Question

Derive the relation between ${{K}_{p}}$ and ${{K}_{c}}$ for a general chemical equilibrium reaction.

Answer 1

We know that${{K}_{p}}$ = The equilibrium constant calculated from the partial pressures of a chemical reaction.
${{K}_{c}}$ = The ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients in a chemical reaction.

Complete step by step answer:
In the question it is given that derive the relation between ${{K}_{p}}$ and ${{K}_{c}}$.
Consider the following reversible chemical reaction,
$aA+bB \leftrightharpoons cC+dD$
The equilibrium constant for the above reaction should be expressed in terms of the concentration and expressed as follows.

${{K}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$

In the above equation, equilibrium is involved between gaseous species, and then the concentrations should be expressed in terms of partial pressures of the gaseous substance involved in the reaction. The equilibrium constant in terms of partial pressures is as follows,

${{K}_{p}}=\dfrac{p_{{C}^{c}}\text{ }p_{{D}^{d}}}{p_{{A}^{a}}\text{ }p_{{B}^{b}}}$

Here ${{p}_{A}}$, ${{p}_{B}}$, ${{p}_{C}}$, ${{p}_{D}}$ represents the partial pressures of the substance A, B, C and D respectively involved in the reaction. Assume that the gases are ideal, then the ideal gas equation is as follows.
PV = nRT
$P=\dfrac{nRT}{V}$
Where P = pressure
n = amount of gas in mol
V = Volume in ${{m}^{3}}$
T = temperature in Kelvin
$\dfrac{n}{V}$ = concentration, C
Thus, at constant temperature, pressure of the gas P is directly proportional to its concentration C,
In general chemical reaction
$aA+bB \leftrightharpoons cC+dD$
The equilibrium constant is

${{K}_{p}}=\dfrac{p_{{C}^{c}}\text{ }p_{{D}^{d}}}{p_{{A}^{a}}\text{ }p_{{B}^{b}}}$

Now, P = CRT
Hence,
${{p}_{A}}$ = [A] RT, where [A] is the molar concentration of A
In the same way
${{p}_{B}}$ = [B] RT
${{p}_{C}}$= [C] RT
${{p}_{D}}$ = [D] RT
Here, [B], [C] and [D] are the molar concentration of B, C and D respectively.
Substituting the above values in the expression for ${{K}_{p}}$
${{K}_{p}}=\dfrac{[{{([C]RT)}^{c}}\text{ }{{([D]RT)}^{d}}]}{[{{([A]RT)}^{a}}\text{ }{{([B]RT)}^{b}}]}$

$\text{=}\dfrac{{{[C]}^{c}}{{[D]}^{d}}{{(RT)}^{c+d}}}{{{[A]}^{a}}{{[B]}^{b}}{{(RT)}^{a+b}}}\text{ }$

$\text{=}\dfrac{{{[C]}^{c}}{{[D]}^{d}}{{(RT)}^{c+d-a+b}}}{{{[A]}^{a}}{{[B]}^{b}}}$

$\text{=}{{\text{K}}_{\text{c}}}{{(RT)}^{c+d-a+b}}$
$\text{=}{{\text{K}}_{\text{c}}}{{(RT)}^{\Delta n}}$
Here ∆n = (c + d) – (a + b) , means number of moles of gaseous products – number of moles of gaseous reactants in a balanced chemical reaction.
Therefore the relation between ${{K}_{p}}$ and ${{K}_{c}}$ is
${{\text{K}}_{\text{p}}}\text{= }{{\text{K}}_{\text{c}}}{{(RT)}^{\Delta n}}$

Note: In a reaction if volume remains constant at equilibrium then ${{K}_{p}}$ is equal to ${{K}_{c}}$
${{K}_{p}}$ = ${{K}_{c}}$
To get the above relation, an equal number of moles of gas should be present on both sides of the chemical reaction.