Question 4.37

[Nasir_Ahmed]

The question says "Comment on the possible values for standard enthalpies of activation, delta(H^o) activated complex, and the standard entropies of activation, delta(S^o)activated complex."

The answer states that the delta(H^o)activated complex will always be positive and are usually much larger than the entropy of activation. Why is that?

[Ivy Kwok 2I]

The standard Gibbs free energy of activation is equal to ΔH°++ - TΔS°++. Because the standard Gibbs free energy of activation is always positive, ΔH°++ must be greater than ΔS°++, thus the standard enthalpy of is usually much larger than the standard entropy of activation.

I'm not 100% sure if this is correct, but I hope it helps!

[Navarro_Bree_1D]

The standard enthalpy of activation will always be positive because bonds are being distorted and extended.

Activation energy = Standard enthalpy of activation + RT.

As we can see from the equation above, the barrier to the transition state is an enthalpic one, that is why the standard enthalpy of activation is greater than the standard entropy of activation.