Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
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) \[{{K}_{c}}={{K}_{p}}{{(RT)}^{-\Delta n}}\]

seo-qna
Last updated date: 20th Jun 2024
Total views: 54k
Views today: 0.54k
Answer
VerifiedVerified
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.