AP Chemistry 7.1-7.2: Introduction to Equilibrium and Reversible Reactions

Dynamic chemical equilibrium

A state in a reversible system where the forward and reverse reaction rates are equal, so the concentrations of reactants and products remain constant over time. The reactions continue microscopically; equilibrium is not a state in which reaction has stopped.

How do reactant and product concentrations change as a reversible reaction approaches equilibrium?

Reactant concentrations generally decrease and product concentrations increase. The forward rate slows as reactants are consumed, while the reverse rate increases as products accumulate, until the rates become equal.

What does a reversible reaction arrow, $\rightleftharpoons$, indicate?

It indicates that the reaction can proceed in both the forward and reverse directions under the stated conditions.

Does equilibrium require equal concentrations of reactants and products?

No. Equilibrium requires equal forward and reverse rates, not equal concentrations. The equilibrium mixture may contain mostly reactants, mostly products, or appreciable amounts of both.

How can a sealed container of liquid bromine establish phase equilibrium?

Bromine initially vaporizes, increasing the gas-phase concentration and therefore the condensation rate. Equilibrium is reached when the rates of vaporization and condensation are equal: $\mathrm{Br_2(l)} \rightleftharpoons \mathrm{Br_2(g)}$.

For the elementary reaction $\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$, what are the forward and reverse rate laws?

The forward rate is $\text{rate}_f=k_f[\mathrm{N_2O_4}]$, and the reverse rate is $\text{rate}_r=k_r[\mathrm{NO_2}]^2$.

What happens to the forward and reverse rates in $\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$ as equilibrium is approached?

The forward rate decreases because $[\mathrm{N_2O_4}]$ decreases, while the reverse rate increases because $[\mathrm{NO_2}]$ increases. At equilibrium, $\text{rate}_f=\text{rate}_r$.

Equilibrium position

The relative amounts of reactants and products present at equilibrium. An equilibrium lying far to the right contains mostly products, whereas one far to the left contains mostly reactants.

What are the initial forward and reverse rates when pure $\mathrm{N_2O_4}$ is used to begin the reaction $\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$?

Initially, $[\mathrm{N_2O_4}]$ is finite and $[\mathrm{NO_2}]=0$. Therefore, the forward reaction proceeds at a finite rate, while the reverse reaction rate is zero.