Question

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/*T*2−1/*T*1) where
*T*1 is the initial temperature and *T*2 is the final
temperature.

Express your answer numerically.

Part B

Calculate the value of ln(*k*1/*k*2) where
*k*1 and *k*2 correspond to the rate constants at the
initial and the final temperatures as defined in part A.

Express your answer numerically.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

Answer #1

**Answer** – Given, rate constant, k1 = 0.100
s^{-1} , rate constant, k2 = 3.10 s^{-1}

T1 =25+273.15 = 298.15 K , T2 = 51 +273.15 = 324.15 K

Part A) In this one we need to calculate (1/T2-1/T1)

= 1/ 324.15 K – 1/ 298.15 K

**= - 0.00027 K**

**Part B)** Calculate the value of
ln(*k*1/*k*2)

= ln 0.100 s^{-1} / 3.10 s^{-1}

**= -3.43**

**Part C)** The activation energy of the reaction-
We know formula

ln(*k*1/*k*2) = Ea/R * (1/T2-1/T1)

-3.43 = Ea/ 8.314J.mol^{-1}.K^{-1} * (-0.00027
K)

Ea = -3.43 *8.314 / -0.00027

= 1.06*10^{5} J/mol

**= 1.06*10 ^{2} kJ/mol**

The rate constant of a chemical reaction increased from 0.100
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:
What is the activation energy of the
reaction?
Express your answer numerically in kilojoules per
mole.
It would be great if you could show all of your work. I've been
trying to figure out this problem for a...

The Arrhenius Equation is typically written as
k=Ae−Ea/RT
However, the following more practical form of this equation also
exists:
lnk2k1=EaR(1T1−1T2)
where k1 and k2 are the rate constants for a
single reaction at two different absolute temperatures
(T1and T2).
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...

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
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
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...

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...

A certain reaction has an activation energy of 31.40 kJ/mol. At
what Kelvin temperature will the reaction proceed 6.50 times faster
than it did at 293 K?
Use the Arrhenius equation
ln (k2/k1) = Ea/R [(1/T1)-(1/T2)]
Where R=8.3145 J/(molxK)

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 reaction has a rate constant of 1.25×10−4 s−1 at 29 ∘C and
0.228 s−1 at 79 ∘C . Part A Determine the activation barrier for
the reaction. Express your answer in units of kilojoules per mole.
Part B What is the value of the rate constant at 18 ∘C ? Express
your answer in units of inverse seconds.

Part A
The activation energy of a certain reaction is 30.5 kJ/mol . At
30 ∘C , the rate constant is 0.0180s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
T2 =
Part B
Given that the initial rate constant is 0.0180s−1 at an initial
temperature of 30 ∘C , what would the rate constant be
at a temperature of 170. ∘C for the same reaction
described in Part A?
Express...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 7 minutes ago

asked 16 minutes ago

asked 29 minutes ago

asked 52 minutes ago

asked 53 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago