JEE Main 2024 · 4 April · morning · Numerical answer · +4 / −1
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.
A + B ⇌(Step 1) C →(Step 2) P
Some details of the above reactions are listed below.
If the overall rate constant of the above transformation (k) is given as k=(k₁ k₂)/k₃ and the overall activation energy (Eₐ) is 400 kJ mol m⁻¹, then the value of Ea₃ is ________ integer)
| Step | Rate constant (sec⁻¹) | Activation energy (kJ mol⁻¹) |
|---|---|---|
| 1 | k₁ | 300 |
| 2 | k₂ | 200 |
| 3 | k₃ | Ea₃ |
Answer and solution
Answer 100
Solution
Chemistry ma’am · 4 steps
Put Arrhenius into every k: multiplying rate constants adds their Ea, and dividing subtracts it.
- Arrhenius for each step. Write each rate constant as kᵢ = Aᵢ e^(-Eₐᵢ/RT). Then substitute all three into k = (k₁ k₂)/k₃.
k = (A₁ A₂)/A₃ e^(-(Eₐ₁ + Eₐ₂ - Eₐ₃)/RT)
- Read off overall Ea. This has the same form as k = A e^(-Eₐ/RT), so the overall activation energy is Eₐ = Eₐ₁ + Eₐ₂ - Eₐ₃. The exponents of the top two add, and the exponent of the bottom one subtracts.
- Put in numbers. 400 = 300 + 200 - Eₐ₃, so Eₐ₃ = 500 - 400 = 100 kJ mol⁻¹.
- So. The activation energy of step 3 is 100 kJ mol⁻¹.
The idea to keep. If the overall k is a product or quotient of step rate constants, add the Ea of every k in the numerator and subtract the Ea of every k in the denominator.
Answer from the NTA final answer key. Solution by PrepWiser.