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.
| Step | Rate constant\(\left(\sec ^{-1}\right)\) | Activation energy\(\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)\) |
| \(1\) | \({k}_1\) | \(300\) |
| \(2\) | \({k}_2\) | \(200\) |
| \(3\) | \({k}_3\) | \(\mathrm{Ea}_3\) |
- A \(70\)
- B \(98\)
- C \(100\)
- D \(90\)
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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