Question 15.13: Finding Equilibrium Concentrations from Initial Concentrati...

1. Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. (In these examples, you must first calculate the concentration of H_2S from the given number of moles and volume.)

H_2S=\frac{1.25 ×10^{-4} \ mol}{0.500 \ L}=2.50 ×10^{-4} \ M

2 H_2S( g) \rightleftharpoons 2 H_2( g) + S_2( g)

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

By inspection, \mathcal{Q}_c = 0; the reaction will proceed to the right.

3. Represent the change in the con-centration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products with respect to x.

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

2 H_2S( g) \rightleftharpoons 2 H_2( g) + S_2( g)

5. Substitute the expressions for the equilibrium concentrations (from Step 4) into the expression for the equilibrium constant. Use  the given value of the equilibrium constant to solve the resulting equation for the variable x. In this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions are not usually simple.
However, since the equilibrium constant is small, you know that the reaction does not proceed very far to the right. Therefore, x will be a small number and can be dropped from any quantities in which it is added to or subtracted from another number (as long as the number itself is not too small).

K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}

=\frac{(2x)^2x}{(2.50 ×10^{-4} – 2x)^2}

1.67 \times 10^{-7}=\frac{4x ^3}{(2.50 ×10^{-4} – 2x)^2}

\frac{4x ^3}{6.25 \times 10^{-4}}

6.25 \times 10^{-4}(1.67 \times 10^{-7}) = 4x^3

x^3=\frac{6.25 \times 10^{-4}(1.67 \times 10^{-7})}{4}

x = 1.38 \times 10^{-5}

Check whether your approxi- mation was valid by comparing the calculated value of x to the number it was added to or sub-tracted from. The ratio of the two numbers should be less than 0.05 (or 5%) for the approximation to be valid. If the approximation is not valid, proceed to Step 5a.

Checking the x is small approximation:

The approximation does not satisfy the <5% rule (although it is close).

5a. If the approximation is not valid, you can either solve the equation exactly (by hand or with your calculator) or use the method of successive approximations. In Example 15.13, use the method of successive approximations.

1.67 \times 10^{-7} = \frac{4x^3}{(2.50 ×10^{-4} – 2.76 ×10^{-5})^2}

Substitute the value obtained for x in Step 5 back into the original cubic equation, but only at the exact spot where x was assumed to be negligible, and then solve the equation for x again. Continue this procedure until the value of x obtained from solving the equation is the same as the one that is substituted into the equation.

If you substitute this value of x back into the cubic equation and solve it, you get x = 1.28 ×10^{-5} , which is nearly identical to 1.27 ×10^{-5}. Therefore, you have arrived at the best approximation for x.

6. Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from Step 4) and calculate the concentrations.

[H_2S] = 2.50 ×10^{-4} – 2(1.28 ×10^{-5})
= 2.24 ×10^{-4} \ M
[H_2] = 2(1.28 \times 10^{-5})
= 2.56\times 10^{-5} \ M
[S_2] = 1.28 \times 10^{-5} \ 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 the approximation method and rounding errors could cause a difference of up to about 5% when comparing values of the equilibrium constant.

K_c=\frac{(2.56 \times 10^{-5})^2(1.28 \times 10^{-5})}{(2.24\times 10^{-4})^2}

= 1.67 \times 10^{-7}

The calculated value of K is equal to the given value. Therefore, the answer is valid.