Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The relationship between ${{K}_{p}}$ and ${{K}_{c}}$ is correctly shown as:This question has multiple correct options(A) ${{K}_{c}}={{K}_{p}}{{(RT)}^{\Delta n}}$(B) ${{K}_{p}}={{K}_{p}}{{(RT)}^{-\Delta n}}$(C) ${{K}_{c}}={{K}_{p}}{{(RT)}^{\Delta n}}$(D) ${{K}_{c}}={{K}_{p}}{{(RT)}^{-\Delta n}}$

Last updated date: 20th Jun 2024
Total views: 54k
Views today: 0.54k
Verified
54k+ views
Hint: Here we know that ${{K}_{c}}$ and ${{K}_{p}}$ are equilibrium constants of gaseous mixture. Here ${{K}_{c}}$ is for molar concentration and ${{K}_{p}}$ is for partial pressure of the gases inside a closed system.

Step by step solution:
${{K}_{c}}$and ${{K}_{p}}$ are the equilibrium constants of gaseous mixtures. Where
${{K}_{c}}$ is defined by molar concentration
${{K}_{p}}$ is defined by partial pressure.
Let’s consider a reversible reaction:
$aA+bB\underset{{}}{\leftrightarrows}cC+dD$
Now equilibrium constant for the reaction expressed in the terms of concentration:
${{K}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$
If the equilibrium reaction involves gaseous species. The equilibrium constant in terms of partial pressures is:
${{K}_{p}}=\dfrac{{{[pC]}^{c}}{{[pD]}^{d}}}{{{[pA]}^{a}}{{[pB]}^{b}}}$
And the ideal gas equation:
$pV=nRT$
By rearrangement:
$p=\dfrac{nRT}{V}=CRT$
So, from the ideal gas equation:
$pA\text{ }=\text{ }\left[ A \right]\text{ }RT$,$\text{ }pB\text{ }=\text{ }\left[ B \right]\text{ }RT$,$\text{ }pC\text{ }=\text{ }\left[ C \right]\text{ }RT$ and $\text{ }pD\text{ }=\text{ }\left[ D \right]\text{ }RT$
Now we will put all these values of partial pressure in the equation of ${{K}_{p}}$:
${{K}_{p}}=\dfrac{{{(\left[ C \right]\text{ }RT)}^{c}}{{(\left[ D \right]\text{ }RT)}^{d}}}{{{(\left[ A \right]\text{ }RT)}^{a}}{{(\left[ B \right]\text{ }RT)}^{b}}}$
By rearranging the equation and putting${{K}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$:
${{K}_{p}}=\dfrac{{{\left[ C \right]}^{c}}{{\text{(}RT)}^{c}}{{\left[ D \right]}^{d}}{{(RT)}^{d}}}{{{\left[ A \right]}^{a}}{{\text{(}RT)}^{a}}{{\left[ B \right]}^{b}}{{\text{( }RT)}^{b}}}$’
${{K}_{p}}={{K}_{c}}\dfrac{{{\text{(}RT)}^{c}}{{(RT)}^{d}}}{{{\text{(}RT)}^{a}}{{\text{( }RT)}^{b}}}$’
${{K}_{p}}={{K}_{c}}{{\text{(}RT)}^{(c+d)-(a+b)}}$’
Let $\Delta n=(c+d)-(a+b)$
Then,
${{K}_{p}}={{K}_{c}}{{\text{(}RT)}^{\Delta n}}$
So, from the above derivation we can say that the correct relationship between ${{K}_{p}}$ and ${{K}_{c}}$: ${{K}_{p}}={{K}_{c}}{{\text{(}RT)}^{\Delta n}}$
And ${{K}_{c}}={{K}_{p}}{{(RT)}^{-\Delta n}}$
Then the correct answer is option “D”.

Note: The equilibrium constants do not include the concentrations of single components such as liquids and solid, and they do not have any units. These constants are only for ideal gases.