7.2 Direction of Reversible

Forward vs Reverse Rates

Set initial concentrations and K, then watch rate_fwd and rate_rev converge on live bars and a rate-vs-time plot. Equilibrium is equal rates, not concentrations

Forward vs ReverseRate BarsDynamic Equilibrium3 Reactions
Topic 7.2

Direction of Reversible Reactions

Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.

EK 7.2.A.1 states the rule directly:

  • If the forward rate exceeds the reverse rate, there is a net conversion of reactants to products.
  • If the reverse rate exceeds the forward rate, there is a net conversion of products to reactants.
  • Equilibrium is reached when these rates are equal.

Why the rates converge. Start with pure reactants: their concentration is high, so the forward rate is high; there is no product, so the reverse rate is zero. As the reaction runs, reactant concentration falls (forward rate drops) while product concentration rises (reverse rate climbs). The two curves must eventually meet, and where they meet, equilibrium begins.

Note this is a statement about net direction — both reactions are always occurring.

Connection to Q and K. This kinetic picture is the mechanistic explanation for the Q-vs-K rule of topic 7.10. Q < K means there is not enough product yet, the forward rate exceeds the reverse rate, and the reaction proceeds forward. Q > K means the reverse. Same physics, two languages.

A caution about reading graphs. Rate-versus-time graphs level off at a common non-zero value at equilibrium. A graph showing both rates dropping to zero represents a reaction that ran to completion, not one at equilibrium.

Key points

  • Net direction is decided by which rate is larger at that moment.
  • Forward rate falls and reverse rate rises as the reaction proceeds; they meet at equilibrium.
  • At equilibrium both rates are equal and non-zero.
  • This kinetic statement is equivalent to the Q-versus-K rule.

Common mistakes

  • Reading a rate graph as if it were a concentration graph. Check the axis label.
  • Expecting rates to reach zero at equilibrium. They reach equality at a non-zero value.
  • Confusing "faster forward reaction" with "more product at equilibrium". Rate and position are different things.

Worked example

A mixture of N₂O₄ and NO₂ is prepared with far more NO₂ than the equilibrium mixture would contain. Describe how the forward and reverse rates change over time for N₂O₄(g) ⇌ 2 NO₂(g).

Initially: [NO₂] is above its equilibrium value while [N₂O₄] is below it. The reverse reaction (2 NO₂ → N₂O₄) depends on [NO₂], so its rate is unusually high. The forward reaction depends on [N₂O₄], so its rate is unusually low. The reverse rate exceeds the forward rate.

Net change: there is a net conversion of NO₂ to N₂O₄ — the system shifts toward reactants.

As time passes: [NO₂] falls, so the reverse rate decreases; [N₂O₄] rises, so the forward rate increases. The two rates converge.

At equilibrium: the rates become equal and both remain constant at a non-zero value. The concentrations stop changing, though both reactions continue.

Full notes for topic 7.2 →