Question

Consider the equilibrium: 2N2O (g) + O2 (g) = 4NO (g) 3.00 moles of NO (g) are introduced into a 1.00 liter evacuated flask. When the system comes to equilibrium, 1.00 moles of N2O (g) has formed. What are the equilibrium concentrations of each substance? What is Kc for the reaction?

Answer 1

The equilibrium concentrations are [N₂O] = 1.00 mol/L; [O₂] = 0.50 mol/L; [NO] = 1.00 mol/L. `K_"c"` = 2.0.

First, write the balanced chemical equation with an ICE table.

2N₂O + O₂ ⇌ 4NO

I/mol·L⁻¹: 0; 0; 3.00
C/mol·L⁻¹: +2`x`; +`x`; -4`x`
E/mol·L⁻¹: 2`x`; `x`; 3.00 - 4`x`

At equilibrium, [N₂O] = 2`x` mol/L= 1.00 mol/L.

So `x` = 0.50

Then [O₂] = `x` mol/L = 0.50 mol/L

and [NO] = (3.00 - 4`x`) mol/L = (3.00 – 4×0.50) mol/L = 1.00 mol/L

`K_"c" = (["NO"]^4)/(["N"_2"O"]^2 ["O"₂]) = 1.00^4/(1.00^2 × 0.50)` = 2.0