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For the reaction at 298 K 2A+B rarr C ...

For the reaction at `298 K`
`2A+B rarr C`
`DeltaH=400 kJ mol^(-1)` and `DeltaS=0.2 kJ K^(-1) mol^(-1)`
At what temperature will the reaction becomes spontaneous considering `DeltaH` and `DeltaS` to be contant over the temperature range.

Text Solution

Verified by Experts

Let us first calculate the temperature at which the reaction will be in equilibrium, i.e., `DeltaG=0`.
Now, `DeltaG=DeltaH-T DeltaS`
`:. 0=DeltaH-T DeltaS`
or `T=(DeltaH)/(DeltaS)=(400 kJ mol^(-1))/(0.2 kJ K^(-1) mol^(-1))=2000 K`
For reaction to be spontaneous, i.e., for `DeltaG` to be -ve `T` should be greater than `2000 K`.
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