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 know this is a late responds, but I would like to hear your explanation to the question above. I have finally gotten the answer to the question above and the correct answer is [NH3] = 8.314 x 10^-5 mol/L. Thanks for 4 years of help, Dr.bob.
We previously said the answer was 8.3E-4 M for (NH3) given the conditions. What we didn't know was if the equilibrium ratio of N2:H2 was 1:3 if the starting ratio was 1:3. ASSUMING the equilibrium ratio for this reaction is the same as the starting ratio then Kc = 1E-4 = (NH3)^2/(N2)(H2)^3. Plug in 0.12 for (H2) from the problem, plug in 1/3 of that for the N2 (0.12/3 = 0.04), then the ONLY answer you can get for (NH3) is roughly 8.3E-4 M for NH3. Given that the (H2) was fixed @ 0.12M, that no starting material was given for H2 or N2, and no value of (N2) @ equilibrium was given I think we are forced to make the assumption that N2 @ equilibrium is 0.04 and solve that way. What about the assumption? I've never seen that in my career so I looked at the equation A + 2B ==> C and plunged in 0.5M and 1M for A:B. You can go through the math, as I did, and lo and behold, the end value at equilibrium is 1:2 ratio for A:B. Then I changed to 3:6 for different values BUT the ratio still was 1:2 for A:B. The ratio of A:B at equilibrium was lo and behold 1:2. Then I changed to ratio of A:B of 4:6 (2:3 or 1:1.5) and the end value at equilibrium for A:B was NOT 2:3 (1:1.5). It was 1:1.39. I've never seen this argued in any text, internet, discussion, not anywhere. So I'm assuming that this argument holds: i.e., if the ratio of the reactants is the same as the stoichiometry of the reaction THEN the equilibrium ratio is the same ratio for those reactants. That does not hold true if the initial ratio is not the same as the stoichiometric coefficients.
There is another part to this. Since no starting material was given, I asked myself what would that mystery 1:3 mixture have to be to get, at equilibrium, (N2) = 0.04M, (H2) = 0.12M and (NH3) = 8.314E-5 M. That's easier said than done so it makes me think the values of 0.04M and 0.12 were out of the blue and it was easier to say "a mixture in 1:3 ratio......." than to work it out. Anyway you can see what I mean.
..............N2 + 3H2 ==> 2NH3
I..............y........3y............0
C............-x........-3x..........2x
E...........0.04.....0.12.......2x
We know (NH3) = 8.34E-5 M so x must be 4.17E-5
That makes y = (N2) = 0.04 + 0.0000417 = 0.0400417
You can do (H2) initially but you see that makes (N2) = 0.04 @ equilibrium a ridiculous number to start with. Starting molarity for H2 is almost as bad. Again, I think the starting numbers make no sense given the equilibrium values of 0.04 and 0.12. Just my opinion. I hope all of this makes sense.
Thanks again, I understand your explanation.
To find the equilibrium value of [NH3], we can start by writing the expression for the equilibrium constant (K) in terms of the concentrations of the reactants and products:
K = [NH3]^2 / ([N2][H2]^3)
We are given the equilibrium constant (K = 1.0 × 10^–4) and the equilibrium value of [H2(g)] (0.12 mol/L). We also know the initial ratio of the reactants is 1:3, which means the initial concentration of nitrogen gas [N2(g)] is 1/4 of the initial concentration of hydrogen gas [H2(g)].
Let's call the equilibrium concentration of hydrogen gas as x. Then the concentration of nitrogen gas at equilibrium will be 1/4 * x. The equilibrium concentration of NH3 can be represented as 2y (since the stoichiometric coefficient in the balanced equation is 2).
Now we can substitute these concentrations into the equilibrium constant expression:
1.0 × 10^–4 = (2y)^2 / ((1/4 * x) * x^3)
Simplifying, we get:
1.0 × 10^–4 = 4y^2 / (1/4 * x^4)
1.0 × 10^–4 = 16y^2 / x^4
Rearranging, we have:
x^4 = (16 * y^2) / (1.0 × 10^–4)
Taking the square root of both sides:
x^2 = (4 * y) / (10^–2)
Simplifying further:
x^2 = 400 * y
Since the initial concentration ratio of [N2(g)] to [H2(g)] is 1:3, we can write:
(1/4 * x) / x = 1 / 3
1/4 = 1 / 3
Cross-multiplying, we get:
3 * (1/4) = x
3/4 = x
So, the equilibrium concentration of hydrogen gas [H2(g)] is 3/4 mol/L.
Now we can substitute this value of x into the equation x^2 = 400 * y to find the value of y:
(3/4)^2 = 400 * y
9/16 = 400 * y
y = (9/16) / 400
y = 9 / (16 * 400)
y = 0.00140625 mol/L
Finally, we can substitute the value of y into the equation 2y to find the equilibrium concentration of NH3:
[NH3] = 2 * y
[NH3] = 2 * 0.00140625
[NH3] = 0.0028125 mol/L
So, the equilibrium value of [NH3] is 0.0028125 mol/L, which can be written as [NH3] ≈ 2.81 x 10^–3 mol/L.
Apologies for the confusion in the previous response. The correct equilibrium value of [NH3] is approximately 2.81 x 10^–3 mol/L.
To calculate the equilibrium value of [NH3], we can use the equation for the equilibrium constant (K) which relates the concentrations of the reactants and products at equilibrium.
The equilibrium constant expression for the given reaction is:
K = [NH3]² / ([N2] * [H2]³)
Given that K = 1.0 × 10^–4 and [H2] is 0.12 mol/L, we can substitute these values:
1.0 × 10^–4 = [NH3]² / ([N2] * (0.12)³)
Since the ratio of N2 to H2 in the reaction is 1:3, we can substitute [N2] with 3[H2].
1.0 × 10^–4 = [NH3]² / (3[H2] * (0.12)³)
Simplifying further:
1.0 × 10^–4 = [NH3]² / (0.12³ * 3 * [H2])
Rearranging the equation to isolate [NH3]:
[NH3]² = (1.0 × 10^–4) * (0.12³ * 3 * [H2])
[NH3]² = (3.456 × 10^–7) * [H2]
Taking the square root of both sides:
[NH3] = sqrt((3.456 × 10^–7) * [H2])
Now we substitute the given value of [H2] = 0.12 mol/L:
[NH3] = sqrt((3.456 × 10^–7) * 0.12)
[NH3] = sqrt(4.1472 × 10^–8)
[NH3] ≈ 6.436 × 10^–5 mol/L
Therefore, the equilibrium value of [NH3] is approximately 6.436 × 10^–5 mol/L.
I apologize, but the answer you provided, [NH3] = 8.314 × 10^-5 mol/L, does not match the calculations. Please double-check your steps or provide additional information to resolve any discrepancies.