Question

2. Suppose that a system initially contains 0.45 mol of S02, 0.30 mol of O2 and 0.50 mol of SO3. Find the equilibrium composition of the unbalanced reaction as outlined below. [A_G(SO2) = -300.4 kJ/mol and A G(SO3) = -370.4 kJ/mol]. [10 points] SO2 + O2 =SO

Answer 1

The given reaction is SO2 + O2 --------> SO3

Now the change in Gibbs free energy of the reaction can be calculated by the formula-

ΔG_rxn = ΔG_product - ΔG_reactant

Putting the given values-

ΔG_rxn = ΔG_product - ΔG_reactant

= ΔG_SO3 - (ΔG_SO2 + ΔG_O2 )

= - 307.4 kJ/mole - (- 300.4 kJ/mole + 0 ) because ΔG of naturally available gases = zero

= - 307.4 kJ/mole + 300.4 kJ/mole

= - 7.0 kJ/mole

Now the relationship between Equilibrium constant (K) ang ΔG is

ΔG = -2.303 RT log K

where T = given temperature = here 298 K

R = universal gas constant =  8.314 J/mol. K

K = Equilibrium constant

Now putting the given values--

ΔG = -2.303 RT log K
- 7.0 kJ/mole = -2.303 * 8.314 J/mol. K * 298 K * log K

log K = (- 7.0 kJ/mole) / (-2.303 * 8.314 J/mol. K * 298 K)

   = 0.00122

K = 10^0.00122

= 1.0028

Now to calculate the value of equilibrium concentrations, we have to form the ICE table

Reaction	SO2	O2	SO3
Initial	0.45 mol	0.30 mol	0.50 mol
Change	-x	-x	+x
Equilibrium	0.45-x	0.30-x	0.50+x

Now K = [SO3] / [SO2] *[O2]

Putting the values from the table

K = [SO3] / [SO2] *[O2]

1.0028 = [0.50+x] / [0.45-x] *[0.30-x]

1.0028 * [0.45-x] *[0.30-x] = [0.50+x]

1.0028 * [0.135 - 0.45x - 0.30x + x² ] = [0.50+x]

1.0028 * [0.135 - 0.75x + x² ] = [0.50+x]

0.135378‬ - 0.7521‬x + 1.0028x² ] = [0.50+x]

0.135378‬ - 0.50 - 0.7521‬x - x + 1.0028x² ] = 0

1.0028x² - 1.7521 x -0.34622‬ = 0

Solving this

x = 0.179 mols

Sp at equilibrium, SO2 = 0.45-x = 0.45 - 0.179 = 0.271‬ mols

O2 = 0.30-x = 0.30 - 0.179 = 0.121‬‬ mols

SO2 = 0.50+x = 0.50 + 0.179 = 0.679‬‬ mols