At `407 K`, the rate constant of a chemical reaction is `9.5xx10^(-5)s^(-1)` and at `420 K`, the rate constant is `1.9xx10^(-4)s^(-1)`. Calculate the Arrhenius parameter of the reaction.

Options

A. `5.4xx10^(-5)s^(-1)`
B. `5.04xx10^(-5)s^(-1)`
C. `5.04xx10^(5)s^(-1)`
D. `5.4xx10^(5)s^(-1)`

Answer

C

Explanation

The Arrhenius equation is,
`log.(k_(2))/(k_(1))=(E_(a))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Given: `k_(a)= 9.5xx10^(-5)s^(-1), k_(2)= 1.9xx10^(-4)s^(-1)`,
`R= 8.314mol^(-1)K^(-1)`,
`T_(1)= 407 K` and `T_(2)= 420K`
Substitution the values in Arrhenius equation
`log.(1.9xx10^(-4))/(9.5xx10^(-5))=(E_(a))/(2.303xx8.314)[(420-407)/(420xx407)]`
`E_(a)= 75782.3J mol^(-1)`
Applying now `log k_(1)= log A-(E_(a))/(2.303RT_(1))`
`log 9.5xx10^(-5)= logA-(75782.3)/(2.303xx8.314xx407)`
or `log.(A)/(9.5xx10^(-5))=(75782.3)/(2.303xx8.314xx407)= 9.7246`
`A=5.04xx10^(5)s^(-1)`
Note: `K` and `A` are expressed in the same units.