The Arrhenius Equation is typically written as

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

However, the following more practical form of this

equation also exists:

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

where $k_1$ and $k_2$ are the rate constants for a single

reaction at two different absolute temperatures

and $T_2).$

The activation energy of a certain reaction is $46\mathrm{.}8k\frac{J}{m}ol.$ At $20C,$ the rate constant

is $0\mathrm{.}0130s^{-1}.$ At what temperature would this reaction go twice as fast?

Express your answer numerically in degrees Celsius

View Available Hint$($s$)$

$T_2=[,]C$

Part B

Given that the initial rate constant is $0\mathrm{.}0130s^{-1}$ at an initial temperature of $20C,$ what

would the rate constant be at a temperature of $100C?$

Express your answer numerically in inverse seconds.

View Available Hint$($s$)$