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Calculate Delta(r)G^(0) for (NH(4)Cl, s)...

Calculate `Delta_(r)G^(0)` for `(NH_(4)Cl, s)` at 310K. Given: `Dleta _(f)H^(0)` for `(NH_(4)Cl, s) = - 314.5` KJ/mol, `Delta_(r ) C_(P) = 0 S_(N_(2)(g))^(0) = 192JK^(-1) mol^(-1), S_(H_(2) (g))^(0) = 130.5 JK^(-1) mol^(-1), S_(Cl_(2)(g))^(0) = 233JK^(-1) mol^(-1), S_(NH_(4)Cl(s))^(0) = 99.5 JK^(-1) mol^(-1)`. All given data at 300K

A

`-198.56kJ//mol`

B

`-426.7kJ//mol`

C

`-202.3kJ//mol`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta_(f)S^(@)(NH_(4)Cl,s)"at 300 K"`
`=S_(NH_(4)Cl(s))^(@)-[(1)/(2)S_(N_(2))^(@)+2S_(H_(2))^(@)+(1)/(2)S_(Cl_(2))^(@)]`
`=-374 " JK"^(-1)//"mol"^(-1)`
`because" "Delta_(r)C_(p)=0`
`therefore" "Delta_(f)S_(310)^(@)=Delta_(f)S_(300)^(@)`
`=-374JK^(-1)mol^(-1)`
`Delta_(f)G_(310)^(@)=Delta_(f)H_(300)^(@)=-314.5`
`Delta_(f)G_(310)^(@)=Delta_(f)H^(@)-310DeltaS^(@)`
`=-314.5-(310(-374))/(1000)`
`=-198.56kJ//mol`
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