Question

We put 10.0 moles of N₂O into a 2.0 L container at some temperature. At equilibrium, 2.20 moles of N2 remain. Calculate the equilibrium constant, Kc, for the reaction at this temperature. 2N_2O(g) → 2N_2(g) + O_2(g)

a) 0.44
b) 0.55
c) 0.66
d) 0.77

Answer 1

The equilibrium constant (Kc) for the reaction 2N₂O(g) → 2N₂(g) + O₂(g) is 1.95. This value does not match any of the multiple-choice options provided.

Step-by-step explanation:

We are tasked to calculate the equilibrium constant, Kc, for the reaction 2N₂O(g) → 2N₂(g) + O₂(g) given that 10.0 moles of N₂O were initially placed into a 2.0 L container and at equilibrium, 2.20 moles of N₂ remain.

To find Kc, we first need to determine the changes in molar concentrations of our reactants and products from the beginning of the reaction to equilibrium.

• Initial mole of N₂: none (since it's a product)
• Initial mole of N₂O: 10.0 moles
• Initial mole of O₂: none (since it's a product)

At equilibrium, 2.20 moles of N₂ remain. That means a total of 10.0 - 2.20 = 7.80 moles of N₂O have reacted.

For every 2 moles of N₂O that react, 2 moles of N₂ and 1 mole of O₂ are produced. Hence, 7.80 moles of N₂ and 3.90 moles of O₂ are produced.

Now we convert these moles to concentrations by dividing by the volume of the container:

• Concentration of N₂ = 2.20 moles / 2.0 L = 1.10 M
• Concentration of N₂O = (10.0 - 7.80) moles / 2.0 L = 1.10 M
• Concentration of O₂ = 3.90 moles / 2.0 L = 1.95 M

Using the expression for the equilibrium constant, Kc = [N₂]²[O₂] / [N₂O]², we can substitute the equilibrium concentrations into this expression:

Kc = (1.10)²(1.95) / (1.10)₂ = 1.95

Therefore, none of the given multiple choice options are correct.