Enter your answer in the provided box. An industrial chemist introduces 2.0 atm of H2 and 2.0 atm of CO2 into a 1.00−L container at 25.0°C and then raises the temperature to 700°C, at which Kc = 0.534: H2(g) + CO2(g) ⇌ H2O (g) + CO(g) How many grams of H2 are present at equilibrium?

0.0944 gram of H2

Explanation:

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

2.0 atm x (973 K / 298 K) = 6.53 atm

Where

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

At equilibriuim, the amounts are

P(H2) = 6.53 - x

P(CO2) = 6.53 - x

P(H2O) = x

P(CO) = x

Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]

Take the square root of each side

(.534)^0.5 = x / (6.53 - x)

x = 0.731 (6.53 - x)

x = 4.77 - 0.731x

1.731x = 4.77

x = 4.77 / 1.731 = 2.76 atm

P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm

P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm

PV = nRT

n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2

0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g

the answer is symmetrical : )