Topic 7.3 Notes – Reaction Quotient and Equilibrium Constant

1. The Reaction Quotient and the Equilibrium Constant

For a general reversible reaction:

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

the law of mass action gives this expression:

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

That structure never changes. What changes is which values you plug in.

• Equilibrium concentrations → this equals $K_c$
• Concentrations at any time t → this equals $Q_c$
• For gases written with partial pressures → $K_p$ or $Q_p$

At equilibrium:

$$Q = K$$

Think of it this way:

• Q = current ratio of products to reactants
• K = target ratio at equilibrium

Q moves toward K as the reaction shifts.

2. Writing Qc and Qp Correctly

General Forms

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

Concentration form (solutions):

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

Pressure form (gases only):

$$Q_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$

Rules you must follow every time:

• Products go on top.
• Reactants go on bottom.
• Coefficients become exponents.
• Use whatever values the problem gives you (initial, after disturbance, etc.).

The AP will not ask you to convert between $K_c$ and $K_p$, but you do need to recognize whether concentrations or partial pressures are being used.

What Does NOT Appear in Q or K

Exclude:

• Pure solids (s)
• Pure liquids (l)

Their concentrations are constant, so they do not affect the ratio.

Include:

• Aqueous species (aq)
• Gases (g)

Example:

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$

$$K_p = P_{\ce{CO2}}$$

The solids disappear from the expression.

Students often lose easy points by including solids. If it says (s) or (l), leave it out.

3. Comparing Q and K to Predict Direction

This is the core skill.

After calculating Q, compare it to K.

Another way to think about it:

$$Q = \frac{\text{products}}{\text{reactants}}$$

• If Q is too small, the system makes more products.
• If Q is too big, the system makes more reactants.

4. Why the Reaction Shifts

Let’s say:

$$Q > K$$

That means the numerator is too large. There are excess products.
The reaction consumes products and forms reactants.
Q decreases until $Q = K$.

If:

$$Q < K$$

There are excess reactants.
The reaction forms more products.
Q increases until it equals K.

Here’s a simple visual way to remember it:

Comparing Q and K and the resulting shift in reaction direction

The system always adjusts to reach the equilibrium ratio, so Q moves toward K from either side.

5. Working Through a Q vs. K Problem

Example:

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

Given:

• $K_c = 50.0$
• $[\ce{H2}] = 0.40\,M$
• $[\ce{I2}] = 0.40\,M$
• $[\ce{HI}] = 0.20\,M$

Step 1: Write Q

$$Q_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$$

Step 2: Plug in

$$Q_c = \frac{(0.20)^2}{(0.40)(0.40)}$$

$$Q_c = \frac{0.040}{0.160} = 0.25$$

Step 3: Compare

$0.25 < 50.0$

So:

• $Q < K$
• Reaction shifts right
• More HI will form

Notice how mechanical this becomes once you write the expression correctly.

Common errors:

• Forgetting to square coefficients.
• Flipping numerator and denominator.
• Comparing backward (always write the inequality explicitly).