For the following reaction at 298 K2A + B → CΔH = 400 kJ mol-1 and

1.	For the following reaction at 298 K2A + B → CΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1. At what temperature will the reaction become spontaneous? Considering ΔH and ΔS to be constant over the temperature range.
Answer» ΔG = ΔH - TΔS For ΔG = 0, ΔH = TΔS or T = ΔH/ΔS T = {400 kJ mol^-1}/{2 kJ K^-1 mol^-1} = 200 K Thus, reaction will be in a state of equilibrium at 200 K and will be spontaneous above this temperature.

For the following reaction at 298 K2A + B → CΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1. At what temperature will the reaction become spontaneous? Considering ΔH and ΔS to be constant over the temperature range.

Answer»

ΔG = ΔH - TΔS

For ΔG = 0, ΔH = TΔS or T = ΔH/ΔS

T = {400 kJ mol^-1}/{2 kJ K^-1 mol^-1} = 200 K

Thus, reaction will be in a state of equilibrium at 200 K and will be spontaneous above this temperature.

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For the following reaction at 298 K2A + B → CΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1. At what temperature will the reaction become spontaneous? Considering ΔH and ΔS to be constant over the temperature range.

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