Chemistry question on equilibrium

Question

Consider the following equilibrium for which H = +23.12 kJ/mol:

NiBr2(s) + H2(g) ⇌ Ni(s) + 2HBr(g)

When 2.00 moles of NiBr2(s) and H2(g) were each placed into an evacuated 10.0 dm3 container at 578.25 K and the reaction reached equilibrium the pressure in the container was 9.38 x 103 torr. The volume of the container was then changed and the reaction was allowed to reach equilibrium for the second time. At the second equilibrium it was found that the concentration of HBr(g) was 0.0855 M. Calculate the total pressure (torr) in the container at the second equilibrium.

Answer 1

For an equilibrium of gases, we can write two different equilibrium expressions, one in terms of concentrations and one in terms of pressures. These two equilibrium constants are related to each other and we can calculate one from the other.

K_c = [HBr]²/[H₂]

K_p=P_HBr²/P_H2

Notice that NiBr₂ and Ni don't appear in either equilibrium expression, because they are solids. Solids do not have concentrations or pressures, if you want me to talk more about why we don't include solids feel free to ask! The other thing I want to point out is that the concentration/pressure of HBr is squared. This is because it has a coefficient of 2. H₂ has a coefficient of 1, so it is raised to the first power.

So we are told we start out with 2 moles of each reactant in a 10 dm³ container. 10 dm³ = 10 L, so [H₂] = 2 mol/10 L = 0.2 M. Let's build an "ICE" reaction table. It will have three lines, for the initial, change, and equilibrium concentrations. The first column for NiBr₂ will be left blank because it is a solid. For the second column, for H₂, we will have

initial: 0.2

change: - x

equilibrium: 0.2 - x

Third column is also a solid, so blank. Fourth columnm for HBr, will be

initial: 0

change: + 2x

equilibrium: 2x

Normally at this point, we would plug the equilibrium concentrations into the expression for K_c, and solve for x.

K_c = [HBr]²/[H₂]

K_c = 4x²/(0.2 - x)

But we don't know K_c! So we'll have to figure x out some other way. We also know the pressure, temperature, and volume of the equilibrium mixture of H₂ and HBr gases. This means I can calculate the number of moles from the ideal gas law. Make sure you are using the correct units that will cancel out with the units of the ideal gas constant R (so convert torr to atm and dm to L).

PV = nRT

n = PV/RT

n = (12.34 atm × 10 L)/(0.08296 L atm/mol K × 578.25 K)

n = 2.601 mol

But concentration would more convenient for us.

n/V = [gas] = 2.601 mol/ 10 L = 0.2601 M

This concentration is the combined concentration of the H₂ and HBr gases.

[gas] = [H₂] + [HBr]

Let's plug in our concentrations from our table that we wrote in terms of x.

0.2601 = 2x + 0.2 - x

0.2601 = x + 0.2

x = 0.0601 M

[HBr] = 2x = 0.1202 M

[H₂] = 0.2 - x = 0.1399 M

So we've answered the first part of the question! For the second question, the volume is changed and then equilibrium is reestablished. We don't know anything about the new equilibrium amounts, so it would be really helpful to know either K_c or K_p. Let's take the concentrations we just calculated and use them to get K_c.

K_c = [HBr]²/[H₂] = 0.1202²/0.1399 = 0.1033

Like I said at the beginning, we can relate K_p and K_c. Knowing K_p would make things easier since the second question gives us a partial pressure and asks for the total pressure. All we need is the ideal gas law.

PV = nRT, solve for concentration, n/V

n/V = P/RT, now plug in P/RT for each concentration

K_c = (P_HBr/RT)² × (RT/P_H2)

When writing this I used that dividing by [H₂] = P_H2/RT is the same as multiplying by the reciprocal. Now one of the RT's cancels out.

K_c = 1/RT × P_HBr²/P_H2

This is the expression we need to solve the rest of the problem, we don't actually need to calculate K_p. But if we did we would have

K_c = K_p/RT

and in general, for any reaction,

K_c = K_p/(RT)^Δn, where Δn = moles of gaseous product - moles of gaseous reactant

But back to the problem.

K_c = 1/RT × P_HBr²/P_H2

We know R and T. They tell us P_HBr. We just calculated K_c. So the unknown is P_H2. Let's solve for it.

P_H2 = P_HBr²/(K_cRT), plug everything in (after converting your HBr pressure from torr to atm so it agrees with R)

P_H2 = 4.057²/(0.1033×0.08206×578.25) = 3.358 atm.

We have both partial pressures. The total pressure is the sum of the partial pressures. Convert back to torr since that is what the question asks for.

P_total = P_H2 + P_HBr = 3.358 + 4.057 = 7.415 atm

7.415 atm × 760 torr/1 atm = 5635 torr

edit: Thanks for pointing out that I gloss over the bit where we get the pressure of HBr, I can go over that now. We'll use the ideal gas law.

PV = nRT

We'll have to solve for pressure.

P = nRT/V

You might be thinking, we can't use this equation! We don't know n or V! But what we DO know is n/V (molarity).

P = (n/V) × RT

The partial pressure you get out of this will be in atm.