Question

Consider the reaction 2H2O2(l)2H2O(l) + O2(g) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.

Answer 1

Given decomposition reaction,

2H₂O₂(l) 2H₂O(l) + O₂(g)

Let's find G^orxn using Standard Gibbs energy of the formation of species ivolved in above decomposition.

We have,

ΔG°_rxn = Σ ΔHG_f (products) - Σ ΔG°_f (reactants)

For above decomposition reaction,

ΔG°_rxn = [2*ΔHG°_f H₂O(l) + ΔHG_f O₂(g)] - [2 * ΔG°_f H₂O₂(l)]

By using thermodynamic table,

ΔG°_rxn = [2* (-237.14 kJ/mol) + (0 kJ/mol)] - [2 * (-120.42 kJ/mol)]

ΔG°_rxn = -233.44 kJ/mol

ΔG°_rxn = -233440 J/mol

Now we know the relation,

ΔG°_rxn = -RT lnK

lnK = -ΔG°_rxn /RT

We have, ΔG°_rxn = -233440 J/mol, R 8.314 J/Kmol, T = 298.15 K

lnK = -[-233440 J/mol / (8.314 J/Kmol*298.15 K)

lnK = +94.17

K = e^94.17

K = 7.9*10⁴⁰.

Value of equilibrium costant is 7.9*10⁴⁰.

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