The Kp for the following equilibrium system is 4.31 x 10‒4 at 375°C : N2(g) + 3H2(g) ↔ 2NH3(g).
If 0.862 atm of N2 and 0.373 atm of H2 were mixed in a constant-volume flask and heated to 375°C, which of the following would be closest to the equilibrium partial pressure of NH3 ?
(1) 0.860 atm (2) 4.40 x 10‒3 atm (3) 2.75 x 10‒4 atm (4) 0.366 atm (5) 1.09 x 10‒2 atm
..............N2 + 3H2 ==> 2NH3
initial....0.862..0.373......0
change.....-p......-3p......+2p
equil...0.862-p..0.373-3p....2p
I looked on the web to find an equation solver for quartics; I'm sure they are there but I didn't find them. I suggest you try substituting the value for NH3 (and the others) into the equation and see which is the closest. That's probably what the author of the problem expected you to do anyway.
Thank You!
To determine the equilibrium partial pressure of NH3, we can use the equation for Kp:
Kp = (P(NH3))² / (P(N2) * P(H2)³)
Given that Kp = 4.31 x 10‒4 and the initial partial pressures of N2 and H2 are 0.862 atm and 0.373 atm respectively, we can substitute these values into the equation:
4.31 x 10‒4 = (P(NH3))² / (0.862 * (0.373)³)
Now we can solve for P(NH3):
(P(NH3))² = 4.31 x 10‒4 * (0.862 * (0.373)³)
(P(NH3))² = 9.95 x 10‒6
P(NH3) ≈ √(9.95 x 10‒6)
P(NH3) ≈ 0.00315 atm
Therefore, the closest value to the equilibrium partial pressure of NH3 is (2) 4.40 x 10‒3 atm.
To find the equilibrium partial pressure of NH3, we can use the formula derived from the equilibrium constant expression:
Kp = (P(NH3))^2 / (P(N2) * (P(H2))^3,
where Kp is the equilibrium constant, and P(NH3), P(N2), P(H2) are the partial pressures of NH3, N2, and H2, respectively.
Given the equilibrium constant Kp = 4.31 x 10‒4 and the initial partial pressures of N2 and H2, we can substitute the values into the equation to find the equilibrium partial pressure of NH3.
Kp = (P(NH3))^2 / (P(N2) * (P(H2))^3
4.31 x 10‒4 = (P(NH3))^2 / (0.862 * (0.373^3))
To solve for P(NH3), we can rearrange the equation:
(P(NH3))^2 = Kp * (P(N2) * (P(H2))^3
P(NH3) = √(Kp * (P(N2) * (P(H2))^3))
Plugging in the values:
P(NH3) = √(4.31 x 10‒4 * (0.862 * (0.373^3)))
P(NH3) ≈ √(1.9223 x 10‒4 * 0.1108)
P(NH3) ≈ √0.00002128
P(NH3) ≈ 0.004613 atm
Out of the given options, the closest answer to 0.004613 atm is option (2) 4.40 x 10‒3 atm.