An equilibrium mixture contains 0.350 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 ) + H 2 O ( g ) − ⇀ ↽ − CO 2 ( g ) + H 2 ( 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?

0.8524 moles have to be added

Explanation:

Step 1: Data given

An equilibrium mixture contains 0.350 mol of each of the products

Each product contains 0.200 moles

Volume = 1.00L

Step 2: The balanced equation

CO(g) + H2O(g) ⇆ CO2(g) + H2(g)

Step 3: Calculate Kc

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

Kc = [0.350][0.350]/[0.200][0.200]

Kc = 3.0625

Step 4: The initial number of moles

nCO2 = 0.200 moles = 0.200 M

nCO = 0.200 moles = 0.200 M

nCO2 = O.350 + n moles =  0.350 + n M

nH2 = 0.350 M

Step 5: the number of moles at the equilibrium

nCO2 = 0.200 moles + X = 0.300 mol

⇒ X = 0.100 moles

nCO = 0.200 moles + 0.100 moles = 0.300 moles

nCO2 = 0.350 + n -0.100 moles =  0.250 + n

nH2 = 0.350 -0.100  = 0.250 moles

Step 6: Calculate Kc

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

3.0625 = (0.250 + n)(0.250) / (0.300)(0.300)

0.2756 = (0.250 + n)(0.250)

1.1024 = 0.250+n

n = 0.8524 moles

0.8524 moles have to be added

a

explanation:

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