Question

write the equilibrium constant expressions for each of the following reactions

3 O2(g) <-----> 2 O2(g)

2 PbS(s) + 3 O2(g) <-----> 2 PbO(s) + 2 SO2(g)

HCOOH(aq) + H2O(l) <-------> H3O+(aq) + HCOO-(aq)


At 1065 degrees C, H2S decomposes to gaseous hydrogen and sulfur according to the equation 2 H2S(g) <---> 2 H2(g) + S2(g)

Answer 1

For reactions in gaseous phases equilibrium constant can be defined in terms of partial pressures

1)3O₂(g) <-> 2O₃(g)

Kp = ${\frac {p_{O_3}^2} {p_{O_2}^3}}$

p - partial pressures, both products and initial substances are gases

2)2PbS(s) + 3O₂(g) <-> 2PbO(s) + 2SO₂(g)

PbS and PbO are solid phases, their concentrations and partial pressures are constant, so their concentrations in equilibrium constant are usually excluded from the equation

Kp = ${\frac {p_{SO_2}^2} {p_{O_2}^3}}$

3)HCOOH(aq) + H₂O(l) <-> H₃O⁺(aq) + HCOO^-(aq)

For reactions occuring in solutions equilibrium constant can be defined in terms of concentrations.

As it's water solution, concentration of water is much greater than concentrations of reagents, so the change of its concentration is usually neglected. Thus, water concentration is almost constant and can be excluded from the equation.

Kc = ${\frac {[H_3O^+][HCOO^-]} {[HCOOH]}}$