Question

The activation energy for a reaction is 9.0 kcal/ а mol. The increase in the rate constant when its temperature is increased from 298 K to 308 K is: [a] 10% [b] 100% (c) 50% [d] 63 %​

Answer 1

Answer:

Correct option is

D

63%

2.303log

K

1

K

2

=

R

E

a

[

T

1

T

2

T

2

−T

1

]

log

K

1

K

2

=

2.303×2

9.0×10

3

[

308×298

308−298

]

K

1

K

2

=1.63;K

2

=1.63K

1

;

K

1

1.63K

1

−K

1

×100=63.0%

Answer 2

Please refer to the attachment:-