Question

Carbon dioxide reacts with ammonia to give ammonium carbamate, a solid. The reverse reaction is also occurs:${\text{CO}}_2\left(g\right){\text{+NH}}_{\text{3}}\left(g\right)\stackrel{}{\rightleftarrows }{\text{NH}}_{\text{4}}{\text{OCOH}}_{\text{2}}$

The forward reaction is first order in CO₂(g) and second order in NH₃(g). Its rate constant is 0.238 atm^-2s^-1 at 0.0°C (expressed in terms of partial pressures rather than concentrations). The reaction in the reverse direction is zero order, and its rate constant, at the same temperature, is 1.60 ×10^-7 atm s^-1 . Experimental studies show that, at all stages in the progress of this reaction, the net rate is equal to the forward rate minus the reverse rate. Calculate the equilibrium constant of this reaction at 0.0°C.

Short answer

The equilbrium constant at O⁰C is $1.49x{10}^6$

Step-by-step solution

Equilbrium constant of reaction  when partial pressures are given:

The equation for the reaction given in the question as

${\text{CO}}_2+2{\text{NH}}_3\underset{}{\leftrightarrows }{\text{NH}}_4{\text{OCH}}_2$

The equilbrium constant k for the reaction when partial pressure is given is expressed as

$$\begin{gathered} \text{K}=\frac{1}{p_{{\text{CO}}_2}{\left(p_{N{H}_3}\right)}^2}........\left(1\right) \\ {\text{Wherep}}_{C{O}_2}\text{isthepartialpressureofcarbondioxide} \\ {\text{p}}_{N{H}_3}\text{isthepartialpartialpressureofammonia} \end{gathered}$$

Calculation of the equilbrium constant

The partial pressure of a gaseous reactant is calculated in the following way:

For forward reaction the rate can be calculated as the product of the concentration of reactants${\text{=k}}_1{\text{P}}_{C{O}_2}{\left({\text{P}}_{N{H}_3}\right)}^2$

In the similar way rate for backward reaction is calculated as product of concendration of products= ${\text{k}}_{-1}$

At the equilibrium condition ,the rate of the forward reaction is equal to the backward reaction, and so

localid="1661401484074" $$\begin{gathered} {\text{k}}_{-1}={\text{k}}_1{\text{P}}_{C{O}_2}{\left({\text{P}}_{N{H}_3}\right)}^2.......\left(2\right) \\ \frac{{\text{k}}_1}{{\text{k}}_{-1}}=\frac{1}{{\text{P}}_{C{O}_2}{\left({\text{P}}_{N{H}_3}\right)}^2}.....\left(3\right) \end{gathered}$$

By comparing equations 1 and 3

localid="1661401419617" $\text{K=}\frac{{\text{k}}_1}{{\text{k}}_{-1}}.........\left(4\right)$

Substituting

localid="1661401443346" $$\begin{gathered} {\text{k}}_1=0.238at{m}^{-2}{s}^{-1} \\ {\text{k}}_{-1}=1.60x{10}^{-7}atm{s}^{-1}\text{inequation-4,thefollowingequationisobtained} \\ \text{K}=\frac{0.238at{m}^{-2}{s}^{-1}}{1.60x{10}^{-7}atm{s}^{-1}} \\ \text{K}=0.1487x{10}^7 \\ \text{K}=1.49x{10}^6 \end{gathered}$$

Therefore, the equilbrium constant at O⁰C is $1.49x{10}^6$