@@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
@@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
So clear, thank you!
why is the gradient -E/R?
The gradient is the derivative of the equation for the line
@@sydneym3251 That's *what* the gradient is but not *why* it has that value, as the original question asked
@@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
@@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
maybe it was practically/physically proven through trial/error (using data)
Thankyou sir it helps
Thank you so much sir
Was activation energy used before arrhenius?
As a matter of fact, Arrhenius himself had introduced this term in 1889
Source : en.wikipedia.org/wiki/Activation_energy
Thank a lot to you sir
Nice video
And how did he know that the gradient is -Ea/R???
The gradient can easily be calculated with the data used for plotting the graph.
maybe it was practically/physically proven using trial/error + raw data
derivation with differential eqn?
Is there one? I’m curious. Please tell me if there is one
Thanks!!