Here you go bro...I attached 3 files, the word file is exactly the same than the pdf file. and there is also an Excel file where i did the graphics to solve the prolbems. If there is anyting that you dont understand just let me know

6.2) During a chemical reaction, the concentration of Species A was measured as a function of time. The
observed concentration of various time intervals is presented below. Determine the reaction order and rate
constant, k. Is Species A being removed or produced?

Time(min)
0
10
20
30
40
50

Concentration of
A(mg/l)
100
80
60
40
20
0

Answer120

Concentration of A mg/L

100
80
y = -2x + 100
R² = 1

60
40
20
0
0

10

20

30

40

50

60

Time, minutes

The reaction order for this chemical reaction is zero because just graphing the concentration vs. time was
enough to adjust a straight line, and evaluating the graphic we can find the equation of the line, the constant k
will be the slope of the line.
𝑚𝑔

So, the rate constant 𝑘 = −2 𝐿∗𝑚𝑖𝑛
Note: This means that there is a removal of Species A

6.4) The concentration of Species D was measured as a function of time during a chemical reaction. Its
observed concentration at various time intervals is presented below. Determine the reaction order and rate
constant, k. IS Species D being removed or produced?
Concentration of
D(mg/l)
200
142
111
90
77
67

Time(hr)
0
1
2
3
4
5
Answer-

250

Concentration of D mg/L

200

150

100

50

0
0

1

2

3

4

5

6

Time, hours

In this case graphing the concentration of D vs time the result is a curve, so this means that the reaction is not
of order zero. Then we have to graph the ln of the concentration of D vs time to verify if we obtain a straight
line.

6
y = -0.2147x + 5.2054
R² = 0.9719

ln(Concentration of D)

5
4
3
2
1
0
0

1

2

3

4

5

6

Time, hours

We can see that the result of the graphic is a straight line so the reaction is of order one and the constant k
will be the slope of the line
𝑚𝑔

So, the rate constant 𝑘 = −0.2147 𝐿∗ℎ𝑟
Note: This means that there is a removal of Species D

6.5) During a chemical reaction, the concentration of Species B was measured as a function of time. The
observed concentration of Species B at various time intervals is presented below. Determine the reaction
order and rate constant, k. Is Species B being removed or produced?

Time(hr)
0
5
10
15
20

Concentration of
B(mg/l)
100
125
150
175
200

Concentration of B mg/L

250
y = 5x + 100
R² = 1

200

150

100

50

0
0

5

10

15

20

25

Time, hours

The reaction order for this chemical reaction is zero because just graphing the concentration vs. time was
enough to adjust a straight line, and evaluating the graphic we can find the equation of the line, the constant k
will be the slope of the line.
𝑚𝑔

So, the rate constant 𝑘 = 5 𝐿∗ℎ𝑟
Note: This means that there is a production of Species B

6.6) If the half-life of a chemical compound is 30 days under anaerobic conditions, determine the first-order
removal-rate constant, k.
AnswerIf it is a first order reaction the time of half-life is calculated by the equation:

𝑡𝑖𝑚𝑒 𝑜𝑓 ℎ𝑎𝑙𝑓 𝑙𝑖𝑓𝑒 =

−ln(0.5) 0.693
=
𝑘
𝑘

In this case we have that the half time of the chemical compound is 30 days under anaerobic conditions.
So, substituting into equ...