Question

Bromine monochloride ,\[BrCl\] decomposes into bromine and chlorine and reaches the equilibrium
$2BrCl(g) \to B{r_2}(g) + C{l_2}(g)$
For which ${K_c} = 32$ at $500K$ .If initially pure \[BrCl\] is present at a concentration of $3.3 \times {10^{ - 3}}mol{L^{ - 1}}$ ,what is it’s molar concentration in the mixture of equilibrium?

Answer 1

Hint: This problem will be solved with the help of the concept that the rate of reversible reactions at which the concentration of the reactants and products do not change with time is known as chemical equilibrium.According to the law of mass action,”Rate of reaction is directly proportional to the raised to the power equal to the respective stoichiometric coefficient ”. Let’s take an example of a reaction that is where A and B reacts and gives C and D as product and this is an reversible reaction,
            $aA + bB \rightleftarrows cC + dD$
Rate of forward reaction $ \propto {[A]^a}{[B]^b}$
Rate of forward reaction $ = {k_f}{[A]^a}{[B]^b}$
Rate of backward reaction $ \propto {[C]^c}{[D]^d}$
Rate of backward reaction $ = {k_b}{[C]^c}{[D]^d}$
At equilibrium , ${k_f}{[A]^a}{[B]^b} = {k_b}{[C]^c}{[D]^d}$
                               ${K_c} = \dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
Where ${K_c} = $ equilibrium constant,${k_f} = $rate constant for forward reaction and ${k_b} = $rate constant for backward reaction.

Complete step by step answer:
Here \[BrCl\] decomposes into bromine and chlorine ,let the amount of $B{r_2}$ and $C{l_2}$ formed be x.Then at equilibrium the following reaction takes place,
                                          $2BrCl(g) \to B{r_2}(g) + C{l_2}(g)$
Initial concentration $3.3 \times {10^{ - 3}}$ 0 0
At equilibrium $3.3 \times {10^{ - 3}}$ -2x x x
Now we will get , ${K_c} = \dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{[B{r_2}][C{l_2}]}}{{{{[BrCl]}^2}}}$

                                      $
   \Rightarrow \dfrac{{(x) \times (x)}}{{{{(3.3 \times {{10}^{ - 3}} - 2x)}^2}}} = 32 \\
   \Rightarrow \dfrac{x}{{3.3 \times {{10}^{ - 3}} - 2x}} = 5.56 \\
   \Rightarrow x = 18.678 \times {10^{ - 3}} - 11.32x \\
   \Rightarrow x = 1.5 \times {10^{ - 3}} \\
 $

Now we know the value of x,therefore at the equilibrium the concentration of $BrCl$,
$[BrCl] = 3.3 \times {10^{ - 3}} - (2 \times 1.5 \times {10^{ - 3}})$
               $ = 0.3 \times {10^{ - 3}}$
               $ = 3.0 \times {10^{ - 4}}mol{L^{ - 1}}$
So here is the answer that $BrCl$ molar concentration at equilibrium will be $3.0 \times {10^{ - 4}}mol{L^{ - 1}}$.

Note : We have approached this problem by applying the concept of equilibrium and this is given in the problem that $BrCl$ have initial concentration $3.3 \times {10^{ - 3}}mol{L^{ - 1}}$ . So we assumed that at equilibrium bromine and chlorine will have $x mol{L^{ - 1}}$ and value of ${K_c}$ is given .we applied mass action law here and solve the equation and we get value of x and after subtracting -2x from the initial molar concentration of bromine chloride we got molar concentration at equilibrium.