George Mason Relative Rate Responses Please help me solve these questions? I also need to show work and figuring out the graph. Experiment 21 Data and Calc

George Mason Relative Rate Responses Please help me solve these questions? I also need to show work and figuring out the graph. Experiment 21
Data and Calculations: Rates of Chemical Reactions,
II. A Clock Reaction
(2)
A. Dependence of Reaction Rate on Concentration
Reaction: 61) + Bro,aq) + 6H(aq) 31. (aq) + Br(+ 3H.O()
In all the reaction mixtures used in this experiment.
rate – 111MBO, TH-
number of moles of Bro, had been used up in the reaction. The color clock” allows you to measure the time
required for the color to change; since in Equation 2 the change in concentration of Bro, ion, Bro,).
nguined for this filmber of roles of Bro,” pracy. The rule of each reaction is determined by the time
is the same in each mixture, the relative rate of each reaction is inversely proportional to the time 1. Since we
being equal to 1000V. Fill in the following table, first calculating the relative reaction rate for each mixture.
are mainly concerned with relative rather than absolute rates, we will for convenience take all relative rates as
ABO,
color change occurred when constant predetermined
Temp.
in (C)
Reactant Concentrations
in Reacting Mixture (M)
[Bro,
(H)
0.0080 0.02e
21
20
0.0080
0.020
0.02
Timer (sec)
Relative Rate
Reaction for Color of Reaction
Mixture to Change
1000/
11″)
1
96 10.4
0.0020
2
41 24.39 0.004
21
3
45 22.22
0.0020
0.016
20
4
27 37.04 0.0020
0.0080
5
14 8.771 0.0016
0.004
The reactant concentrations in the reaction mixture are nor those of the stock solutions, since the reagents
were diluted by the other solutions. The final volume of the reaction mixture is 50 mL in all cases. Since the
number of moles of reactant does not change on dilution we can say, for example, for I ion, that
moles of r = [1]stock X Vstock = [1°misture X Vanixture
For Reaction Mixture I.
0.04
0.03
20
[1stock = 0.010 M.
Vstock = 10 mL
Vimixture
50 mL
Therefore,
11-mare
0.010 Mx10 mL
= 0.0020 M
50 mL
Calculate the rest of the concentrations in the table using the same approach.
Determination of the Orders of the Reaction
Given the data in the table, the problem is to find the order for each reactant and the rate constant for the reac-
tion. Since we are dealing with relative rates, we can modify Equation 2 to read as follows:
relative rate = k1151″[Br0,1″(H
(5)
(continued on following page)
168 Experiment 2+ Rates of Chemical Reactions, I. A Clock Reaction
We need to determine the relative rate constant k’ and the orders a, #, and p in such a way as to be consistent
with the data in the table.
The solution to this problem is quite simple, once you make a few observations on the reaction mixtures.
Each mixture (2 to 4) differs from Reaction Mixture in the concentration of only one species (see table),
This means that for any pair of mixtures that includes Reaction Mixture 1. there is only one concentration that
changes. From the ratio of the relative rates for such a pair of mixtures we can find the order for the reactant
whose concentration was changed. Proceed as follows:
Write Equations below for Reaction Mixtures 1 and 2, substituting the relative rates and the concentra-
tions of r. Bros. and Hions from the table you have just completed.
Relative Rate 10
– K’I 0.0020 | 0.0080 II 0.020 V
Relative Rate 2 = 24
lo.colo l’ 0.020 P
Divide the first equation by the second, noting that nearly all the terms cancel out. The result is simply
Relative Rate 1
Relative Rate 2
= kl 0.004
If you have done this properly, you will have an equation involving only m as an unknown. Solve this equation
for m. the order of the reaction with respect to lion.
m = 7.7 (nearest integer)
Applying the same approach to Reaction Mixtures 1 and 3, find the value of n, the order of the reaction with
respect to Bro, ion.
Relative Rate 1 =
10
=
klo 11.000 1’1 .Ozo
Relative Rate 3 =
=k’l o.orzo 10.016
1″ 0.002
Dividing one equation by the other:
n =
1.2=1
Now that you have the idea, apply the method once again, this time to Reaction Mixtures 1 and 4, and
find p, the order with respect to H ion.
Relative Rate 4 =
= kl .000 0.00301″ 0.04 IP
Dividing the equation for Relative Rate I by that for Relative Rate 4, we get
p=
0.40
Having found m. n, and p (nearest integers), the relative rate constant, k’. can be calculated by substitution of m.
n, p, and the known rates and reactant concentrations into Equation 5. Evaluate k’ for Reaction Mixtures 1 to 4.
Mixture
1
2
3
4
k’
man
Standard deviation in k’
(See Appendix VIII)
Why should k’ have nearly the same value for each of the above reactions?
Using kucu in Equation 5, predict the relative rate and time, I prad. for Reaction Mixture 5. Use the concen-
trations in the table.
Relative ratepeed
“peed
lots
Experiment 2+ Rates of Chemical reactions. Alex Resin
159
B. Effect of Temperature on Reaction Rate: The Activation Energy
To find the activation energy for the reaction it will be helpful to complete the following table.
The dependence of the rate constant. k’, for a reaction is given by Equation 4:
In k’ =-
E
RT
+ constant
Since the reactions at the different temperatures all involve the same reactant concentrations, the rate
constants, K. for two different mixtures will have the same ratio as the reaction rates themselves for the two
mixtures. This means that in the calculation of Eg. we can use the observed relative rates instead of rate con-
stants. Proceeding as before, calculate the relative rates of reaction in each of the mixtures and enter these
values in (c). Take the natural logarithm of the relative rate for each mixture and enter these values in (d). To
set up the terms in 1/T, fill in (b), (e), and (1) in the table.
Approximate Temperature in °C
20
40
10
0
(a) Time 1 in seconds for color to appear
96
??
llo
(b) Temperature of the reaction mixture in °C
21
41
10
(c) Relative rate = 1000/t
(d) In of relative rate
(e) Temperature T in K
(1) 1/T, K-
To evaluate Eg, make a graph of In relative rate vs. 1/T, using Excel or the graph paper provided. (See
Appendix V.)
Find the slope of the line obtained by drawing the best straight line through the experimental points.
Slope
?
The slope of the line equals -EJR, where R
culate the activation energy, Ey, for the reaction.
(8.31 joules/mole K) if E, is to be in joules per mole. Cal-
Ex =
joules/mo
Optional C. Effect of a Catalyst on Reaction Rate
Reaction 1
Catalyzed
Reaction 1
Section
Name
Experiment 21
Data and Calculationst Rates of Chemical Reactions,
II. A Clock Reaction

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