An $8 L$ $8 L$ container holds $6$ $6$ mol and $5$ $5$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . How much does the pressure change by?

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An $8 L$ $8 L$ container holds $6$ $6$ mol and $5$ $5$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . How much does the pressure change by?

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Answer 1 · 2018-03-31T03:20:51

The final pressure will be 0.767 times the initial pressure.

Explanation:

The overall strategy for this problem is to determine the change in the number of moles present and then use the ideal gas law to calculate the pressure change.

We have a reaction where

A+ 2B -> C

Let $n_{A}$ $n_A$ represent the number of moles of A after the reaction and $n_{A}^{0} = 6$ $n_A^0=6$ represent the number of moles of A before the reaction.

Let $n_{B}$ $n_B$ represent the number of moles of B after the reaction and $n_{B}^{0} = 5$ $n_B^0=5$ represent the number of moles of B before the reaction.

Let $n_{C}$ $n_C$ represent the number of moles of C after the reaction and $n_{C}^{0} = 0$ $n_C^0=0$ represent the number of moles of C before the reaction.

If all 5 moles of B react, then

$n_{A} = n_{A}^{0} - \frac{5}{2} = 6 - \frac{5}{2} = \frac{7}{2}$ $n_A=n_A^0-5/2=6-5/2=7/2$

$n_{B} = n_{B}^{0} - 5 = 5 - 5 = 0$ $n_B=n_B^0-5=5-5=0$

$n_{C} = n_{C}^{0} + \frac{5}{2} = 0 + \frac{5}{2} = \frac{5}{2}$ $n_C=n_C^0+5/2=0+5/2=5/2$

The total number of moles, $n$ $n$ , after the reaction will be

$n = n_{A} + n_{B} + n_{C} = \frac{7}{2} + 0 + \frac{5}{2} = 6$ $n=n_A+n_B+n_C=7/2+0+5/2=6$ moles.

Initially there were $n_{0} = 6 + 5 + 0 = 11$ $n_0=6+5+0=11$ total moles.

Now we can use the Ideal Gas Law to calculate the ratio of the final pressure, $P$ $P$ to the initial pressure, $P_{0}$ $P_0$ . Let $T_{0} = 320 K$ $T_0=320 K$ be the temperature before the reaction and let $T = 450 K$ $T=450K$ be the temperature after the reaction. The Ideal Gas Law tells us that

$\frac{P}{P_{0}} = \frac{n}{n_{0}} \cdot \frac{T}{T_{0}} = \frac{6}{11} \cdot \frac{450}{320} = \frac{135}{176} \approx 0.767 .$ $P/P_0=n/n_0*T/T_0=6/11*450/320=135/176~~0.767.$

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An $8 L$ $8 L$ container holds $6$ $6$ mol and $5$ $5$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . How much does the pressure change by?

1 Answer

Explanation:

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An 8L8 L container holds 66 mol and 55 mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 320oK320^oK to 450oK450 ^oK. How much does the pressure change by?

1 Answer

Explanation:

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An $8 L$ $8 L$ container holds $6$ $6$ mol and $5$ $5$ mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from $320^{o} K$ $320^oK$ to $450^{o} K$ $450 ^oK$ . How much does the pressure change by?