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?

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 $2 \frac{2}{3}$ mol of B will remain, unreacted.

So, the reaction will result in the container now holding $2 \frac{2}{3}$ mol of B. Plus, there will be $\frac{5}{3}$ 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 $\frac{5}{3}$ of a mole out.

Next, we apply the ideal gas law:

$\frac{{P}_{1} {V}_{1}}{{n}_{1} {T}_{1}} = \frac{{P}_{2} {V}_{2}}{{n}_{2} {T}_{2}}$

In this problem, ${V}_{1} = {V}_{2} = 6 L$ and we leave this out.

${n}_{1} = 11$ mol

${n}_{2} = 4 \frac{1}{3}$ mol (or $2 \frac{2}{3}$ mol of you leave out the product.)

${T}_{1} = 480 K$

${T}_{2} = 420 K$

We will determine the value of ${P}_{2} / {P}_{1}$ (the fractional change in P)

From the ideal gas law:

${P}_{2} / {P}_{1} = \frac{{n}_{2} {T}_{2}}{{n}_{1} {T}_{1}} = \frac{4 \frac{1}{3} \cdot 420}{11 \cdot 480} = 0.345$

The pressure drops to 34.5 % of its initial value.

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

${P}_{2} / {P}_{1} = \frac{{n}_{2} {T}_{2}}{{n}_{1} {T}_{1}} = \frac{2 \frac{2}{3} \cdot 420}{11 \cdot 480} = 0.212$

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