Question

A mixture of 0.47 mol of H2 and 3.59 mol of Cl2 in one-liter flask is heated to 2800 oC. Calculate the equilibrium partial pressures of H2, Cl2 and HCl at equilibrium if total pressure is 2.00 atmos. Kp = 193 at 2800 oC

Answer 1

Let's write the overall equation:

H₂ + Cl₂ <--------> 2HCl

We have the value of Kp, the temperature and R, so we can calculate Kc, and then, the equilibrium concentrations:

Kp = Kc(RT)ⁿ

n = 2 - (1+1) = 0

Kp = Kc(RT)⁰

Kp = Kc

The value of Kc is equal to Kp so, let's write an ICE chart:

r: H₂ + Cl₂ <--------> 2HCl

i: 0.47 3.59 0

e: 0.47-x 3.59-x 2x

193 = (2x)² / (0.47-x)(3.59-x)

193 = 4x² / (1.6873-4.06x+x²)

193(1.6873-4.06x+x²) = 4x²

325.65 - 783.58x + 193x² = 4x²

189x² - 783.58x + 325.65 = 0

Using the quadratic formula, and solving for x, we have the following values:

x₁ = 0.4577 M

x₂ = 3.688 M

For obvious reason, the value of x is 0.4577 M; then the concentration in equilibrium:

[H2] = 0.47-0.4577 = 0.0123 M

[Cl2] = 3.59-0.4577 = 3.1323 M

[HCl] = 2*0.4577 = 0.9194 M

As we know, the volume is 1 L, so concentration is equal to moles. The total moles:

0.0123 + 3.1323 + 0.9194 = 4.064 moles

Now, the molar fractions:

XH = 0.0123 / 4.064 = 0.003

XCl = 3.1323 / 4.064 = 0.771

XHCl = 0.9194 / 4.064 = 0.226

Finally the pressures:

PH = 0.003 * 2 = 0.006 atm

PCl = 0.771 * 2 = 1.542 atm

PHCl = 0.226*2 = 0.452 atm

Hope this helps