Question 12.EP.6: When hydrogen gas reacts with iodine gas at elevated tempera...
When hydrogen gas reacts with iodine gas at elevated temperatures, the following equilibrium is established:
H_{2}(g)+I_{2}(g) \rightleftarrows 2 \ HI(g)
A student measured the equilibrium constant as 59.3 at 400°C. If one trial begins with a mixture that includes 0.050 M hydrogen and 0.050 M iodine, what will be the equilibrium concentrations of reactants and products?
Strategy
It may help to begin by summarizing the information from the problem.
Given: [H_{2}]_{initial}= [I_{2}]_{initial} = 0.050 M; K = 59.3 Find: [H_{2}]_{eq}, [I_{2}]_{aq}, and \ [HI]_{aq}
We’ll use (and clarify) the three point strategy outlined above. The chemical equation for the equilibrium is given in the problem. From that equation, we can write the equilibrium expression easily enough:
K=\frac{[HI]^{2}}{[H_{2}][I_{2}]}=59.3
Next we construct the table of concentrations, with places to enter the initial concentrations, the changes in concentration, and the final (equilibrium) concentrations for the three species. So far, only the given initial concentrations are known. (Because only H_{2} and I_{2} are present initially, the initial concentration of HI is zero.)
The rest of the table will be filled in as we work our way through the solution. At this point, the problem seems daunting; we have three unknown equilibrium concentrations, and only one real equation to work with—the equilibrium expression. We will need to use the stoichiometry of the reaction to relate the changes in concentration to one another.
| H_{2} | I_{2} | HI | |||
| Initial Concentration | 0.050 M | 0.050 M | 0 M | ||
| Change in Concentration | |||||
| Final Concentration | |||||
Because only H_{2} and I_{2} are present initially, we know that their concentrations must decrease (and the HI concentration must increase) as the system goes to equilibrium. We can also see from the balanced equation that each H_{2} molecule must react with one I_{2} molecule to form two HI molecules. This gives us the relationship we need to express the changes in concentration in terms of a single variable. Let’s call the change in [H_{2}] “−x.” (The negative sign shows that the concentration decreases.) The 1:1 mole ratio between H_{2} and I_{2} means that the change in [I_{2}] must also be −x, and the 1:2 mole ratio between H_{2} and HI means that the change in [HI] must be +2x. This gives us the second row in the table .
Next we can fill in the third row. To do this, we use the simple relationship that the concentration of each species at equilibrium will be its initial concentration plus the change in concentration. So we just need to sum each column to get expressions for the third row.
Of course, our aim is to find actual values for those three equilibrium concentrations, so we still need to find a way to solve for the variable (x). To do this, we return to the equilibrium expression. If we insert the expressions from the third row of our table into the equilibrium expression, we’ll get the following equation:
K=\frac{[HI]^{2}}{[H_{2}][I_{2}]}=\frac{(2x)^{2}}{(0.050-x)(0.050-x)}=59.3
or
\frac{(2x)^{2}}{(0.050-x)^{2}}=59.3
This equation has only one unknown, so we should be able to solve it for x and then use that result to obtain values for the final concentrations. We could multiply the denominator out and solve using the quadratic formula. Or we might save a little algebra by realizing that the left hand side is a perfect square. This lets us take the square root of both sides to get a linear equation, which we can solve more easily.
\frac{2x}{0.050-x}=7.70
0.39 = 9.70x
x = 0.040
Finally, we use this value of x to calculate all equilibrium concentrations.
[H_{2}] = [I_{2}] = 0.050 − 0.040 = 0.010 M
\ [HI] = 2x = 0.080 M
Analyze Your Answer
We can compare this result to the initial amounts to make sure it seems reasonable. First, we know that x can’t be bigger that 0.05 because that would make [H_{2}] and [I_{2}] negative. Looking more closely, we can also see that because K is noticeably greater than one, we expect most of the original H_{2} and I_{2} to end up on the product side as HI. This is consistent with our answer.
| H_{2} | I_{2} | HI | |||
| Initial Concentration | 0.050 M | 0.050 M | 0 M | ||
| Change in Concentration | -x | -x | +2x | ||
| Final Concentration | |||||
| H_{2} | I_{2} | HI | |||
| Initial Concentration | 0.050 M | 0.050 M | 0 M | ||
| Change in Concentration | -x | -x | +2x | ||
| Final Concentration | 0.050-x | 0.050-x | 2x | ||