Question

. The following data is given at a certain temperature: (1) 2N2 (g) + O2 (g) 2N2O (g) K1 = 1.2 x 10-35 (2) N2O4 (g) 2NO2 (g) K2 = 4.6 x 10-3 (3) ½N2 (g) + O2 (g) NO2 (g) K3 = 4.1 x 10-9 Calculate the molar equilibrium constant, K for the reaction: 2N2O (g) + 3O2 (g) 2N2O4 (g)

Answer 1

To calculate the molar equilibrium constant (K) for the reaction

2N2O (g) + 3O2 (g) ⇌ 2N2O4 (g)

We can use the concept of the equilibrium constant expression and the relationship between the equilibrium constants of individual reactions.

The reaction can be broken down into two steps:

Step 1: 2N2 (g) + O2 (g) ⇌ 2N2O (g) (K1 = 1.2 x 10^(-35))
Step 2: N2O (g) + O2 (g) ⇌ N2O4 (g) (K2 = 4.6 x 10^(-3))

To find the equilibrium constant for the overall reaction, we multiply the equilibrium constants of the individual steps:

K = K1 * K2

K = (1.2 x 10^(-35)) * (4.6 x 10^(-3))

K = 5.52 x 10^(-38)

Therefore, the molar equilibrium constant (K) for the reaction 2N2O (g) + 3O2 (g) ⇌ 2N2O4 (g) is 5.52 x 10^(-38)