Q16.119 CP - Chemistry: Molecular Nature Of Matter And Change

Question

The growth of Pseudomonas bacteria is modelled as a first-order process with k = 0.035 $\mathit{m}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}$  at 37°C. The initial Pseudomonas population density is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$cells/L. 

(a) What is the population density after two h? 

(b) What is the time required for the population to go from $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{t}\mathit{o}\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$cells/ L?

Step-by-Step Solution

Answer

(a) Populated density after 2hrs is  $7.0\times {10}^{4}cells/L$

(b) The time required is 20min

Step 1: Calculation of Population density after 2hr

As the reaction follows first-order kinetics. The rate of the reaction is:

$ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0$ 

Given, 

K=0.035$mi{n}^{-1}$  

The initial concentration or initial density  ${\left[A\right]}_0=1.0\times {10}^{-3}cells/L$

Time =2hr

In minutes it becomes 120 min

So, 

$$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ ln\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=Kt \\ ln\frac{{\left[A\right]}_t}{1.0\times {10}^3}=0.035\left(120\right) \\ {\left[A\right]}_t=7.0\times {10}^{4}cells/L \end{gathered}$$ 

Hence, the populated density after 2hrs is $7.0\times {10}^{4}cells/L$

 

Step 2: Calculation of time required for population

As the reaction follows first-order kinetics. The rate of the reaction is:

 $ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0$

Given, 

K=0.035$mi{n}^{-1}$  

The initial concentration or initial density  ${\left[A\right]}_0=1.0\times {10}^{-3}cells/L$

Final concentration or density ${\left[A\right]}_t=2\times {10}^{3}cells/L$ 

So, 

 $$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ ln\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=Kt \\ ln\frac{2.0\times {10}^3}{1.0\times {10}^3}=0.035\left(t\right)l \\ ln2.0=0.032t \\ t=20min \end{gathered}$$

Hence, the time required is 20min