Question

An equilibrium mixture contains 0.20 moles of hydrogen gas, 0.80 moles of carbon dioxide, 0.10 moles of carbon monoxide, and 0.40 moles of water vapor in a 1.00-liter 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 0.20 moles? The equation for the reaction is:

CO_(g)+H₂O_(g) <=> CO_{2 (g)}+H_{2 (g)}

Answer 1

Answer – Given, at equilibrium

moles of H₂ = 0.20 moles , moles of CO₂ = 0.80 moles , moles of CO = 0.10 mole

moles of H₂O = 0.40 moles , volume = 1.0 L

so concentration is same with moles for each, since volume = 1.0 L

First we need to calculate the Kc for this reaction

Kc = [CO₂][ H₂] / [CO] [H₂O]

      = (0.80 M * 0.20 M) / (0.10 M* 0.40 M)

      = 4.00

So, we need to increase the moles of CO by 0.20 moles

So we need to do the reverse reaction

    CO_{2 (g)} + H_{2 (g)} <------> CO_(g) + H₂O_(g) , Kc = 0.25

I     Y           0.20                0.1        0.40

C   -x          -x                     +x          +x

E Y-x       0.20-x            0.20          0.4+x

So, 0.1+x = 0.20

So, x = 0.10

So, at equilibrium, [CO] = 0.20 M

[H₂O] = 0.40+0.1 = 0.50

[CO₂] = Y -0.10

[H₂] = 0.20-x = 0.20-0.1 = 0.1 M

So, Kc = [CO] [H₂O] / [CO₂][ H₂]

0.25 = 0.20 * 0.50 / (Y-0.10) *0.1

So, 0.25 = 0.1 / 0.1Y – 0.01

So, 0.25 (0.1Y-0.01) = 0.1

0.025Y – 0.0025 = 0.1

0.025 Y = 0.1+0.0025

0.025 Y = 0.1025

So, Y = 0.1025 / 0.025

          = 4.1

So, Y = 4.1 – 0.80 = 3.3 M

So moles CO₂ = 3.3 M * 1.0 L = 3.3 moles

So 3.3 moles of CO₂would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.20 moles.