Another Equilibrium Question!
The following reaction is at equilibrium in a closed 20.0-L container. At equilibrium, the pressures of NH3 and HCl are both equal to 0.00552 atm, and the mass of NH4Cl (s) is 35.4 g. Calculate the equilibrium pressure of NH3 after 11.3 g of NH4Cl (s) is added to the container
NH4Cl(s) -> HCl(g) + NH3(g)
Not sure how to approach this problem, I've tried several methods..
And get stuck on them
I'm not sure if I should solve Kp first, which should be .00552*2 so Kp = .01104, but I get confused on how to use the mass of NH4Cl and when 11.3g of NH4Cl is added.
3 answers
Kp, if you wish to calculate it is p^2 = 0.00552^2 but I'm not sure you need it. The problem is confusing to me too, I wonder if this is a trick question. Since NH4Cl is a solid, it does not enter into the equilibrium as long as there is a speck of NH4Cl solid present. Therefore, the pressures of the NH3 and HCl in the problem are those that exist over a bed of solid NH4Cl at the prescribed temperature. Adding more NH4Cl has no effect on the equilibrium EXCEPT for the volume in the 20.0 L container taken up by the added solid. If that takes up considerable volume, that will increase P and it would shift the equilibrium to the left. BUT, since there is no density given (to calculate a volume of the solid),and with no temperature listed you can't use pv = nRT to try to arrive at the number of moles present BEFORE and AFTER the 11.3 g has been added. So my thinking is that we ignore the small amount of volume taken up by the added solid and give the pressures as 0.00552 atm. Do you have a data base you key these answers to. If so, I'd like to know how this one turns out.
Yes i think this was a trick question, the solid just needs to be completely ignored. This question isn't part of a database unfortunately, just a practice question for exam
what is the mass in kilograms of 750.0mL of mercury at 25.0 C(d=13.5g/mL)
1 answer