A $4 L$ $4 L$ container holds $19$ $19$ mol and $20$ $20$ 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 $210^{o} K$ $210^oK$ to $120^{o} K$ $120^oK$ . How much does the pressure change?

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A $4 L$ $4 L$ container holds $19$ $19$ mol and $20$ $20$ 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 $210^{o} K$ $210^oK$ to $120^{o} K$ $120^oK$ . How much does the pressure change?

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Answer 1 · 2018-04-28T05:35:28

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

$n_{0}^{A}$ $n_0^A$ = the initial number of moles of gas A = 19.

$n_{0}^{B}$ $n_0^B$ = the initial number of moles of gas B = 20.

$n_{0}$ $n_0$ = 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!)

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

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

$3 A + 4 B \to C$ $3A+4BrarrC$

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

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

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

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

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

Because of the reaction stoichiometry

$n^{A} = n_{0}^{A} - (\frac{3}{4}) x = 19 - (\frac{3}{4}) 20 = 4$ $n^A=n_0^A-(3/4)x=19-(3/4)20=4$ moles of gas A.

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

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

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

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

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

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A $4 L$ $4 L$ container holds $19$ $19$ mol and $20$ $20$ 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 $210^{o} K$ $210^oK$ to $120^{o} K$ $120^oK$ . How much does the pressure change?

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A 4L4 L container holds 1919 mol and 2020 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 210oK210^oK to 120oK120^oK. How much does the pressure change?

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A $4 L$ $4 L$ container holds $19$ $19$ mol and $20$ $20$ 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 $210^{o} K$ $210^oK$ to $120^{o} K$ $120^oK$ . How much does the pressure change?