Description
Hello,
Exercises:
6.2, 6.4, 6.5, 6.6, 6.8, 6.9, 6.10
Please do not copy anything. I do not want you to copy other student's homework and give it to me please. This happened a lot to me.
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Explanation & Answer

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...
