Le Chatelier's Principle

Question

A system of R -> <- P is at equilibrium. You add more R to the system. (The reaction is reversible)

R = reactant, P = product

Does adding R change the rate of the reverse reaction?

Answer 1

Adding more R to a system at equilibrium will indeed affect the rate of the reverse reaction. However, it's important to note that equilibrium is a dynamic state where the rates of the forward and reverse reactions are equal. By adding more R, you are disturbing the equilibrium and shifting it to the right (towards the products) in order to re-establish equilibrium.

When you add more R, the concentration of R increases, causing an increase in the rate of the forward reaction. As the forward reaction proceeds at a higher rate, the concentration of P (product) increases, and the concentration of R decreases. This change in concentrations will ultimately lead to an increase in the rate of the reverse reaction as well.

The system will continue to adjust until a new equilibrium is reached. The forward and reverse reactions will then occur at equal rates again, but with altered concentrations of R and P due to the addition of more R.

Answer 2

When a system is at equilibrium the rate of the forward and reverse reactions are equal. When the system is disturbed, the rates will change in order to bring the system back to equilibrium. Adding reactant will shift the reaction to the right towards product formation, meaning the rate of the forward reaction will increase while the rate of the reverse reaction decreases.