Answer
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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.
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