This example problem demonstrates how to calculate the equilibrium concentrations from initial conditions and the reaction's equilibrium constant. This equilibrium constant example concerns a reaction with a "large" equilibrium constant.

### Problem:

15.7 g of H_{2} gas is mixed with 294 g of I_{2} gas in a 5.00 L tank at 25 °C. The two gasses react to form hydrogen iodide gas by the reaction

H_{2}(g) + I_{2}(g) ↔ 2 HI(g).

What are the equilibrium concentrations of each gas?

Given: K = 710 at 25 °C

### Solution:

**Step 1 - Find initial concentrations**

[H_{2}]_{o} = 15.7 g/5.00 L x 1 mol/2.016 g

[H_{2}]_{o} = 1.56 M

[I_{2}]_{o} = 294 g/5.00 L x 1 mol/253.2 g

[I_{2}]_{o} = 0.232 M

[HI]_{0} = 0 M**Step 2 - Find final concentration if reaction goes to completion**

If the reaction is allowed to go to completion, all of the reactants will be consumed to produce the products. This allows us to determine the maximum value of the concentration of products.

From the reaction, we see for every mole of H_{2} and I_{2}, 2 moles of HI will be formed. This indicates the limiting reactant will be the I_{2} and some small amount of H_{2} gas will be left over. At the end of the reaction, the final concentrations would be

[H_{2}]_{f} = 1.56 M - 0.232 M

[H_{2}]_{f} = 1.328 M

[I_{2}]_{f} = 0 M

[HI]_{f} = 2(0.232 M)

[HI]_{f} = 0.464 M**Step 3 - Find equilibrium concentrations**

We know the reaction will reach an equilibrium condition.

Some amount of I_{2} gas will not be consumed. If the amount not consumed is "X" then the concentrations at equilibrium are

[H_{2}] = [H_{2}]_{f} + X

[H_{2}] = 1.328 + X

[I_{2}] = X

[HI] = [HI]_{f} - 2X

[HI] = 0.464 - 2X

The equilibrium constant for this reaction is

K = [HI]^{2}/[H_{2}][I_{2}]

Substitute the values from above

710 = (0.464-2X)^{2}/(1.328 + X)(X)

710(1.328 + X)(X) = (0.464-2X)^{2}

710(1.32X + X^{2}) = 4X^{2} - 1.856X + 0.215

937.2x + 710X^{2} = 4X^{2} - 1.856X + 0.215

(710-4)X^{2} + (937.2 + 1.856)X - 0.215 = 0

706X^{2} + (939.056)X - 0.215 = 0

Solve for X using the quadratic formula

for ax^{2} + bx + c

x = [-b ± (b^{2} - 4ac)^{½}]/2a

X = 2.3 x 10^{-4} and -1.3303

The negative X value is not used as this implies a negative concentration.

Therefore

X = 2.3 x 10^{-4} M

Use this to find the equilibrium concentrations

[H_{2}] = 1.328 + X

[H_{2}] = 1.328 + 2.3 x 10^{-4}

[H_{2}] = 1.328 M

[I_{2}] = X

[I_{2}] = 2.3 x 10^{-4} M

[HI] = 0.464 - 2X

[HI] = 0.464 - 2(2.3 x 10^{-4} M)

[HI] = 0.464 - 4.6 x 10^{-4} M)

[HI] = 0.464 M

### Answer:

The equilibrium concentrations of the reaction are

[H_{2}] = 1.328 M

[I_{2}] = 2.3 x 10^{-4} M

[HI] = 0.464 M

### Alternate Method:

You may notice the equilibrium concentrations are very close to the maximum values found in Step 2. This is generally true of reactions with large equilibrium constants. Since K is the ratio of products to reactants, the larger the K, the more products you would expect at equilibrium.

This reasoning would allow us to make some changes to Step 3. For large values of K, we would expect the value of X to be insignificant compared to the maximum concentrations. We could then assume the change in concentration would be zero. This would make the equilibrium concentrations equal to

[H_{2}] = [H_{2}]_{f} + X

[H_{2}] = 1.328 + 0 = 1.328

[I_{2}] = X

[HI] = [HI]_{f} - 2X

[HI] = 0.464 - 2(0)

[HI] = 0.464

This would make the equilibrium constant equal to

K = [HI]^{2}/[H_{2}][I_{2}]

K = (0.464)^{2}/(1.328)X

720 = 0.215/(1.328)X

956.16X = 0.215

X = 2.25 X 10^{-4}

This value of X, based on the assumption is very close to the actual value calculated above.

This assumption works only when the value of X is within 5% of the other concentrations in the reaction.