Many reactions can run backward too. In a sealed flask, the products of colliding reactants collide with each other and re-form the reactants. What happens when both directions run at once?
Key idea
A reversible reaction reaches equilibrium when the forward and reverse reactions run at equal rates. From then on the amounts of reactants and products stay constant, though both reactions keep going. Constant does not mean equal: the amounts can be very different.
Why it works
At the start only reactants are present, so the forward reaction is fast and the reverse reaction cannot run. As reactants are used up, the forward rate falls; as products build up, the reverse rate rises. The two rates meet, and after that each molecule of product that forms is matched by one that turns back. It is like a crowded shop where people enter and leave at the same rate: the number inside holds steady while the people keep changing.
Worked example: Not equal, just constant
A flask starts with $1.00\ \text{M}$ of A and no B. At equilibrium, $[\mathrm{A}] = 0.25\ \text{M}$ for $\mathrm{A} \rightleftharpoons \mathrm{B}$. What is $[\mathrm{B}]$?
- Each A that reacts becomes one B.
- $[\mathrm{B}] = 1.00 - 0.25 = 0.75\ \text{M}$.
- The two stay at $0.25$ and $0.75\ \text{M}$ while molecules keep switching back and forth.
Answer: $0.75\ \text{M}$.
Worked example: Using the coefficients
A flask starts with $1.00\ \text{M}$ of $\mathrm{NO}_{2}$. At equilibrium in $$2\mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}$$ $[\mathrm{N_2O_4}] = 0.30\ \text{M}$. What is $[\mathrm{NO_2}]$?
- Each $\mathrm{N}_{2}\mathrm{O}_{4}$ formed used two $\mathrm{NO}_{2}$: $2 \times 0.30 = 0.60\ \text{M}$ used.
- $[\mathrm{NO_2}] = 1.00 - 0.60 = 0.40\ \text{M}$.
Answer: $0.40\ \text{M}$.
Common mistake
Thinking the reactions stop at equilibrium, or that the reactants and products must be present in equal amounts. Both reactions continue; only the rates are equal. Check: "equilibrium" is about rates being equal, not amounts.