मराठी

According to Arrhenius equation rate constant k is equal to Ae-EaRT. Which of the following options represents the graph of ln k vs 1T? - Chemistry

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प्रश्न

According to Arrhenius equation rate constant k is equal to `A e^((-Ea)/(RT)`. Which of the following options represents the graph of ln k vs `1/T`?

पर्याय

MCQ

उत्तर


Explanation:

To arrhenius equation `k = A e^((-Ea)/(RT)` 

Taking log on both sides in `k` = In `A.e^(-(E_a)/(RT))`

In `k` = In `A - (E_a)/(RT)`

In `k = - (-E_a)/R xx 1/T` + In A

y = mx + c

This equation can be related to equation of straight line.

From the graph, it is very clear that slope of the plot = `(-E_a)/R` and intercept = In A.

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पाठ 4: Chemical Kinetics - Exercises [पृष्ठ ४८]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 12
पाठ 4 Chemical Kinetics
Exercises | Q I. 6. | पृष्ठ ४८
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