NH3 is normally encountered as a gas with a pungent odor. It is
formed from hydrogen and nitrogen (equation given below). A
chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0
litre reaction vessel in 1:3 ratio. The equilibrium constant for the
reaction is 1.0 × 10^–4, and the equilibrium value of [H2(g)] is 0.12 mol/L. Calculate the equilibrium value of [NH3].
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
I got 2 different answers and I don't know which one is right.
1) [N2(g)] : [H2(g)] = 1:3
Thus [N2] = 0.04 mol/L, since [H2] is 0.12 mol/L and has a ratio of 1:3 when compared to N2.
Then Kc = ([NH3]^2) / ([N2] [H2]^3)
1.0 × 10^–4 = ([NH3]^2) / (0.04) (0.12)^3
[NH3]^2 = 6.912 x 10^9
[NH3] = 8.314 x 10^-5 mol/L
oops clicked submit by accident
2) I used an ICE table
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
I : 1 mol 3 mol 0
C : -x -3x +2x
E : 1-x mol 3-3xmol 2x mol
since [H2]eq = 0.12 mol/L, then 3-3x = 0.12, therefore x = 0.96
to get [NH3]eq = 2 x 0.96 = 1.92
which one is right? Dr.bob ?
I have missed you. Frankly, I think something is missing like a number for EITHER N2 or H2 initially. In your first trial you have assume that if the ratio is 1:3 initially that it will be 1:3 at equilibrium. Is that true?
For the second trial, if you substitute 1 and 3 initially for N2 and H2 then your answers look good (except that next to last step is a 10^-9). But plug those values into th K expression and you don't get anywhere near the K listed in the problem. So then substitute 2 and 6 for N2 and H2 (still a ratio of 1:3) and you get a different value for NH3 and K is not the same NOR is it the value listed in the problem. My opinion is that you need a value for N2 to start or H2 to start and you don't have that. I found that same problem posted on the web several places with solutions (for a fee of course). I was able to read some of them but the English grammar was so poor I couldn't read them. The ones with better grammar were off limits to me. If you find the answer, plug that into the ICE chart and see if you get the K of 10^-4. By the way, I looked up K for that reaction and it was listed as 120; however, this problem has no temperature listed so that discrepancy doesn't mean anything. Sorry I failed you. Please let me know when you get the answer from the teacher. I still think something is missing.
Hello again C.P. -- I've had another day to thing about this and I've hardened by position somewhat. I don't retract anything I posted earlier but I want to add this. I think this is an ICE problem through and through; it just needs a starting amount for N2 or for H2 then the "other" one will be in the ratio of N2:H2 of 1:3. The fact that starting with 1 for N2 and 3 mole/L for H2 does not give the Kc listed of 10^-4 means one of two things. Either the Kc listed in the problem is not correct or that the starting amounts of 1M and 3M are not correct. They give a Qc > Kc; therefore, the 1M and 3M are too high. So I think there is some combination that will give Kc, whatever the Kc is, and still be in the ratio of 1:3 of reacting materials. There still could be some "trick" to the problem that I don't know about but I thought I knew all the tricks by this stage of my life. I really would like to know the final outcome so when you learn how to get the answer the teacher wants please post it to my attention OR just as another chemistry question for I read all of them unless I'm not at home. Good luck in your chemistry endeavors.
OK. I have the answer but I came about it in a cockeyed manner. I think the correct answer is the one you solved with the Kc and came up with 8.3E-5 M for (NH3). I'm having trouble understanding why that gives the correct answer. Here is what I went through.
1) .......................N2 + 3H2 ==> 2NH3
Look through what you did for N2 start @ 1M and H2 start @ 3M. See my comments earlier about the K value. It's huge and not even close to 10^-4.
2) It is obvious that H2 must be > 0.12 M to start.
3) Start with H2 at 0.13, make N2 start at 0.13/3 = ? and go through the calculations. I won't put them here but I ended up with
Qc = (NH3)^2/(N2)(H2)^3 = (0.06667)^2/(0.04)(0.12)^3 = 0.642, too large!
4) Starting material must be larger than 0.12 and smaller than 0.13 so try 0.125.
You can go through the calculations and
Qc = (0.01333)^2/(0.04)(0.12)^3 = 0.16. still too large but slightly closer.
5) Getting ready to try something between 0.12 and 0.125 when I noticed THAT THE DENOMINATOR DID NOT CHANGE. Looked back over calculations from yesterday and the denominator was 0.04 and (0.12)^3 there too. So we ask ourselves "What would the numerator have to be to make that fraction equal to Kc and that answer is
10^-4 = (X)^2/(0.04)(0.12)^3 and X must be 8.3E-5 M. What is X in the formula? It's (NH3) and that format is Kc = (NH3)^2/(N2)(H2)^3 which is what you started with. So the answer to your original question is the Kc calculation is the correct answer. I guess I'm finished with the problem; however, I'm still trying to understand WHY the ICE procedure doesn't work and WHY I had to resort to a trial and error and finally to the Kc approach. I won't rest until I come up with an explanation. I'm going to bed; it's past my bedtime.
OK, five days later I have it. I expect you're through with this and probably you won't read this response; however, if you would like an explanation of why you get two answers I can tell you. Just post on Jiskha to let me know and I'll provide a thorough answer.
To calculate the equilibrium value of [NH3], we can use the equation you provided and the given equilibrium constant (Kc = 1.0 × 10^–4). Let's go through the steps again to verify the correct answer.
1) Find the initial concentrations of N2 and H2 based on the given ratio of 1:3. Assume the initial concentration of N2 is x, so the initial concentration of H2 would be 3x. However, in this case, we are given the concentration of H2, which is 0.12 mol/L. Therefore, we can obtain the value of x by dividing 0.12 by 3.
[H2] = 3x
0.12 = 3x
x = 0.04 mol/L
So, the initial concentration of N2 is 0.04 mol/L.
2) Use the equilibrium constant expression and the known values to find the equilibrium concentration of NH3 (let's call it y).
Kc = ([NH3]^2) / ([N2] [H2]^3)
1.0 × 10^–4 = (y^2) / (0.04) (0.12)^3
1.0 × 10^–4 = (y^2) / (0.04) (0.001728)
1.0 × 10^–4 = (y^2) / 6.912 x 10^9
Multiply both sides of the equation by 6.912 x 10^9 to isolate y^2:
(y^2) = (1.0 × 10^–4) (6.912 x 10^9)
(y^2) = 6.912 x 10^5
Take the square root to find y:
y = √(6.912 x 10^5)
y ≈ 831.4
Since the concentration cannot be negative, the equilibrium value of [NH3] is positive:
[NH3] ≈ 831.4 mol/L
Therefore, the equilibrium value of [NH3] is approximately 831.4 mol/L.