Question

Given the following UNBALANCED reaction: NH3 (g) <--> N2 (g) + H2 (g) If 1

we start with 2.2 mol NH3 AND 1.1 mol N2 in a 0.95L container and are left
with 1.4mol of NH3 at equilibrium, calculate k. *
Options: a) 1.29 b)1.32 c) 1.35 and d) 1.46

Answer 1

C. 1.35

Step-by-step explanation:

2NH3 (g) <--> N2 (g) + 3H2 (g)

Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0

change in concentration 2x x 3x

-0.84 M +0.42M +1.26M

Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M

K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M