# 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

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

So, the reaction will result in the container now holding

Note: If the problem meant that this new product was not a gas, then change my answer by leaving these

Next, we apply the ideal gas law:

In this problem,

We will determine the value of

From the ideal gas law:

The pressure drops to 34.5 % of its initial value.

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

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