An equilibrium mixture contains 0400 mol of each of the prod

An equilibrium mixture contains 0.400 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. 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? Number mol

Solution

             CO(g) + H2O(g) <----> CO2(g) + H2(g)
first equil 0.2 M   0.2 M          0.4 M   0.4 M

added         +y      -              -        -

change        -x       -x             +x     +x

re-est
equilibrium   0.3     0.2-x          0.4+x   0.4+x

Kc = [h2][co2]/[co][h2o]

   = (0.4*0.4)/(0.2*0.2)

   = 4
after re-est equilibrium,

4 = (0.4+x)^2/((0.3*(0.2-x))

x = 0.03923

so that,

[co] = 0.2+y-x = 0.3

     0.3 = 0.2+y-0.03923

y = no of mol of CO added = 0.13923 mol

answer: 0.13923 mol