Question

1 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

Answer: The equilibrium constant for this reaction is
`1.95* 10^(-10)`

Step-by-step explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

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

At eqm. conc. (2.00) M ( 0.00250 ) M (0.0500) M

The expression for equilibrium constant for this reaction will be,

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

Now put all the given values in this expression, we get :

`K_c=((0.00250)^2* (0.0500)^3)/((2.00)^2)`

By solving we get :

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

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