Reaction Quotient: Predicting the Direction of Chemical Change
Analogy
- 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: $$aA + bB \rightleftharpoons xX + yY$$
- 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 ($\text{mol dm}^{-3}$).
Tip
Ensure that all concentrations used in the $Q$ calculation are expressed in the same units, typically $\text{mol dm}^{-3}$
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: $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$
- At 475 K, the equilibrium constant $K$ is 0.59. Suppose the reaction mixture contains the following concentrations: $$[\text{N}_2] = 0.50 \, \text{mol dm}^{-3}, \, [\text{H}_2] = 0.50 \, \text{mol dm}^{-3}, \, [\text{NH}_3] = 0.50 \, \text{mol dm}^{-3}$$
- Step 1: Write the expression for $Q$: $$Q = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$
- 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 $K$: $$Q = 2.0 \qquad K = 0.59$$
- Since $Q > K$, the reverse reaction is favored.
- The system will shift toward the reactants to decrease the concentration of $\text{NH}_3$ and increase the concentrations of $\text{N}_2$ and $\text{H}_2$.
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.
Example question
Using $Q$ to Predict Reaction Behavior
Consider the following equilibrium:
$$2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g), \, K = 11.5 \, \text{at } 283 \, \text{K}$$
A reaction mixture contains: $$[\text{NO}_2] = 0.025 \, \text{mol dm}^{-3}, \, [\text{N}_2\text{O}_4] = 0.10 \, \text{mol dm}^{-3}$$
Determine the direction of the spontaneous reaction.
Solution
- Write the expression for $Q$: $$Q = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}$$
- Substitute the given concentrations: $$Q = \frac{0.10}{(0.025)^2} = \frac{0.10}{0.000625} = 160$$
- Compare $Q$ to $K$: $$Q = 160 \qquad K = 11.5$$
- Since $Q > K$, the reverse reaction is favored.
- The system will shift toward the reactants, increasing $[\text{NO}_2]$ and decreasing $[\text{N}_2\text{O}_4]$.
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 $\text{H}_2 + I_2 \rightleftharpoons 2\text{HI}$, calculate $Q$ if $[\text{H}_2] = 0.10 \, \text{mol dm}^{-3}, [\text{I}_2] = 0.20 \, \text{mol dm}^{-3}, [\text{HI}] = 0.30 \, \text{mol dm}^{-3}$. Predict the direction of the reaction if $K = 50$.
