Question

At ${\mathbf{25}}^o\mathbf{C}$ , ${\mathbf{K}}_p=\mathbf{2}.\mathbf{9}\times {\mathbf{10}}^{-\mathbf{3}}$ for the reaction

${\mathbf{NH}}_{\mathbf{4}}{\mathbf{OCONH}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{2}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

In an experiment carried out at ${\mathbf{25}}^o\mathbf{C}$, a certain amount of ${\mathbf{NH}}_4{\mathbf{OCONH}}_2$ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

Short answer

The total pressure in the container at equilibrium is 0.27 atm.

Step-by-step solution

The expression for equilibrium constant

For the given balanced equation, the expression for the equilibrium constant, K_p in terms of partial pressures is written below.

${\text{K}}_{\text{p}}\text{=}{\left({\text{P}}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}\left({\text{P}}_{{\text{CO}}_{\text{2}}}\right)$ …(1)

Since the partial pressures of pure liquids and solids are ignored.

The equilibrium constant value for the reaction is given as, ${\text{K}}_{\text{P}}{\text{=2.9×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}3}}$.

ICE table

Let x be the change in partial pressure.

Setup the ICE table as:

$\begin{array}{cccccc}& {\text{NH}}_{\text{4}}{\text{OCONH}}_{\text{2}}\text{(s)}& \stackrel{}{\rightleftharpoons }& {\text{2NH}}_{\text{3}}\text{(g)}& \text{+}& {\text{CO}}_{\text{2}}\text{(g)}\\ \text{Iinitial}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}}& & \text{0}& & \text{0}\\ \text{Change}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}}& & \text{+2x}& & \text{+x}\\ \text{Equilibrium}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}}& & \text{2x}& & \text{x}\end{array}$

The value of “x”

Insert the K_p value and the terms of the equilibrium partial pressures of the gases from ICE table into equation (1) to find the value of x.

$\begin{array}{l}{\text{2.9×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}3}}\text{=}{\left(\text{2x}\right)}^{\text{2}}\left(\text{x}\right)\\ {\text{2.9×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}3}}{\text{=4x}}^{\text{3}}\\ {\text{x}}^{\text{3}}\text{=}\frac{{\text{2.9×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}3}}}{\text{4}}\end{array}$

Solve for x.

$\text{x=0.090}$

Hence, the value of x is 0.090 atm.

Partial pressure of each gas at equilibrium

The equilibrium partial pressure of ammonia is,

$\begin{array}{c}{\text{P}}_{{\text{NH}}_{\text{3}}}\text{=2x}\\ \text{=2×0.090}\\ \text{=0.18\hspace{0.17em}atm}\end{array}$

Similarly, the equilibrium partial pressure of carbon dioxide is,

$\begin{array}{c}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=x}\\ \text{=0.090\hspace{0.17em}atm}\end{array}$

The total pressure, ${\text{P}}_{\text{total}}$at equilibrium is the sum of the partial pressures of ammonia and carbon dioxide.

$\begin{array}{c}{\text{P}}_{\text{total}}{\text{=P}}_{{\text{NH}}_{\text{3}}}{\text{+P}}_{{\text{CO}}_{\text{2}}}\\ \text{=0.18\hspace{0.17em}atm+0.090\hspace{0.17em}atm}\\ \text{=0.27\hspace{0.17em}atm}\end{array}$

Therefore, the total pressure at equilibrium is 0.27 atm.