Question

An industrial chemist introduces 3.4 atm H₂ and 3.4 atm CO₂ into a 1.00-L container at 25.0°C and then raises the temperature to 700.0°C, at which K_eq = 0.534:
H₂(g) + CO₂(g) ⇔ H₂O(g) + CO(g)
How many grams of H₂ are present after equilibrium is established?

Answer 1

Change to concentration:

[H2] = P/(RT) = 3.4/(0.082*298) = 0.1391

[CO2] = P/(RT) = 3.4/(0.082*298) = 0.1391

initially

K = [H2O][CO]/ [H2][CO2]

in equilibrium

[H2O] = 0 +x

[CO] = 0+x

[H2] = 0.1391-x

[CO2] = 0.1391-x

subs

0.534 = x*x / (0.1391-x)^2

sqrt(0.534) = x/(0.1391-x)

0.7307(0.1391-x) = x

0.10164037 - 0.7307x = x

1.7307x = 0.10164037

x = 0.10164037 /1.7307 = 0.05872

[H2O] = 0 +x = 0.05872

[CO] = 0+x = 0.05872

[H2] = 0.1391-0.05872 = 0.08038 M

[CO2] = 0.1391-0.05872 = 0.08038