) "for" (NH_(4)Cl,s)` at 310 K.
Given : `Delta_(f)H^(@)(NH_(4)Cl,s)=-314.5 KJ//"mol," "Delta_(r)C_(p)=0`
`S_(N_(2)(g))^(@)=192 JK^(-1), " "S_(H_(2)(g))^(@)=130.5 JK^(-1)mol^(-1),`
`S_(Cl_(2))^(@)(g) =233 JK^(-1)"mol"^(-1)," "S^(@)NH_(4)Cl(s)=99.5 JK_(1)"mol"^(-1)`
All given data are at 300K.
A. `-198.56 KJ//"mol"`
B. `-426.7 KJ//"mol"`
C. `-202.3 KJ//"mol"`
D. None of these
1 Answers
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)^(@)`
`=-374 JK^(-1)mol^(-1)`
`Delta_(f)H+_(310)^(@)=Delta_(f)H_(300)^(@)=-314.5`
`Delta_(f)G_(310)^(@)=Delta_(f)H^(@)-310DeltaS^(@) =- 314.5-(310(-374))/(1000) =-198.56 KJ//"mol"` .