Question

At equilibrium, a reaction vessel contains 4.50 atm of Br2 and 1.10 atm of NBr3. According to the reaction: 2 NBr3 (g) ⇌ N2 (g) + 3 Br2 (g) with Kp = 4.8, determine the equilibrium partial pressure of N2.

Answer 1

Answer:According to the balanced equation:

2 NBr3 (g) ⇌ N2 (g) + 3 Br2 (g)

We are given the initial partial pressures of Br2 as 4.50 atm and NBr3 as 1.10 atm.

Let's assume that the equilibrium partial pressure of N2 is x atm.

Using the equilibrium constant expression, we have:

Kp = (partial pressure of N2)^1 * (partial pressure of Br2)^3 / (partial pressure of NBr3)^2

Substituting the given values, we get:

4.8 = x^1 * (4.50 atm)^3 / (1.10 atm)^2

Simplifying the equation:

4.8 = x * 91.125 / 1.21

Cross multiplying:

4.8 * 1.21 = x * 91.125

5.808 = 91.125x

Dividing both sides by 91.125:

x = 5.808 / 91.125

x ≈ 0.0637 atm

Therefore, the equilibrium partial pressure of N2 is approximately 0.0637 atm.