Question

For the reaction${\mathbf{N}}_2{\mathbf{O}}_4\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{NO}}_2\mathbf{\left(}\mathbf{g}\mathbf{\right)}{\mathbf{K}}_p\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}$ at a certain temperature. If 0.040 atm of${\mathbf{N}}_2{\mathbf{O}}_4$ is reacted initially, calculate the equilibrium partial pressures of${\mathbf{NO}}_2\left(\mathbf{g}\right)$and.${\mathbf{N}}_2{\mathbf{O}}_4\left(\mathbf{g}\right)$

Short answer

The partial pressures of ${\text{NO}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$are0.056 and 0.028 respectively.

Step-by-step solution

Define chemical equilibrium

The equilibrium of a reaction is set on the basis of the rate of the reaction. When the rates of products and reactants attain equilibrium values, then the reaction is said to be in equilibrium. As per the usual norms when the equilibrium constant is given, the method followed to find out the individual concentrations at equilibrium for the reactants and products is ICE (initial, change equilibrium)

Calculation

The equilibrium reaction is as follows:

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\rightleftharpoons {\text{2NO}}_{\text{2}}\left(\text{g}\right)$

The dissociation constant or the reaction is

$K_p=\frac{{\left(P_{N{O}_2}\right)}^2}{\left(P_{N_2{O}_4}\right)}$

The ICE table for the reaction is

$\begin{array}{ccccc}& & {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)& \rightleftharpoons & {\text{2NO}}_{\text{2}}\left(\text{g}\right)\\ Initial\left(atm\right)& & 0.040& & 0\\ Change\left(atm\right)& & -x& & +2x\\ AtEquilibrium\left(atm\right)& & 0.040-x& & 2x\end{array}$

On substituting the given values, we get

$\begin{array}{c}{K}_p=\frac{{\left(P_{N{O}_2}\right)}^2}{\left(P_{N_2{O}_4}\right)}\\ 0.25=\frac{{\left(2x\right)}^2}{(0.040-x)}\\ 0.25\times 0.040-0.25x=4x^2\\ 4x^2+0.25x-0.01=0\end{array}$

On applying quadratic equation, we get

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$;

wherea=4, b=+0.25 and c=-0.01.

On substituting the values and simplifying the given quadratic equation, we get

x=0.028

The partial pressure of${\text{NO}}_{\text{2}}$is calculated as follows:

$\begin{array}{c}{P}_{N{O}_2}=2x\\ =2\times 0.028\\ =0.056\end{array}$

The partial pressure of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$is calculated as follows:

$\begin{array}{c}{P}_{N_2{O}_4}=x\\ =0.028\end{array}$

Thus, the partial pressures of ${\text{NO}}_{\text{2}}$and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$are0.056 and 0.028 respectively.