The Arrhenius equation shows the relationship

between the rate constant $k$ and the temperature $T$

in kelvins and is typically written as

$k=A{e}^{-\frac{E_a}{R}T}$

where $R$ is the gas constant $(8\mathrm{.}314\frac{J}{m}ol*K),$

$A$ 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

$ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

which is mathmatically equivalent to

$ln\frac{k_1}{k_2}=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

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 $46\mathrm{.}0k\frac{J}{m}ol.$ At $23C,$ the rate constant is $0\mathrm{.}0120s^{-1}.$ At what

temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

View Available Hint$($s$)$

$T_2=$

Part B

Given that the initial rate constant is $0\mathrm{.}0120s^{-1}$ at an initial temperature of $23C,$ what would the rate constant

be at a temperature of $200.C$ for the same reaction described in Part A$?$

Express your answer with the appropriate units.

View Available Hint$($s$)$

$k_2=$