The given reaction is SO2 + O2 --------> SO3
Now the change in Gibbs free energy of the reaction can be calculated by the formula-
ΔGrxn = ΔGproduct - ΔGreactant
Putting the given values-
ΔGrxn = ΔGproduct - ΔGreactant
= ΔGSO3 - (ΔGSO2 + ΔGO2 )
= - 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 = 100.00122
= 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
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