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The Arrhenius equation shows the relationship between the rate constant

`k`

and the temperature

`T`

in kelvins and is typically written as

`k=Ae^(-(E_(a))/(R)T)`

where

`R`

is the gas constant

`(8.314(J)/(m)ol*K)`

,

`A`

is a constant called the frequency factor, and

`E_(a)`

is the activation energy for the reaction. However, a more practical form of this equation is

`ln((k_(2))/(k_(1)))=(E_(2))/(R)((1)/(T_(1))-(1)/(T_(2)))`

which is mathmatically equivalent to

`ln((k_(1))/(k_(2)))=(E_(a))/(R)((1)/(T_(2))-(1)/(T_(1)))`

where

`k_(1)`

and

`k_(2)`

are the rate constants for a single reaction at two different absolute temperatures (

`T_(1)`

and

`T_(2)`

). The activation energy of a certain reaction is

`47.4k(J)/(m)ol`

. At

`26\deg C`

, the rate constant is

`0.0120s^(-1)`

. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. View Available Hint(s)

`?`

`q,`

Part B Given that the initial rate constant is

`0.0120s^(-1)`

at an initial temperature of

`26\deg C`

, what would the rate constant be at a temperature of 200 .

`\deg C`

for the same reaction described in Part A? Express your answer with the appropriate units. View Available Hint(s)

`k_(2)=`

`?`