Ncert Exercises & Solutions

Q.17. For the reaction at 298 K,
2A + B → C
∆H = 400 kJ $m o l^{- 1}$ and ∆S = 0.2 kJ $k^{- 1}$ $m o l^{- 1}$
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

NEETprep Answer:

From the expression,
∆G = ∆H – T∆S
Assuming the reaction at equilibrium, ∆T for the reaction would be:

T = (∆ H - ∆ G) \frac{1}{∆ S} = \frac{∆ H}{∆ S} (∆ G = 0 a t e q u i l i b r i u m) T = \frac{400 k J m o l^{- 1}}{0.2 k J K^{- 1} m o l^{- 1}} T = 2000 K

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

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