Great Deal! Get Instant $10 FREE in Account on First Order + 10% Cashback on Every Order Order Now

1. Imagine two competing first-order reactions with rate constants k1 > k2. In other words, the first reaction is faster. Show mathematically that, as you lower the temperature, the selectivity will...

1 answer below »
1. Imagine two competing first-order reactions with rate constants k1 >
k2. In other words, the first reaction is faster. Show mathematically that,
as you lower the temperature, the selectivity will normally increase. Hint:
Selectivity can be measured as the ratio between two rate constants. A
larger ratio of rate constants almost always gives a more selective
reaction. You can use the relationship between k and ?G‡ to make this
proof. Which equation defines the relationship between k and ?G‡?
Important Q: Why is this only an approximation? Under what
conditions can the selectivityremain temperature-independent?
For many of the next questions, I recommend using Excel or some
other mathematical software package to, e.g., calculate the slope of
lines.
Last question:
In a way similar to how we proved kH / kD = 7 as the theoretical
limit for primary KIEs in class, now use the same equations to
prove that the theoretical limit for secondary KIEs is kH / kD =
1.4. You will need to use the change in bond bending frequency
shown on our handout and on page 429 of the textbook.
Page 429
Document Preview:

1. Imagine two competing first-order reactions with rate constants k1 > k2. In other words, the first reaction is faster. Show mathematically that, as you lower the temperature, the selectivity will normally increase. Hint: Selectivity can be measured as the ratio between two rate constants. A larger ratio of rate constants almost always gives a more selective ‡ reaction. You can use the relationship between k and ?G to make this ‡ proof. Which equation defines the relationship between k and ?G ? Important Q: Why is this only an approximation? Under what conditions can the selectivityremain temperature-independent? For many of the next questions, I recommend using Excel or some other mathematical software package to, e.g., calculate the slope of lines.

Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
130 Votes
TTs170213_75297_6
Question1
Answer
See the following parallel reaction
The relative rate is given by
There fore for being more selective k1>k2
By Eyring equation,
Therefore, we can understand that k and ΔG# and inversely related. For k to be high, ΔG#
has to be low.
We need k1 to be higher than k2. Therefore, it is clear that we need ΔG# of forward reaction
to be lesser than that of the reverse reaction.
Important Question-
Answer
It is only an approximation. This is based the values of k1 and k2. When we vary T, these k
values will change. The change will not be proportional for k1 and k2. But here we assume
that they changes proportionally, hence it is an approximation.
Selectivity will be temperature independent is the system is near equili
ium.
Question 3.3 Problem
Tempt
Temp
Kelvin K delta G 1/T K
2.9 276.06 16.9
-
6490.314 0.003622 16.9
11.8 284.96 11
0.003509 11
18.1 291.26 8.4
-
5154.518 0.003433 8.4
21.9 295.06 7.9
-
5071.194 0.003389 7.9
29.3 302.46 6.5 -4707.78 0.003306 6.5
32 305.16 6.1
-
4588.637 0.003277 6.1
34.9 308.06 5.7
-
4458.504 0.003246 5.7
37.2 310.36 5.3
-
4304.014 0.003222 5.3
42.5 315.66 4.6 -4005.7 0.003168 4.6
By vant Hoff’s isochore equation
From graph slope =24715
Delta H= -Rx slope
=-8.314x24715
=
Question 11
time conc delta(conc) delta t rate
0 0.0165
600 0.0124 -0.0041 600 -6.83333E-06
1200 0.0093 -0.0031 600 -5.16667E-06
1800 0.0071 -0.0022 600 -3.66667E-06
2400 0.0053 -0.0018 600 -0.000003
3000 0.0039 -0.0014 600 -2.33333E-06
3600 0.0029 -0.001 600 -1.66667E-06
Rate= change in cocn/change in time
Rate1=K(conc)
n eqn1
Rate2=K(conc)
n eqn2
-6.88x10-6 =k(0.0123)^n
-5.1667x10-6 =k(0.0093)^n where n is the order of the reaction
Eqn 1/eqn 2
1.322= (1.32)^n
Therefore n=1 reaction is of first orde
Question 3.4
Answer
By Eyring equation
y = 24715x - 74.815
0
5
10
15
20
0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037
K2
K2
Linear (K2)

The...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here