Arrhenius Equation Derivation

[@ghassansiregar1376]

So clear, thank you!

[@humanppplus4306]

why is the gradient -E/R?

[@sydneym3251]

The gradient is the derivative of the equation for the line

[@timotejbernat462]

@@sydneym3251 That's *what* the gradient is but not *why* it has that value, as the original question asked

[@sydneym3251]

@@timotejbernat462 Explaining why taking the derivative of an equation yields an answer is interesting. I suppose I could explain what a derivative is. Maybe that would help

[@timotejbernat462]

@@sydneym3251 The question is about the physical basis for why has the particular value of -E/R, not the mathematical basis underlying the derivative, you seem to have misunderstood that

[@sleepydog9968]

maybe it was practically/physically proven through trial/error (using data)

[@SimranKaur-zf1zf]

Thankyou sir it helps

[@Updesh001]

Thank you so much sir

[@MikeEnergy_]

Was activation energy used before arrhenius?

[@nirmalshibu9769]

As a matter of fact, Arrhenius himself had introduced this term in 1889
Source : en.wikipedia.org/wiki/Activation_energy

[@aakashmeena3805]

Thank a lot to you sir

[@MikeEnergy_]

Nice video

[@efrainm4649]

And how did he know that the gradient is -Ea/R???

[@aashirwadmishra3706]

The gradient can easily be calculated with the data used for plotting the graph.

[@sleepydog9968]

maybe it was practically/physically proven using trial/error + raw data

[@kronosai4555]

derivation with differential eqn?

[@linkytunes2133]

Is there one? I’m curious. Please tell me if there is one

[@romin_1122]

Thanks!!