JEE MAIN - Chemistry (2024 - 4th April Morning Shift - No. 23)
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.
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Some details of the above reactions are listed below.
| Step | Rate constant (sec$$^{-1}$$) | Activation energy (kJ mol$$^{-1}$$) |
|---|---|---|
| 1 | $$\mathrm{k_1}$$ | 300 |
| 2 | $$\mathrm{k_2}$$ | 200 |
| 3 | $$\mathrm{k_3}$$ | $$\mathrm{Ea_3}$$ |
If the overall rate constant of the above transformation (k) is given as $$\mathrm{k=\frac{k_1 k_2}{k_3}}$$ and the overall activation energy $$(\mathrm{E}_{\mathrm{a}})$$ is $$400 \mathrm{~kJ} \mathrm{~mol} \mathrm{~m}^{-1}$$, then the value of $$\mathrm{Ea}_3$$ is ________ integer)
Answer
100
Explanation
$$\begin{aligned} & k=\frac{k_1 k_2}{k_3} \\ & E_{a_{e f f}}=E_{a_1}+E_{a_2}-E_{a_3} \end{aligned}$$
$$\begin{aligned} & 400=300+200-E_{\mathrm{a}_3} \\ & 400=500-E_{\mathrm{a}_3} \\ & E_{\mathrm{a}_3}=100 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned}$$
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