Kaplan Practice Test 2: #47 of Physical Sciences

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Gauss44

Full Member
10+ Year Member
Joined
Oct 28, 2012
Messages
3,185
Reaction score
416
Passage:

A researcher investigated the equilibrium between CO2, C, and CO as a function of temperature. The equation is given below.

CO2(g) + C(s) ⇌ 2CO(g)

Carbon dioxide, at 298 K and 1 atm, and an excess of powdered carbon were introduced into a furnace, which was then sealed so that the pressure would increase as the temperature rose. The furnace was heated to and held constant at a predetermined temperature. The pressure within the furnace chamber was recorded after it had remained unchanged for one hour. The table below shows the pressures recorded for a series of temperatures together with the pressures expected if no reaction had taken place.

I don't think the table (directly below) is relevant, so please ignore any formatting problems.

T(K) Pr (P recorded after reaction in atm) Pe (P expected without reaction in atm)
900 3.4 3.0
950 3.8 3.2
1000 4.3 3.4
1050 5.0 3.5
1200 7.2 4.0

#47
Question:


When the system stabilized at 1200 K, a sample of helium was injected into the furnace. What should happen to the amount of carbon dioxide in the system?

A. It should increase
B. It should decrease
C. It should be completely converted to carbon monoxide
D. It will remain the same

Highlight for correct answer and Kaplan's explanation: Correct answer is D. You should know that helium, a noble gas, is very unreactive and would almost certainly not react with any of the species in the furnace. Because the helium does not react with any of the species that participate in the equilibrium, the equilibrium is unaffected by the addition of helium. Even though it increases the total pressure inside the system, the partial pressures of the reacting gases are unchanged (Dalton's law), and therefore, they keep on behaving as the helium weren't present. My question is, why isn't A correct per le chatelier's principle? There are more moles of gas on the right side of the equation, so why wouldn't it shift left?

Edit (highlight to read): #50 then says, "Which of the following is NOT necessarily true about the equilibrium reaction among CO2, C, and CO?
A. The standard entropy change is positive.
B. A decrease in pressure at constant temperature would shift the equilibrium to the right.
C. Addition of CO will shift the equilibrium to the left.
D. The standard Gibbs free energy change is negative.
Correct answer is D.
Answer key says: ...Choice A is wrong (that is the statement is certainly true) because the entropy change is positive. Choice B is wrong because, according to le chatelier's principle, a decrease in pressure will favor the side of the reaction with more moles of gas. In this case, it is the products side, and the equilibrium is said to have shifted to the right. Choice C is wrong because, again according to le chateliers principle, addition of a product will shift the equilbrium to the left; that is, the reactants side...


I'm starting to think it's an error.

Members don't see this ad.
 
Last edited:
Awesome question. I struggled with this in GChem2 for a while.

So right, Le Chatelier's states that when a stress causes a shift in the equilibrium, the system will act to counteract the effect of the disturbance. Now the equilibrium expression for that reaction looks like:

K = (PCO)^2 / (PCO2)

So when we add a stress that changes that, we shift back to equilibrium. When we add the helium, it increases the total pressure by the same amount of pressure as if we had it in a container alone. So if our total pressure was 1 atm, and that consisted of 0.66 atm CO2 and 0.33 CO, then we added 5 atm, of helium, the total pressure would be 6 atm, but the pressure due to CO2 would still be 0.66 atm and that due to CO would be 0.33. Since the equilibrium isn't changed, nothing happens.

I think the case you're thinking of is if they said they introduced helium in such a way that total pressure was not affected. Well now, total pressure is the same, but you're adding more gas. That means that the pressure contributed due to the gases already there is less what it was before. It's like saying x + y = 10. Then you all of a sudden say that x + y + z = 10. Well in the second case, in order for those all to equal 10, x and y have to be less than what they were in the first case.
 
Top