R2.3.5 Reaction quotient (Q) (Higher Level Only)

Reaction Quotient: Predicting the Direction of Chemical Change

You are observing a chemical reaction in progress.
You know the equilibrium constant $K$ , which represents the ratio of products to reactants when the system is at equilibrium.
But what if the reaction hasn’t yet reached equilibrium? How can you predict whether the reaction will move forward to produce more products or backward to regenerate reactants?

This is where the reaction quotient, $Q$ , becomes an invaluable tool. By comparing $Q$ to $K$ , you can determine the direction of the reaction and understand how the system will adjust to reach equilibrium.

The Reaction Quotient: Definition and Calculation

Definition

Reaction quotient

The reaction quotient, $Q$ , measures the relative concentrations of products and reactants in a chemical reaction at a given moment in time, whether or not the system is at equilibrium

It is calculated using the same formula as the equilibrium constant $K$ , but with non-equilibrium concentrations of the reacting species.

For a general reaction:

$a A + b B ⇌ x X + y Y$

The expression for $Q$ is:

$Q = \frac{[X]^{x} [Y]^{y}}{[A]^{a} [B]^{b}}$

Products appear in the numerator, raised to the power of their stoichiometric coefficients.
Reactants appear in the denominator, also raised to the power of their stoichiometric coefficients.
Square brackets $[. . .]$ denote the concentrations of the species in moles per cubic decimeter ( ${mol dm}^{- 3}$ ).

Tip

Ensure that all concentrations used in the $Q$ calculation are expressed in the same units, typically ${mol dm}^{- 3}$

Comparing $Q$ and $K$ : What It Tells Us

The value of $Q$ in relation to $K$ reveals the direction in which the reaction will proceed to achieve equilibrium:

If $Q < K$ :
- The concentration of reactants is too high (or the concentration of products is too low) compared to the equilibrium state.
- The forward reaction is favored, and the system will shift toward the products.
If $Q > K$ :
- The concentration of products is too high (or the concentration of reactants is too low) compared to the equilibrium state.
- The reverse reaction is favored, and the system will shift toward the reactants.
If $Q = K$ :
- The system is at equilibrium.
- The rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of reactants or products.

Example

Determining the Direction of Reaction

Consider the reaction:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

At 475 K, the equilibrium constant $K$ is 0.59. Suppose the reaction mixture contains the following concentrations:

$[N_{2}] = 0.50 {mol dm}^{- 3}, [H_{2}] = 0.50 {mol dm}^{- 3}, [{NH}_{3}] = 0.50 {mol dm}^{- 3} .$

Step 1: Write the expression for $Q$ :

$Q = \frac{[{NH}_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}$

Step 2: Substitute the given concentrations:

$Q = \frac{(0.50)^{2}}{(0.50) (0.50)^{3}} = \frac{0.25}{0.125} = 2.0$

Step 3: Compare $Q$ to $K$ :

$Q = 2.0$ , $K = 0.59$ .

Since $Q > K$ , the reverse reaction is favored. The system will shift toward the reactants to decrease the concentration of ${NH}_{3}$ and increase the concentrations of $N_{2}$ and $H_{2}$ .

Common Mistake

Many students confuse $Q$ with $K$ . Remember, $K$ is calculated only at equilibrium, while $Q$ can be calculated at any point during the reaction.

Relating reaction quotient and equilibrium constant depending on time.

Example question

Using $Q$ to Predict Reaction Behavior

Consider the following equilibrium:

$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g), K = 11.5 at 283 K .$

A reaction mixture contains: $[{NO}_{2}] = 0.025 {mol dm}^{- 3}, [N_{2} O_{4}] = 0.10 {mol dm}^{- 3} .$

Determine the direction of the spontaneous reaction.

Solution

Write the expression for $Q$ : $Q = \frac{[N_{2} O_{4}]}{[{NO}_{2}]^{2}}$
Substitute the given concentrations: $Q = \frac{0.10}{(0.025)^{2}} = \frac{0.10}{0.000625} = 160$
- Compare $Q$ to $K$ :
- $Q = 160$ , $K = 11.5$ .
Since $Q > K$ , the reverse reaction is favored. The system will shift toward the reactants, increasing $[{NO}_{2}]$ and decreasing $[N_{2} O_{4}]$ .

Reflection and Broader Implications

Self review

What is the key difference between $Q$ and $K$ ?
If $Q = 0$ , what does this tell you about the reaction mixture?
For the reaction $H_{2} + I_{2} ⇌ 2 HI$ , calculate $Q$ if $[H_{2}] = 0.10 {mol dm}^{- 3}, [I_{2}] = 0.20 {mol dm}^{- 3}, [HI] = 0.30 {mol dm}^{- 3}$ . Predict the direction of the reaction if $K = 50$ .

Theory of Knowledge

How does the concept of equilibrium and $Q$ reflect the balance we observe in natural systems, such as ecosystems or the global carbon cycle?
Can these ideas help us better understand sustainability?

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