Question

A first-order reaction has a rate constant of 3.82 × 10-5 s-1 at 292 K. If the activation energy is 60 kJ/mol, what temperature (in K) is needed to double the rate of the reaction?

Answer 1

Step-by-step explanation:

Substituting k

1

​

=2×10

−2

,k

2

​

=4×10

−2

,T

1

​

=300K,T

2

​

=310K,R=8.314JK

−1

mol

−1

in the expression

log

k

2

​

k

1

​

​

=

2.303×R

E

a

​

​

(

T

1

​

T

2

​

T

2

​

−T

1

​

​

) , we get

log

2×10

−2

4×10

−2

​

=

2.303×8.314

E

a

​

​

(

300×310

310−300

​

)

log2=

19.147

E

a

​

​

×

300×310

10

​

E

a

​

=0.3010×19.147×300×31=53598Jmol

−1

E

a

​

=53.598kJmol

−1