Question

# An equilibrium mixture contains 0.650 mol of each of the products (carbon dioxide and hydrogen gas)...

An equilibrium mixture contains 0.650 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container.

CO(g) + H2O(g) <-----> CO2(g) + H2(g)

How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

For the given reaction, Kc = [CO][H2]/[CO2][H2O]

= (0.650x0.650)/(0.200x0.200)

= 10.5625

CO(g) + H2O(g) <-----> CO2(g) + H2(g)

Initially at equilibrium,

Concentration in mol/l                       0.200         0.200                        0.650                   0.650

Concentration at equilibrium          0.300         0.300                             0.550+n          0.550

after change

where n is the added moles of CO2.

So, after reestablishment of equilibrium,

Kc = 10.5625 = (0.550+n)0.550/(0.300x0.300)

n = 1.1784

Thus, 1.1784 moles of CO2 would have to be added.

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