Q16.113 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Nitrification is a biological process for removing $\mathit{N}{\mathit{H}}_{\mathbf{3}}$ from wastewater as $\mathit{N}{{\mathbf{H}}_{\mathbf{4}}}^{\mathbf{+}}$: 

 $\mathbf{N}{{\mathbf{H}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{2}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }\mathbf{N}{\mathbf{O}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{ }\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{ }\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

The first-order rate constant is given as

 ${\mathit{K}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{47}{\mathit{e}}^{\mathbf{0}\mathbf{.}\mathbf{095}\left(T°C-15°C\right)}$

 Where ${\mathit{K}}_{\mathbf{1}}$ is in $\mathit{d}\mathit{a}{\mathit{y}}^{\mathbf{-}\mathbf{1}}$ and T is in °C

1. If the initial concentration $\mathit{N}{\mathit{H}}_{\mathbf{3}}$ is 3.0 $\mathit{m}\mathit{o}\mathit{l}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$, how long will it take to reduce the concentration to 0.35 $\mathit{m}\mathit{o}\mathit{l}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$ in the spring (T =20°C)? 
2.  In the winter (T = 10°C)?
3.  Using your answer to part (a), what is the rate of ${\mathit{O}}_{\mathbf{2}}$ consumption?

Step-by-Step Solution

Answer

1. 3days
2. 7days
3. Rate=  $-\frac{1}{2}\frac{\delta O_2}{\delta t}$

Step 1: Calculation of time in spring

The first-order rate constant is given at T=20°C as:

$$\begin{gathered} K_1=0.47e^{0.095\left(T°C-15°C\right)} \\ {K}_1=0.47e^{0.095\left(20°C-15°C\right)} \\ {K}_1=0.47e^{0.095\left(5°C\right)} \\ {K}_1=0.47e^{0.475} \\ {K}_1=0.755da{y}^{-1} \end{gathered}$$

The first-order kinetics followed as:

Given,

 $$\begin{gathered} {\left[A\right]}_0=3.0mol/m^3 \\ {\left[A\right]}_t=0.35mol/m^3 \end{gathered}$$

Therefore,

$$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ Kt=ln\frac{{\left[A\right]}_0}{{\left[A\right]}_t} \\ 0.755t=ln\frac{3}{0.35} \\ t=3days \end{gathered}$$ 

  

Step 2: Calculation of Rate of consumption

For a given reaction rate of a reactant is written as:

 $N{H_4}^{+} + 2O_2\to N{O}_3^{-} +2H^{+} +H_{2}O$

The rate of oxygen is:

Rate=  $-\frac{1}{2}\frac{\delta O_2}{\delta t}$

Step 3: Calculation of time in winter

The first-order rate constant is given at T=10°C as:

$$\begin{gathered} K_1=0.47e^{0.095\left(T°C-15°C\right)} \\ {K}_1=0.47e^{0.095\left(10°C-15°C\right)} \\ {K}_1=0.47e^{0.095\left(-5°C\right)} \\ {K}_1=0.47e^{-0.475} \\ {K}_1=0.299da{y}^{-1} \end{gathered}$$

The first-order kinetics followed as:

Given,

$$\begin{gathered} {\left[A\right]}_0=3.0mol/m^3 \\ {\left[A\right]}_t=0.35mol/m^3 \end{gathered}$$

Therefore,

 $$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ Kt=ln\frac{{\left[A\right]}_0}{{\left[A\right]}_t} \\ 0.299t=ln\frac{3}{0.35} \\ t=7days \end{gathered}$$