Topic 7.1 Notes – Introduction to Equilibrium

1. Reversible Processes and Dynamic Equilibrium

A reversible process can proceed in both directions under the same conditions. In chemical equations, that’s shown with a double arrow:

$$\ce{A <=> B}$$

This means reactants form products and products can reform reactants.

Everyday reversible processes

You’ve already seen these:

• Physical changes
  • $\ce{H2O(l) <=> H2O(g)}$ (evaporation and condensation)
  • $\ce{CO2(aq) <=> CO2(g)}$ (gas dissolving and escaping)
  • $\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)}$ (dissolution and precipitation)
• Chemical changes
  • Acid-base reactions (proton transfer), like $\ce{NH3 + H2O <=> NH4^{+} + OH^{-}}$
  • Redox reactions (electron transfer), where oxidation and reduction can reverse under the right conditions

When a reversible process occurs in a closed system (no matter enters or leaves), it can reach equilibrium.

What equilibrium actually means

At equilibrium:

• The forward rate = reverse rate
• Concentrations (or partial pressures) are constant
• Reactants and products are present

Constant does not mean equal. You can have much more product than reactant at equilibrium. What’s equal are the rates.

That distinction shows up all the time in multiple-choice questions.

2. What Equilibrium Looks Like Experimentally

If you’re observing a system at equilibrium:

• No visible change (color, pressure, etc.)
• Measurable quantities stop changing over time
• The reaction is still happening

Students often think “no change” means “no reaction.” It means no net change.

Why the system must be closed

In a closed container:

• Concentrations only change because of the forward and reverse reactions.
• The system can self-adjust until rates become equal.

In an open system:

• Gas can escape.
• A product can be removed.
• Matter changes for reasons other than the reaction.

That prevents true equilibrium. If a question describes gas bubbling out into the air, that’s not an equilibrium situation.

3. How Equilibrium Is Established Over Time

Take a simple reaction:

$$\ce{A <=> B}$$

Imagine starting with only A.

• At the beginning:
  • $[A]$ is high → forward rate is fast
  • $[B]$ is near zero → reverse rate is slow
• As B forms:
  • $[A]$ decreases → forward rate slows
  • $[B]$ increases → reverse rate speeds up
• Eventually:
  • Forward rate = reverse rate → equilibrium

Concentration vs. Time Graph

Here’s what that looks like on a concentration-time graph.

Concentration vs. time for a reversible reaction

This example shows $\ce{N2O4 <=> 2NO2}$, but the shape is the same for any reversible reaction.

Notice:

• Lines become flat at equilibrium.
• Final concentrations are different.
• The “flat” region means constant concentration, not zero rate.

You may be asked to explain why a line levels off. The correct reasoning is that the forward and reverse rates are equal.

Rate vs. Time Graph

Rates tell the same story from a different angle.

Forward and reverse rates approaching equilibrium

At equilibrium:

• Rates are equal and constant.
• They are usually not zero.

If a question asks what happens before equilibrium, talk about changing concentrations affecting rate.

4. The Dynamic Nature of Equilibrium

Equilibrium is dynamic.

At equilibrium:

• Particles are still colliding.
• Bonds are still breaking and forming.
• A is turning into B.
• B is turning back into A.

There is no net change because both happen at the same rate.

Think of it like water flowing into and out of a tank at the same speed. The water level stays constant, even though water is moving.

This idea is huge on the AP exam. If you ever write that “the reaction stops at equilibrium,” you will lose points.

5. Connecting Rates to the Equilibrium Constant

For a simple reaction $\ce{A <=> B}$:

Forward rate:
$$\text{rate}_f = k_f[A]$$

Reverse rate:
$$\text{rate}_r = k_r[B]$$

At equilibrium:

$$k_f[A] = k_r[B]$$

Rearranging gives a constant ratio:

$$\frac{[B]}{[A]} = \frac{k_f}{k_r} = K$$

That ratio is the equilibrium constant, K.

Important conceptual points:

• $K$ reflects how far the reaction proceeds toward products.
• $K$ is constant at a given temperature.
• Changing initial concentrations does not change $K$.
• Changing temperature does change $K$.

For this topic, focus on the idea that equal rates lead to a constant ratio of products to reactants.