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

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.

& {{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{aligned}$$ **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.