Complete Question
The complete question is shown on the first uploaded image
Answer:
Part A
activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]
Part B
The frequency plot is [tex]A = 2.4*10^{13} s^{-1}[/tex]
Explanation:
From the question we are told that
at [tex]T_1 = 300 \ K[/tex] [tex]k_1 = 5.70 *10^{-2}[/tex]
and at [tex]T_2 = 310 \ K[/tex] [tex]k_2 = 0.169[/tex]
The Arrhenius plot is mathematically represented as
[tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]
Where [tex]E_a[/tex] is the activation barrier for the reaction
R is the gas constant with a value of [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]
Substituting values
[tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]
=> [tex]E_a = 84 .0 \ KJ/mol[/tex]
The Arrhenius plot can also be mathematically represented as
[tex]k = A * e^{-\frac{E_a}{RT} }[/tex]
Here we can use any value of k from the data table with there corresponding temperature let take [tex]k_2 \ and \ T_2[/tex]
So substituting values
[tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]
=> [tex]A = 2.4*10^{13} s^{-1}[/tex]
