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Chemical Kinetics question

2015 · Q86
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Chemical Kinetics question

2015 · Q86

NEETChemistryChemical KineticsMCQ+4 / −1
The activation energy of a reaction can be determined from the slope of which of the following graphs?
  1. A
    ln k vs. 1T{1 \over T}T1​
  2. B
    Tln⁡ k{T \over {\ln \,k}}lnkT​ vs. 1T{1 \over T}T1​
  3. C
    ln k vs. TTT
  4. D
    ln⁡kT{{\ln k} \over T}Tlnk​ vs. TTT
View written solutionFree

Correct answer: A

According to Arrhenius equation,

k=Ae−EaETk = A{e^{ - {{{E_a}} \over {ET}}}}k=Ae−ETEa​​

Taking natural log on both the sides we get,

ln k = ln A −EaET{ - {{{E_a}} \over {ET}}}−ETEa​​ ...........(1)

Comparing (1) with standard form of equation of line

y = mx + C

We get Slope, m = −EaR{ - {{{E_a}} \over R}}−REa​​

Hence, if ln k is plotted against 1/T, slope of the line will be −EaR{ - {{{E_a}} \over R}}−REa​​.

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