For the reaction at 298 K,
2A + B → C
ΔH = 400 kJ mol-1and ΔS = 0.2 kJ K-1mol-1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
From the expression,
ΔG = ΔH - TΔS
Assuming the reaction at equilibrium, ΔTfor the reaction would be:
T = (ΔH - ΔG) / ΔS
T = ΔH / ΔS (ΔG = 0 at equilibrium)
= 400 kJ mol-1 / 0.2 kJ K-1mol-1
T= 2000 K
For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 17: For the reaction at 298 K, 2A + B → C ΔH = 400 kJ mol-1and ΔS = 0.2 kJ K-1mol....
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