A $3 L$ $3 L$ container holds $9$ $9$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $160$ $160$ $K$ $K$ to $360$ $360$ $K$ $K$ . How much does the pressure change?

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A $3 L$ $3 L$ container holds $9$ $9$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $160$ $160$ $K$ $K$ to $360$ $360$ $K$ $K$ . How much does the pressure change?

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Answer 1 · 2016-04-04T03:34:28

The pressure will be reduced by the chemical reaction but increased by the temperature change. The reaction prevails, so the final pressure is approximately $0.54$ $0.54$ times the initial pressure.

Explanation:

We need to assume that A and B are monatomic initially, because we are not told.

The balanced equation will be:

$2 A + 3 B \to A_{2} B_{3}$ $2A + 3B to A_2B_3$

Gas B will be the limiting reagent, since the reaction requires $3$ $3$ $m o l$ $mol$ of B for every $2$ $2$ $m o l$ $mol$ of A, and 12 is less than $\frac{3}{2}$ $3/2$ times 9.

That means all $12$ $12$ $m o l$ $mol$ of B will react, with $8$ $8$ $m o l$ $mol$ of A, producing $4$ $4$ $m o l$ $mol$ of $A_{2} B_{3}$ $A_2B_3$ . There will still be $1$ $1$ $m o l$ $mol$ of unreacted A, for a total of $5$ $5$ $m o l$ $mol$ of gas.

The volume does not change, so there are two steps in considering the change in pressure:

We have gone from $21$ $21$ $m o l$ $mol$ of gas to $5$ $5$ $m o l$ $mol$ of gas in the same volume, to the pressure will be $\frac{5}{21}$ $5/21$ of its initial value.
The temperature has changed from $160$ $160$ $K$ $K$ to $360$ $360$ $K$ $K$ , which means the pressure will be $\frac{360}{160}$ $360/160$ of the initial value.

Combining these two things, the final pressure will be $\frac{5}{21} \times \frac{360}{160} = \frac{1800}{3360} \approx 0.54$ $5/21xx360/160=1800/3360~~0.54$ times as great.

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A $3 L$ $3 L$ container holds $9$ $9$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $160$ $160$ $K$ $K$ to $360$ $360$ $K$ $K$ . How much does the pressure change?

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Explanation:

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A 3 L3L3 L container holds 9 99 mol and 12 1212 mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 160160160 KKK to 360360360 KKK. How much does the pressure change?

1 Answer

Explanation:

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A $3 L$ $3 L$ container holds $9$ $9$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from $160$ $160$ $K$ $K$ to $360$ $360$ $K$ $K$ . How much does the pressure change?