Question

At a certain temperature, 0.900 mol SO3 is placed in a 5.00 L container.

2SO3(g)−⇀↽−2SO2(g)+O2(g)

At equilibrium, 0.110 mol O2 is present. Calculate Kc.

Answer 1

Answer:

Step 1: Explanation

Kc is defined as the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature

Step 2: Write the balanced the chemical equation

2SO₃ (g) <--------------> 2SO₂ (g)+O₂(g)

Step 3: Calculate the Concentrations by dividing the given moles by volume of container.

Given,

Initial SO₃ = moles / volume = (0.900 moles / 5 litre ) = 0.18 M

as well the concentration of O₂ at equilibrium = ( 0.110 moles / 5 L ) = 0.022 M

Step 3: Write the ICE table

2SO₃ (g) <--------------> 2SO₂ (g) + O₂(g)

Initial amount 0.18 0 0

Change amount    -2x +2x +x

Equilibrium amount 0.18-2x 2x x

Since the value of O2 at equlibrium is given [O2] = 0.022 M = [x]

hence value of x = 0.022 M

Hence all value at equilibrium is

[SO₃] = 0.18 -2x = 0.18 -(2 × 0.022 ) =0.136

[SO₂] = 2x = 2 × 0.022 = 0.044

[O₂] = x = 0.022

Step 4: Calculate the Kc

2SO₃ (g) <--------------> 2SO₂ (g) + O₂(g)

The Kc expression will be

Kc = [SO2]² × [O2] / [SO3]²

on substituting the value

Kc= ( 0.044)²×(0.022) / (0.136)² = 2.3 ×10^-3

Hence, the value of Kc = 2.3 ×10^-3