Question

An equilibrium mixture contains 0.550 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?

Answer 1

We’re given the following equilibrium:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

At equilibrium, we have a 1 L container with 0.200 mol of reactants ( CO and H2O) and 0.55 mol of (CO2 and H2)

Molarity = Number of moles of solute / Volume of solution in Litre

This means that the concentration are:

[CO] = [H2O] = 0.2 M

[CO2] = [H2] = 0.55 M

We will now calculate the Kc

Kc = [Products] / [Reactants]

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

= [0.55 X 0.55] / [0.2X0.2]

= 7.56

Now, in order for the CO concentration to be 0.300 M at the new equilibrium point, it must have increased by 0.100 M, which means that the other concentrations must also have increased proportionally based on this stoichiometry.

H₂O + co - H₂ + 0.55 M initial 0.2M 0.2M CO2 0.55M + z -(0.1M) change tool toolM - (0.1m) 0.3M 0.45 M equiliboium 0-3M 0.45m + Z Ko [H2) (CO2) 0.45M] [.45 m + z] [H₂o] [co] [0.3m] [0.3M] This is equal to 7.56 [ke]

Now, We will form an equation [0.45 m] (.45M + 2] = 7.56 [0.3 m] [O.3 MJ On solving we get, 0.45) [0.45 + 2] = 7.56] (0.3] [0-3] [0.45][0-45 + 2] = 0.6804 0.45 +2= 0.6804 = 0.45 0.45 +2= 1.512 Z= 1.512 - 0.450 or Z-1.062

So, in the given 1L vessel, 1.06 moles of CO2 must be added to reach the new equilibrium conditions.