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For a first-order reaction A rarr P, the...

For a first-order reaction `A rarr P`, the temperature (T) dependent rate constant (k) was found to follow the equation, log k `= -(2000)(1)/(T) + 6.0`.
The pre-expontential factor A and activation energy `E_(a)`, respectively, are

A

`1.0 xx 10^(6)s^(-1)` and `9.2 kJ mol^(-1)`

B

`6.0 s^(-1)` and `16.6 kJ mol^(-1)`

C

`1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`

D

`1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Using Arrhenius theory, k` = A e^(-E_(a)//RT)`
It is given that log k `= 6 - (2000)/(T)` …(1)
Also log k = log A `-(E_(a))/(2.303 RT)` …..(2)
Comparing both Eqs. (1) and (2) and solving , we get
`6 - (2000)/(T) -= log A - (E_(a))/(2.303 RT) log A = 6 rArr A` = antilog (6) `= 10^(6)s^(-1)`
`(E_(a))/(2.303 RT) = (2000)/(T) rArr E_(0) = 2000 xx 2.303 xx 8.314 = 38.3 kJ mol^(-1)`
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