Question

2NO2 (g) <-> N2O4 (g) is a reversible reaction. At equilibrium, a 0.4 L flask contains 0.35 moles of N2O4 (g) and 0.26 moles No2 (g). What is the Keq for this reaction? Answer choices 11.4 .47 2.1 3.84

Answer 1

According to the explanation given in the previous session about Keq, we will follow the same step by step in order to find this Keq as well, starting by setting up the reaction:

2 NO2 <--> N2O4

We have:

0.4 L

0.35 moles of N2O4

0.26 moles of NO2

Finding the concentration of each compound:

M = 0.35/0.4

M = 0.875 for N2O4

M = 0.26/0.4

M = 0.65 for NO2

Now finding the Keq:

Keq = [0.875]/[0.65]^2

Keq = 2.1, letter C