A $6 L$ $6 L$ container holds $5$ $5$ mol and $6$ $6$ 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 $480^{o} K$ $480^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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A $6 L$ $6 L$ container holds $5$ $5$ mol and $6$ $6$ 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 $480^{o} K$ $480^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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Answer 1 · 2017-10-17T21:21:46

Depending on how you interpret the question, the answer is that the pressure drops to either 34.5% or to 21.2% of its starting value.
See the explanation below.

Explanation:

If I am understanding the question properly, it says that 3 molecules of A and 2 molecules of B bind together to form a single molecule. What I'm not clear on is whether that new molecule is one of gas, or not. Assuming it is, here's what must be done.

First, we have a limiting reagent problem to determine how many molecules of B remain once all of A has reacted.

Because it takes 3 molecules of A to react with only 2 of B, the 5 moles of A present will react with $5 mol of A \times \frac{2 mol of B}{3 mol of A} = 3 \frac{1}{3}$ $5" mol of A"xx(2" mol of B")/(3" mol of A") = 3 1/3$ mol of B will be used up and $2 \frac{2}{3}$ $2 2/3$ mol of B will remain, unreacted.

So, the reaction will result in the container now holding $2 \frac{2}{3}$ $2 2/3$ mol of B. Plus, there will be $\frac{5}{3}$ $5/3$ of a mol of the new product present, because all of A is used, at a rate of three molecules per one molecule of product formed.

Note: If the problem meant that this new product was not a gas, then change my answer by leaving these $\frac{5}{3}$ $5/3$ of a mole out.

Next, we apply the ideal gas law:

$\frac{P_{1} V_{1}}{n_{1} T_{1}} = \frac{P_{2} V_{2}}{n_{2} T_{2}}$ $(P_1V_1)/(n_1T_1)=(P_2V_2)/(n_2T_2)$

In this problem, $V_{1} = V_{2} = 6 L$ $V_1=V_2=6L$ and we leave this out.

$n_{1} = 11$ $n_1=11$ mol

$n_{2} = 4 \frac{1}{3}$ $n_2=4 1/3$ mol (or $2 \frac{2}{3}$ $2 2/3$ mol of you leave out the product.)

$T_{1} = 480 K$ $T_1=480K$

$T_{2} = 420 K$ $T_2=420K$

We will determine the value of $\frac{P_{2}}{P_{1}}$ $P_2/P_1$ (the fractional change in P)

From the ideal gas law:

$\frac{P_{2}}{P_{1}} = \frac{n_{2} T_{2}}{n_{1} T_{1}} = \frac{4 \frac{1}{3} \cdot 420}{11 \cdot 480} = 0.345$ $P_2/P_1=(n_2T_2)/(n_1T_1) = (4 1/3 * 420)/(11*480)=0.345$

The pressure drops to 34.5 % of its initial value.

If you leave out the product (assume it to be solid), the answer is

$\frac{P_{2}}{P_{1}} = \frac{n_{2} T_{2}}{n_{1} T_{1}} = \frac{2 \frac{2}{3} \cdot 420}{11 \cdot 480} = 0.212$ $P_2/P_1=(n_2T_2)/(n_1T_1) = (2 2/3 * 420)/(11*480)=0.212$

The pressure drops to 21.2% of its original value in this case.

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A $6 L$ $6 L$ container holds $5$ $5$ mol and $6$ $6$ 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 $480^{o} K$ $480^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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Explanation:

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A 6L6 L container holds 55 mol and 66 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 480oK480^oK to 420oK420^oK. How much does the pressure change?

1 Answer

Explanation:

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A $6 L$ $6 L$ container holds $5$ $5$ mol and $6$ $6$ 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 $480^{o} K$ $480^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?