Question

Bromine monochloride, $BrCl$ decomposes into bromine and chlorine and reaches the equilibrium.
 $2BrC{l}_{(g)}\rightleftarrows B{r}_{2(g)}+C{l}_{2(g)}$ for which $K_c=32$ at $500K$ .
If initially pure $BrCl$ is present at a concentration of $3.30\times {10}^{-3}mollitr{e}^{-1}$ ; what is its molar
concentration in the mixture at equilibrium?

Answer 1

Hint :To calculate an equilibrium concentration from an equilibrium constant, an understanding of the concept of equilibrium and how to write an equilibrium constant is required. Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.

Complete Step By Step Answer:
For knowing the value of molar concentration $BrCl$ we need to first calculate the degree at which it gets decomposed, that is the degree of dissociation of $BrCl$ into its constituent ions.
let the degree of dissociation be $\alpha$ .
We observe the decomposition reaction at initial concentration and at equilibrium.
 $2BrC{l}_{(g)}\rightleftarrows B{r}_{2(g)}+C{l}_{2(g)}$
  $3.30\times {10}^{-3}$ 0 0
  $3.3\times {10}^{-3}-2\alpha$ $\alpha$ $\alpha$
At the initial stage only $BrCl$ is present and its concentration is $3.30\times {10}^{-3}mollitr{e}^{-1}$ , but the concentration of ions will be zero.
At the equilibrium dissociation occurs and equal molecules of ions are formed that is $\alpha$ and 2 $\alpha$ amount is reduced from the concentration of $BrCl$ .
Now $K_c$ of the reaction at equilibrium is given by:
 $K_c={\displaystyle \frac{\left[B{r}_2\right]\left[C{l}_2\right]}{{\left[BrC{l}_2\right]}^2}}$
Substituting the respective concentration in the above equations:
 $K_c={\displaystyle \frac{\alpha \times \alpha }{{(3.3\times {10}^{-3}-2\alpha )}^2}}$
 $K_c={\displaystyle \frac{{\alpha }^2}{{(3.3\times {10}^{-3}-2\alpha )}^2}}$
 $K_c=32$ (given)
 ${\displaystyle \frac{{\alpha }^2}{{(3.3\times {10}^{-3}-2\alpha )}^2}}=32$
Taking square root both sides we get:
 ${\displaystyle \frac{\alpha }{3.3\times {10}^{-3}-2\alpha }}=4\sqrt{2}$
 $\alpha =18.67\times {10}^{-3}-8\sqrt{2}\alpha$
 $\alpha +8\sqrt{2}\alpha =18.67\times {10}^{-3}$
 $\left(1+8\sqrt{2}\right)\alpha =18.67\times {10}^{-3}$
 $\alpha =1.5162\times {10}^{-3}$
Hence now we have the value of degree of dissociation.
So, Molar Concentration of $BrCl$ at equation:
That is given by
 $3.3\times {10}^{-3}-2\alpha$
 $3.3\times {10}^{-3}-2\times 1.5162\times {10}^{-3}$ = $0.267\times {10}^{-3}M$
Hence the required answer is: $0.267\times {10}^{-3}M$ .

Note :
Molar concentration (also called molarity, amount concentration or substance concentration) is a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.