Question

At a given temperature, a 5.0 L glass container contains the following reaction mixture at equilibrium:

SO2(g) + N2O(g) ⇌ SO3(g) + N2(g)

If the masses of each of the gases are 3.5 g (SO3), 4.6 g (SO2), 18.5 g (N2), and 0.98 g (N2O), what is the equilibrium constant, Kc, for this reaction at the given temperature?

Answer 1

Step 1: calculate concentration of SO3

Molar mass of SO3,

MM = 1*MM(S) + 3*MM(O)

= 1*32.07 + 3*16.0

= 80.07 g/mol

mass(SO3)= 3.5 g

use:

number of mol of SO3,

n = mass of SO3/molar mass of SO3

=(3.5 g)/(80.07 g/mol)

= 4.371*10^-2 mol

volume , V = 5 L

use:

Molarity,

M = number of mol / volume in L

= 4.371*10^-2/5

= 8.742*10^-3 M

Step 2: calculate concentration of SO2

Molar mass of SO2,

MM = 1*MM(S) + 2*MM(O)

= 1*32.07 + 2*16.0

= 64.07 g/mol

mass(SO2)= 4.6 g

use:

number of mol of SO2,

n = mass of SO2/molar mass of SO2

=(4.6 g)/(64.07 g/mol)

= 7.18*10^-2 mol

volume , V = 5 L

use:

Molarity,

M = number of mol / volume in L

= 7.18*10^-2/5

= 1.436*10^-2 M

Step 3: calculate concentration of N2

Molar mass of N2 = 28.02 g/mol

mass(N2)= 18.5 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(18.5 g)/(28.02 g/mol)

= 0.6602 mol

volume , V = 5 L

use:

Molarity,

M = number of mol / volume in L

= 0.6602/5

= 0.132 M

Step 4: calculate concentration of N2O

Molar mass of N2O,

MM = 2*MM(N) + 1*MM(O)

= 2*14.01 + 1*16.0

= 44.02 g/mol

mass(N2O)= 0.98 g

use:

number of mol of N2O,

n = mass of N2O/molar mass of N2O

=(0.98 g)/(44.02 g/mol)

= 2.226*10^-2 mol

volume , V = 5 L

use:

Molarity,

M = number of mol / volume in L

= 2.226*10^-2/5

= 4.453*10^-3 M

Step 5: calculate Kc

Kc = [SO3][N2]/[SO2][N2O]

= (8.742*10^-3)*(0.132)/(1.436*10^-2)*(4.453*10^-3)

= 18.0

Answer: 18.0