Question

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ${\text{ICl}}$ was $0.78{\text{M}}$?
$2{\text{ICl}}\left( {\text{g}} \right) \rightleftharpoons {{\text{I}}_2}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_2}\left( {\text{g}} \right);{{\text{K}}_{\text{c}}} = 0.14$

Answer 1

Equilibrium is a state in which there are no observable changes as time goes by. Equilibrium constant ${{\text{K}}_{\text{c}}}$ is a number equal to the ratio of equilibrium concentrations of products to the equilibrium concentrations of reactants each raised to the power of its stoichiometric coefficient.

Complete step by step answer:
For a reversible reaction at equilibrium and constant temperature, a certain ratio of reactant and product concentrations has a constant value.
Consider the equilibrium reaction below:
$2{\text{ICl}}\left( {\text{g}} \right) \rightleftharpoons {{\text{I}}_2}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_2}\left( {\text{g}} \right)$
Let the equilibrium concentrations of ${{\text{I}}_2}$ and ${\text{C}}{{\text{l}}_2}$ are ${\text{x}}.{\text{M}}$.
Initially the concentrations of $${\text{ICl,}}$$${{\text{I}}_2}$and${\text{C}}{{\text{l}}2}$are$0.78{\text{M}}$,$0{\text{M}}$and$0{\text{M}}$. At equilibrium, the concentrations are$\left( {0.78 - 2{\text{x}}} \right)$,${\text{x}}.{\text{M}}$and${\text{x}}.{\text{M}}$. Therefore equilibrium constant can be expressed as:${{\text{K}}{\text{c}}} = \dfrac{{\left[ {{{\text{I}}_2}} \right]\left[ {{\text{C}}{{\text{l}}_2}} \right]}}{{{{\left[ {{\text{ICl}}} \right]}^2}}}$Substituting the values, we get$0.14 = \dfrac{{{\text{x}} \times {\text{x}}}}{{{{\left( {0.78 - 2{\text{x}}} \right)}^2}}}$Taking root on both sides, we get$\sqrt {0.14} = \sqrt {\dfrac{{{{\text{x}}^2}}}{{{{\left( {0.78 - 2{\text{x}}} \right)}^2}}}} \\
0.374 = \dfrac{{\text{x}}}{{\left( {0.78 - 2{\text{x}}} \right)}} \\ $Cross-multiplying,$ 0.374\left( {0.78 - 2{\text{x}}} \right) = {\text{x}} \\
{\text{0}}{\text{.291 - 0}}{\text{.748x = x}} \\
{\text{0}}{\text{.291 = 1}}{\text{.748x}} \\ $Therefore,${\text{x}} = \dfrac{{0.291}}{{1.748}} = 0.166$Hence we can say that when the concentration of${\text{ICl}}$is$0.78{\text{M}}$, then the concentration of${{\text{I}}_2}$and${\text{C}}{{\text{l}}_2}$will be$0.166{\text{M}}$. At equilibrium, the concentration of${\text{ICl}}$$ = 0.78{\text{M}} - 2{\text{x}}$Substituting the value of${\text{x}}$,$\left[ {{\text{ICl}}} \right]{\text{ = 0}}{\text{.78M - }}\left( {2 \times 0.166{\text{M}}} \right) = 0.78{\text{M}} - 0.332{\text{M}} = 0.448{\text{M}}$$\left[ {{{\text{I}}_2}} \right] = 0.166{\text{M}} \\
\left[ {{\text{C}}{{\text{l}}_2}} \right] = 0.166{\text{M}} \\ $

Additional information:
The magnitude of ${{\text{K}}_{\text{c}}}$ will tell about whether the equilibrium reaction favors the reactants or products. If equilibrium constant is greater than one, it favors products. If equilibrium constant is less than one, it favors reactants.

Note:
The steps involved in finding equilibrium concentrations are:
-Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown ${\text{x}}$ which represents the change in concentration.
-Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of equilibrium constant, solve for ${\text{x}}$.
-Then calculate equilibrium concentrations of all species.