# A 4 L container holds 19  mol and 20  mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from 210^oK to 120^oK. How much does the pressure change?

Apr 28, 2018

The pressure drops from 168 atm to 22.2 atm.

#### Explanation:

I'll answer a question for a fellow Wisconsinite!

The trickiest part of this problem is determining the change in the number of moles from the reaction.

Let

${n}_{0}^{A}$ = the initial number of moles of gas A = 19.

${n}_{0}^{B}$ = the initial number of moles of gas B = 20.

${n}_{0}$ = the total number of moles of gas present before the reaction = 19 + 20 = 39.

We are given all the information we need to calculate the initial pressure if we make the assumption that gasses A and B are ideal. (In reality this is not often the case at 120K!)

${P}^{0} = \frac{{n}^{0} R {T}^{0}}{V} ^ 0 = \frac{39 \left(0.082057\right) \left(210\right)}{4} \approx 168$ atm

When the two types of gasses react, they form a new gas C. The problem states that the reaction goes as follows:

$3 A + 4 B \rightarrow C$

We know that the limiting reagent here is gas B since we require 33% more B than A for the reaction yet we only have about 5% more B than A before the reaction. We will assume that this reaction goes to completion so that there is no gas B after the reaction.

Let

$x$ = the number of moles of gas B that reacts = 20.

${n}^{A}$ = the number of moles of A left after the reaction.

${n}^{B}$ = the number of moles of B left after the reaction = 0.

${n}^{C}$ = the number of moles of C formed by the reaction.

Because of the reaction stoichiometry

${n}^{A} = {n}_{0}^{A} - \left(\frac{3}{4}\right) x = 19 - \left(\frac{3}{4}\right) 20 = 4$ moles of gas A.

${n}^{C} = \frac{x}{4} = \frac{20}{4} = 5$ moles of gas C

The total number of moles of gas after the reaction, $n$, is

$n = {n}^{A} + {n}^{B} + {n}^{C} = 4 + 0 + 5 = 9$ moles of gas.

Again assuming the ideal gas law holds we can calculate the pressure after the reaction.

$P = \frac{n R T}{V} = \frac{9 \left(0.082057\right) \left(120\right)}{4} \approx 22.2$ atm