2CO(g) + O2(g) -> 2CO2(g)

Question

2.0 mol of CO(g) and 2.0 mol of O2(g) are pumped into a rigid, evacuated 4.0L container, where they react to form CO2(g), as shown above. Which of the following values does NOT represent a potential set of concentrations for each gas at a given point during the reaction?

CO O2 CO2

A) 0.5 0.5 0

B) 0 0.25 0.5

C) 0.25 0.25 0.5

D) 0.25 0.38 0.25

The correct answer is C, but it would be really helpful if someone can further explain it. Thank you!

Answer 1

The balanced equation shows that for every mole of CO2 produced, one mole of CO has reacted, and 0.5 mole of O2 has reacted. So, if we designate the moles of CO2 produced at any point in the reaction as X, then X moles of CO has reacted, and 0.5X moles of O2 has reacted.

The number of moles of CO not reacted is 2.0 - X, and the number of moles of O2 not reacted is 2.0 - X.

The concentrations at any time are the moles of each divided by 4 liters.

CO O2 CO2

2.0-X/4.0 2.0-0.5X /4.0 X/4.0

Substitute the number of moles of CO2 for X that will give you the concentrations specified in the problem and calculate the corresponding concentrations predicted for the other reactants.

Fox example, row B specified that the concentration of CO2 is 0.5, this means that X must be 2.0 moles (2.0/4.0 = 0.5). Now,, calculate the expected concentrations for CO and O2. For CO, (2.0 - 2.0/)4.0 = 0, and for O2 (2.0 -0.5*2.0)/4.0 = 0.25. This conforms to the values given in the problem.

Do this for each row, and you will find that the row C results do not conform to the given values.

Answer 2

Suppose x moles of O2 reacts.

Then 2x moles of CO will react, and 2x moles of CO2 will be formed.

So we will have 2-2x moles of CO, 2-x moles of O2 and 2x moles of CO2

As the container is 4l, the concentrations in moles per liter will be

(2-2x)/4, (2-x)/4, (2x/4)

So for each of A,B,C,D we need to be able to solve for x, so that these three values fit with the table values

For (C)

(2x/4) = 0.5

So x = 1

(2-2x)/4 = 0 so we should have 0 moles of CO left and we have 0.25 from the table so

C doesnt work.