# 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
Jul 3, 2016

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

#### Explanation:

Before reaction, we have:

• ${V}_{1} = 7 \text{ L}$
• ${n}_{\text{A1" = 21 " mol}}$, ${n}_{\text{B1" = 12 " mol}}$
• ${T}_{1} = 350 \text{ K}$

Applying the law:

${P}_{1} {V}_{1} = {n}_{1} R {T}_{1} \rightarrow$

$\rightarrow {P}_{1} = \frac{{n}_{1} R {T}_{1}}{{V}_{1}} = \left\{39 \text{ mol" cdot 0,082 " atm·L/K·mol" cdot 350 " K"}/{7 " L}\right\}$
$= 159.9 \text{ atm}$

Where we have used that ${n}_{1} = {n}_{\text{A1" + n_"B1" = 21 + 12 " mol}}$.

Now, let us write the reaction: 3 molecules of $\text{B}$ + 5 molecules of $\text{A}$ form ${\text{A"_5 "B}}_{3}$.

$5 {\text{A" + 3 "B" rightarrow "A"_5 "B}}_{3}$

With our initial conditions, taking in account that we need $\frac{5}{3} = 1.67$ times more $\text{A}$ than $\text{B}$:

1. We would need, for 21 mol of $\text{A}$, $\frac{3}{5} \cdot 21 = 12.6$ mol of $\text{B}$.
2. We would need, for 12 mol of $\text{B}$, $\frac{5}{3} \cdot 12 = 20$ mol of $\text{A}$.

Given that we just have 12 mol of $\text{B}$, the option 1 is not possible, so we choose option 2 ($\text{A}$ is limiting reactant, and $\text{B}$ is excess reactant).

Thus, reaction is:

$20 \text{A" + 12 "B" rightarrow 4 "A"_5 "B"_3 + 1 "B}$

So we'll have in the end: n_2 = n_{"A"5"B"3} + n_"B" = 4 + 1 = 5 " mol".

By applying ideal gas law again:

P_2 = {5 " mol" cdot 0,082 " atm·L/K·mol" cdot 420 " K"}/{7 " L"} = 24.6 " atm"

And the change on pressure is:

$\Delta P = {P}_{2} - {P}_{1} = 24.6 - 159.9 = - 135.3 \text{ atm}$