Question

# Discuss the relation between ${{K}_{p}}\text{ } and \text{ }{{K}_{c}}$ for this equilibrium:$2{{H}_{2}}O(g)+2C{{l}_{2}}(g)\rightleftharpoons 4HCl(g)+{{O}_{2}}(g)$[A]${{K}_{p}}={{K}_{c}}$[B]${{K}_{p}}>{{K}_{c}}$[C]${{K}_{p}}<{{K}_{c}}$[D]${{K}_{p}}=\frac{1}{{{K}_{c}}}$

Hint: To solve this, first we have to write the equilibrium constants in terms of pressure and concentration. Then we can use the ideal gas equation to find the pressure for each gas in terms of concentration and then use it in the ${{K}_{p}}$ equation.

Complete Step by Step Solution: We use the terms ${{K}_{p}}\text{ }and\text{ }{{K}_{c}}$ to explain the equilibrium constants for gases. ${{K}_{p}}$ is the equilibrium constant in terms of partial pressure and ${{K}_{c}}$ is the equilibrium constant in terms of concentration.
As we express the equilibrium constant in terms of the reactant and the product, similarly we can express ${{K}_{p}} and {{K}_{c}}$ in terms of the partial pressure of the reactants and the products and the concentration of the reactants and the products respectively.
For the given equilibrium,$2{{H}_{2}}O(g)+2C{{l}_{2}}(g)\rightleftharpoons 4HCl(g)+{{O}_{2}}(g)$, we can write that-
${{K}_{c}}=\frac{{{\left[ HCl \right]}^{4}}{{\left[ {{O}_{2}} \right]}^{1}}}{{{\left[ {{H}_{2}}O \right]}^{2}}{{\left[ C{{l}_{2}} \right]}^{2}}}$
${{K}_{p}}=\frac{P_{HCl}^{4}\times P_{{{O}_{2}}}^{1}}{P_{{{H}_{2}}O}^{2}\times P_{C{{l}_{2}}}^{2}}$
We know that, for an ideal gas-
PV = nRT
Or, P =$\frac{n}{V}$RT
We know that the number of moles divided by the total volume gives us the concentration. Therefore, we can write that-
P = CRT
Where, P is the pressure, C is the concentration of the gases, R is the universal gas constant and T is the temperature.
Now, using this, we can find the pressure for each gas as-
\begin{align} & {{P}_{HCl}}=\left[ HCl \right]RT \\ & {{P}_{{{O}_{2}}}}=\left[ {{O}_{2}} \right]RT \\ & {{P}_{{{H}_{2}}O}}=\left[ {{H}_{2}}O \right]RT \\ & {{P}_{C{{l}_{2}}}}=\left[ C{{l}_{2}} \right]RT \\ \end{align}
Putting the values of pressure in the equilibrium equation, we will get
${{K}_{p}}=\frac{{{\left[ HCl \right]}^{4}}{{\left( RT \right)}^{4}}\times {{\left[ {{O}_{2}} \right]}^{1}}{{\left( RT \right)}^{1}}}{{{\left[ {{H}_{2}}O \right]}^{2}}{{\left( RT \right)}^{2}}\times {{\left[ C{{l}_{2}} \right]}^{2}}{{\left( RT \right)}^{2}}}$
Rearranging the above equation we can write that,
${{K}_{p}}=\frac{{{\left[ HCl \right]}^{4}}\times {{\left[ {{O}_{2}} \right]}^{1}}}{{{\left[ {{H}_{2}}O \right]}^{2}}\times {{\left[ C{{l}_{2}} \right]}^{2}}}{{\left( RT \right)}^{(4+1)-(2+2)}}$
As we calculated above, ${{K}_{c}}=\frac{{{\left[ HCl \right]}^{4}}{{\left[ {{O}_{2}} \right]}^{1}}}{{{\left[ {{H}_{2}}O \right]}^{2}}{{\left[ C{{l}_{2}} \right]}^{2}}}$
Therefore, we can write the above equation as-
${{K}_{p}}={{K}_{c}}RT$
As R, the universal gas constant has a positive value and temperature will have a positive value too, we can write that ${{K}_{p}}>{{K}_{c}}$
Therefore, the correct answer is option [B] ${{K}_{p}}>{{K}_{c}}$

Note: It is important to remember here that the value of ${{K}_{p}}\text{ } and \text{ }{{K}_{c}}$ will be different for different reactions. The general relation between ${{K}_{p}}\text{ } and \text{ }{{K}_{c}}$ can be written as ${{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}$ where the terms have their usual meanings and $\Delta n$ is the difference between the total number of moles of reactants and total number of moles in the product.