How do you use entropy and enthalpy to explain how chemical reactions come to equilibrium?

Question

Let's say I have a reversible chemical reaction between A and B that will come to equilibrium in a closed system. How do I use the concepts of enthalpy, entropy and naturally then, Gibbs free energy, to explain how the system will eventually come to dynamic equilibrium if I started with all A? The only formula I know in this area is $∆G = ∆H - T∆S$ and I want to know if I could just based on this formula. I know that equilibrium occurs when ∆G = 0 and that ∆H should be pretty constant as the reaction progresses. So I was thinking about how ∆S changes but I'm a bit confused on how it changes and especially then further implications. For example, if ∆H and ∆S share opposite signs, then the sign of ∆S will have to reverse in order for ∆G = 0. But how can, say, A becoming B reverse in entropic favourability purely based off the amounts of A and B? I feel like I have a lot of fundamental misconceptions about entropy, enthalpy and equilibrium and could someone help clear things up?

Answer 1

[OP] But how can, say, A becoming B reverse in entropic favourability purely based off the amounts of A and B?

The entropy of reaction is dependent on the reaction coefficient Q (so on the activities or concentrations of the reactants and products, $\frac{[\ce{B}]}{[\ce{A}]}$ for the example given by the OP).

As the reaction proceeds in the forward direction, the entropy of reaction decreases (and the Gibbs energy of reaction increases) until the reaction reaches equilibrium. No matter what the standard entropy of reaction $\Delta_r S^\circ$, the actual entropy of reaction $\Delta_r S$ will be positive in the absence of products (at the “beginning” of the reaction) and negative when a reactant runs out (if the reaction were to go to completion).

$$\Delta_r S = \Delta_r S^\circ - R \ln{Q}$$

If you plug this into the defining equation for the Gibbs energy, you will find that the Gibbs energy of reaction is also dependent on Q.

$$\Delta_r G = \Delta_r G^\circ + R T \ln{Q}$$

This makes sense because otherwise, we could not explain how a reaction reaches equilibrium as concentrations change.

Finally, for those who have learned about the relationship of $Q$ and $K$ to determine in which direction equilibrium lies, you can express $\Delta_r G^\circ$ as $ - R T \ln{K}$ to get the following expression for the Gibbs energy of reaction:

$$\Delta_r G = R T \ln{\frac{Q}{K}}$$

This is a direct way of relating the ratio of $Q$ and $K$ to the Gibbs energy of reaction. The expression yields zero when $Q$ is equal to $K$, and it shows the concentration-dependence (because $Q$ is part of the expression).

Answer 2

For a reaction (A = B + heat) to reach equilibrium at constant temperature energy must be transferred to the environs. This means that the system Delta H is negative and, since energy is transferred out of the system, the system Delta S is also negative. For the reverse reaction Delta H is positive and Delta S is also positive. At equilibrium Delta G, (Delta H minus TDelta S) equals ZERO. Therefore, at equilibrium Delta H = TDelta S. This means that a reaction that reaches equilibrium has the same sign of Delta H and Delta S for the system. Attainment of equilibrium in one direction is driven by energy and entropy loss from the system and energy and entropy gain of the environment; in the reverse direction the reaction is driven by energy and entropy loss of the environment and energy and entropy gain of the system. Consequences of this are that the equilibrium can be attained from either direction and the reaction is spontaneous from either direction to equilibrium. The position of equilibrium is determined by the values of the enthalpy and entropy changes

The standard Delta G[0] is the free energy change from the Standard States to EQUILIBRIUM! This means that the standard reaction is in only one direction and the equation as written and the expression of Keq determines the sign [not the value] of Delta G[0]. The Standard State is a displacement from equilibrium and the reaction is spontaneous in one direction to equilibrium. Displacement in the other direction from the standard state is not spontaneous but that is not what reversing the equation means. Reversing the equation describes the reaction from Equilibrium to the Standard State.

In the spontaneous direction: Standard State to Equilibrium; Delta G[0] = -RTlnKeq. Reversing the equation is the equation from equilibrium to the Standard State [not the attainment of equilibrium from the reverse direction]. Because the reaction is spontaneous in only one direction from the standard state, [unless the standard state happens to be equilibrium such as a phase change for a pure substance] there is no STANDARD free energy change for attainment of equilibrium for the reverse reaction to equilibrium; the free energy for non-standard activities is determined by the value of Q.

In the reverse direction: Equilibrium to Standard State; Delta G[0] = -RTln(Keq). this Keq is the reciprocal of the forward reaction so the standard free energy is of opposite sign. Keq>1 means a positive ln, Keq<1 one means a negative ln. the value of Keq determines the sign of Delta G[0].

In order to measure conditions away from the standard state the RTlnQ factor is evaluated to study the actual activities involved.