17.3 Equilibrium Constants
Learning Objectives
- Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
- Relate the magnitude of an equilibrium constant to properties of the chemical system
Now that we have a symbol (⇌) to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:
We can write the reaction quotient (Q) for this equation. When evaluated using concentrations, it is called Qc. We use brackets to indicate molar concentrations of reactants and products.
The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction [latex]2\text{NO}_2(g)\;{\rightleftharpoons}\;\text{N}_2\text{O}_4(g)[/latex] is given by this expression:
Example 17.3a
Writing Reaction Quotient Expressions
Write the expression for the reaction quotient for each of the following reactions:
- [latex]3\text{O}_2(g)\;{\rightleftharpoons}\;2\text{O}_3(g)[/latex]
- [latex]\text{N}_2(g)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(g)[/latex]
- [latex]4\text{NH}_3(g)\;+\;7\text{O}_2(g)\;{\rightleftharpoons}\;4\text{NO}_2(g)\;+\;6\text{H}_2\text{O}(g)[/latex]
Solution
- [latex]Q_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}[/latex]
- [latex]Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}[/latex]
- [latex]Q_c = \frac{[\text{NO}_2]^4[\text{H}_2\text{O}]^6}{[\text{NH}_3]^4[\text{O}_2]^7}[/latex]
Exercise 17.3a
Write the expression for the reaction quotient for each of the following reactions:
- [latex]2\text{SO}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{SO}_3(g)[/latex]
- [latex]\text{C}_4\text{H}_8(g)\;{\rightleftharpoons}\;2\text{C}_2\text{H}_4(g)[/latex]
- [latex]2\text{C}_4\text{H}_{10}(g)\;+\;13\text{O}_2(g)\;{\rightleftharpoons}\;8\text{CO}_2(g)\;+\;10\text{H}_2\text{O}(g)[/latex]
Check Your Answer[1]
The numeric value of Qc for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Qc is determined. When pure reactants are mixed, Qc is initially zero because there are no products present at that point. As the reaction proceeds, the value of Qc increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease (Figure 17.3a). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.
When a mixture of reactants and products of a reaction reaches equilibrium at a given temp