Chemistry, asked by parvesh9457kumar, 9 months ago

Half life of a first order chemical
reaction is 69 hr at 300 K Also, rate of this reaction is doubled
as temperature is increased from 300 K to 310 K. Determine Activation energy and pre-exponential factor for this reaction.​

Answers

Answered by anujdiwakr334
0

Answer:

Explanation:

Your tool of choice here will be the Arrhenius equation, which looks like this

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

k

=

A

exp

(

E

a

R

T

)

a

a

−−−−−−−−−−−−−−−−−−−−−−−−

, where

k

- the rate constant for a given reaction

A

- the pre-exponential factor, specific to a given reaction

E

a

- the activation energy of the reaction

R

- the universal gas constant, useful here as

8.314

J mol

1

K

1

T

- the absolute temperature at which the reaction takes place

Notice that he problem provides you with the pre-exponential factor, the activation energy, and the rate constant, and asks for the temperature in degrees Celsius at which the reaction has that particular rate constant.

Your strategy here will be to use the Arrhenius equation to solve for

T

, the absolute temperature at which the reaction takes place, then use the conversion factor

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

T

[

K

]

=

t

[

C

]

+

273.15

a

a

−−−−−−−−−−−−−−−−−−−−−−−−

to go from Kelvin to degrees Celsius.

So, rearrange the Arrhenius equation to isolate the exponential term on one side of the equation

k

A

=

exp

(

E

a

R

T

)

Take the natural log of both sides of the equation

ln

(

k

A

)

=

ln

[

exp

(

E

a

R

T

)

]

This will be equivalent to

ln

(

k

A

)

=

E

a

R

T

Rearrange to isolate

T

on one side of the equation

T

=

E

a

R

1

ln

(

k

A

)

Plug in your values to get - do not forget to convert the activation energy from kilojoules to joules

T

=

62.0

10

3

J

8.314

J

mol

1

K

1

1

ln

(

0.434

s

1

9.9

10

10

s

1

)

T

=

285.14 K

Expressed in degrees Celsius, the temperature will be

t

[

C

]

=

285.14 K

273.15

=

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

12

C

a

a

−−−−−−−−−

The answer is rounded to two sig figs, the number of sig figs you have for the pre-exponential factor.

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