# An 8 L container holds 5  mol and 9  mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 320^oK to 450 ^oK. How much does the pressure change by?

Jun 22, 2016

The pressure decreases by $2316.4 \text{ ""kPa}$

#### Explanation:

The key to this problem is to find the number of moles of gas before and after the reaction.

We can then use the ideal gas equation to find the pressure in each case and hence the pressure change.

Before the gases react we have a total number of moles of 5 + 9 = 14.

We can write the equation as:

${A}_{\left(g\right)} + 2 {B}_{\left(g\right)} \rightarrow A {B}_{2 \left(g\right)}$

This tells us that 1 mole of $A$ reacts with 2 moles of $B$ to give 1 mole of $A {B}_{2}$.

So 9 moles of $B$ will react with 4.5 moles of $A$ to give 4.5 moles of $A {B}_{2}$.

This means that there must be an excess 0.5 moles of $A$ which have not reacted.

So the total moles of gas after reaction = 0.5 + 4.5 = 5.

The ideal gas equation gives us :

$P V = n R T$

$P$ is the pressure

$V$ is the volume

$n$ is the number of moles

$R$ is the gas constant with the value $8.31 \text{ ""J/K/mol}$

$T$ is the absolute temperature

Initial:

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

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

${P}_{1} = \frac{14 \times 8.31 \times 320}{0.008} = 4653600 {\text{ ""N/m}}^{2}$

(note I have converted $\text{L}$ to ${\text{m}}^{3}$)

Final:

${P}_{2} = \frac{{n}_{2} R {T}_{2}}{V}$

$\therefore {P}_{2} = \frac{5 \times 8.31 \times 450}{0.008} = 2337187.5 {\text{ ""N/m}}^{2}$

$\therefore \Delta P = {P}_{1} - {P}_{2} = 4653600 - 2337187.5 = 2316412.5 {\text{ ""N/m}}^{2}$

$\Delta P = 2316.4 \text{ ""kPa}$

($1 \text{N/m"^2=1"Pa}$)

So you can see that the pressure has dropped by $2316.4 \text{ ""kPa}$

Although the temperature has increased this is more than offset by the decrease in the number of particles brought about by the reaction, hence the reduction in pressure.