I keep going through this problem in my homework, but I cannot get the right ans

Question

I keep going through this problem in my homework, but I cannot get the right answer. I believe I am messing up on a small step. Any help?

14. An industrial chemist introduces 1.5 atm H2 and 1.5 atm CO2 into a 1.00-L container at 25.0°C and then raises the temperature to 700.0°C, at which Keq = 0.534:
H2(g) + CO2(g) H2O(g) + CO(g)
How many grams of H2 are present after equilibrium is established?

Explanation / Answer

H2(g) + CO2(g) H2O(g) + CO(g)

Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:

1.5 atm x (973 K / 298 K) = 4.89 atm

Kc = Kp because the moles of product equals the moles of reactants.

At equilibriuim, the amounts are
P(H2) = 4.89 - x
P(CO2) = 4.89 - x
P(H2O) = x
P(CO) = x

Kc = Kp = .534 = (x)(x) / [(4.89 - x)(4.89 - x)]
Take the square root of each side
(.534)^0.5 = x / (4.89 - x)
x = 0.731 (4.89 - x)
x = 3.56 - 0.731 x
1.731x = 3.56
x = 3.56 / 1.731 = 2.05 atm

P(H2) at equilibriuim = 4.89 - 2.05 = 2.83 atm
P(CO2) at equilibrium = 4.89 - 2.05 = 2.83 atm

PV = nRT
n = PV/RT = [(2.83 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0354 mol H2

0.0354 mol H2 x (2.00 g / 1.00 mol) = 0.07 g H2 ................... Answer

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