# A 5 L container holds 5  mol and 10  mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from 320^oK to 450 ^oK. By how much does the pressure change?

Feb 24, 2018

There is a change of $- 56.649$ atmospheres.

#### Explanation:

To solve for $\Delta P$, we need two values: ${P}_{\text{initial}}$ and ${P}_{\text{final}}$.

First, we calculate the initial pressure using the Ideal Gas Law:

${P}_{i} V = {n}_{i} R {T}_{i}$

Here,

$V = 5 \text{L}$

${n}_{i} = 15 \text{mol}$

${T}_{i} = 320 \text{K}$

$R = 0.0821 {\text{L atm K"^-1"mol}}^{_} 1$

To solve for ${P}_{i}$, we rearrange:

${P}_{i} = \frac{{n}_{i} R {T}_{i}}{V}$

And input:

${P}_{i} = \frac{15 \cdot 0.0821 \cdot 320}{5}$

${P}_{i} = 78.816 \text{atm}$

Now, we need the equation that took place. It is:

$5 \text{A"+10"B"rarr2"A"_2"B"_5+"A}$. Here, one mole of $\text{A}$ is left unreacted.

In the products side, we find we now have $3$ moles of gas. Use the Ideal Gas Law again:

${P}_{f} V = {n}_{f} R {T}_{f}$

Here,

${n}_{f} = 3 \text{mol}$

${T}_{f} = 450 \text{K}$

Rearranging to solve for ${P}_{f}$:

${P}_{f} = \frac{{n}_{f} R {T}_{f}}{V}$ and inputting:

${P}_{f} = \frac{3 \cdot 0.0821 \cdot 450}{5}$

${P}_{f} = 22.167 \text{atm}$

We know that $\Delta P = {P}_{f} - {P}_{i}$. Inputting:

$\Delta P = 22.167 - 78.816$

$\Delta P = - 56.549 \text{atm}$