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A graph of ln k vs. 1/T is plotted and the linear trendline has an equation of y = –5773x + b. Please calculate the activation energy in units of kJ/mol.


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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.

Arrhenius equation and kinetics

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

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