"Synthesis gas" is a mixture of carbon monoxide and water vapor. At high temperature synthesis gas will form carbon dioxide and hydrogen, and in fact this reaction is one of the ways hydrogen is made industrially.

A chemical engineer studying this reaction fills a 200.mL flask at 12.°C with 1.3atm of carbon monoxide gas and 3.6atm of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 0.82atm of carbon monoxide gas, 3.12atm of water vapor and 0.48atm of carbon dioxide.

The engineer then adds another 0.43atm of carbon monoxide, and allows the mixture to come to equilibrium again. Calculate the pressure of hydrogen after equilibrium is reached the second time. Round your answer to 2 significant digits.

The equilibrium occurs when, in a reversible reaction, the velocity of the formation of the products is equal to the velocity of the formation of the reactants. In this scenario, the partial pressures and the concentration of the components remains constant.

The equilibrium can be characterized by the equilibrium constant, which can be calculated by the concentration (Kc), or by the partial pressure (Kp). In the expression of Kc, solids and pure liquids are not put, and in the expression of Kp, only gases are considered.

The constant is calculated by the product of the concentration, or pressure, of the products, elevated by their coefficients, divided by the product of the concentration, or pressure, of the reactants, elevated by their coefficients. Its value only changes with the temperature.

So, for the reaction given:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

1.3 atm 3.6 atm 0 0 Initial

-x -x +x +x Reacts (stoichiometry is 1:1:1:1)

1.3-x 3.6-x x x Equilibrium

At equilibrium:

pCO = 1.3 - x = 0.82

x = 1.3 - 0.82 = 0.48 atm

pH₂O = 3.12 atm

pCO₂ = 0.48 atm

pH₂ = 0.48 atm

Thus,

Kp = pCO₂*pH₂/(pCO*pH₂O)

Kp = 0.48*0.48/(0.82*3.12)

Kp = 0.090

When more carbon monoxide (CO) is added, the equilibrium will shift to the right, and more products will be formed, is order to reestablish the equilibrium (Le Chatelier's pricniple), so:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

0.82 atm 3.12 atm 0.48 atm 0.48 atm 1st equilibrium

1.25 3.12 0.48 0.48 After the CO addition

-x -x +x +x Reacts

1.25-x 3.12-x 0.48 +x 0.48+x New equilibrium

Because the temperature is the same, Kp = 0.090

0.090 = (0.48+x)*(0.48+x)/[(1.25-x)*(3.12-x)]

0.090 = (0.2304 + 0.96x + x²)/(3.9 - 4.37x + x²)

0.2304 + 0.96x + x² = 0.351 - 0.3933x + 0.09x²

0.91x² + 1.3533x - 0.1206 = 0

Solving this 2nd grade equation at a graphic calculator, for x > 0 and x < 1.25

x = 0.084 atm

pH₂ = 0.48 + 0.084

pH₂ = 0.56 atm