Question

The Arrhenius Equation is typically written as

*k*=*A**e*−*E*a/*R**T*

However, the following more practical form of this equation also exists:

ln*k*2*k*1=*E*a*R*(1*T*1−1*T*2)

where *k*1 and *k*2 are the rate constants for a
single reaction at two different absolute temperatures
(*T*1and *T*2).

**Part A**

The activation energy of a certain reaction is 32.1 kJ/mol . At 20 ∘C, the rate constant is 0.0130 s−1. At what temperature would this reaction go twice as fast?

Express your answer numerically in degrees Celsius

**Part B**

Given that the initial rate constant is 0.0130 s−1 at an initial temperature of 20 ∘C, what would the rate constant be at a temperature of 100 ∘C?

Express your answer numerically in inverse seconds.

Answer #1

A)

we have:

T1 = 20 oC

=(20+273)K

= 293 K

K2/K1 = 2/1

Ea = 32.1 KJ/mol

= 32100 J/mol

we have below equation to be used:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(2/1) = (32100.0/8.314)*(1/293.0 - 1/T2)

0.6931 = 3860.9574*(1/293.0 - 1/T2)

T2 = 309 K

= (309-273) oC

= 36 oC

Answer: 36 oC

B)

we have:

T1 = 20 oC

=(20+273)K

= 293 K

T2 = 100 oC

=(100+273)K

= 373 K

K1 = 1.3*10^-2 s-1

Ea = 32.1 KJ/mol

= 32100 J/mol

we have below equation to be used:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/1.3*10^-2) = (32100.0/8.314)*(1/293.0 - 1/373.0)

ln(K2/1.3*10^-2) = 3861*(7.32*10^-4)

K2 = 0.2195 s-1

Answer: 0.220 s-1

± The Arrhenius Equation
The Arrhenius equation shows the relationship between the rate
constant k and the temperature T in kelvins and
is typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), A
is a constant called the frequency factor, and Ea
is the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathmatically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a
single reaction...

± The Arrhenius Equation
The Arrhenius equation shows the relationship between the rate
constant k and the temperature T in kelvins and
is typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), A is
a constant called the frequency factor, and Eais
the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathmatically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a
single reaction at...

The
Arrhenius equation shows the relationship between the rate constant
k and the temperature Tin kelvins and is
typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), Ais
a constant called the frequency factor, and Ea is
the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathmatically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a
single reaction at two different absolute temperatures
(T1and...

The Arrhenius equation shows the relationship between the rate
constant k and the temperature T in kelvins and is typically
written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K),
A is a constant called the frequency factor, and Ea is the
activation energy for the reaction. However, a more practical form
of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically
equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate
constants for a single reaction at two different absolute...

A. The Arrhenius
equation shows the relationship between the rate constant
k and the temperature T in kelvins
and is typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), A
is a constant called the frequency factor, and Ea
is the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathematically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a
single reaction at two different...

Explain the Arrhenius Equation, k = Ae-Ea/RT, in your own words.
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The rate constant of a chemical reaction increased from 0.100
s−1 to 3.10 s−1 upon raising the temperature from 25.0 ∘C to 51.0
∘C .
Part A
Calculate the value of (1/T2−1/T1) where
T1 is the initial temperature and T2 is the final
temperature.
Express your answer numerically.
Part B
Calculate the value of ln(k1/k2) where
k1 and k2 correspond to the rate constants at the
initial and the final temperatures as defined in part A.
Express your answer numerically....

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s−1 to 2.60 s−1 upon raising the temperature from 25.0 ∘C to 45.0
∘C .
I solved
(1T2−1T1) =
−2.11×10−4
K−1
In (k1/k2) = -3.26
But, I'm having problems on this question:
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Express your answer numerically in kilojoules per
mole.
It would be great if you could show all of your work. I've been
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There are several factors that affect the rate of a reaction.
These factors include temperature, activation energy, steric
factors (orientation), and also collision frequency, which changes
with concentration and phase. All the factors that affect reaction
rate can be summarized in an equation called the Arrhenius
equation:
k=Ae−Ea/RT
where k is the rate constant, A is the
frequency factor, Ea is the activation energy,
R=8.314 J/(mol⋅K) is the universal gas constant, and
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A certain...

Calculate Ea for the uncatalyzed reactions. Use equation 9 and
two uncatalyzed reactions that are identical in every aspect except
for temperature.
Eq. 9: ln k2/k1 = (Ea/R)(1/T1 - 1/T2)
Reaction
[H2O2]M
[I-]M
Δ[I2]
Reaction Time, Δt (s)
Reaction Temp. (K)
RATE = Δ[I2]/Δt (M/s)
Apparent rate, constant, k'
ln k'
1/T
1
0.008
0.015
1.25E-04
126
295.15
9.92E-07
8.27E-03
-4.79546
0.003388
2
0.012
0.015
1.25E-04
136
295.15
9.19E-07
5.11E-03
-5.2773
0.003388

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