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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. \(\mathrm{A}+\mathrm{B} \underset{\text { Step } 3}{\text { Step } 1} \mathrm{C} \xrightarrow{\text { Step } 2} \mathrm{P}\) Some details of the above reaction are listed below.
StepRate constant\(\left(\sec ^{-1}\right)\)Activation energy\(\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)\)
\(1\)\({k}_1\)\(300\)
\(2\)\({k}_2\)\(200\)
\(3\)\({k}_3\)\(\mathrm{Ea}_3\)
If the overall rate constant of the above transformation (k) is given as \(\mathrm{k}=\frac{\mathrm{k}_1 \mathrm{k}_2}{\mathrm{k}_3}\) and the overall activation energy \(\left(E_2\right)\) is \(400 \mathrm{~kJ} \mathrm{~mol}^{-1}\), then the value of \(\mathrm{Ea}_3\) is _______ \(\mathrm{kJ} \mathrm{mol}^{-1}\). (nearest integer)

  1. A \(70\)
  2. B \(98\)
  3. C \(100\)
  4. D \(90\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(100\)

Step-by-step Solution

Detailed explanation

\(\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}\)…
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