Question

Calculate $$K_c$$ and $K_p$ for the given reaction at 295K, if the equilibrium concentrations are $\left[N_2{O}_4\right]=0.75\mathrm{M}$ and $\left[N{O}_2\right]=0.062\mathrm{M}$, $R=0.08206\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$.
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-
$$K_p=K_c{\left(RT\right)}^{\mathrm{\Delta }n}$$

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
$$K_c={\displaystyle \frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}}$$
We can define $$K_c$$ for the given reaction in the same way,
$$K_c={\displaystyle \frac{{\left[N{O}_2\right]}^2}{\left[N_2{O}_4\right]}}$$
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,
$$K_c={\displaystyle \frac{{\left[N{O}_2\right]}^2}{\left[N_2{O}_4\right]}}={\displaystyle \frac{{\left(0.062\right)}^2}{0.75}}=0.00512$$
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,
$$K_p={\displaystyle \frac{{\left[P_C\right]}^c{\left[P_D\right]}^d}{{\left[P_A\right]}^a{\left[P_B\right]}^b}}$$
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:
$$K_p=K_c{\left(RT\right)}^{\mathrm{\Delta }n}$$
Where, “R” is the universal gas constant; “T” is the temperature at which the equilibrium is maintained and $\mathrm{\Delta }n$is the difference in the number of moles of products and reactants.
As mentioned above, the formula for $\mathrm{\Delta }n$ is,
$$\mathrm{\Delta }n=\text{No}\text{. of moles of products}-\text{No}\text{. of moles of reactants}$$
Applying the above formula, we find $\mathrm{\Delta }n$ 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.
 $$\begin{array}{rl}& K_p=K_c{\left(RT\right)}^{\mathrm{\Delta }n}=0.00512\times (0.08206\times 295)\\ & \Rightarrow K_p=0.1239\approx 0.124\end{array}$$

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 $\mathrm{\Delta }n$. 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.