Question:
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?

Answer:

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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Question:
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?

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Question: 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?

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Question:
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?