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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

`-426.7 kJ//mol`

C

`2-2.3 kJ//mol`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(2) N_(2) + 2H_(2) + (1)/(2) Cl_(2) rarr NH_(4) Cl`
`Delta G_(gamma)^(0) = Delta H_(gamma)^(0) - T Delta S_(gamma)^(0)`
`Delta H_(310)^(0) = Delta H_(300)^(0) = - 314.5`
`Delta S_(gamma)^(0) = 99.5 - ((1)/(2) xx 192 + 2 xx 130.5 + (1)/(2) xx 233)`
`= - 374`
`Delta G^(0) = - 314.5 - (310 xx (-374))/(1000)`
`= - 198.56` Kj/mol
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