A 6 L container holds 5 mol and 6 mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 480^oK to 420^oK. How much does the pressure change?

1 Answer
Oct 17, 2017

Depending on how you interpret the question, the answer is that the pressure drops to either 34.5% or to 21.2% of its starting value.
See the explanation below.

Explanation:

If I am understanding the question properly, it says that 3 molecules of A and 2 molecules of B bind together to form a single molecule. What I'm not clear on is whether that new molecule is one of gas, or not. Assuming it is, here's what must be done.

First, we have a limiting reagent problem to determine how many molecules of B remain once all of A has reacted.

Because it takes 3 molecules of A to react with only 2 of B, the 5 moles of A present will react with 5" mol of A"xx(2" mol of B")/(3" mol of A") = 3 1/3 mol of B will be used up and 2 2/3 mol of B will remain, unreacted.

So, the reaction will result in the container now holding 2 2/3 mol of B. Plus, there will be 5/3 of a mol of the new product present, because all of A is used, at a rate of three molecules per one molecule of product formed.

Note: If the problem meant that this new product was not a gas, then change my answer by leaving these 5/3 of a mole out.

Next, we apply the ideal gas law:

(P_1V_1)/(n_1T_1)=(P_2V_2)/(n_2T_2)

In this problem, V_1=V_2=6L and we leave this out.

n_1=11 mol

n_2=4 1/3 mol (or 2 2/3 mol of you leave out the product.)

T_1=480K

T_2=420K

We will determine the value of P_2/P_1 (the fractional change in P)

From the ideal gas law:

P_2/P_1=(n_2T_2)/(n_1T_1) = (4 1/3 * 420)/(11*480)=0.345

The pressure drops to 34.5 % of its initial value.

If you leave out the product (assume it to be solid), the answer is

P_2/P_1=(n_2T_2)/(n_1T_1) = (2 2/3 * 420)/(11*480)=0.212

The pressure drops to 21.2% of its original value in this case.