Question

For the reaction 2NOBr (g) ↔ 2NO (g) + Br2 (g), when the concentrations are [NOBr] = 0.10 m, [NO] = 0.010 M, and [Br2] = 0.0050 M, what is the equilibrium constant of this reaction? A. 5.0 × 10–5 M B. 0.000005 M C. 5.0 × 105 M D. 5.0 M E. None of the Above

Answer 1

A. Remember that K=[Reactants]/[products]

Answer 2

Answer: B. 0.000005

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,

`NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)`

At eqm. conc. 0.10 M 0.010 M 0.0050 M

The expression for equilibrium constant for this reaction will be,

`K_c=([NO]^2[Br_2])/([NOBr])`

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

`K_c=((0.010)^2* (0.0050))/((0.10))`

`K_c=5* 10^(-6)=0.000005`

Thus the value of the equilibrium constant is 0.000005 M .