# Understanding Reaction Rates and equilibria

As a high school student grappling with the interplay of reaction rates, Le Chatelier's principle, and equilibria, I have a question about a system that hasn't yet achieved equilibrium.

Specifically, if the reaction rate to the right exceeds that to the left, should I expect more products or more reactants to be formed as the system approaches equilibrium? On one hand, the higher reaction rate to the right suggests more products are being produced, but on the other hand, doesn't the reaction rate to the right need to decrease for equilibrium to be reached?

• Note that as soon as reactions starts, two processes start to occur. Conversion of reactants into products and conversion of products back into reactants. Sep 17 at 12:46
• It's not that reverse reaction starts occuring only after equilibrium point. It is already occuring with the forward reaction. Past the equilibrium point, the reverse reaction rate is more than forward reaction rate contrary to before equilibrium where forward rate is more than reverse rate. Sep 17 at 12:47
• You cannot always expect more product in the right side of equilibrium. If the rate if independent of concentration of reactants and products, such as in zero order reaction, conc. of product need not be more than reactants. It's just their respective rates. Sep 17 at 12:51
• In the equilibrium reaction is A⇌B, "if the reaction rate to the right exceeds that to the left" then B is produced from A at a greater rate than A is produced than B at that moment, so yes, "the higher reaction rate to the right suggests more products are being produced." It is also correct that the reaction rate to the right (relative to the rate to the left) needs to decrease for equilibrium to be reached because after equilibrium is reached, the two rates will be the same. Sep 17 at 14:15

At $$t=0$$ the forward rate $$A\to B$$ is at a maximum, as $$\ce{[B] = 0}$$ (assuming only A at the start) and after a second or so some B is present and rate $$A\to B$$ has reduced a bit as there is less $$A$$ and the rate of $$B\to A$$ has increased but is still small. Some A is, of course reformed, but that is always accounted for in the forwards rate. As time passes the rate of $$A\to B$$ falls further and $$B\to A$$ continues to rise until they are equal and equilibrium has ben reached. No further change in rates occurs, unless of course you change something, such as temperature etc. The position of equilibrium (amount of A vs B) depends on the ratio of rate constants.

Specifically, if the reaction rate to the right exceeds that to the left, should I expect more products or more reactants to be formed as the system approaches equilibrium?

Let's assume that at the beginning there are no products. Thus at the beginning of the reaction there can't be any reverse reaction. Also at the beginning the forward reaction rate is at its peak, since the reactants are at their maximum concentration.

On one hand, the higher reaction rate to the right suggests more products are being produced, but on the other hand, doesn't the reaction rate to the right need to decrease for equilibrium to be reached?

Now as the forward reactions starts two things happen. First the forward reaction rate, not the rate constant but the overall reaction rate, starts to slow because the reactants are being used up. The second thing that happens is the reverse reaction starts. Since equilibrium hasn't yet been reached, the forward reaction rate is faster than the reverse reaction rate.

But as the forward reaction drives the system to equilibrium the forward reaction rate continues to slow because the reactant concentrations are decreasing and the reverse reaction rate continues to speed up because the product concentrations are increasing. Equilibrium is reached when the two reaction rates are finally equal.

Note that if there are no products initially then the reverse reaction rate can never be faster than the forward reaction rate. Thus the system is always going towards equilibrium.