
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) \[{{K}_{c}}={{K}_{p}}{{(RT)}^{-\Delta n}}\]
Answer
165.9k+ 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.
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.
Recently Updated Pages
Classification of Elements and Periodicity in Properties | Trends, Notes & FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Hess Law of Constant Heat Summation: Definition, Formula & Applications

Disproportionation Reaction: Definition, Example & JEE Guide

Environmental Chemistry Chapter for JEE Main Chemistry

Trending doubts
Combination of Capacitors - In Parallel and Series for JEE

Electrical Field of Charged Spherical Shell - JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Thermodynamics Class 11 Notes: CBSE Chapter 5

JEE Advanced 2025 Notes

Redox reaction Class 11 Notes: CBSE Chapter 7

Electrochemistry JEE Advanced 2025 Notes

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models
