excel kinetics lab

8/11/2019 Excel Kinetics Lab

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Katherine Chan

25 February 2014

Chem Lab

Chemical Kinetics Lab Questions:

The reaction in consideration here is A B.

1. Construct a spreadsheet for the 0thorder rate law with k= 1.0 and [A]0= 1000.

Generate data [A]t and time using integrated rate law and increments of t= 10s. Use

the data generated, construct a plot of [A] vs. time. Attach your graph and label it

Graph 1a. (extend time increments until you reach 500s)

a.

What is the shape of your graph?

The line is linear because the slope decreases steadily.

b.

After how many seconds will [A] drop to 600?

After 400 seconds, [A] will drop to 600 mol/L.

c.

Now change your [A]0value to 2000. How does the new graph differ from the

old one? Attach the graph and label it Graph 1b.

Not only the initial value start at 2000 mol/L in the second graph, but

the slope is more negative (or "steeper) in the first graph.

0

200

400

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0 100 200 300 400 500 600

Concentration(mol/L)

Time (s)

Graph 1a


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d.

Now change your kvalue from 1.0 to 0.5 and [A]0back to 1000. How does

this graph differ from Graph 1a and 1b? Attach the graph and label it Graph1c.

Compared to the Graph 1a and 1b, this graph has a smaller slope in

that the other graph had a bigger change in concentration over change

in time.

2. Construct a spreadsheet for the 1storder rate law with k= 1.0 and [A]0= 1000.Use increments of t= 0.10s. Use the data generated, construct plots of [A] vs. time

and [B] vs. time on the same graph. Attach your graph and label it Graph 2a. (extend

the time increments until you reach 5.00s)

0

500

1000

1500

2000

2500

0 100 200 300 400 500 600

Concentration(m

ol/L)

Time (s)

Graph 1b

0

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400

600

800

1000

1200

0 100 200 300 400 500 600


Time (s)

Graph 1c


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a.

After how many seconds will [A] and [B] be equal?

Between 0.7 and 0.8 seconds will they be equal.

b.



The new graph differs from the original because the concentration of

A is the same as the concentration of B sooner at 0.0 seconds. Though

both graphs show a variation of a logarithmic equation, the beginning

of Graph 2a shows more of a rapid increase of [B] as [A] rapidly

decreases. Graph 2b's slopes do not increase or decrease as much as

2a's.

0

200

400

600

800

1000

1200

0 1 2 3 4 5 6


Time (s)

Graph 2a

[A] [B]

0

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600

800

1000

1200

0 1 2 3 4 5 6


Time (s)

Graph 2b

[A] [B]


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c.

Now change your kvalue from 1.0 to 0.5 and [A]0back to 1000. How does

this graph differ from Graph 2a and 2b? Attach the graph and label it Graph

2c.

Compared to the graph of 2a and 2b, the concentration of A and

concentration of B are the same later at 1.4 seconds. Furthermore, the

rate does not plateau as much as time increases in Graph 2c comparedto Graph 2a and Graph 2b; the rate begins to plateau at ~3.0 s in

Graph 2a and ~2.5 s in 2b. Graph 2c shows that it's barely starting to

plateau at 5 s.

d.

Repeat part c by increasing the value of kto 1.5. At what point will [A] = 0?

Attach the graph and label it Graph 2d.

When k=1.5, the concentration of A and B are the same between 0.5

and 0.6 seconds. The graph also plateaus sooner around 2.0 seconds.

The slope of the concentration of [B] also increases more rapidly in

Graph 2d than the previous graphs.

0

200

400

600

800

1000

1200

0 1 2 3 4 5 6

Concentration

(mol/L)

Time (s)

Graph 2c

[A] [B]


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e.

Change your kvalue back to 1.0. Construct a graph using integrated rate law

to show a linear relationship between [A] and time. Attach the graph and

label it Graph 2e.

3. Construct a spreadsheet for the 2ndorder rate law with k= 1.0 and [A]0= 1000.

Use time increments of t= 0.00010s. Use the data generated, construct plots of [A]

vs. time and [B] vs. time on the same graph. Attach your graph and label it Graph 3a.

(extend the time increments until you reach 0.0050s)

0

200

400

600

800

1000

1200

0 1 2 3 4 5 6


Time (s)

Graph 2d

[A] [B]

0

200

400

600

800

1000

1200

0 1 2 3 4 5 6


Time (s)

Graph 2e

[A] [B]


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a.

At the point when [A] = [A]0, what is the rateof the reaction?

When the concentration is 500 M (half of the original concentration),

the rate of reaction is 500,000 M/s.

b.



Compared to 3a, the concentration starts at 500 M at 0.0 seconds.

Also, at the beginning of Graph 3a between 0.0 to 0.0001 seconds, the

slope of both lines are steep; in Graph 3b, the slope doesn't vastly

increase or decrease exponentially as much.

0

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600

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1000

1200

0 0.001 0.002 0.003 0.004 0.005 0.006


Time (s)

Graph 3a

1/[A] [B]

0

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400

500

600

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800

900

0 0.001 0.002 0.003 0.004 0.005 0.006


Time (s)

Graph 3b

1/[A] [B]


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c.

Now change your kvalue from 1.0 to 0.05 and [A]0back to 1000. How does this

graph differ from Graph 3a and 3b? Attach the graph and label it Graph 3c.

Graph 3c is very different from Graph 3a and 3b. While all of the

previous graphs have shown the concentrations were equivalent at

one point, the concentration in Graph 3c are always different. The

concentration of B is a lot more than the concentration of A at anygiven time. The shape of both lines are also more linear.

d. Repeat part c by increase the value of kto 1.5. Attach the graph and label it

Graph 3d.

0

200

400

600

800

1000

1200

0 0.001 0.002 0.003 0.004 0.005 0.006


Time (s)

Graph 3c

1/[A] [B]

0

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600

800

1000

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0 0.001 0.002 0.003 0.004 0.005 0.006


Time (s)

Graph 3d

1/[A] [B]


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e.

Change your kvalue back to 1.0. Construct a graph using integrated rate law

to show a linear relationship between [A] and time. Attach the graph and

label it Graph 3e.

4. The rate law of a chemical reaction is determined to be Rate = k[A]2[B]. What is

the unit of k?

5. Using graphs generated above, determine the half life for all three rate laws. Using

k=1.0 and [A]0=1000

6. True or false: the rate at which the reaction is occurring is independent of initial

concentration. Explain your answer.

7. A reaction is in progress. One of the reactants is orange in color. Can you

determine when the reaction has completed? If so, how?

8. Using the same rate constant, k, and the initial reactant concentration, [A]0,

different increments of time had to be used to generate the rate data, depending on

the order of reaction. Briefly explain why.

0

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0 0.001 0.002 0.003 0.004 0.005 0.006


Time (s)

Graph 3e

1/[A] [B]

excel kinetics lab

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