Question

Part A - Calculating K from Initial and Equilibrium Concentrations When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibrium, the concentration of N2O4 is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction below at 400 K? N2O4(g) ⇌ 2 NO2(g)

Answer 1

Solution :-

Lwts first calculate the moles of the N2O4

Moles= mass / molar mass

Moles of N2O4 = 9.2 g /92.011 g per mol = 0.10 mol N2O4

Now lets calculate the initial concentration of the N2O4

Initial concentration of the N2O4 = 0.10 mol / 0.5 L = 0.20 M

N2O4(g)   --------- > 2 NO2(g)

0.20 M                         0

-x                                  +2x

0.20-x                          2x

0.20 –x = 0.057 M

X= 0.20 M – 0.057 M = 0.143 M

so lets calculate the equilibrium concetration of the NO2

[NO2] eq = 2x = 2*0.143 M = 0.286 M

Now lets calculate the Kc

Kc= [NO2]^2/[N2O4]

Kc = [0.286]^2/[0.057]

Kc =1.44

So the Kc = 1.44