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A graph plotted between $\log K$ vs $1 / T$ for calculating activation energy is shown by
Options:
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
A graph plotted between $\log k \mathrm{Vs} \frac{1}{T}$ for calculating activation energy is shown as from Arrhenius equation
$\log k=\log A-\frac{E_A}{2.303 R T}$
$\log k=\log A-\frac{E_A}{2.303 R T}$
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