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Chemical Kinetics and Nuclear Chemistry question

2024 · 29 Jan · Shift 1 · Q26
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Chemical Kinetics and Nuclear Chemistry question

2024 · 29 Jan · Shift 1 · Q26

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For a reaction taking place in three steps at same temperature, overall rate constant K=K1 K2 K3\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}K= K3​K1​ K2​​. If Ea1,Ea2\mathrm{Ea}_1, \mathrm{Ea}_2Ea1​,Ea2​ and Ea3\mathrm{Ea}_3Ea3​ are 40, 50 and 60 kJ/mol60 \mathrm{~kJ} / \mathrm{mol}60 kJ/mol respectively, the overall Ea\mathrm{Ea}Ea is ‾\underline{\hspace{2cm}}​kJ/mol\mathrm{kJ} / \mathrm{mol}kJ/mol.
Numerical answer
View written solutionFree

Correct answer: 30

  1. Use the Arrhenius equation for each step:

ki=Aie−Ea,i/RTk_i = A_i e^{-E_{a,i}/RT}ki​=Ai​e−Ea,i​/RT

Given,

K=K1K2K3K = \frac{K_1 K_2}{K_3}K=K3​K1​K2​​

Substitute the Arrhenius forms:

K=A1e−Ea1/RT⋅A2e−Ea2/RTA3e−Ea3/RTK = \frac{A_1 e^{-E_{a1}/RT} \cdot A_2 e^{-E_{a2}/RT}}{A_3 e^{-E_{a3}/RT}}K=A3​e−Ea3​/RTA1​e−Ea1​/RT⋅A2​e−Ea2​/RT​

  1. Simplify the exponential part:

K=A1A2A3  e−Ea1/RTe−Ea2/RTe+Ea3/RTK = \frac{A_1A_2}{A_3} \; e^{-E_{a1}/RT} e^{-E_{a2}/RT} e^{+E_{a3}/RT}K=A3​A1​A2​​e−Ea1​/RTe−Ea2​/RTe+Ea3​/RT

K=A1A2A3  e−(Ea1+Ea2−Ea3)/RTK = \frac{A_1A_2}{A_3} \; e^{-(E_{a1}+E_{a2}-E_{a3})/RT}K=A3​A1​A2​​e−(Ea1​+Ea2​−Ea3​)/RT

Comparing with overall Arrhenius form,

K=A e−Ea/RTK = A\, e^{-E_a/RT}K=Ae−Ea​/RT

So,

Ea=Ea1+Ea2−Ea3E_a = E_{a1}+E_{a2}-E_{a3}Ea​=Ea1​+Ea2​−Ea3​

  1. Substitute the values:

Ea=40+50−60=30 kJ/molE_a = 40 + 50 - 60 = 30 \text{ kJ/mol}Ea​=40+50−60=30 kJ/mol

  1. Final answer:

30\boxed{30}30​

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