suppose that you place exactly 100 bacteria into a flask containing nutrients for the bacteria and that you find the following data at 37 °C:

t (min):0 15.0 30.0 45.0 60.0
Number of bacteria: 100 200 400 800 1600

What is the order of the rate of production of the bacteria?

How many bacteria will be present after 1.20 × 102 min?

What is the rate constant for the process?

please help! I'm not sure where to start!

This is a first order process since rate = k(A)

how do i find the rate constant?

Does it make sense to you that the bacteria started with 100, is doubling every 15 minutes and you HAVE LESS THAN YOU STARTED WITH after 120 minutes? Doesn't make sense to me.

k is 0.693/t1/2

Look at this table from your post, then extend it to 120 minutes.
min....count
0........100
15.......200
30.......400
45.......800
60......1600
75........?
90........?
105.......?
120.......?

THANKS. I figured it out.

To determine the order of the rate of production of the bacteria, we need to analyze the relationship between the number of bacteria and time. If the rate is constant, it would imply a first-order reaction. If the rate changes with the concentration of the reactants or products, it would indicate a higher order reaction.

To identify the reaction order, we can examine the data and plot it on a graph. The number of bacteria is plotted on the y-axis, while time is plotted on the x-axis. Once we have the graph, we can determine the best-fit line or curve.

Let's start with the first question: What is the order of the rate of production of the bacteria?

1. Plot the data on a graph with time (t) on the x-axis and the number of bacteria on the y-axis.
2. Check the shape of the graph. Does it form a straight line or a curve?
3. If the graph is a straight line, it indicates a first-order reaction; if it forms a curve, it could be a higher-order reaction.
4. Determine the equation that best fits the data points on the graph.
5. Based on the equation, identify the order of the rate of production of the bacteria.

Now, let's move on to the second question: How many bacteria will be present after 1.20 × 10^2 min?

To determine the number of bacteria, we can use the equation we found in the previous step. Since we haven't found the equation yet, we'll have to wait until the order of the reaction is determined.

Lastly, the third question: What is the rate constant for the process?

The rate constant can be derived once we know the reaction order and have the equation that describes the relationship between time and the number of bacteria.

So, let's start by plotting the data and analyzing the graph to determine the order of the rate of production of the bacteria.

The rate is doubling every 15.0 minutes so the half life is 15.0 min.

k = 0.693/t1/2. Solve for k and substitute into the below equation.
ln(No/N) = kt
No = 100
N = ?
k = from above
t = 1.2E2 min
Solve for N.

I pulled the k = 0.693/t1/2 our of a hat but you can get it this way.
ln(No/N) = kt
Pick out No = 200 and
N = 100
then t = 15 min and solve for k. You will get the same k as in the above formula BECAUSE
ln 200/100 = k(15)
0.693/15 = ?.
I don't pay any attention the the signs; if your prof does the k is a negative sign meaning the bacteria are expanding instead of decomposing.

i got N=0.83 does the seem right?