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Answer:
The order of the production of bacteria is 1.
12,787 bacteria will be present after 112 min.
[tex]0.04331 min^{-1}[/tex] is the rate constant for the process.
Explanation:
Rate of formation of bacteria when there 100 bacteria: R
Rate of formation of bacteria when there 200 bacteria: R' =2R
Rate of formation of bacteria is directly proportional to the number of bacteria:
[tex]\frac{R}{R'}=\frac{k[200]^x}{k[400]^x}[/tex]
[tex]\frac{R}{2R}=(\frac{200}{400})^x[/tex]
x = 1
The order of the production of bacteria is 1.
The half life of the process = [tex]t_{1/2}=16 minutes[/tex]
Rate constant of the process = k
For first order process k is only dependent on half life:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{16 min}=0.04331 min^{-1}[/tex]
For the first order process the increment numbers is given ;
[tex]N=N_o\times e^{kt}[/tex]
Where:
[tex]N_o[/tex]= Initial population
N = Population after time t.
We have ;
[tex]N_o=100[/tex]
t = 112 minutes
[tex]N=100\times e^{0.04331 min^{-1}\times 112 min}[/tex]
[tex]N = 12,786.82\approx 12,787[/tex]
12,787 bacteria will be present after 112 min.
[tex]0.04331 min^{-1}[/tex] is the rate constant for the process.
After 112 minutes, there are 12769 bacterial cells in the culture.
We know that bacterial growth always follows first order kinetics. The half life of the bacteria growth is the time taken for the number of bacteria cells in the culture to double.
We can see that the half life of the bacterial cells from the question is 16 mins. Now, we have to obtain the rate constant of the bacterial growth as follows;
k = 0.693/t1/2
k = 0.693/16 mins
k = 0.0433 min-1
Using the relation;
N = Noe^kt
No = 100
N = 100e^ 0.0433(112)
N = 12769 bacterial cells
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