Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium $2BrCl(g) \rightleftharpoons Br_{2}(g)+Cl_{2}(g)$ for which Kc = 32 at 500K. If initially pure BrCl is present at a concentration of 3.3 × $10^{-3}$ mol $L^{-1}$, what is its molar concentration in the mixture at equilibrium ?

Answer 1

Given,

Initial concentration of the mixture $=\quad 3.30\quad \times \quad { 10 }^{ -3 }\quad { mol }/{ L }$

$2Br{ Cl }_{ (g) }\quad \rightarrow \quad { Br }_{ 2(g) }\quad +\quad { Cl }_{ 2(g) }$

At equilibrium, $\left( 3.30\quad \times \quad { 10 }^{ -3 }\quad -\quad \left( x \right) \right) \quad { mol }/{ L }$

${ K }_{ c }\quad =\quad \frac { \left( \frac { x }{ 2 } \right) \left( \frac { x }{ 2 } \right) }{ { \left( 3.30\quad \times \quad { 10 }^{ -3 }\quad -\quad \left( x \right) \right) }^{ 2 } }$

Where, given ${ K }_{ c }$ at 500K is 32

So, $32\quad =\quad \frac { { x }^{ 2 } }{ \left( 4\quad \times \quad \left( 3.30\quad \times \quad { 10 }^{ -3 } \right) \right) \quad -\quad x }$

By simplifying the above equation, we get 5.60

$x\quad =\quad 11.32\quad \left( \left( 3.30\quad \times \quad { 10 }^{ -3 } \right) \quad -\quad x \right)$

$x\quad =\quad 3.0\quad \times \quad { 10 }^{ -3 }$

At equilibrium, $[BrCl] =\quad \left( 3.30\quad \times \quad { 10 }^{ -3 } \right) \quad -\quad \left( 3.0\quad \times \quad { 10 }^{ -3 } \right)$

$=\quad \left( 0.30\quad \times \quad { 10 }^{ -3 } \right)$

$=\quad 3.0\quad \times \quad { 10 }^{ -4 }\quad { mol }/{ L }$