Question 14.10: Finding Equilibrium Concentrations from Initial Concentratio...
Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant
Consider the reaction.
N_{2}O_{4}(g) \xrightleftharpoons[]{} 2 NO_{2}(g)K_{c} = 0.36 (at 2000 °C)
A reaction mixture at 100 °C initially contains [NO_{2}] = 0.100 M. Find the equilibrium concentrations of NO_{2} and N_{2}O_{4} at this temperature.
1. Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products.Leave room in the table for the changes in concentrations and for the equilibrium concentrations.
2. Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations. Compare Q to K and predict the direction in which the reaction will proceed.
3. Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products in terms of x. It is usually most convenient to let x represent the change in concentration of the reactant or product with the smallest stoichiometric coefficient.
4. Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.
5. Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x. In some cases, such as Example 14.9, you can take the square root of both sides of the expression to solve for x. In other cases, such as Example 14.10, you must solve a quadratic equation to find x.Remember the quadratic formula:
ax² + bx + c = 0
x =\frac{ -b ±\sqrt{2b2 – 4ac}}{2a}
6. Substitute x into the expressions for the equilibrium concentrations of the reactants and
products (from step 4) and calculate the concentrations. In cases where you solved a quadratic and have two values for x, choose the value for x that gives a physically realistic answer. For example, reject the value of xthat results in any negative concentrations.
7. Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that rounding errors could cause a difference in the least significant digit when comparing values of the equilibrium constant.
| [N_{2}O_{4}] | [NO_{2}] | |
| Initial | 0.00 | 0.100 |
| Change | ||
| Equil |
Q_{c} =\frac{ [NO_{2}]^{2}}{[N_{2}O_{4}]} =\frac{ (0.100)^{2}}{0.00}
= ∞
Q > K; therefore, the reaction will proceed to the left.
| [N_{2}O_{4}] | [NO_{2}] | |
| Initial | 0.00 | 0.100 |
| Change | +x | -2x |
| Equil |
| [N_{2}O_{4}] | [NO_{2}] | |
| Initial | 0.00 | 0.100 |
| Change | +x | -2x |
| Equil | x | 0.100-2x |
K_{c} = \frac{[NO]^{2}}{[N_{2}O_{4}]}
= \frac{(0.100 – 2x)^{2}}{x}
0.36 = \frac{0.0100 – 0.400x + 4x^{2}}{x}
0.36x = 0.0100 – 0.400x + 4x²
4x² – 0.76x + 0.0100 = 0 (quadratic)
x =\frac{ -b ±\sqrt{2b2 – 4ac}}{2a}
= \frac{-(-0.76) ± \sqrt{(-0.76)2 – 4(4)(0.0100)}}{2(4)}
= \frac{0.76 ± 0.65}{8}
x = 0.176 or x = 0.014
We reject the root x = 0.176 because it gives a negative concentration for NO_{2}.Using x = 0.014, we get the following concentrations:
[NO_{2}] = 0.100 – 2x
= 0.100 – 2(0.014) = 0.072 M
[N_{2}O_{4}] = x
= 0.014 M
K_{c} = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]} = \frac{0.072^{2}}{0.014}
= 0.37
Since the calculated value of K_{c} matches the given value (to within one digit in the least significant figure), the answer is valid.