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.

$M=\frac{n}{V}$

Putting the values,

$M=\frac{0.134 mol}{250 mL(\frac{10^{-3}L}{1 mL})}=0.536 M$

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:

$k_{c}=\frac{[b]}{[a]}$

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

Putting the values,

$2.36=\frac{x}{0.536-x}$

On rearranging,

$x=\frac{1.265}{3.36}=0.376$

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