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Calculate the magnitude of standard free...

Calculate the magnitude of standard free energy of formation of ammonium chloride at `25^(@)C` (approximate integer in Kcal`mol^(-1)`), the equation showing the formation of `NH_(4)CI` from its elements is
`1/2NH_(4)(g) + 2H_(2)(g) + 1/2CI_(2)(g) rarr NH_(4)CI(s)`
For `NH_(4)CI, DeltaH_(f)^(@)` is `-313 KJ mol^(01)`, Also given that
`" "S_(N_(2))^(@) = 191.5 JK^(-1) mol^(-1)" "S_(N_(2))^(@) = 130.6 JK^(-1) mol^(-1)`
`" "S_(CI_(2))^(@) = 223.0 JK^(-1) mol^(-1) " "S_(NH_(4)CI)^(@) = 94.6 JK^(-1)mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
48
For the formation reaction above
`DeltaS_(f)^(@) = S_(NH_(4)CI)^(@) -((1)/(2)S_(N_(2))^(@) + 2S_(H_(2))^(@) + (1)/(2)S_("CI_(2))^(@))`
`=94.6-[1/2(191.5) + 2(130.6) + 1/2(223.0)]`
`=-373.85 JK^(-1) mol^(-1)`
Now that we have `DeltaH` and `DeltaS` , we can find `DeltaG`,
`DeltaG_(f)^(@) = DeltaH_(f)^(@) - TDeltaS_(f)^(@)`
`=-313 xx 10^(3) J mol^(-1) - (298 K) xx (-373.85 Jk^(-1) mol^(-1))`
`=-201.6 xx 10^(3) Jmol^(-1) "or" -201.6 KJ mol^(-1)`
`=-48 Kcal mol^(-1).`
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