Question

The value of K for the reaction
`O_(3)(g)+OH(g) hArr H(g)+2O_(2)(g)`
Changed from `0.096` at `298 K` to `1.4` at `373 K`. Above what temperature will the reaction become thermodynamically spontaneous in the forward direction assuming that `DeltaH^(ɵ)` and `DeltaS^(ɵ)` values for the reaction do not change with change in temperature? Given that `DeltaS_(298)^(ɵ)=10.296 J K^(-1)`.

Answer 1

We have
`"log" K_(2)/K_(1)=(DeltaH^(ɵ))/(2.303 R)((T_(2)-T_(1)))/(T_(1)T_(2))`
`"log" 1.4/0.096=(DeltaH^(ɵ))/(2.303xx8.314)((373-298)/(373-298))`
`DeltaH^(ɵ)=33025 J`
Now the temperature above which the forward reaction will be spontaneous is actually the temperature at which the reaction attains equilibrium, that is, when `K=1` or `log K=0`
`:. DeltaG^(ɵ)=-2.303RT log K=-2.303RT log 1.0=0`
From thermodynamics, we get
`DeltaG^(ɵ)=DeltaH^(ɵ)-T DeltaS^(ɵ)`
`0=33025-Txx10.296`
or `T=320.75 K`