Question 14.12: Finding Equilibrium Concentrations from Initial Concentratio...
Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant
Consider the reaction for the decom-position of hydrogen disulfide.
2 H_{2}S(g)\xrightleftharpoons[]{} 2 H_{2}(g) + S_{2}(g)
K_{c} = 1.67 \times 10^{-7} at 800 °C
A 0.500-L reaction vessel initially contains 0.0125 mol of H_{2}S at 800 °C. Find the equilibrium concentrations of H_{2} and S_{2}.
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_{2}S from the given number of moles and volume.)
2. Use the initial concentrations to calculate the reaction quotient (Q). 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 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.
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 expression for the variable x.In this case, the resulting equation is cubic in x. Although cubic equations can be solved, determining the solutions is not usually easy. 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 you can drop it from any quantities in which it is added to or subtracted from another number (as long as the number itself is not too small).
Check whether your approximation is valid by comparing the calculated value of x to the number it was added to or subtracted 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.
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 this example, the method of successive approximations is used.
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.
6. Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations.
7. Check your answer by substituting the calculated equilibrium concentrations 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.
[H_{2}S] = \frac{0.0125 mol}{0.500 L} = 0.0250 M
2 H_{2}S(g)\xrightleftharpoons[]{} 2 H_{2}(g) + S_{2}(g)| [H_{2}S] | [H_{2}] | [S_{2}] | |
| Initial | 0.0250 | 0.00 | 0.00 |
| Change | |||
| Equil |
By inspection, Q = 0; the reaction will proceed to the right.
2 H_{2}S(g)\xrightleftharpoons[]{} 2 H_{2}(g) + S_{2}(g)| [H_{2}S] | [H_{2}] | [S_{2}] | |
| Initial | 0.0250 | 0.00 | 0.00 |
| Change | -2x | +2x | +x |
| Equil |
| [H_{2}S] | [H_{2}] | [S_{2}] | |
| Initial | 0.0250 | 0.00 | 0.00 |
| Change | -2x | +2x | +x |
| Equil | 0.0250-2x | 2x | x |
K_{c} = \frac{[H_{2}]^{2}[S_{2}]}{[H_{2}S]^{2}}
=\frac{ (2x)^{2}x}{(0.0250 – 2x)^{2}}
1.67 \times 10^{-7} =\frac{ 4x^{3}}{(0.0250 – 2x)^{2}}
. 
1.67 × 10^{−7} =\frac{4x^{3}}{ (0.0250 − \cancel{2}x)^{2}}
1.67 ×10^{-7} = \frac{4x^{3}}{6.25 \times 10^{-4}}
6.25 ×10^{-4}(1.67 × 10^{-7}) = 4x^{3}
x³ =\frac{ 6.25 \times 10^{-4}(1.67 \times 10^{-7})}{4}
x = 2.97 × 10^{-4}
Check the x is small approximation.
\frac{2.97 \times 10^{-4}}{0.0250}× 100% = 1.19%
The x is small approximation is valid; proceed to step 6.
[H_{2}S] = 0.0250 – 2(2.97 × 10^{-4})
= 0.0244 M
[H_{2}] = 2(2.97 × 10^{-4})
= 5.94 × 10^{-4} M
[S_{2}] = 2.97 × 10^{-4} M
K_{c} = \frac{(5.94 \times 10^{-4})^{2}(2.97 \times 10^{-4})}{(0.0244)^{2}}
= 1.76×10^{-7}
The calculated value of K is close enough to the given value when you consider the uncertainty introduced by the approximation. Therefore the answer is valid.