Question

What's the new pressure of a mixture of `CO` and `H_2`after the volume is halved?

A mixture of `CO(g)` and `H_2(g`) is pumped into a previously evacuated 2.0 L reaction vessel. The total pressure of the reaction system is 1.2 atm at equilibrium. What will be the total pressure of the system if the volume of the reaction vessel is reduced to 1.0 L at constant temperature?

(A) Less than 1.2 atm
(B) Greater than 1.2 atm but less than 2.4 atm
(C) 2.4 atm
(D) Greater than 2.4 atm

Answer 1

C) 2.4 atm

Explanation:

Both of the species in the container are in a gaseous state, so we will use the Ideal Gas Law to solve this question:

`PV = nRT`

Since the question doesn't mention any reactions, we can assume that `n`, the number of moles and `T`, the temperature remain constant (`R` of course staying constant because it is a constant). That leaves `V` being changed, in this case halved. Everything on the right side of the equation stayed constant, which means `P`, pressure, must double in order to balance the decrease in volume:

`1.2` atm * `2 = 2.4` atm

Answer 2

`"B")` `"Greater than 1.2 atm but less than 2.4 atm."`

Explanation:

Having done some research online, this question comes from the 2014 AP Chemistry exam. It refers to an equilibrium:

`"CO(g) + H"_2("g")` `rightleftharpoons` `"CH"_3"OH(g)"`

Therefore, Le Chatelier's principle is involved:

If a system at equilibrium is subjected to a change of pressure, temperature, or the number of moles of a component, there will be a tendency for a net reaction in the direction that reduces the effect of this change.
http://www.chem1.com/acad/webtext/chemeq/Eq-02.html

So, using Boyle's law, we expect the new pressure to be `"2.4 atm"`, but because we are dealing with an equilibrium, Le Chatelier's principle applies, and we end up with a pressure that is greater than `"1.2 atm"` but less than `"2.4 atm"`.