Stumped on chemical equilibrium problem

Question

Suppose 93.0g of HI(g) is placed in a glass vessel and heated to 1107 K. At this temperature, equilibrium is quickly established between HI(g) and its decomposition products, H2(g) and I2(g):

2HI(g) ⇌ H2(g) + I2(g)

The equilibrium constant at 1107 K is 0.0259, and the total pressure at equilibrium is observed to equal 6.45 atm.

(a) Calculate the equilibrium partial pressures of HI(g), H2(g), and I2(g).

(b) Calculate the volume of the vessel.

I am stumped because I do not know of a way to find the initial pressure of HI(g). I have tried to assume the volume of the vessel, but that does not work out. We are not given relative moles or any equilibrium partial pressures, just the total equilibrium pressure. Mole fractions do not seem to work since I do not know the total pressure prior to equilibrium.

I know that I use 93.0g HI and divide it by the molar mass of HI (127.91245 g/mol) to get the moles of HI. Typically I would then use pV=nRT and set up the partial pressure of the chemical as p=RT*n/V where V is the volume of the vessel. However, that is not possible here since the volume of the vessel is not given. We also do not know the ratios of moles or grams between chemicals at equilibrium.

Once the partial pressure of HI(g) is found, it is smooth sailing from there on with the ICE table. Since we are given the total pressure I just need to find the equilibrium partial pressures, and then solve for the change in pressure. However, that is not possible without knowledge of the initial HI(g) partial pressure.

Any help with the process of doing this problem would be very much appreciated. Thank you.

(The answer for a is 0.785 atm for the equilibrium partial pressures of H2(g) and I2(g), and 4.88 atm for the partial pressure of HI. The answer for b is 10.2 L)

Answer 1

Hope the following helps. You were on the right track, but just made the problem harder than it actually is.

93.0 g HI x 1 mol / 127.9 g = 0.727 mols HI(g)

Set up an ICE table and determine moles of each species at equilibrium

2HI(g) ⇌ H2(g) + I2(g)

0.727..........0..........0............Initial

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

0.727-2x......x..........x...........Equilibrium

Kp = (H2)(I2) / (HI)²

0.0259 = (x)(x) / (0.727-2x)²

Using the quadratic equation to solve for x

x = 0.0886 mols

Use x to solve for moles of each @ equilibrium

H2 = 0.0886 mols

I2 = 0.0886 mols

HI = 0.727 - 0.177 = 0.550

Total moles = 0.727 mols

Next find mole fraction of each and multiply by total pressure to find partial pressure

X_H2 = 0.0886 mol / 0.727 mol = 0.219 x 6.45 atm = 0.786 atm

X_I2 = 0.0886 mol / 0.727 mol = 0.219 x 6.45 atm = 0.786 atm

X_HI = 0.550 mol / 0.727 mol = 0.757 x 6.45 atm = 4.88 atm

To find the volume, use PV = nRT and solve for V

V = nRT / P = (0.727 mol)(0.0821 Latm/Kmol)(1107) / 6.45 atm

V = 10.2 L