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An equilibrium mixture contains 0.500 mol of each of the products (carbon dioxide and hydrogen gas)...

An equilibrium mixture contains 0.500 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) +H20 (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?

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Answer #1

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


Kc = (0.500)(0.500) / (0.200)(0.200) = 6.25

. . . . . . .. . . .CO + H2O <--> CO2 . .+. . . H2


initial : . . . . .0.200 . 0.200 . . .0.500+n . . .0.500


change: . . +0.100 +0.100 . . .-0.100 . .   . -0.100
equilibrium: 0.300 . 0.3300 . . .0.400+n . . . 0.400

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


6.25= (0.400 + n)(0.400) / (0.300)(0.300)

0.5625=(0.400 + n)(0.400)

0.05625=(0.400 + n)=

= 1.40625
n = 1.40625-0.400 =1.006 M

moles of carbon dioxide would have to be added = 1.011 mole

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