Question

g Potassium chlorate decomposes to product potassium chloride and oxygen gas. 2KClO3(s) ⇔ 2KCl(s) + 3O2(g) When this reaction was run at room temperature, the following equilbrium concentrations were measured: [O2] = 0.0500 M; [KCl] = 0.00250 M; [KClO3] = 2.00 M What is the equilibrium constant for this reaction?

Answer 1

`1.95* 10^(-10)` is the equilibrium constant for this reaction.

Step-by-step explanation:

Equilibrium concentrations were measured:

`[O_2] = 0.0500 M`

`[KCl] = 0.00250 M`

`[KClO_3] = 2.00 M`

`2KClO_3(s)\rightleftharpoons 2KCl(s) + 3O_2(g)`

The expression of an equilibrium constant is given by :

`K_c=([KCl]^2[O_2]^3)/([KClO_3]^2)`

`=([0.00250 M]^2[0.0500 M]^3)/([2.00 M]^2)`

`K_c=1.95* 10^(-10)`

`1.95* 10^(-10)` is the equilibrium constant for this reaction.