Equilibrium Constant?

Question

Im kind of confused on how to work this problem. Any help would be much appreciated.

Calculate the equilibrium constant for the decomposition of liquid water into hydrogen and oxygen gas at 25 degrees C.

Does anybody know the formula or how to work the problem?

Answer 1

Firstly, a quick review of the Equilibrium Law...

For the general reaction:
aA + bB => cC + dD
(Reactants) (Products)

The formula for the equilibrium constant, Kc
= [Products] / [Reactants]
= ([C]^c x [D]^d) / [([A]^a + [B]^b)

with [ ] being the concentration of the substances in mol dm^-3.

Right, so we're all set. Take note that this question provided the information whereby the temperature given is 25 degrees C. At 25 C, the absolute pressure is 101kPA (with reference to the standard conditions and their values).

The information you gave:
H2O => H2 + O2

Balance the equation, and you get:
2H2O => 2H2 + O2

Work out the partial pressures for H2O, H2 and O2 respectively with the following formula:

The partial pressure, p, of a gas
= (no. of moles of gases / total number of moles) x total pressure

To find the individual no. of moles of gases:
Reaction: 2H2O => 2H2 + O2
Initial no: x 0 0
Final no: x - 0.2 0.2x 0.1x
= 0.8x

Substitute the value of total pressure = 101kPa into the equation.

Total number of moles of gases
= 0.8x + 0.2x + 0.1x
= 1.1x

p[H2O], partial pressure of water
= (0.8x / 1.1x) x 101kPa
= 73.5

p[H2], partial pressure of hydrogen
= ( 0.2x / 1.1x) x 101kPa
= 18.4

p[O2], partial pressure of oxygen
= (0.1x / 1.1x) x 101kPa
= 9.2

Therefore, Kc
= ([partial pressure of hydrogen]^2 x [partial pressure of oxygen]) / [partial pressure of water]^2

i.e.

= (p^2[H2] x p[O2]) / p^2[H2O]
= ([18.4]^2 x [9.2]) / [73.5]^2

Kc = 0.577

Answer 2

Consult a good textbook for Equilibrium. The concentration of the water is required.