# A 5 L container holds 9  mol and 12  mol of gasses A and B, respectively. Every three of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from 320^oK to 210 ^oK. By how much does the pressure change?

May 7, 2018

The Pressure within the container decreases by
$\Delta P = 9.43 \cdot {10}^{6} \textcolor{w h i t e}{l} \text{Pa}$

#### Explanation:

Number of moles of gaseous particles before the reaction:
${n}_{1} = 9 + 12 = 21 \textcolor{w h i t e}{l} \text{mol}$

Gas A is in excess.
It takes $9 \cdot \frac{3}{2} = 13.5 \textcolor{w h i t e}{l} \text{mol">12 color(white)(l)"mol}$ of gas B to consume all gas A and $12 \cdot \frac{2}{3} = 8 \textcolor{w h i t e}{l} \text{mol"<9 color(white)(l)"mol}$ vice versa.
$9 - 8 = 1 \textcolor{w h i t e}{l} \text{mol}$ of gas A would be in excess.

Assuming that every two molecules of A and three molecules of B combine to yield a single gaseous product molecule, the number of moles of gas particles present in the container after the reaction would equal to
$\textcolor{\mathrm{da} r k b l u e}{{n}_{2}} = 12 \cdot \frac{\textcolor{\mathrm{da} r k b l u e}{1}}{3} + 1 = \textcolor{\mathrm{da} r k b l u e}{5} \textcolor{w h i t e}{l} \text{mol}$

The volume of the container in the appropriate SI unit would be
$V = 5 \textcolor{w h i t e}{l} {\text{dm}}^{3} = 5 \cdot {10}^{- 3} \textcolor{w h i t e}{l} {m}^{3}$

The temperature drops from ${T}_{1} = 320 \textcolor{w h i t e}{l} \text{K}$ to ${T}_{2} = 210 \textcolor{w h i t e}{l} \text{K}$ during the reaction. Apply the ideal gas law with

$R = 8.314 \textcolor{w h i t e}{l} {\text{m" ^3* "Pa"* "mol" ^(-1)* "K}}^{- 1}$

gives

${P}_{1} = \frac{{n}_{1} \cdot R \cdot {T}_{1}}{V} = 1.12 \cdot {10}^{7} \textcolor{w h i t e}{l} \text{Pa}$

$\textcolor{\mathrm{da} r k b l u e}{{P}_{2}} = \frac{\textcolor{\mathrm{da} r k b l u e}{{n}_{2}} \cdot R \cdot {T}_{1}}{V} = \textcolor{\mathrm{da} r k b l u e}{1.75 \cdot {10}^{6}} \textcolor{w h i t e}{l} \text{Pa}$

$\textcolor{\mathrm{da} r k b l u e}{\Delta P} = \textcolor{\mathrm{da} r k b l u e}{{P}_{2}} - {P}_{1} = \textcolor{\mathrm{da} r k b l u e}{- 9.43 \cdot {10}^{6}} \textcolor{w h i t e}{l} \text{Pa}$

Hence the pressure decrease by $\textcolor{\mathrm{da} r k b l u e}{9.43 \cdot {10}^{6}} \textcolor{w h i t e}{l} \text{Pa}$.

Note that the figures in dark blue are dependent on the assumption that every two moles of gas A and three moles of gas B combine to form one mole of product, which is also a gas. See if you can find the right $\Delta P$ based on further information given in the question.

Reference
 The Ideal Gas Law, http://www.science.uwaterloo.ca/~cchieh/cact/c120/idealgas.html