Question

Consider the reaction below: a (aq) ↔ b (aq) kc = 2.36 if the reaction is started by placing 0.134 mol of a into 250.0 ml of solution, what will the concentration of a be at equilibrium?

Answer 1

The equilibrium reaction is as follows:

`a(aq)\rightleftharpoons b(aq)`

The initial number of moles of a (aq) and volume of solution is given 0.134 mol and 250 mL.

First calculate the molarity of solution which is defined as number of moles of solute in 1 L of solution.

`M=(n)/(V)`

Putting the values,

`M=(0.134 mol)/(250 mL((10^(-3)L)/(1 mL)))=0.536 M`

Thus, initial concentration of a(aq) will be 0.536 M.

At equilibrium, let the change in concentration be x, thus, concentration of a(aq) will be 0.536-x and that of b(aq) will be x.

The expression for
`k_(c)` for the equilibrium reaction is:

`k_(c)=([b])/([a])`

Here [a] and [b] are equilibrium concentration of reactant a and product b respectively.

Putting the values,

`2.36=(x)/(0.536-x)`

On rearranging,

`x=(1.265)/(3.36)=0.376`

Now, equilibrium concentration of a (aq) will be:

`[a]=0.536-x=0.536-0.376=0.16 M`

Therefore, equilibrium concentration of a(aq) is 0.16 M