Question

A sample of ammonium carbamate placed in a glass vessel at 25°C undergoes the reaction:

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

The total pressure of gases in equilibrium with the solid is found to be 0.115 atm.

(a) Calculate the partial pressures of NH₃andCO₂.

(b) Calculate the equilibrium constant at 25°C.

Short answer

The partial pressure of${\mathrm{NH}}_3$is 0.076 atm.

The partial pressure of ${\mathrm{CO}}_2$is 0.0383 atm.

The equilibrium partial pressure for the reaction is$2.21\times {10}^{-4}$.

Step-by-step solution

Equilibrium Partial Pressure

The chemical constant that determines the concentration of product and reactant in a chemical reaction is known as the equilibrium constant. K_Pis the term that is used for the equilibrium constant for partial pressure.

Calculation

The total pressure of the gases at equilibrium is 0.115 atm.

The equilibrium reaction is as shown:

${\mathrm{NH}}_4{\mathrm{OCONH}}_2\left(\mathrm{s}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

The ICE table for the reaction is as shown:

The total pressure for the gases can be written as:

$\begin{array}{rcl}2\mathrm{x}+\mathrm{x}& =& 0.115 \mathrm{atm}\\ 3\mathrm{x}& =& 0.115\mathrm{atm}\\ \mathrm{x}& =& \frac{0.115\mathrm{atm}}{3}\\ & =& 0.0383 \mathrm{atm}\\ & & \end{array}$

The partial pressure of NH₃ at equilibrium is 2x, it partial pressure is calculated as:

$\begin{array}{rcl}{\mathrm{P}}_{{\mathrm{NH}}_3}& =& \left(2\mathrm{x}\right)\mathrm{atm}\\ & =& 2\times 0.0383 \mathrm{atm}\\ & =& 0.076\mathrm{atm}\end{array}$

Thus, the partial pressure ofNH₃ is 0.076 atm.

The equilibrium expression for the reaction is as shown:

$\mathrm{K}={\left({\mathrm{P}}_{{\mathrm{NH}}_3}\right)}^2\left({\mathrm{P}}_{{\mathrm{CO}}_2}\right)$

On substituting the given values, we get

$\begin{array}{rcl}{K}_p& =& {\left(P_{N{H}_3}\right)}^2\left(P_{C{O}_2}\right)\\ & =& {\left(0.076\right)}^2\left(0.0383\right)\\ & =& 2.21\times {10}^{-4}\end{array}$

Thus, the equilibrium partial pressure for the reaction is$\begin{array}{rcl}& & 2.21\times {10}^{-4}\end{array}$ .