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?

1 Answer
Aug 19, 2014

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