A $7 L$ $7 L$ container holds $21$ $21$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from $350^{o} K$ $350^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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A $7 L$ $7 L$ container holds $21$ $21$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from $350^{o} K$ $350^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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Answer 1 · 2016-07-03T16:37:37

Let us assume that both gases are ideal gases, so we may use the ideal gas law.

Explanation:

Before reaction, we have:

$V_{1} = 7 L$ $V_1 = 7 " L"$
$n_{A1} = 21 mol$ $n_"A1" = 21 " mol"$ , $n_{B1} = 12 mol$ $n_"B1" = 12 " mol"$
$T_{1} = 350 K$ $T_1 = 350 " K"$

Applying the law:

$P_{1} V_{1} = n_{1} R T_{1} \to$ $P_1 V_1 = n_1 R T_1 rightarrow$

$\to P_{1} = \frac{n_{1} R T_{1}}{V_{1}} = \frac{39 mol \cdot 0, 082 atm\cdotL/K\cdotmol \cdot 350 K}{7 L}$ $rightarrow P_1 = {n_1 R T_1}/{V_1} = {39 " mol" cdot 0,082 " atm·L/K·mol" cdot 350 " K"}/{7 " L"}$
$= 159.9 atm$ $= 159.9 " atm"$

Where we have used that $n_{1} = n_{A1} + n_{B1} = 21 + 12 mol$ $n_1 = n_"A1" + n_"B1" = 21 + 12 " mol"$ .

Now, let us write the reaction: 3 molecules of $B$ $"B"$ + 5 molecules of $A$ $"A"$ form $A_{5} B_{3}$ $"A"_5 "B"_3$ .

$5 A + 3 B \to A_{5} B_{3}$ $5 "A" + 3 "B" rightarrow "A"_5 "B"_3$

With our initial conditions, taking in account that we need $\frac{5}{3} = 1.67$ $5/3 = 1.67$ times more $A$ $"A"$ than $B$ $"B"$ :

We would need, for 21 mol of $A$ $"A"$ , $\frac{3}{5} \cdot 21 = 12.6$ $3/5 cdot 21 = 12.6$ mol of $B$ $"B"$ .
We would need, for 12 mol of $B$ $"B"$ , $\frac{5}{3} \cdot 12 = 20$ $5/3 cdot 12 = 20$ mol of $A$ $"A"$ .

Given that we just have 12 mol of $B$ $"B"$ , the option 1 is not possible, so we choose option 2 ( $A$ $"A"$ is limiting reactant, and $B$ $"B"$ is excess reactant).

Thus, reaction is:

$20 A + 12 B \to 4 A_{5} B_{3} + 1 B$ $20 "A" + 12 "B" rightarrow 4 "A"_5 "B"_3 + 1 "B"$

So we'll have in the end: $n_{2} = n_{A 5 B 3} + n_{B} = 4 + 1 = 5 mol$ $n_2 = n_{"A"5"B"3} + n_"B" = 4 + 1 = 5 " mol"$ .

By applying ideal gas law again:

$P_{2} = \frac{5 mol \cdot 0, 082 atm\cdotL/K\cdotmol \cdot 420 K}{7 L} = 24.6 atm$ $P_2 = {5 " mol" cdot 0,082 " atm·L/K·mol" cdot 420 " K"}/{7 " L"} = 24.6 " atm"$

And the change on pressure is:

$Delta P = P_2 - P_1 = 24.6 - 159.9 = -135.3 " atm"$

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A $7 L$ $7 L$ container holds $21$ $21$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from $350^{o} K$ $350^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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Explanation:

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A $7 L$ $7 L$ container holds $21$ $21$ mol and $12$ $12$ mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from $350^{o} K$ $350^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?