Chemistry 2 - Partial Pressure

Question

At a certain temperature, the equilibrium constant, 𝐾,

for the given reaction is 53.3.

H2(g)+I2(g)↽−−⇀2HI(g)𝐾=53.3

At this temperature, the reactants were placed in a container to react. If the partial pressures of H2

and I2 were each 0.300 bar initially, what is the partial pressure of HI at equilibrium?

𝑃HI= ? bar

Answer 1

Here's another approach which dodges quadratics - It's specific to this problem, while looking at this kind of problem like Anthony T. does will get you there more consistently (if you can do quadratics). Well, here goes:

In an equilibrium equation the reaction constant 'K' is set equal to the concentrations (or partial pressures) of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients, as follows:

For the reaction H₂(g)+I₂(g) ↔ 2HI(g), we get K_eq = 53.3 = [HI]²/([I₂]*[H₂])

We are also given the initial partial pressures of I2 and H2 which is 0.300 bar each. The total pressure of a system is the sum of the partial pressures, and in our reaction 2 moles of reactant gas are used up to produce 2 moles of product, so the total pressure should remain the same at equilibria as it is at the start.

Initial partial pressure will be the sum of [I₂] and [H₂]: 0.600 bar.

The final pressure must be the same, so at equilibrium: 0.600 = [HI]+[H₂]+[I₂].

Also because one mole of [H₂] is used for each mole of [I₂] consumed, and because they start at the same starting pressures we can simplify both the partial pressure and the equilibria equations as follows, substituting H₂ for I_{2 - because the partial pressures start and must remain identical}:

The partial pressure equation becomes: 0.600 = [HI]+[H₂]+[H₂] = [HI]+2[H₂].

And the equilibrium equation becomes: 53.3 = [HI]²/([H₂]*[H₂]) = [HI]²/[H₂]².

A fun algebra fact helps us - for positive values of x and y, x²/y² = (x/y)², so 53.3 = ([HI]/[H₂])²

So √53.3 = 7.30 = [HI]/[H₂] ...

Solving for H₂ gives us [H₂] = [HI]/7.30

Plug that into our revised partial pressure equation and we get 0.600 = [HI]+2([HI]/7.30).

2/7.30 is 0.274: 0.600 = [HI]+0.247[HI] = 1.247[HI]...

...Divide both sides by 1.247, and we get [HI] =0.471 bar.

Answer 2

As the partial pressures of the gases involved are proportional to the number of moles of the gases, we can write and ICE table as follows:

                   H2     I2        HI

Initial         0.3     0.3      0

Change       -X      -X      +2X

Equil.        0.3-X   0.3-X   2X

The equilibrium expression is K = (2X)² / (0.3 – X) (0.3- X)

Or 53.3 = 4X² / (0.09 -0.6X +X²)

Simplifying leads to a quadratic equation in X which can be solved using the quadratic formula.

I will leave that up to you as it takes up a lot of space, but you should get two answers:

X1 = 0.413 and X2 = 0.236. As the partial pressure of the HI is 2X, this results in partial pressures of 0.825 bar and 0.471 bar. We can eliminate 0.825 bar since if the reaction went all the way to completion, the partial pressure of HI would be only 0.6 bar. So the answer is 0.471 bar.