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InterviewSolution
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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)`. |
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Answer» 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` |
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