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There is a container that has a volume of $5~\mathrm{L}$ on one side and $3~\mathrm{L}$ on the other and is separated in the middle. There are $2~\mathrm{mol}$ of $\ce{N2}$ on one side at $0~^\circ\mathrm{C}$ and $4~\mathrm{mol}$ of $\ce{H2}$ on the other at $0~^\circ\mathrm{C}$. When the valve is turned and the gases react, what will the total pressure be?

The balanced reaction is $$\ce{N2 + 3H2 <=> 2NH3}$$ My thinking was that the reaction occurs at STP so the pressure would stay at 1 atm?
Or that you could use \begin{align} pV &= nRT\\ p(8~\mathrm{L}) &=6(.082)(273.15)\\ p &= 0.06 \end{align} The $8~\mathrm{L}$ comes from adding the two volumes, and the 6 from adding the number moles together.

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    $\begingroup$ Welcome to Chemistry.SE. Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that ‎we aren't doing your homework for you. $\endgroup$ – user15489 Jun 9 '15 at 2:08
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    $\begingroup$ You are neglecting the fact, that these gasses react with each other. Also the temperature is given as $0~^\circ\mathrm{C}$, so STP it is not. I feel like there is an equilibrium constant missing in the question. $\endgroup$ – Martin - マーチン Jun 9 '15 at 3:56
  • $\begingroup$ Thank you for adding your attempt! This makes it much easier to help you. $\endgroup$ – user15489 Jun 9 '15 at 4:13
  • $\begingroup$ That is all the information that was given for the problem. And I thought that STP was 1 atm and 0 degrees C, thats why I said that. $\endgroup$ – Shaggy Jun 9 '15 at 10:13
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Note: Not all the reactants will react to form the products, but rather exist in equilibrium. However since the question doesn't mention anything about the equilibrium constant, I will assume that the reaction goes to completion as otherwise it will be impossible to do this question.

If you are aware of the various gas laws, doing this question is actually really simple and doesn't involved many calculations.

Step 1: Finding Initial Pressure

Firstly, the question doesn't specify what the initial pressure of the container is. Therefore you can't just assume that it is $1\ \mathrm{atm}$. So the first step would be finding the initial pressure: $$p = \frac{nRT}{V} = \frac{6\ \mathrm{mol}\times 8.3125\ \mathrm{L\ kPa\ mol^{-1}\ K^{-1}}\times273\ \mathrm{K}}{8\ \mathrm{L}} = 1702.4\ \mathrm{kPa} = 16.8\ \mathrm{atm}$$

Step 2: Find Moles of Products

Looking at the stoichiometry of the chemical equation, $4$ moles of gas react to form $2$ moles of gas. Therefore the ratio is $2:1$. In this case, we originally had $6$ moles of gas. Therefore at the end of the reaction, there should be $3$ moles of gas.

Step 3: Find Final Pressure

Now we know that the total number of moles of gas has halved (from $6$ moles to $3$ moles). One of the gas laws state that the number of moles of gas is directly proportional to the pressure of the system. This is clearly obviously by looking at the ideal gas equation. Therefore, by halving the original amount of moles of gas should result in halving of the original pressure of the system.

Therefore the final pressure will be: $8.4\ \mathrm{atm}$

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