Question

A sample of ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\left(\mathbf{g}\right)$is placed in an empty cylinder at ${\mathbf{25}}^{\mathbf{°}}\mathbf{C}$. After equilibrium is reached, the total pressure is 1.5 atm, and 16% (by moles) of the original ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\left(\mathbf{g}\right)$has dissociated to ${\mathbf{NO}}_{\mathbf{2}}\left(\mathbf{g}\right)$.

a. Calculate the value of ${\mathbf{K}}_{\mathbf{p}}$for this dissociation reaction at${\mathbf{25}}^{\mathbf{°}}\mathit{C}$.

b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\left(\mathbf{g}\right)$and ${\mathbf{NO}}_{\mathbf{2}}\left(\mathbf{g}\right)$.

c. What percentage (by moles) of the original ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\left(\mathbf{g}\right)$is dissociated at the new equilibrium position (total pressure 5 1.00 atm).

Short answer

1. Thus, the value of $K_p$is 0.14.
2. $P_{N_2{O}_4}=0.67atm$and $P_{N{O}_2}=0.33atm$
3. Thus, the percentage of dissociated ${\mathrm{N}}_2{\mathrm{O}}_4$is 20 %.

Step-by-step solution

Law of mass action

The rate of reaction which is proportional to the concentration of each reactant is referred to as the law of mass action. At a state of chemical equilibrium, the ratio of reactant concentration and product concentration is constant.

Subpart (a)

The equilibrium reaction is as shown:

${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\to 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$

The ICE table for the reaction is as:

$$\begin{gathered} {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ Initial\left(atm\right)\mathrm{x}0 \\ Change\left(atm\right)-0.16\mathrm{x}+0.32\mathrm{x} \\ AtEquilibrium\left(atm\right)0.84\mathrm{x}0.32\mathrm{x} \end{gathered}$$

From the given data the total pressure is 1.5 atm. Thus, we can write as:

$\begin{array}{c}\left(0.84\right)x+\left(0.32\right)x=1.5atm\\ 1.16x=1.5atm\\ x=\frac{1.5 atm}{1.16}\\ =1.29\end{array}$

$K_p$for the reaction is written as:

$\begin{array}{c}{K}_p=\frac{{\left(P_{N{O}_2}\right)}^2}{\left(P_{N_2{O}_4}\right)}\\ =\frac{{\left(0.32\times 1.29\right)}^2}{\left(0.84\times 1.29\right)}\\ =\frac{0.16}{1.1}\\ =0.14\end{array}$

Thus, the value of $K_p$is 0.14.

Subpart (b)

The total pressure of the gases at equilibrium is 1.0 atm. If$P_{N_2{O}_4}$and $P_{N{O}_2}$are the partial pressure of the respective gases, then we can write

$P_{N_2{O}_4}+P_{N{O}_2}=1.0 atm.....\left(I\right)$

We also know that$K_p$is 0.14.

$\begin{array}{c}{K}_p=\frac{{\left(P_{N{O}_2}\right)}^2}{\left(P_{N_2{O}_4}\right)}\\ 0.14=\frac{{\left(P_{N{O}_2}\right)}^2}{\left(P_{N_2{O}_4}\right)}......\left(II\right)\end{array}$

If we solve equation 1 and 2 we get the following partial pressures of gases at equilibrium as:

$P_{N_2{O}_4}=0.67atm$and $P_{N{O}_2}=0.33atm$

Subpart (c)

If ${\mathrm{P}}_{°}$is the initial concentration of $N_2{O}_4$and x is the change in concentration of ${\mathrm{N}}_2{\mathrm{O}}_4$, we can look at the changes in concentration of all species as follows:

$$\begin{gathered} {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ Initial\left(atm\right){\mathrm{P}}_{°}0 \\ Change\left(atm\right)-\mathrm{x}+2\mathrm{x} \\ AtEquilibrium\left(atm\right)0.670.33 \end{gathered}$$

Now we need to calculate the amount of dissociated${\mathrm{N}}_2{\mathrm{O}}_4$as follows:

$\begin{array}{c}2x=0.33atm\\ x=\frac{0.33}{2}\\ =0.165atm\end{array}$

$\begin{array}{c}{\mathrm{P}}_{°}=0.67+x\\ =0.67+0.165\\ =0.84atm\end{array}$

The percent dissociation of${\mathrm{N}}_2{\mathrm{O}}_4$is calculated as follows:

$$\begin{gathered} \%N_2{O}_{4}dissociated=\frac{\mathrm{x}}{{\mathrm{P}}_{°}}\times 100\% \\ \begin{array}{c}=\frac{0.165}{0.84}\times 100\%\\ =20\%\end{array} \end{gathered}$$

Thus, the percentage of dissociated ${\mathrm{N}}_2{\mathrm{O}}_4$is 20 %.