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}}mol{\text{ }}litr{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} = \dfrac{{\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} = \dfrac{{\alpha \times \alpha }}{{{{(3.3 \times {{10}^{ - 3}} - 2\alpha )}^2}}} $
 $ {K_c} = \dfrac{{{\alpha ^2}}}{{{{(3.3 \times {{10}^{ - 3}} - 2\alpha )}^2}}} $
 $ {K_c} = 32 $ (given)
 $ \dfrac{{{\alpha ^2}}}{{{{(3.3 \times {{10}^{ - 3}} - 2\alpha )}^2}}} = 32 $
Taking square root both sides we get:
 $ \dfrac{\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.