.
10 points
X
For the first-order reaction
1/2 N204(8) ► NO2(g); AH = 28.6 kJ
the rate constant is k = 7.62 * 10% s-1 at 8 °C, and the activation energy is 27.0 kJ/mol. What is the rate constant at 23 °C?
S
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12.5 – Activation Energy Calculations
Useful Equations
Eq= R
=
In(kz) –In(ki)
1 / 4 1 / 2
()
R= 8.314
J
mole.K
TK = Tc + 273.15
=
exty
: exey ex-y
=
eln(x) = 2
=
ey
In (x + y) = ln (x) + In (y) In ()
2 •
=
= ln (2) – In (y)
In (x) = y ln (2)
In (ex) = x
=
5 points
The rate constant for a first-order reaction is 6.10 * 10-75-1 at 385 K and 8.79 * 10-25-1 at 705 K. What is the activation energy? (R = 8.31 J/(mol · K)).
*
S
.
Tyne your answer
1
0
5 points
The rate constant for a first-order reaction is 3.68 * 10-85-1 at 492 K and 7.50 * 10-45-1 at 813 K. What is the activation energy? (R= 8.31 J/(mol · K)).
=
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Next
2
10 points
-100
For the first-order reaction
1/2 N204(8)
> NO2(g); AH= 28.6 kJ
the rate constant is k 7.71 * 104 s-1 at -36 °C, and the activation energy is 20.2 kJ/mol. What is the rate constant at 16 °C?
S
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1
5 points
0:
The rate constant for a first-order reaction is 5.02 * 10-75-1 at 461 K and 3.91 * 10-35-1 at 612 K. What is the activation energy? (R = 8.31 J/(mol · K)).
*
Type your answer…
Ne
2
10 points
B
For the first-order reaction
1/2 N204(g)
· NO2(g); AH= 28.6 kJ
the rate constant is k = 4.02 * 103 s-1 at -39 °C, and the activation energy is 45.3 kJ/mol. What is the rate constant at 19 °C?
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2
0/10 points
For the first-order reaction
1/2 N204(8) ► NO2(g); AH = 28.6 kJ
the rate constant is k 4.02 * 103 s-1 at -39 °C, and the activation energy is 45.3 kJ/mol. What is the rate constant at 19 °C?
=
х
18,629
Correct Answer: 407,894
Feedback
Based on answering incorrectly
Ok, there are several things that could have gone wrong here:
1. You forgot to convert your temperature to K
1. we always have to convert to K, it’s a function of the kinetic energy being defined as KE = 0 at 0 kelvin. It also saves us from funky negatives
2. You forgot to convert your activation energy to J
1. You should always write your units. Our value of Ris in J, since our activation energy ends up living in an exponential function it’s not immediate obvious that
you’re missing a factor of 1000. Please write your units, it’s self checking and you put a ton of time into this calculation.
3. You got tricked by the change in enthalpy listed in the question and went down the wrong path with the equation you used to solve this.
1. Reasonable mistake. We don’t need the enthalpy here, it’s already included in our activation energy. The activation energy is defined as the minimum energy
required to form a product during a collision between reactants.
2. The change in enthalpy is going to be the energy we have to put in to break the bonds minus the energy we get out when we form the new bonds. Since the
reaction is endothermic ((Delta H>0)) the activation energy is going to be the enthalpy plus the energy required to get from our reactants to the the transition
state, or the energy required to form the activated complex. This graph,
Transition state
Ea
Energy
A + B
AH
C+D
Extent of reaction
illustrates the energy diagram for an exothermic reaction. In the
exothermic case the activation energy is just the energy required to reach the transition state – as the energy of the reactants is lower than the energy of the
products. Activation energy would still be required for this reaction as we have to get from the reactants to the products (releasing energy in the process) we first
have to get to the transition state and form the activated complex.
4. Things just got really awful with the natural log and exponential functions and the math got a bit weird, lets do some math
We can express the activation energy in terms of k1, k2, T1 and T2 using:
E.
===
-R
(
In(k2) –In(kı)
(-)
We’re trying to find k2, or the rate constant at T2.
We have EQ, R, k1, T1 and T2 so lets get all of the terms without ką to the other side of the equation
We have Eq, R, k1, T1 and T2 so lets get all of the terms without k2 to the other side of the equation
a:
– (1 – 1) + In (ki) = ln (kz
🙂 In
E
R
Mathematically this is getting pretty awful, so lets say A = –
(1-1)
Now we can rewrite the equation as
R
T2
Ti
A + In (kı) = ln (k2)
We need to get the k2 out of the natural log function and to do so we need to plug the natural log into an exponential function. We’re allowed to do whatever we want
mathematically as long as we do it to both sides
e(A+In(kı)) = e(ln(k2))
— е
=
Since the exponential of a natural log function just gives back the argument in the natural log (eln(z)
going to need to use another identity of the exponential function (ea+b = e.eb.
x) we get back k2 on the right hand side. On the left hand side we’re
A.eln(ki)
eln(k) + A.kı = k2
E
R
1 (1-1)
T2
So we’ve got everything but A. We defined A as A
temperatures in kelvin.
From here on out we just need to plug in Eą, R, T2 and T1, being careful to express En in J, and our
1
5 points
The rate constant for a first-order reaction is 7.41 * 10-95-1 at 418 K and 1.72 * 10-45-1 at 659 K. What is the activation energy? (R = 8.31 J/(mol · K)).
Type your answer…
10 points
For the first-order reaction
1/2 N204(g) → NO2(g); AH= 28.6 kJ
the rate constant is k 7.80 * 102 s-1 at 6 °C, and the activation energy is 33.5 kJ/mol. What is the rate constant at 37 °C?
Type your answer…
2 A (g) + B (9)
be
Let
> 2 c (g)
the experimental rate equation
Rate = K[A]*[B]P
where
K is the
rate constant.
for experiment 1
Rate = 12 [A] = 0.10
[B] = 0.10
SO
Rate = K [0.10] [o. 10] &
12 = K (0-10)& co. 109
for experimenty2
Rate = 24, [A] = 0.10
[B] = 0.20
24 = K (0.10)*(0-20)
So,
Rate = 36
SO
5
For experiment a
[A] = 0.10, LB) = 0.30
3)
36 = K (0.109 (0.30)
for experiment 44
Rate = 48 [A] = 0.20
[6] -0.10
SO,
48 = *(0.2094 (0.1038
for experiment = 5
Rate =
, [A] = 0.30 , (B) = 0.40
108 – K (0.3039 (0.10) – ©
108
equation
We
get
divided
by equation
CO. 10)
K
(oroja
24
12
x
X
(0.201.me
(0.40)
or,
2 = 2
Oy
21
or ßa 1
We get –
equation ® divided equation ④
48
(6.2004 (0.10 )
12
(0.104 (0.10 )
x
x
K
B
4 =
27
or,
or 28 =
22
OY,
x=2
SO
d = 2
and
ß = 1
Now,
the experimental rate equation is –
K [A] ² [B]
Rate =
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