Question

Consider the reaction to produce ammonia shown below.
N2 (g) + 3H2 (g) * ) 2NH3 (g) KP = 44 at 473 K
Initially a mixture of N2, H2, and NH3 have the following pressures.
PN2 = 10atm PH2 = 30atm PNH3 = 3atm
What is the pressure of each gas once equilibrium is established?

Answer 1

Consider the reaction to produce ammonia shown below.

N2 (g) + 3H2 (g) * ) 2NH3 (g) KP = 44 at 473 K

Initially a mixture of N2, H2, and NH3 have the following pressures.

PN2 = 10atm PH2 = 30atm PNH3 = 3atm

What is the pressure of each gas once equilibrium is established?

Solution

$$N _ {2} (g) + 3 H _ {2} (g) = 2 N H _ {3} (g)$$ $$K p = \frac {P _ {N H 3} {} ^ {2}}{P _ {N 2} \cdot P _ {H 2} {} ^ {3}};$$

where $P_{NH3}, P_{N2}, P_{H2}$ pressures of $NH_3, N_2$ and $H_2$ when equilibrium is established.

Let $x$ be the change in pressure for $N_2$ . Then the change in pressure for $H_2$ is $3x$ (we can see from the chemical equation that the coefficient ratio of $N_2$ and $H_2$ is 1:3), the change in pressure of $NH_3$ is $2x$ (we can see from the chemical equation that the coefficient ratio of $N_2$ and $NH_3$ is 1:2).

Let's express pressures of gases when equilibrium is established in the form of table:

Comment: for $\mathsf{N}_2$ and $\mathsf{H}_2$ change in pressure is negative as pressures of these gases decrease, for $\mathsf{NH}_3$ change in pressure is positive as it's pressure increase.

Solve the equation and find x:

$$K p = \frac {P _ {N H 3} {} ^ {2}}{P _ {N 2} \cdot P _ {H 2} {} ^ {3}};$$ $$4 4 = \frac {(3 + 2 x) ^ {2}}{(1 0 - x) \cdot (3 0 - 3 x) ^ {3}};$$ $$4 4 (1 0 - x) \cdot (3 0 - 3 x) ^ {3} = (3 + 2 x) ^ {2};$$ $$4 4 (1 0 - x) \cdot 3 ^ {3} (1 0 - x) ^ {3} = (3 + 2 x) ^ {2};$$ $$1 1 8 8 (1 0 - x) ^ {4} - (3 + 2 x) ^ {2} = 0;$$ $$3 4. 4 7 ^ {2} (1 0 - x) ^ {4} - (3 + 2 x) ^ {2} = 0;$$ $$(3 4. 4 7 (1 0 - x) ^ {2} - (3 + 2 x)) \cdot (3 4. 4 7 (1 0 - x) ^ {2} + (3 + 2 x)) = 0;$$

Answer provided by AssignmentExpert.com

34.47(10-x)² - (3+2x) = 0 or 34.47(10-x)² + (3+2x) = 0.

Find roots of the first equation:

34.47(10-x)² - (3+2x) = 0;

34.47(100-20x+x²) - 3-2x = 0;

3447-689.4x+34.47x²-3-2x=0

34.47x² - 691.4x + 3444 = 0

D = (-691.4)² - 4·34.47·3444 = 478033.96 - 474858.72 = 3175.24 = 56.35²;

x₁ = (691.4 - 56.35) / 2 · 34.47 = 9.21;

x₂ = (691.4 + 56.35) / 2 · 34.47 = 10.85.

Find roots of the second equation:

34.47(10-x)² + (3+2x) = 0;

34.47(100-20x+x²) + 3+2x = 0;

3447-689.4x+34.47x²+3+2x=0

34.47x² - 687.4x + 3450 = 0

D = (-687.4)² - 4·34.47·3450 = 472518.76 - 475686 = -3167.24;

D<0, the second equation has no roots.

Analysis of the roots: x₁ = 9.21; x₂ = 10.85. Initial pressure of N₂ 10 atm, then P_N2 when equilibrium is established should be positive: P_N2 = 10-x = 10-9.21 = 0.79 (atm) (compare: 10-10.85 = -0.85). We should use x = 9.21.

P_H2 = 30-3x = 30-3·9.21 = 2.37 (atm).

P_NH3 = 3+2x = 3+2·9.21 = 21.42 (atm).

Answer: P_N2 = 0.79 atm, P_H2 = 2.37 atm, P_NH3 = 21.42 atm.