Question

# The equilibrium constant for the gas phase reaction of carbon monoxide with water to form carbon...

The equilibrium constant for the gas phase reaction of carbon monoxide with water to form carbon dioxide and molecular hydrogen is 0.58 at 1000 degrees C. If a mixture of 0.0200 molar CO, 0.0100 molar H2O, and 0.0050 molar CO2 is allowed to come to equilibrium, what will the equilibrium concentrations of all four species be?

Do the units matter if you use 20 mM CO, 10 mM H2O, and 5 mM CO2, for the inital concentrations?

Please show all work on this Problem, thank you!!!

CO(g) + H2O(g) <---> CO2(g) + H2(g)
Kc = [CO2][H2]/[CO][H2O] = 0.58

Since you have no H2 initially, you know that you must form some to get to an equilibrium. Since all of the substances are in a 1:1:1:1 molar ratio, let x = molar concentration of H2 at equilibrium. Then, the equilibrium concentration of everything will be:
[CO] = 0.0200 -x
[H2O] = 0.0100 - x
[CO2] = 0.050+x
[H2] = x

Plug these into the expression for Kc to get:

0.58 = (0.005+x)(x)/(0.020-x)(0.010-x)

Now, you have a whale of an algebra problem. Just multiply everything through and rearrange it to the form
ax^2 + bx+c = 0. Then use the quadratic equation to solve for x. One of your two roots will make sense. With that, you can calculate the equilibrium concentrations of everything