Using the Arrhenius equation | Kinetics | Chemistry | Khan Academy
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- Опубликовано: 15 ноя 2014
- How to use the Arrhenius equation to calculate the activation energy.
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Are you who makes videos for the Channel Question? You have the same voice! Like literally!
very clearly explained! Thanks for sharing :)
how do get the slope of points with using stat on calculator. ie can you use Y2-Y1 / X2-X1.. for points on the graph
I extract iron from iron ore and know i will calculate the arrhenius equation. So my experiments under temperature are 343K ,353K and 363K , 373k and 383k for each temperature time is 10,20,30,40 and 50 mints .So how can i find arrhenius factor there activation energy and rate constant.
Equation is k= A × e^Ea/RT
R is gas constant 8.315 J.K^-1 mol^-1.
Actually i want to calculate these data to make a graph lnk verses 1/T.
As i done kinetics model equations. In which 1-(1-x)^1/3 equation is best fit because the highest R² values.
From where i can get this rate constant K.
Hello i have question.
how did u get the slope plz i need an explanation
first you plot the points on a coordinate plain ln k as "y" and 1/T as "x" you see that the points aren't so well in line ,and there's jumpiness ,that they are scattered,which I guess, is the case in real life experiment, but non the less you could spot a common trend. When we see a relationship in a scatterplot, we can use a line to summarize the relationship in the data.
There are more advanced ways to fit a line to data(I think that is why he was using a calculator), but in general, we want the line to go through the "middle" of the points.(you simply put a straight line through the scattered points at the area you think is the average place or the middle). This is called linear regreation.Once you have the line(wich represents ln k = -Ea/R(1/T)+ln A In our case) you use m = y2-y1/x2-x1 to find the slope(wich should be -Ea/R in our case), by choosing any two arbitrary point on the line.if its unclear, khan academy has videos on linear regression and also coordinates geometry,images would be more helpful in understanding this.otherwise it's a realy easy concept.