Question

Calculating an equilibrium composition after a prior equilibrium determines K "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 75.0 L tank with 31. mol of carbon monoxide gas and 30. mol of water vapor. When the mixture has come to equilibrium he determines that it contains 9.0 mol of carbon monoxide gas, 8.0 mol of water vapor and 22. mol of hydrogen gas. The engineer then adds another 7.5 mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of carbon dioxide after equilibrium is reached the second time. Round your answer to 2 significant digits. mol

Answer 1

To solve this problem, we need to use the concept of chemical equilibrium. At equilibrium, the rates of the forward and reverse reactions are equal, so the concentrations of the reactants and products are constant.

In the first equilibrium, the reaction is:

CO + H2O <-> CO2 + H2

We are given that the initial amounts of CO and H2O are 31 mol and 30 mol, respectively, and the final amounts are 9 mol and 8 mol, respectively. We can use these values to calculate the equilibrium constant, K, for the reaction:

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

Substituting the given values, we get:

K = (22 mol) / (9 mol)(8 mol) = 3.11

In the second equilibrium, we have added 7.5 mol of water, so the amount of CO and H2O at equilibrium will be slightly different. Let's call the new amounts x and y, respectively. The reaction at equilibrium is still:

CO + H2O <-> CO2 + H2

The equilibrium constant expression remains the same, so we can set up the following equation:

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

Substituting the known values and solving for [CO2], we get:

[CO2] = K[CO][H2O] / [H2]
= (3.11)(9 mol)(y mol) / (22 mol)

We know that the total number of moles of CO and H2O is the sum of the initial amount and the amount added, or 31 mol + 7.5 mol = 38.5 mol. We also know that the total number of moles of CO and H2O at equilibrium must be equal to the total number of moles initially, or 38.5 mol. Therefore, we can set up the following equation:

x mol + y mol = 38.5 mol

Substituting this equation into the equation for [CO2] and solving for [CO2], we get:

[CO2] = (3.11)(9 mol)(38.5 mol - x mol) / (22 mol)

We can't solve this equation directly, because we don't know the value of x. However, we do know that the equilibrium concentration of CO must be equal to the initial concentration minus the concentration of CO2, or 9 mol - [CO2] mol. We can substitute this expression into the equation for [CO2] and solve for [CO2]:

[CO2] = (3.11)(9 mol)(38.5 mol - (9 mol - [CO2] mol)) / (22 mol)

Solving this equation, we find that [CO2] = 16.9 mol.

Therefore, the final concentration of CO2 after the second equilibrium is reached is 16.9 mol.