Question

1. The following results were collected for two experiments involving the reaction at 600 oC between gaseous sulfur dioxide and oxygen to from gaseous sulfur trioxide. Experiment 1 Experiment 2 Initial Equilibrium Initial Equilibrium [SO2]o = 2.00 M [SO2] = 1.50 M [SO2]o = 0.500 M [SO2] = 0.590 M [O2]o = 1.50 M [O2] = 1.25 M [O2]o = 0 M [O2] = 0.0450 M [SO3]o = 3.00 M [SO3] = 3.50 M [SO3]o = 0.350M [SO3] = 0.260 M a. Write the balanced chemical equation. b. Write the mass action expression. c. Calculate the equilibrium constant for each experiment. Comment on your results. What will cause a change in the equilibrium constant?

Answer 1

A. balanced equation

2SO+O₂ ---------> 2SO₃

B. Mass action expression = product concentration/reactant concentration

                                       = [SO₃]²/[SO₂]²[O₂]

C.Kequ1 = [3]²/[2]²[1.5]=1.5

   K2 = [3.5]²/[1.5]²[1.25] = 4.3

   K3 = [0.35]²/[0.5][0] = 0.12

   k4 = [0.26]²/[0.59]²[0.045] = 4.50

C. When Kequ is larger than 1 it would means that the equilibrium is starting to favour the product

    When Kequ is less thn 1 the reaction is reactant favored at equlibrium.