Question 15.11: A 1.000 L flask is filled with 1.000 mol of H2(g) and 2.000 ...
A 1.000 L flask is filled with 1.000 mol of H_{2} (g) and 2.000 mol of I_{2} (g) at 448 °C. The value of the equilibrium constant K_{c} for the reaction
H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)at 448 °C is 50.5. What are the equilibrium concentrations of H_{2}, I_{2} , and HI in moles per liter?
Analyze We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species.
Plan In this case, we are not given any of the equilibrium concentrations. We must develop some relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.8, where we calculated an equilibrium constant using initial concentrations.
Solve
(1) We note the initial concentrations of H_{2} and I_{2}:
(2) We construct a table in which we tabulate the initial concentrations:
\hspace{95 pt}H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)| Initial concentration (M) | 1.000 | 2.000 | 0 |
| Change in concentration (M) | |||
| Equilibrium concentration (M) |
(3) We use the stoichiometry of the reaction to determine the changes in concentration that occur as the reaction proceeds to equilibrium. The H_{2} and I_{2} concentrations will decrease as equilibrium is established and that of HI will increase. Let’s represent the change in concentration of H_{2} by x. The balanced chemical equation tells us that for each x mol of H_{2} that reacts, x mol of I_{2} are consumed and 2x mol of HI are produced:
\hspace{95 pt}H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)
| Initial concentration (M) | 1.000 | 2.000 | 0 |
| Change in concentration (M) | -x | -x | +2x |
| Equilibrium concentration (M) |
(4) We use initial concentrations and changes in concentrations, as dictated by stoichiometry, to express the equilibrium concentrations. With all our entries, our table now looks like this:
\hspace{95 pt}H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)
| Initial concentration (M) | 1.000 | 2.000 | 0 |
| Change in concentration (M) | -x | -x | +2x |
| Equilibrium concentration (M) | 1.000 – x | 2.000 – x | 2x |
(5) We substitute the equilibrium concentrations into the equilibrium expression and solve for x:
K_{c} =\frac{[HI]²}{[H_{2}][I_{2}] }=\frac{(2x)²}{(1.000 – x)(2.000 – x)}= 50.5
If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x:
4x² = 50.5(x² – 3.000x + 2.000)
46.5x² – 151.5x + 101.0 = 0
Solving the quadratic equation (Appendix A.3) leads to two solutions for x:
x =\frac{- b \pm \sqrt{b^{2} – 4ac} }{2a} (Appendix A.3)
x =\frac{- (-151.5) \pm \sqrt{(-151.5)^{2} – 4(46.5)(101.0)} }{2 (46.5)}= 2.323 or 0.935
When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative concentrations of H_{2} and I_{2}. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = 0.935 to find the equilibrium concentrations:
[H_{2}] = 1.000 – x = 0.065 M
[I_{2}] = 2.000 – x = 1.065 M
[HI] = 2x = 1.87 M
Check We can check our solution by putting these numbers into the equilibrium expression to assure that we correctly calculate the equilibrium constant:
K_{c} =\frac{[HI]²}{[H_{2}][I_{2}] }=\frac{(1.87)²}{(0.065)(1.065)}= 51
Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not chemically meaningful. Reject this solution to the quadratic equation.