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Question: Relationship between \( {K_p} \) and \( {K_c} \) is \( {K_p} = {K_c}{(RT)^{\Delta n}} \) . Tell whet...

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

Explanation

Solution

Kp{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 Kc{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.

Complete answer:
Now we are given the reaction Kp=Kc(RT)Δn{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. Kp{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 Kc{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+bBcC+dDaA + bB \to cC + dD
Let pa,pb,pcandpd{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 Kp{K_p} , defined as
Kp=pccpddpaapbb{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=nRTpV = nRT
Now we also can say that p=nVRTp=CRTp = \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 pa=CaRT{p_a} = {C_a}RT
Now let’s see the formula for the equilibrium concentration in terms of molar concentration, that is Kc{K_c}
Kc=CccCddCaaCbb{K_c} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}}
Now putting the value of pressure from pa=CaRT{p_a} = {C_a}RT in the equation Kp=pccpddpaapbb{K_p} = \dfrac{{{p_c}^c{p_d}^d}}{{{p_a}^a{p_b}^b}}
We get,
Kp=(CcRT)c(CdRT)d(CaRT)a(CbRT)b{K_p} = \dfrac{{{{({C_c}RT)}^c}{{({C_d}RT)}^d}}}{{{{({C_a}RT)}^a}{{({C_b}RT)}^b}}}
Kp=CccCddCaaCbb×(RT)c+dab\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 Kc=CccCddCaaCbb{K_c} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}} and putting the value in the above equation
Kp=CccCddCaaCbb×(RT)c+dab{K_p} = \dfrac{{{C_c}^c{C_d}^d}}{{{C_a}^a{C_b}^b}} \times {(RT)^{c + d - a - b}}
Kp=Kc×(RT)Δn\Rightarrow {K_p} = {K_c} \times {(RT)^{\Delta n}}
Thus, the relation given above is correct where Δn\Delta n represents the change in column.
The relation Kp=Kc(RT)Δn{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.