This problem is providing us with the trendline resulting from the linearization of the Arrhenius equation as y = –5773x + b. In such a way, the activation energy in kJ/mol is required and found to be 48.0kJ/mol.
In chemistry, the foundations of chemical kinetics are studied via the Arrhenius equation, as way for us to relate the rate constant, temperature and activation energy in a chemical reaction, therefore it can be written as:
[tex]k=k_0*exp(-\frac{Ea}{RT} )[/tex]
Thus, when one tabulates k vs T data, one can linearize this equation in order to find the activation energy as follows:
[tex]ln(k)=ln(k_0*exp(-\frac{Ea}{RT} ))[/tex]
Which can be reduced to:
[tex]ln(k)=ln(k_0)+ln(exp(-\frac{Ea}{RT} ))\\\\ln(k)=ln(k_0)-\frac{Ea}{RT}[/tex]
And resembles:
[tex]y=b+mx[/tex]
This means the slope is:
[tex]m=-\frac{Ea}{R}[/tex]
And we know its value as -5773 according to the given trendline. Thereby, we can solve for the activation energy, in the required units as follows:
[tex]Ea=-m*R=-(-5773K)*8.148.3145\frac{J}{mol*K}*\frac{1kJ}{1000J} \\\\Ea=48.0kJ/mol[/tex]
Learn more about chemical kinetics: brainly.com/question/26351746
