Question

15.7 g of H₂ gas is mixed with 294 g of I₂ gas in a 5.00 L tank at 25 ºC. The two gasses react to form hydrogen iodide gas by the reaction H₂(g)+I₂<->2HI(g). Given K=710 at ºC, what are the equilibrium concentrations of each gas? (Answers are listed in order of [H₂], [I₂] and [HI].)

a.1.560, 0.230, 0.000 b.0.000, 1.560, 3.120 c.1.330, 0.000, 0.460 d.1.328, 0.000230, 0.464

Answer 1

Step 1 - We are to find initial concentrations
[H₂]_o = 15.7 g/5.00 L x 1 mol/2.016 g
[H₂]_o = 1.56 M
[I₂]_o = 294 g/5.00 L x 1 mol/253.2 g
[I₂]_o = 0.232 M
[HI]₀ = 0 M

Step 2 - In second step we are to 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 H₂ and I₂, 2 moles of HI will be formed.

This indicates the limiting reactant will be the I₂ and some small amount of H₂ gas will be left over. At the end of the reaction, the final concentrations would be
[H₂]_f = 1.56 M - 0.232 M
[H₂]_f = 1.328 M
[I₂]_f = 0 M
[HI]_f = 2(0.232 M)
[HI]_f = 0.464 M

Step 3 - In third step we are to find equilibrium concentrations

We know the reaction will reach an equilibrium condition. Some amount of I₂ gas will not be consumed. If the amount not consumed is "X" then the concentrations at equilibrium are
[H₂] = [H₂]_f + X
[H₂] = 1.328 + X
[I₂] = X
[HI] = [HI]_f - 2X
[HI] = 0.464 - 2X
The equilibrium constant for this reaction is

K = [HI]²/[H₂][I₂]

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 + X²) = 4X² - 1.856X + 0.215
937.2x + 710X² = 4X² - 1.856X + 0.215
(710-4)X² + (937.2 + 1.856)X - 0.215 = 0
706X² + (939.056)X - 0.215 = 0

Solve for X using the quadratic formula

for ax² + bx + c
x = [-b ± (b² - 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

[H₂] = 1.328 + X
[H₂] = 1.328 + 2.3 x 10^-4
[H₂] = 1.328 M

[I₂] = X
[I₂] = 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

[H₂] = 1.328 M
[I₂] = 2.3 x 10^-4 M
[HI] = 0.464 M