Question 15.10: Finding Equilibrium Concentrations from Initial Concentratio...

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.

N_2O_4(g) \rightleftharpoons 2 NO_2(g)

2. Use the initial concentrations to calculate the reaction quotient (\mathcal{Q} ) for the initial concentrations. Compare \mathcal{Q} to K to predict the direction in which the reactionwill proceed.

\mathcal{Q_c} =\frac{[NO_2]^2}{[N_2][N_2O_4]} =\frac{(0.100)^2}{0.00}

=∞

\mathcal{Q}>K ; therefore, the reaction will proceed to the left.

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.

N_2O_4(g)  \rightleftharpoons 2 \ NO_2(g)

4. Sum each column for each reactant and each product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

N_2O_4(g)  \rightleftharpoons 2 \ NO_2(g)

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 15.9, you can take the square root of both sides of the expression to solve for x. In other cases, such as Example 15.10, you must solve a quadratic equation to find x.
Remember the quadratic formula:

ax^2 + bx + c = 0

x=\frac{-b\pm \sqrt{b^2 – 4ac} }{2a}

K_c=\frac{[NO_2]^2}{[N_2][N_2O_4]}

=\frac{(0.100 – 2x)^2}{x}

0.36=\frac{0.0100 – 0.400x + 4x^2}{x}

0.36x =0.0100 – 0.400x + 4x^2

4x^2 – 0.76x + 0.0100 = 0 (quadratic)

x=\frac{-b\pm \sqrt{b^2 – 4ac} }{2a}

x=\frac{-(-0.76)\pm \sqrt{(-0.76)^2 – 4(4)(0.0100)} }{2(4)}

=\frac{+0.76\pm 0.65 }{8}

x = 0.176 or x = 0.014

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 x that results in any negative concentrations.

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_2O_4] = x
= 0.014 M

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.

K_c=\frac{[NO_2]^2}{[N_2][N_2O_4]}

=\frac{(0.072)^2}{0.014}= 0.37

Since the calculated value of Kc matches the given value (to within one digit in the least significant figure), the answer is valid.