# A 7 L container holds 7  mol and 18  mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from 320^oK to 330 ^oK. By how much does the pressure change?

Oct 14, 2017

Pressure changes directly with amount of matter and with temperature. We'll do this in two steps.

#### Explanation:

If the reaction is of the form $A + 3 B \to C$ where $C = A {B}_{3}$:

(1)
In the reaction 18 mol of B will bind to 6 mol of A to form 6 mol of C (assuming C is also a gas!).
1 mol of A will be left. So now we have 7 mol in stead of 25 mol.
${p}_{1} = {p}_{0} \times \frac{7}{25}$

(2)
Temperature goes from 320 K to 330 K:
${p}_{e n d} = {p}_{1} \times \frac{330}{320}$

Total change:
${p}_{e n d} = {p}_{0} \times \frac{7}{25} \times \frac{330}{320} \approx 0.29 \times {p}_{0}$

Note:
More likely the reaction is of the form:
${A}_{2} + 3 {B}_{2} \to 2 C$ where $C = A {B}_{3}$

In this case the number of moles goes from:
$7 A + 18 B \to 1 A + 12 C$
And the fraction in step (1) will be $\frac{13}{25}$