Question

For which of the following reactions does the equilibrium constant depend on the units of concentration?
A. $NO\left( g \right)\rightleftharpoons \dfrac{1}{2}{{N}_{2}}\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)$
B. ${{C}_{2}}{{H}_{5}}OH\left( l \right)+C{{H}_{3}}COOH\left( l \right)\rightleftharpoons C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\left( l \right)+{{H}_{2}}O\left( l \right)$
C. $2H\left( g \right)\rightleftharpoons {{H}_{2}}\left( g \right)+{{I}_{2}}\left( g \right)$
D. $COC{{l}_{2}}\left( g \right)\rightleftharpoons CO\left( g \right)+C{{l}_{2}}\left( g \right)$

Answer 1

Rate constant is basically a proportionality constant, which indicates the relation between the molar concentration of reactants and the rate of reaction.

-We will discuss the expression for equilibrium constant for all the options for which the equilibrium constant depends on the units of concentration.
-In first option, that is$NO\left( g \right)\rightleftharpoons \dfrac{1}{2}{{N}_{2}}\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)$ the value of equilibrium constant will be:
${{k}_{C}}=\dfrac{{{\left[ {{N}_{2}} \right]}^{\dfrac{1}{2}}}{{\left[ {{O}_{2}} \right]}^{\dfrac{1}{2}}}}{\left[ NO \right]}$
Now, we will put the values of unit in above formula:

& {{k}_{C}}=\dfrac{{{\left[ mol/l \right]}^{\dfrac{1}{2}}}{{\left[ mol/l \right]}^{\dfrac{1}{2}}}}{\left[ mol/l \right]} \\\ & {{k}_{C}}=no\text{ }unit \\\ \end{aligned}$$ -Here we can see that the equilibrium constant is having no unit. So, the equilibrium constant doesn’t depend on the units of concentration. -In second option, that is ${{C}_{2}}{{H}_{5}}OH\left( l \right)+C{{H}_{3}}COOH\left( l \right)\rightleftharpoons C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\left( l \right)+{{H}_{2}}O\left( l \right)$ the value of equilibrium constant will be $$\begin{aligned} & {{k}_{C}}=\dfrac{\left[ Cu \right]\left[ Z{{n}^{2+}} \right]}{\left[ Zn \right]\left[ C{{u}^{2+}} \right]} \\\ & {{k}_{C}}=\dfrac{\left[ mol/l \right]\left[ mol/l \right]}{\left[ mol/l \right]\left[ mol/l \right]} \\\ & {{k}_{C}}=no\text{ }unit \\\ \end{aligned}$$ -Here we can see that the equilibrium constant is having no unit. So, the equilibrium constant doesn’t depend on the units of concentration. -In third option, that is $2H\left( g \right)\rightleftharpoons {{H}_{2}}\left( g \right)+{{I}_{2}}\left( g \right)$ the value of equilibrium constant will be $$\begin{aligned} & {{k}_{C}}=\dfrac{\left[ C{{H}_{3}}COO{{C}_{2}}{{H}_{5}} \right]\left[ {{H}_{2}}O \right]}{\left[ {{C}_{2}}{{H}_{2}}OH \right]\left[ C{{H}_{3}}COOH \right]} \\\ & {{k}_{C}}=\dfrac{\left[ mol/l \right]\left[ mol/l \right]}{\left[ mol/l \right]\left[ mol/l \right]} \\\ & {{k}_{C}}=no\text{ }unit \\\ \end{aligned}$$ -Here we can see that the equilibrium constant is having no unit. So, the equilibrium constant doesn’t depend on the units of concentration. -In fourth option, that is $COC{{l}_{2}}\left( g \right)\rightleftharpoons CO\left( g \right)+C{{l}_{2}}\left( g \right)$ ,the value of equilibrium constant will be $$\begin{aligned} & {{k}_{C}}=\dfrac{\left[ CO \right]\left[ C{{I}_{2}} \right]}{\left[ COC{{I}_{2}} \right]} \\\ & {{k}_{C}}=\dfrac{\left[ mol/l \right]\left[ mol/l \right]}{\left[ mol/l \right]} \\\ & {{k}_{C}}=mol/l \\\ \end{aligned}$$ -Here we can see that the equilibrium constant is having a mol/l unit. So, it depends on units of concentration, hence this option is correct. **Hence, we can conclude that the correct option is (D), that is for the reaction $COC{{l}_{2}}\left( g \right)\rightleftharpoons CO\left( g \right)+C{{l}_{2}}\left( g \right)$, the equilibrium constant depend on the units of concentration.** **Note:** We should not get confused in terms ${{k}_{C}}$and${{Q}_{C}}$. As ${{k}_{C}}$is the equilibrium constant, that is the ratio of concentrations of products and reactants when the reaction is at equilibrium. While ${{Q}_{C}}$ is the reaction quotient, which is used to determine in which direction a reaction will proceed.