Question

If the following reaction has Kp= 4.5x10⁻⁵ at 720K. N₂(g)+3H₂(g) 2NH₃(g). If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH₃)= 93 atm, P(N₂)=48 atm, and P(H₂)=52 atm.

Answer 1

The reaction N2(g) + 3H2(g) ⇒ 2NH3(g) will shift to the left to reach equilibrium, as the reaction quotient (Q) is greater than the equilibrium constant (Kp), indicating excess NH3.

Step-by-step explanation:

To determine the direction in which the reaction will shift to reach equilibrium, we use the reaction quotient, Q, and compare it to the equilibrium constant, Kp. The reaction in question is the synthesis of ammonia:

N2(g) + 3H2(g) ⇒ 2NH3(g)

The equilibrium constant expression at 720K is given by:

Kp = [P(NH3)]2 / ([P(N2)] × [P(H2)]3)

Given:

• Kp = 4.5 × 10−5
• P(NH3) = 93 atm
• P(N2) = 48 atm
• P(H2) = 52 atm

We can calculate the reaction quotient, Q:

Q = (93 atm)2 / (48 atm × (52 atm)3)

Q = 8689 atm2 / (48 atm × 140608 atm3)

Q = 8689 atm2 / 6753152 atm4

Q ≈ 1.286 × 10−3

Since Q > Kp, the reaction will shift to the left to reach equilibrium, converting NH3 into N2 and H2.