Question

A 0.10-mol sample of pure ozone, O3, is placed in a sealed 1.0-L container and the reaction 2O3 (g) -> 3O2(g) is allowed to reach equilibrium. A 0.50-mol sample of pure ozone is placed in a second 1.0-L container at the same temperature and allowed to reach equilibrium. Without doing any calculations, predict which of the following will be different in the two containers at equilibrium. Which will be the same? Briefly explain your answers.

a) amount of O2

b) concentration of O2

c) ratio of [O2]/[O3]

d) the ratio [O2]^3/[O3]^2

e) the ratio [O3]^2/[O2]^3

Answer 1

Answer :

A given equilibrium reaction.

3 O₂ (g) <-----> 2 O₃ (g).

In first container, Initial concentration of O₃ = # of moles / Volume in L = 0.1moles/1.0 L = 0.1 M

In second container, Initial concentration of O₃ = 0.5/1.0 = 0.5 M

As initial concentrations of O₃ are unequal in both container.

a) Amount of O₂ formed at equilibrium will be different.

b) [O₂] will also be different.

c) [O₂]/[O₃] will be different as this is not the rate expression for the equilibrium constant.

d) & e): Ratios [O₂]³/[O₃]² and [O₃]²/[O₂]³ are expressions for rate constants for reaction in a specified direction or in reverse direction. And hence they will be same in both vessels.

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