[thegoodaaron]

Hello all. I'm having a hard time with the following problem. More specifically, I am having a hard time figuring out the partial pressure of ammonia. I overheard several classmates talking about how nitrogen was the limiting reactant and, as such, once the reaction goes to completion, there would still be H2 in the container contributing to the pressure, but I'm not sure how to use that information (assuming it is even correct). Any suggestions? I appreciate any help you can provide!

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Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia gas (NH3). You have nitrogen and hydrogen gases in a 14.0 L container fitted with a movable piston (the piston allows the container volume to change so as to keep the pressure constant inside the container). Initially the partial pressure of each reactant gas is 1.03 atm. Assume the temperature is constant and the reaction goes to completion.

(a) Calculate the partial pressure of ammonia in the container after the reaction has reached completion.

(b) Calculate the volume of the container after the reaction has reached completion.

[Yggdrasil]

Always start with a balanced chemical reaction.

[thegoodaaron]

N2 + 3H2 --> 2NH3

I'm not sure what to do with this information.

[Yggdrasil]

Do you know how to figure out the limiting reagent from this information?  If so, can you figure out how much ammonia is produced when the reaction goes to completion?  (Note: if it makes it easier, you can assume the container is at 273K since the temperature will end up not mattering in the final answer).

[thegoodaaron]

Hmm, I guess I don't know how. I recognize that each gas has the same partial pressure,meaning that there number of moles of each gas is the same. But up until now I have only computed limiting reactants when specific quantities are given to me. As far as I can tell, this problem only tells me that the amounts of each are equal. How can I use this information to determine the limiting reactant and, from there, determine the amount of ammonia produced when the reaction goes to completion?

[Yggdrasil]

You are absolutely correct that there is an equal amount of each gas.  Perhaps the question will make more sense if we can work with actual numbers.  Since the problem doesn't specify a temperature, you can try assuming that temperature doesn't matter.  So, lets choose an arbitrary temperature to work at, say 300K.

Now from the ideal gas law, you can calculate the number of moles of each reactant.  PV = nRT rearranges to give n = PV/RT = 0.586 moles of each gas (P = 1.03 atm, V = 14.0L, T = 300K)

Given that you start with 0.586 moles of N₂ and 0.586 moles of H₂ can you find out how many moles of NH₃ will be present after the reaction is complete?  Can you figure out how many moles of the non-limiting reagent are left after the reaction is complete?  Can you convert these quantities to pressures?

[thegoodaaron]

Wonderful, thank you!

.586mol N2 x (3mol H2/1mol N2) = 1.758 mol H2 needed; not available; limiting
.586mol H2 x (1mol N2/3mol H2) = 0.195 mol N2 needed; available

amount N2 left after reaction = amount available - amount needed = .586 - .195 = .391mol N2

.586mol H2 x (2mol NH3/3mol H2) = 0.391mol NH3

Thus, once the reaction has gone to completion, the amounts of ammonia and nitrogen is equal. And because the total pressure on the container will always be 2.06 because of the piston, that means that the partial pressure of ammonia is also 1.03.

Then, to determine the new volume, all you need to do is multiply 14.0L by 2/3 because the new partial pressures are 2/3 the size of the initial partial pressures. Thank you!!

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I have one more question. You are given the value of K for the reaction 2 NO(g) <----->  N2(g) + O2(g) and are asked, for a given set of partial pressures, which direction will the reaction shift (or is it at equilibrium). I know how to solve these kinds of problems with initial concentrations by computing Q and comparing its value to the value of K. But how can this be done with partial pressures?

EDIT: I figured it out. Thank you so much for your help, this time and the last time you helped me.