Topic 7.6 Notes – Properties of the Equilibrium Constant

1. The Equilibrium Constant and Its Mathematical Form

At equilibrium, the forward rate equals the reverse rate, so concentrations stop changing. They are constant, but not necessarily equal.

For a general reaction:

$$\ce{aA + bB <=> cC + dD}$$

The equilibrium constant is:

$$K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

A few things you must remember:

• Only gases (g) and aqueous (aq) species appear in $K$.
• Pure solids (s) and pure liquids (l) are omitted.
• Coefficients in the balanced equation become exponents.
• $K$ depends only on temperature.
• The reaction quotient $Q$ has the same mathematical form as $K$, but uses non‑equilibrium concentrations.

Because $K$ and $Q$ have identical structure, anything algebraically valid for $K$ also works for $Q$. That idea becomes important later.

2. The Three Algebra Rules for Manipulating K

These rules are the equilibrium version of Hess’s Law. Instead of adding ΔH values, you’ll be multiplying or exponentiating $K$ values.

Reversing a Reaction

If you flip a reaction, the equilibrium constant becomes its reciprocal.

Example:

$$\ce{SO2(g) + 1/2 O2(g) <=> SO3(g)} \quad K = 4.0$$

Reverse it:

$$\ce{SO3(g) <=> SO2(g) + 1/2 O2(g)}$$

New constant:

$$K_{\text{new}} = \frac{1}{4.0} = 0.25$$

Why this works: when you reverse the reaction, the numerator and denominator in the equilibrium expression switch places.

Rule: Reverse reaction → use $1/K$.

Multiplying the Coefficients by a Factor c

If every coefficient is multiplied by some number $c$, then:

$$K_{\text{new}} = K^c$$

Example:

$$\ce{H2(g) + I2(g) <=> 2HI(g)} \quad K = 50$$

Multiply entire equation by 2:

$$\ce{2H2(g) + 2I2(g) <=> 4HI(g)}$$

Now:

$$K_{\text{new}} = 50^{2} = 2500$$

If you multiplied by 1/2, you would take the square root of $K$.

Why this works: coefficients become exponents in the equilibrium expression, so scaling coefficients scales exponents.

Rule: Multiply coefficients by $c$ → raise $K$ to the power $c$.

Adding Reactions Together

When reactions are added, their equilibrium constants are multiplied.

If:

Reaction 1 → $K_{1}$
Reaction 2 → $K_{2}$

Overall reaction → $K_{\text{overall}} = K_{1} \times K_{2}$

This works because multiplying the individual equilibrium expressions gives the algebraic form of the overall reaction.

Rule: Add reactions → multiply their $K$ values.

Never add $K$ values. That’s a common mistake.

3. Combining the Rules to Build an Overall K

Most test questions mix all three rules.

Here’s how it plays out logically:

1. Write the target overall reaction.
2. Compare it to the given reactions.
3. Decide which reactions must be:
   • Reversed
   • Multiplied
4. Perform those changes.
5. Add reactions and cancel intermediates.
6. Adjust the $K$ values:
   • Reverse → invert $K$
   • Multiply coefficients → raise $K$
   • Add reactions → multiply $K$ values

This is structurally identical to Hess’s Law, except:

• Hess’s Law → add ΔH
• Equilibrium → multiply $K$

The AP loves giving two or three reactions where one needs reversing and another needs scaling. The final $K$ must match the final balanced equation exactly. If the equation isn’t correct, neither is the constant.

4. Connection to Q and Why This Matters

Since $Q$ has the same mathematical form as $K$:

• Reverse reaction → $Q$ becomes $1/Q$
• Multiply coefficients → $Q^c$
• Add reactions → multiply $Q$ expressions

Conceptually, a large $K$ (≫1) means products are favored. A small $K$ (≪1) means reactants are favored. When you manipulate the equation, you are not changing the chemistry. You are changing how you describe it mathematically.

You are viewing the same equilibrium from a different perspective.