For the reaction:N2 (g) + 3H2 (g) → 2NH3(g)ΔH = -95.4 kJ mol-1 and

1.	For the reaction:N2 (g) + 3H2 (g) → 2NH3(g)ΔH = -95.4 kJ mol-1 andΔS = -198.3 JK-1 mol-1Calculate the temperature at which Gibbs energy change (∆G) is equal to zero. Predict the nature of the reaction at this temperature and above it?
Answer» ∆G = ∆H − T∆S When ∆G = 0 ∆H = T∆S T = \(\frac{\Delta H}{\Delta S}\) = \(\frac{-95.4\times1000\,J\,mol^{-1}}{-198.3\,J\,K}\) = 481 K At this temperature, the reaction would be in equilibrium. With increase in temperature, the opposing factor T∆S would become more and hence ∆G would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperature below 481 K.

For the reaction:N2 (g) + 3H2 (g) → 2NH3(g)ΔH = -95.4 kJ mol-1 andΔS = -198.3 JK-1 mol-1Calculate the temperature at which Gibbs energy change (∆G) is equal to zero. Predict the nature of the reaction at this temperature and above it?

Answer»

∆G = ∆H − T∆S

When ∆G = 0

∆H = T∆S

T = \(\frac{\Delta H}{\Delta S}\)

= \(\frac{-95.4\times1000\,J\,mol^{-1}}{-198.3\,J\,K}\)

= 481 K

At this temperature, the reaction would be in equilibrium. With increase in temperature, the opposing factor T∆S would become more and hence ∆G would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperature below 481 K.

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For the reaction:N2 (g) + 3H2 (g) → 2NH3(g)ΔH = -95.4 kJ mol-1 andΔS = -198.3 JK-1 mol-1Calculate the temperature at which Gibbs energy change (∆G) is equal to zero. Predict the nature of the reaction at this temperature and above it?

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