Keq, Equilibrium Calculations and Given Initial Conditions

Question

0.00400 mol/L of NO and 0.00300 mol/L of O2 are reacted together: 

2 NO(g) + O2(g) --> 2 NO2(g) 

 Initially, there is no NO2 in the flask. At equilibrium, it is determined that [NO2] = 0.00350 mol/L. What is the value of Keq for the reaction? 

Answer 1

When given initial conditions and asked to find equilibrium concentrations, it's easiest to set up an ICE table.

2NO(g) + O₂(g) <===> 2NO₂(g)

0.004......0.003.................0........Initial

-2x.........-x......................+2x......Change

0.004-2x..0.003-x...........2x........Equilibrium

Since we are told that [NO2] at equilibrium = 0.0035 M, we now know that 2x = 0.0035 M and x = 0.00175 M

We can now fill in the ICE table and find equilibrium concentrations of the reactants and then solve for Keq.

[NO] = 0.004 - 0.0035 = 0.0005 M

[O₂] = 0.003 - 0.00175 = 0.00125

[NO₂] = 0.0035

Keq = [NO₂]² / [NO]²[O₂] = (0.0035)² / (0.0005)²(0.00125)

Keq = 3.92x10⁴