The reaction contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 x 10-5. Suppose in a case [N2] = 0.80 mol L-1 and [...

Question

The reaction contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 x 10^-5. Suppose in a case [N₂] = 0.80 mol L^-1 and [O₂] = 0.20 mol L^-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.

Philoid · Accepted Answer

Let x amount has been decayed from reactant to product. So,

N₂(g) + O₂(g) 2NO(g)

At t=0 0 0.8 0.20

At equilibrium, 0.8-x 0.2-x 2x

Given, equilibrium constant K = 10^-5.

We know, K =

∴ K =

Putting the concentrations of N₂, O₂, NO at equilibrium,

10^-5 = .

If we consider the value of x to be very small for simplicity of calculations, then,

0.8-x ≈ 0.8

0.2-x ≈ 0.2

So, 10^-5 =

or, 4x² = 16×10^-6.

or, x = 2×10^-3 [M]

Now, the amount of reactants and products at equilibrium are,

For N₂ it was 0.8-x = 0.8-0.002 = 0.798[M]

For O₂ it was 0.2-x = 0.2-0.002 = 0.198[M]

For NO, it was 2x = 2×0.002 = 0.004[M]