Question

Calculate \[{{K}_{c}}\] and ${{K}_{p}}$ for the given reaction at 295K, if the equilibrium concentrations are $\left[ {{N}_{2}}{{O}_{4}} \right]=0.75\operatorname{M}$ and $\left[ N{{O}_{2}} \right]=0.062\operatorname{M}$, $R=0.08206\operatorname{L}\operatorname{atm}{{\operatorname{K}}^{-1}}{{\operatorname{mol}}^{-1}}$.
Reaction:\[{{N}_{2}}{{O}_{4}}\left( g \right)\rightleftharpoons 2N{{O}_{2}}\left( g \right)\]

Answer 1

Hint: Find \[{{K}_{c}}\] by using the equilibrium concentrations already given in the question. To find ${{K}_{p}}$ use the formula-
\[{{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}\]

Complete step-by-step answer:
\[{{K}_{c}}\] is known as the equilibrium constant when the concentration of the reactants and products are given in moles per litre. Let’s take an example of the following reaction-
\[aA+bB\rightleftharpoons cC+dD\]
The reaction is in equilibrium, so
\[{{K}_{c}}=\dfrac{{{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}}{{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}}\]
We can define \[{{K}_{c}}\] for the given reaction in the same way,
\[{{K}_{c}}=\dfrac{{{\left[ N{{O}_{2}} \right]}^{2}}}{\left[ {{N}_{2}}{{O}_{4}} \right]}\]
As you can see, the values required for calculating \[{{K}_{c}}\] are already given in the question. Putting the values in their respective places we get,
\[{{K}_{c}}=\dfrac{{{\left[ N{{O}_{2}} \right]}^{2}}}{\left[ {{N}_{2}}{{O}_{4}} \right]}=\dfrac{{{\left( 0.062 \right)}^{2}}}{0.75}=0.00512\]
So, we get \[{{K}_{c}}\] for the given reaction as $5.12\times {{10}^{-3}}$.
Let us move on to find ${{K}_{p}}$. This is also an equilibrium constant but is only defined when the partial pressures of reactants and products are given rather than their molar concentrations. As partial pressure is involved, ${{K}_{p}}$ is most often defined for gaseous reactions. To calculate this constant, we simply substitute the molar concentrations in the formula for \[{{K}_{c}}\] with their respective partial pressures. So,
\[{{K}_{p}}=\dfrac{{{\left[ {{P}_{C}} \right]}^{c}}{{\left[ {{P}_{D}} \right]}^{d}}}{{{\left[ {{P}_{A}} \right]}^{a}}{{\left[ {{P}_{B}} \right]}^{b}}}\]
Where, ${{P}_{A}}$ is the partial pressure of the gaseous reactant “A” and the others are defined in a similar way.
But, here we have not been provided with the individual partial pressures of reactants and products. We have to use \[{{K}_{c}}\] to find ${{K}_{p}}$ and they are related as follows:
\[{{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}\]
Where, “R” is the universal gas constant; “T” is the temperature at which the equilibrium is maintained and $\Delta n$is the difference in the number of moles of products and reactants.
As mentioned above, the formula for $\Delta n$ is,
\[\Delta n=\text{No}\text{. of moles of products}-\text{No}\text{. of moles of reactants}\]
Applying the above formula, we find $\Delta n$ is 1. The universal gas constant and temperature of the reaction is already given. We can proceed to find the ${{K}_{p}}$ of this reaction.
 \[\begin{align}
  & {{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}=0.00512\times \left( 0.08206\times 295 \right) \\
 & \Rightarrow {{K}_{p}}=0.1239\approx 0.124 \\
\end{align}\]

Therefore, the ${{K}_{p}}$ of this reaction is $1.24\times {{10}^{-1}}$.

Notes: You should always subtract the number of moles of reactants from the products in order to gain $\Delta n$. This is a common mistake as students sometimes do the opposite and end up with a wrong answer. The value of R should be used in accordance with the units of other given values of the question.