Chemistry, asked by rajeshwaris210, 10 months ago

68.
b) 212 kJ
c) 35 kj
d) 12110
The E, of a non-catalyzed reaction at 27 C is 300 kcal/mol and E of the same reaction catalyzed
by an enzyme is 10 kcal mol.Calculate the ratio of rate constants for catalyzed and non-catalyzed
reaction.​

Answers

Answered by kumarisangita
0

Answer:

The Arrhenius equation 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

T

- the absolute temperature at which the reaction takes place

In essence, the Arrhenius equation establishes a relationship between the rate constant of a reaction and the absolute temperature at which the reaction takes place.

In other words, this equation allows you to figure out how a change in temperature will ultimately affect the rate of the reaction.

The two temperatures at which the reaction takes place can be calculated using the conversion factor

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

a

a

T

[

K

]

=

t

[

C

]

+

273.15

a

a

−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you will have

T

1

=

100

C

+

273.15

=

373.15 K

T

2

=

120

C

+

273.15

=

393.15 K

If you take

k

1

to be the rate constant of the reaction at

T

1

, you can say that

k

1

=

A

exp

(

E

a

R

T

1

)

(

1

)

Similarly, if you take

k

2

to be the rate constant of the reaction at

T

2

, you will have

k

2

=

A

exp

(

E

a

R

T

2

)

(

2

)

Now, let's assume that your reaction is

n

order with respect to a reactant

A

n

A

products

The differential rate law for this generic reaction would look like this

rate

=

k

[

A

]

n

Assuming that you'll perform the reaction at

T

1

and at

T

2

using the same concentration for the reactant, you can say that you have

rate

1

=

k

1

[

A

]

n

and

rate

2

=

k

2

[

A

]

n

Your goal here will be to find the ratio that exists between the rate of the reaction at

T

2

and the rate of the reaction at

T

1

. This comes down to finding

rate

2

rate

1

=

k

2

[

A

]

n

k

1

[

A

]

n

=

k

2

k

1

=

?

Now, divide equations

(

2

)

and

(

1

)

to get

k

2

k

1

=

A

exp

(

E

a

R

T

2

)

A

exp

(

E

a

R

T

1

)

This will be equivalent to

k

2

k

1

=

exp

[

E

a

R

(

1

T

1

1

T

2

)

]

Before plugging in your values, make sure that you do not forget to convert the activation energy from kcal per mole to cal per mole by using the conversion factor

1 kcal

=

10

3

cal

You will have

k

2

k

1

=

exp

[

15

10

3

cal

mol

1

1.987

cal

mol

1

K

1

(

1

373.15

1

393.15

)

K

1

]

k

2

k

1

=

2.799

I'll leave the answer rounded to two sig figs

rate

2

rate

1

=

¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

2.8

a

a

−−−−−−−

Therefore, the reaction will proceed

2.8

times faster if the temperature is increased by

20

C

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