Question

A student ran the following reaction in the laboratory at 692 K:

When she introduced 0. 0393 moles of and 0. 0702 moles of into a 1. 00 liter container, she found the equilibrium concentration of to be 0. 00197 M. Calculate the equilibrium constant, , she obtained for this reaction

Answer 1

To calculate the equilibrium constant (K) for the given reaction, we need to use the concentrations of the reactants and products at equilibrium. From the information provided:

Initial moles of N₂: 0.0393 moles
Initial moles of O₂: 0.0702 moles
Initial volume of the container: 1.00 liter
Equilibrium concentration of NO₂: 0.00197 M

The balanced chemical equation for the reaction is:
2NO₂(g) ⇌ N₂(g) + 2O₂(g)

The equilibrium constant expression for this reaction is:
K = [N₂] * [O₂]² / [NO₂]²

Now let's calculate the equilibrium constant (K):

[N₂] = (0.0393 moles) / (1.00 liter) = 0.0393 M
[O₂] = (0.0702 moles) / (1.00 liter) = 0.0702 M
[NO₂] = 0.00197 M

K = (0.0393 M) * (0.0702 M)² / (0.00197 M)²

K ≈ 985.41

Therefore, the equilibrium constant (K) obtained for this reaction at 692 K is approximately 985.41.