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Calculate the dissociation constant of N...

Calculate the dissociation constant of `NH_(4)OH` at `298k`, if `DeltaH^(Theta)` and `DeltaS^(Theta)` for the given changes are as follows:-
`NH_(3) + H^(o+) hArr overset(o+)NH_(4)`,
`DeltaH^(Theta) =- 52.2 KJ mol^(-1), DeltaS^(Theta) = 1.67 J K^(-1)mol^(-1)`
`H_(2)O hArr H^(o+) + overset(Theta)OH, DeltaH^(Theta) = 56.6 kJ mol^(-1)`.
`DeltaS^(Theta) =- 76.53 JK^(-1)mol^(-1)`

Text Solution

Verified by Experts

`NH_(3)+H^(o)+ hArr NH_(4)^(o+) , DeltaH^(Theta) = - 52.5`
Adding, `H_(2)O hArr H^(o+) + overset(Theta)OH, DeltaH^(Theta) = 56.6`
`ulbar(NH_(3)+H_(2)O hArr NH_(4)^(o+) + overset(Theta)OH, DeltaH^(Theta) = 4.4 kJ mol^(-1))`
Similarly, `DeltaS^(@)` for the change `=- 76.53 JK^(-1) mol^(-1)`
or for the change
`NH_(4)OH hArr NH_(4)^(o+) + overset(Theta)OH, DeltaH^(Theta) = 4.4 kJ mol^(-1)`
and `DeltaS^(Theta) =- 76.53 JK^(-1) mol^(-1)`
Now we have `DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`:. DeltaG^(Theta) = 4.4 - (-76.53 xx 10^(-3)) xx 298 = 27.2 KJ mol^(-1)`
Also, `DeltaG^(Theta) =- 2.303 RT log K_(b)`
`27.21 =- 2.303 xx 8.314 xx 10^(-3) xx 298 log K_(b)`
`:. K_(b) = 1.7 xx 10^(-5)`
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