7. The equilibrium constant Kc for the reaction H2 (g) + I2 (g) ⇌ 2 HI (g) is 54.3 at 430°C. At the start of the reaction there are 0.714 moles of hydrogen, 0.984 moles of iodine, and 0.886 moles of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

[HI] = 0.83 M

[H₂] = 0.070 M

[I₂] = 0.18 M

Explanation:

We have the equilibrium:

H₂ (g) + I₂ (g) ⇄ 2 HI (g)

for which

Kc = 54.3 = [HI]² / ([H₂][I₂]

The concentrations of the gases initially are:

[H₂]₀ = 0.714 mol / 2.40 L = 0.2975 M

[I₂]₀ = 0.984 mol / 2.40 L = 0.4100 M

[HI]₀ = 0.886 mol / 2.40 L = 0.3692 M

We should use Q since the question states we are not at equilibrium:

Q = [HI]² / ([H₂][I₂] = 0.369² / (0.298 x 0.410) = 1.15

Q is less than Kc (1.15 vs 54.3), so the system will shift to the right to attain equilibrium, that is the concentrations of H₂ and I₂ will decrease to favor HI

Lets call x the concentration decrease of H₂ and I₂, we can then write the equation for the equilibrium:

Kc = 54.3 = (0.369 + 2x) ² / (0.298 - x) (0.410 - x)

A quadratic equation results which we can solve from our calculator or making use of software:

0.369² + 2 (0.369 x 2x) + (2x) ² = 54.3

x² - 0.708 x + 0.122

0.136 + 1.476x + 4x² = 54.3 (x² - 0.708 x + 0.122)

0.136 + 1.476x + 4x² = 54.3 x² - 38.444 + 6.625

0 = 50.3 x² - 39.920 x + 6.489

The roots are X₁ = 0.566 and X₂ = 0.228

The first is impossible since it will give negative concentrations at equilibrium for H₂ and I₂.

So using X₂ = 0.228

[HI] = 0.369 + 2 (0.228) = 0.83 M

[H₂] = 0.298 - 0.228 = 0.070 M

[I₂] = 0.410 - 0.228 = 0.18 M

After all this work we can check our answer to see if it agrees with Kc

0.83² / (0.070 x 0.18) = 54.7

There is some rounding errors due to the nature of the small numbers involved, but the answer is correct.

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] = 0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q = [n (HI) / V]² / [n (H2) / V][n (I2) / V]

Q = (n (HI) ²) / (nH2 * nI2)

Q = 0.886² / (0.714*0.984)

Q = 1.117

Q
Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 - X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n (HI) / V]² / [n (H2) / V][n (I2) / V]

Kc = (n (HI) ²) / (nH2 * nI2)

KC = 54.3 = (0.886+2X) ² / ((0.714 - X) * (0.984 - X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) / 2.40 = 0.826 M