# AHL: Reaction quotient

> IB Chemistry HL · R2: How much / how fast / how far?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-reaction-quotient/

This subtopic introduces the reaction quotient ($Q$), a tool that describes the relative amounts of reactants and products at any point in a reversible reaction, not just at equilibrium. You will learn to calculate $Q$ and use it to predict reaction direction.

**Prerequisites:** [Equilibrium constant ($K_c$)](https://www.owlsprep.com/study/ib-chemistry-hl-u5-equilibrium-constant/); [Dynamic equilibrium](https://www.owlsprep.com/study/ib-chemistry-hl-u5-dynamic-equilibrium/)

## Learning objectives

- Distinguish between the reaction quotient ($Q$) and equilibrium constant ($K$)
- Calculate $Q$ from measured concentration or partial pressure values
- Compare $Q$ and $K$ to predict the direction of reaction to reach equilibrium
- Relate changes in $Q$ to Le Chatelier's principle for disturbed equilibrium systems

## Definition and Expression of the Reaction Quotient

**Reaction Quotient** — A dimensionless quantity that describes the ratio of product to reactant concentrations (or partial pressures) at any given point in a reversible reaction, regardless of whether equilibrium has been reached.

*Notation:* $Q$, or $Q_c$ for concentration, $Q_p$ for partial pressure

*Example:* For the general reaction $aA + bB \rightleftharpoons cC + dD$, $Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ where $[X]$ = concentration of $X$ at time $t$.

The reaction quotient has the exact same form as the equilibrium constant $K_c$. The only fundamental difference is that $K_c$ is only calculated from concentrations at equilibrium, while $Q$ can be calculated at any time during the reaction.

**Worked example:** Write the reaction quotient expression $Q_c$ for the reversible reaction: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$

1. Recall the general form of $Q_c$:
2. $$Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
3. Where exponents equal the stoichiometric coefficients of each species, and products go in the numerator, reactants in the denominator.
4. Substitute the species and coefficients from the given reaction:
5. $$Q_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$

> **Exam tip:** Always match exponents to the balanced equation's stoichiometric coefficients, never use 1 for all exponents by default.

## Predicting Reaction Direction from $Q$ vs $K$

The primary application of the reaction quotient is to predict which direction a reaction will proceed to reach equilibrium. By comparing the calculated $Q$ to the known equilibrium constant $K$ at the same temperature, we get three clear outcomes:

- If $Q < K$: The product concentration term is too small. The reaction proceeds **forward** to make more products, increasing $Q$ until $Q = K$.
- If $Q > K$: The product concentration term is too large. The reaction proceeds **reverse** to make more reactants, decreasing $Q$ until $Q = K$.
- If $Q = K$: The reaction is already at equilibrium, with no net change in concentrations.

**Worked example:** For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_c = 0.064$ at 450°C. A reaction mixture has $[N_2] = 1.00$ M, $[H_2] = 0.50$ M, $[NH_3] = 0.25$ M. Calculate $Q_c$ and predict the direction of reaction.

1. Write the correct $Q_c$ expression from the balanced equation:
2. $$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$
3. Substitute the given concentration values into the expression:
4. $$Q_c = \frac{(0.25)^2}{(1.00)(0.50)^3} = \frac{0.0625}{0.125} = 0.5$$
5. Compare $Q_c$ to $K_c$: $0.5 > 0.064$, so $Q_c > K_c$. When $Q > K$, the reaction proceeds in the reverse direction to produce more reactants.

> **Exam tip:** Double-check arithmetic for exponents, as a small calculation error can reverse your prediction.

## $Q$ and Disturbances to Equilibrium

When a system at equilibrium is disturbed by a change in concentration, pressure, or volume (at constant temperature), the value of $Q$ changes immediately, while $K$ remains unchanged. Comparing the new $Q$ to the original $K$ confirms the direction of shift predicted by Le Chatelier's principle.

> **tip**
>
> Only temperature changes alter the value of $K$. All other equilibrium disturbances only change $Q$, not $K$.

**Worked example:** The reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ is at equilibrium, so $Q = K_c$. Additional $H_2$ is added to the vessel at constant temperature. Use $Q$ to predict the direction of shift.

1. At original equilibrium:
2. $$Q_{original} = \frac{[HI]^2}{[H_2]_{original}[I_2]} = K_c$$
3. After adding $H_2$, $[H_2]$ immediately increases, so the denominator of $Q$ becomes larger:
4. $$Q_{new} = \frac{[HI]^2}{[H_2]_{new}[I_2]}$$
5. A larger denominator means $Q_{new} < K_c$. Temperature is constant so $K_c$ does not change. Since $Q < K$, the reaction shifts forward to produce more $HI$, which matches Le Chatelier's prediction.

## Common pitfalls

- **Wrong:** Claiming reaction proceeds forward when $Q > K$, reverse when $Q < K$.
  - Why it fails: Confusing which term is too large: $Q > K$ means product concentration is too high, not too low.
  - Correct: $Q < K$ → forward direction; $Q > K$ → reverse direction.
- **Wrong:** Using experimental rate law exponents instead of stoichiometric coefficients for $Q$.
  - Why it fails: Mixing up the form of rate laws and equilibrium expressions.
  - Correct: $Q$ exponents always match the stoichiometric coefficients from the balanced equation.
- **Wrong:** Changing the value of $K$ after a concentration change, and comparing new $K$ to old $Q$.
  - Why it fails: Confusing the factors that change $Q$ vs $K$.
  - Correct: Only temperature changes $K$. All other disturbances change $Q$, not $K$.
- **Wrong:** Putting reactants in the numerator and products in the denominator of the $Q$ expression.
  - Why it fails: Reversing the order from memory, especially in exam pressure.
  - Correct: Always products over reactants, raised to their stoichiometric coefficients.

## Cheatsheet

| Condition | Reaction Direction | Key Note |
| --- | --- | --- |
| $Q < K$ | Forward (→) | More products needed, $Q$ increases to $K$ |
| $Q > K$ | Reverse (←) | More reactants needed, $Q$ decreases to $K$ |
| $Q = K$ | At equilibrium | No net change in concentrations |
| Same expression as $K$ | All cases | $Q$ = any time, $K$ = only equilibrium |
| Only $T$ changes $K$ | All disturbances | Concentration/pressure change $Q$ only |

## What's next

Mastering the reaction quotient is a critical foundation for all advanced equilibrium topics in IB Chemistry HL. The ability to predict reaction direction from $Q$ is used across acid-base equilibria, solubility equilibria, and Gibbs free energy calculations for non-spontaneous reactions. It also provides a quantitative confirmation of Le Chatelier's qualitative predictions for equilibrium shifts, which is often tested in extended response questions. Your next steps build directly on this concept to solve more complex equilibrium problems.

- [R3: What are the mechanisms of chemical change?](https://www.owlsprep.com/study/ib-chemistry-hl-u6-overview/)
- [Introduction to organic chemistry: functional groups](https://www.owlsprep.com/study/ib-chemistry-hl-u6-introduction-to-organic-chemistry-functional/)
- [Nomenclature of organic compounds](https://www.owlsprep.com/study/ib-chemistry-hl-u6-nomenclature-of-organic-compounds/)

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