Understanding Gibbs free energy and enthalpy

Question

Let's say that a reaction releases -50kJ/mol of Gibbs free energy (delta G), and -75kJ/mol of enthalpy (we know that -25kJ/mol went to make sure that the total entropy of universe increased).

Now let's say we reversed the reaction.

Book says that the minimum energy required to make the reaction occur spontaneously in the reverse direction is the negative of the original change in Gibbs free energy (50kJ/mol).

What bothers me here is that I would think that you would need 75kJ/mol to reverse the reaction - the total energy the original reaction emitted needs to be put back in order to bring the system back to it's original potential energy level. It seems as though enthalpy shows that one amount of energy is needed to reverse the reaction, while Gibbs potential energy indicates a different amount of required energy.

I have no problem solving any of the computational problems relating to these concepts, but I'm curious about how to more fully understand them conceptually. I've noticed that a lot of people are great at solving technical problems related to these concepts, but are not too worried about how to interpret them conceptually. I guess for me, on the other hand, it's more important to understand than to compute...

Thank you for your help!

Answer 1

The equation for Gibbs Free Energy is:

$$\Delta G = \Delta H - T \Delta S$$

The change in Gibbs Free Energy already includes the amount of energy associated with the change in energy. So to reverse the reaction, the total change in energy is given by $-\Delta G$.

Now the question is how much energy do you NEED to get the reaction to go backwards. The minimum amount of energy for it to go backwards spontaneously is $-\Delta G$ because that puts the product on the same energy level as the reactants so in principle they are energetically equivalent. However, what you should really be concerned about was the activation energy of the forward reaction, $E_a$, because this is the mountain that the reaction needs to climb. There is a diagram below, but it is based on the $\Delta H$ values. You can convert this to a graph with $\Delta G$ as well.

Answer 2

Thank you for your answer IT Tsoi.

I wanted to write a comment, but it turned out too long, so I put it as an answer.

It looks like the to have the reaction reach the activation energy you need to give it is more energy than the enthalpy reversal and Gibbs reversal (75kJ and 50kJ in my example). So from the graph you attached, we can see that in practice you may need much more energy than either of these values (especially if you first need to break a bond before the second bond can form).

I was thinking why in principle Gibbs energy input can be less than 75kJ, and the reaction can still, in an idealized circumstance, reverse when given 50kJ.

I think maybe it's like this for reaction in question where the forward is spontaneous (ΔG = -50), and the reverse is not spontaneous (ΔG = +50).

1) You first give it negative ΔG, the entropy change is then bigger than zero, and the reverse reaction can go on.

2) At that point in an idealized circumstance the reaction can suck by itself the remaining 25kJ from the environment to bridge the gap between how much energy you gave it (ΔG) and how much energy is the potential energy difference between the reactants and products of the reverse reaction (ΔH). It can do so because even after taking an extra 25kJ from the environment, the change in entropy of the universe is greater or equal to zero (It's like water evaporating in a dark room by taking energy from the environment because total entropy change is greater than 0).

3) But it can't take more energy from the surroundings than ΔH - ΔG (75 - 50), because by taking more, the entropy of the surroundings will decrease too much making the total change of the entropy of the universe negative. So the (reverse) reaction couldn't take as much energy as is the activation energy, but can by itself get only up to 25kJ from the surroundings (after it's given 50kJ - the negative of ΔG of the forward reaction). To get to the full activation energy it needs an additional energy input, so in practice even after the reaction got 50kJ and then was able to acquire 25kJ more from the surrounding by itself (to a total of ΔH), it still needs more energy to reach the Ea.

So (maybe) in summary: if you give it ΔG, it can suck heat by itself to get to ΔH, but it can't go over ΔH by itself (otherwise ΔS univ will be negative), so to get to Ea it probably needs more energy inputed than ΔG.