Consider the following reaction at a high temperature. br2(g) ⇆ 2br(g) when 1.35 moles of br2 are put in a 0.780−l flask, 3.60 percent of the br2 undergoes dissociation. calculate the equilibrium constant kc for the reaction.

Answer : The equilibrium constant for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of .

Now we have to calculate the dissociated concentration of .

The balanced equilibrium reaction is,

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of = = 1.2 %

So, the dissociate concentration of =

The value of x = 0.2077 M

Now we have to calculate the concentration of at equilibrium.

Concentration of = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

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

Therefore, the equilibrium constant for the reaction is, 0.1133

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.