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# Relationship between ${K_p}$ and ${K_c}$ is ${K_p} = {K_c}{(RT)^{\Delta n}}$ . Tell whether it’s true or false.

Last updated date: 15th Sep 2024
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Hint: ${K_p}$ is defined as the equilibrium constant calculated when the partial pressures of the different reactants and products are taken of a given chemical reaction while ${K_c}$ is defined as the equilibrium constant calculated when the different molar concentrations of the different reactants and the products are taken of a given chemical reaction.

Now we are given the reaction ${K_p} = {K_c}{(RT)^{\Delta n}}$. We first need to define and see what are the terms present on both sides of the equation. ${K_p}$ is defined as the equilibrium constant calculated when the partial pressures of the different reactants and products are taken of a given chemical reaction while ${K_c}$ is defined as the equilibrium constant calculated when the different molar concentrations of the different reactants and the products are taken of a given chemical reaction.
Now to prove the above relation we should take a reaction for example:
$aA + bB \to cC + dD$
Let ${p_a},\,{p_b},\,{p_c}\,and\,{p_d}$ be the partial pressures of the two reactants and the two products given in the reaction.
Now the equilibrium constant of partial pressure is ${K_p}$ , defined as
${K_p} = \dfrac{{{p_c}^c{p_d}^d}}{{{p_a}^a{p_b}^b}}$
Now the partial pressure or pressure of an agas in a mixture of ideal gases always follows the principle $pV = nRT$
Now we also can say that $p = \dfrac{n}{V}RT \Rightarrow p = CRT$
Here the constant that replaces moles divided by volume is called the concentration of the molecule or the molar concentration.
Therefore, partial pressure of a gas A is equal to ${p_a} = {C_a}RT$
Now let’s see the formula for the equilibrium concentration in terms of molar concentration, that is ${K_c}$
${K_c} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}}$
Now putting the value of pressure from ${p_a} = {C_a}RT$ in the equation ${K_p} = \dfrac{{{p_c}^c{p_d}^d}}{{{p_a}^a{p_b}^b}}$
We get,
${K_p} = \dfrac{{{{({C_c}RT)}^c}{{({C_d}RT)}^d}}}{{{{({C_a}RT)}^a}{{({C_b}RT)}^b}}}$
$\Rightarrow {K_p} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}} \times {(RT)^{c + d - a - b}}$
Using the equation ${K_c} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}}$ and putting the value in the above equation
${K_p} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}} \times {(RT)^{c + d - a - b}}$
$\Rightarrow {K_p} = {K_c} \times {(RT)^{\Delta n}}$
Thus, the relation given above is correct where $\Delta n$ represents the change in column.
The relation ${K_p} = {K_c}{(RT)^{\Delta n}}$ given is true.

Note:
The equilibrium constant is a very important criterion for a reaction to happen, that is if the equilibrium constant is greater than one then the reaction occurs and the products are formed else the reaction doesn’t occur. The equilibrium constant can be manipulated with the help of temperature.