An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxid

Question

An equilibrium mixture contains 0.750 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)

An equilibrium mixture contains 0.750 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?

Explanation / Answer

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

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

   = 0.75^2/0.2^2

   = 14.06

After addition of CO2,

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

14.06 = ((0.75+x-0.1)*(0.75-0.1))/(0.3*0.3)

x = 1.3

NO of mol of co2 must add = 1.3 mol

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