Equlibrium for a non-spontaneous reaction

Question

Consider a reversible reaction $\ce{A -> B}$, at standard state wher $\ce{A}$ and $\ce{B}$ are at $\pu{1 atm}$. If the free energy of products is greater than the free energy of reactants, $\Delta_{\text{R}} G^\circ > 0$, and since $Q = 1$, the overall free energy for this reaction is $\Delta_{\text{R}} G^\circ > 0$ and this tells me the reaction is non-spontaneous. Yet, being at standard state does not mean that the reaction is at equilibrium, in fact $Q = 1$ could be greater or less than $K$, the products are not at equilibrium, and hence amount of $\ce{A}$ and $\ce{B}$ will adjust to a point where $Q = K$. Wouldn't that have just mean that the reaction is not at equilibrium and hence would spontaneously adjust to reach equilibrium. But yet $\Delta_{\text{R}} G^\circ$ tells me that it would not be spontaneous. Wouldn't this two concepts be conflicting against each other?

Answer 1

Yet, being at standard state does not mean that the reaction is at equilibrium, in fact Q=1 could be greater or less than K.

$\Delta_{\text{R}} G^\circ $ and K are linked. If $\Delta_{\text{R}} G^\circ $ is positive, K will be less than one. Consequently, Q > K and the reaction will go backwards.

But yet $\Delta_{\text{R}} G^\circ $ tells me that it would not be spontaneous. Wouldn't this two concepts be conflicting against each other?

No, there is no conflict. If the reaction is spontaneous, equilibrium is reached in the forward direction. If the reaction is non-spontaneous, equilibrium is reached in the reverse direction.

These rules only work for starting with concentrations at standard state. No matter what the value of the Gibbs energy of reaction, you can always choose an initial state such that the reaction will go forward or reverse. In some cases, however, it will very quickly reach equilibrium, so there is very little reaction.

Answer 2

If delta G is positive, that means if you start a reaction with both A and B in a container already and each has a partial pressure of 1 atm, the reaction is not at equilibrium. forward reaction is not spontaneous. Reverse reaction is. So it means the reaction will start with B changes to A to reach equilibrium. At equilibrium, B partial pressure will be greater than 1 and A partial pressure will be less than 1 atm. This reaction has a k less than 1. It means this reaction favors the reactants.

Most kids think that a non spontaneous reaction means if I start with just A in the container, A won’t proceed to change to B. As you learn from equilibrium chapter, if you start with just A and no B, the forward reaction will proceed to change to B. The delta G (products free energy - reactants free energy) at this condition is negative since there is no products yet so The delta G at standard state is for when A and B both are 1 atm. Also remember if we say a reaction is nonspontaneous, we mean the forward reaction is non spontaneous, it also means the reverse reaction is automatically spontaneous.

Answer 3

Note: This first part isn't actually that related to the specific scenario you are giving, but I think it's a misconception that you had that led you to ask this question. If you want a direct answer, go to the last paragraph.

Just because it is non-spontaneous, it doesn't mean that no reaction will occur. In fact, the reaction will always occur to some extent, even if only 0.001% of the particles react.

Consider this formula, which I'm sure you have seen before: $$\Delta G=-RT\ce{ln}K$$ This makes sense because if the reaction is spontaneous ($\Delta G<0$), $K>0$, so the forward reaction, which is spontaneous is favored over the reverse reaction; vice versa.

If we, for example, say $\Delta G = +100$ kJ, at 298 K, $K$ would equal $3.4\cdot10^{-17}$. Thus, the forward reaction pretty much completely dominates the reverse reaction, so we say that the reaction is essentially irreversible. However, the reverse reaction is still happening, to a very small extent.

If we happened to make $\Delta G = +10$ kJ, $K$ would equal only about $56$, so the reverse reaction becomes visible. The terms spontaneous and non-spontaneous are only generalizations; it doesn't mean that a non-spontaneous reaction doesn't occur at all. There will always be some particles in that substance that will overcome the activation energy and react.

In terms of your question, since the reaction $\ce{A->B}$ is non-spontaneous, the reaction $\ce{B->A}$ is actually spontaneous (think about their equilibrium constants). Thus, the reverse reaction will be favored until the system reaches equilibrium, or when $Q=K$.