Calculate the equilibrium concentrations of each gas at 430 ∘C.

Question

A mixture of 0.435 M H2,0.435 M H2, 0.357 M I2,0.357 M I2, and 0.874 M HI0.874 M HI is enclosed in a vessel and heated to 430 °C.

H2(g)+I2(g)↽−−⇀2HI(g)𝐾c=54.3 at 430 ∘CH2(g)+I2(g)↽−−⇀2HI(g)Kc=54.3 at 430 ∘C

Calculate the equilibrium concentrations of each gas at 430 ∘C.

Answer 1

H₂(g) + I₂(g) ↽−−⇀ 2HI(g) 𝐾c=54.3

0.435....0.357...........0.874........Initial

Q = [HI]² / [H₂][I₂] = (0.874)² / (0.435)(0.357)

Q = 4.91 which is less than Kc of 54.3, so the reaction will proceed toward the product side

H₂(g) + I₂(g) ↽−−⇀ 2HI(g) 𝐾c=54.3

0.435....0.357...........0.874........Initial

-x..........-x................+2x............Change

0.435-x...0.357-x......0.874+2x...Equilibrium

54.3 = (0.874+2x)² / (0.435-x)(0.357-x)

x = 0.289 (check the calculations. Used a quadratic calculator)

Equilibrium concentrations:

[H₂] = 0.435 - 0.289 = 0.146 M

[I₂] = 0.357 - 0.289 = 0.068 M

[HI] = 0.874 + (2x0.289) = 1.45 M

Answer 2

You do the ice diagram to work out the concentrations at equilibrium in terms of the extent of reaction taking place. We can use concentrations for the ₂extent instead of moles because volume is constant. You can think of the extent as the H₂ molarity reacted (x)..

Equilibrium concentrations after reaction:

H₂: .435 - x I₂: .357 - x and HI: .874 - 2x (note if x turns out negative, it means reaction went the other way)

Now use the equilibrium values in the K expression and equate to K_C

54.3 = (.874-x)²/((.435-x)(.357-x)) You can solve this with quadratic equation or numerically

I get x = .310 and ..464 (thrown out, > .435)

You can now substitute x into the expression for the equil. conc.

Please consider a tutor. Take care.