# A #7 L# container holds #21 # mol and #12 # mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from #350^oK# to #420^oK#. How much does the pressure change?

##### 1 Answer

Let us assume that both gases are ideal gases, so we may use the ideal gas law.

#### Explanation:

Before reaction, we have:

#V_1 = 7 " L"# #n_"A1" = 21 " mol"# ,#n_"B1" = 12 " mol"# #T_1 = 350 " K"#

Applying the law:

Where we have used that

Now, let us write the reaction: 3 molecules of

With our initial conditions, taking in account that we need

- We would need, for 21 mol of
#"A"# ,#3/5 cdot 21 = 12.6# mol of#"B"# . - We would need, for 12 mol of
#"B"# ,#5/3 cdot 12 = 20# mol of#"A"# .

Given that we just have 12 mol of **limiting reactant**, and **excess reactant**).

Thus, reaction is:

So we'll have in the end:

By applying ideal gas law again:

And the change on pressure is: