Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: T on July 21, 2015, 05:42:43 AM
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Hello,
The question is:
1 mole of nitrogen gas and 3 moles of hydrogen gas were added to a rigid 1 L container at 298 K and left to react. At equilibrium it was found that the concentrations of nitrogen, hydrogen and ammonia were 0.116 mol Lā1, 0.348 mol Lā1 and 1.768 mol Lā1 and the pressure inside the container was 55 bar.
1 mole of hydrogen chloride gas was subsequently added to the container and the system was left to re-establish equilibrium (hydrogen chloride gas reacts with ammonia gas to produce solid ammonium chloride).
What are the concentrations (or range of concentrations) of nitrogen, hydrogen, ammonia and hydrogen chloride once equilibrium has been reached for the second time?
I have attached the multi-choice answers.
So the 2 equations after HCl was added is:
N2 + 3H2 ::equil:: 2NH3
HCl + NH3 ::equil:: NH4Cl
So after HCl is added the ammonia will react with it creating NH4Cl. So therefore the N2 and H2 will increase forward reaction to balance the equilibrium. So A and C are ruled out. The ammonia will react to form NH4Cl so B is also ruled out. Now I am left with D and E, I choose E since some NH4Cl will dissociate back into HCl + NH3. However, the answer is D. I can't think of any reason the HCl concentration would be 0, if HCl is 0 then the equilibrium constant for HCl and NH3 is impossible.
[tex]Kc=\frac{1}{[HCl] [NH3]}[/tex]
Can someone give me a hint as to why the HCl concentration would be 0?
Thanks
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some NH4Cl will dissociate back into HCl + NH3
Not significantly at 298 K - I guess this this why D is considered correct.
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Ammonia chloride is solid at this conditions, so there will be no HCl left.
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Line D and E should be a normal ICE table (second change and second equilibrium). Sign at x (+ or - for different reagents) you can deduce from Kc calculated from line A and Q(uotient) calculated from line C. Some concentration of ammonia exists at the new beginning and other phase (solid ammonium chloride) can be neglected.
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Not significantly at 298 K - I guess this this why D is considered correct.
But if there is still some dissociation shouldn't the answer still be E?
Ammonia chloride is solid at this conditions, so there will be no HCl left.
Why? Won't some ammonium chloride dissociate back?
Line D and E should be a normal ICE table (second change and second equilibrium). Sign at x (+ or - for different reagents) you can deduce from Kc calculated from line A and Q(uotient) calculated from line C. Some concentration of ammonia exists at the new beginning and other phase (solid ammonium chloride) can be neglected.
I looked on line and ICE table: I = initial concentration, C = change, E = equilibrium concentration. Are you suggesting D is change and E is equilibrium? I don't really understand what you are suggesting and how it will help me answer the question, could you please tell me?
Thanks to everyone who replied
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Why? Won't some ammonium chloride dissociate back?
It can only dissociate if it is in gas phase. At 298 K both is not the case.
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Why can't ammonium chloride dissociate back if it is solid?
NH4Cl :rarrow: NH3 + HCl
Thanks
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The answer is that E is technically correct, as there must always be some pressure of HCl, however small, if there is equilibrium. However, it is, in practice, extremely small.
From tabulated free energy of formation data, I estimate the dissociation constant K = PHCl*PNH3 ā 10-16 atm2 at 298K. As you can see, you have a substantial pressure of excess ammonia, so the HCl pressure is going to be tiny - for practical purposes, indistinguishable from zero.