Question

If the rate constant k of a reaction doubles when the temperature increases from 121 °C to 279 °C, what is the activation energy of the reaction in units of kJ/mol? Do not enter units with your numerical answer. Do not use scientific notation.

Answer #1

A reaction has a rate constant of 0.393 at 291 K and 1.41 at 345
K. Calculate the activation energy of this reaction in kJ/mol.

Find the ratio between rate constant at 300 K and 320
k when the activation energy of this reaction is 50 KJ/mol.

± 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...

If the rate constant increases from 0.40 M
–1 s –1 at 25°C to 0.80 M
–1 s –1 at 35°C, what is the activation
energy in kJ/mol for this reaction?

The activation energy, Ea for a particular reaction is
13.6 kj/mol. If the rate constant at 754 degrees celsius is
24.5/min at egat temperature in celsius will the rate constant be
12.7/min? r= 8.314j/mol • K

Reaction rates increase with temperature because as the
temperature increases:
The equilibrium constant increases
The activation energy increases
The activation energy decreases
The molecules have higher average kinetic energy
The molecules have lower average kinetic energy

Part A
The activation energy of a certain reaction is 36.4 kJ/mol . At
21 ∘C , the rate constant is 0.0160s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
Hints
Part B
Given that the initial rate constant is 0.0160s−1 at an initial
temperature of 21 ∘C , what would the rate constant be
at a temperature of 130. ∘C for the same reaction
described in Part A?
Express your...

The rate constant of a first-order reaction is 0.0032 x
10-4 L/mol *s at 640 K. If the activation energy is
176,406J/mol, calculate the temperature at which is rate constant
is 0.0039 x10 -4 L/mol*s.
Show your work please.

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...

Part A The activation energy of a certain reaction is 30.6
kJ/mol . At 22 ∘C , the rate constant is 0.0110s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast? Express your answer with the appropriate units. Hints T2 = 39
∘C SubmitMy AnswersGive Up Correct Part B Given that the initial
rate constant is 0.0110s−1 at an initial temperature of 22 ∘C ,
what would the rate constant be at a temperature of 170....

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