JEE Chemical Kinetics 2

Chemistry Level 2

A reaction takes place in various steps. The rate constants for the first, second, third and fifth steps are k 1 , k 2 , k 3 k_1,k_2,k_3 and k 5 , k_5, respectively. The overall rate constant is given by k = k 2 k 3 ( k 1 k 5 ) 0.5 . k=\dfrac{k_2}{k_3}\left (\dfrac{k_1}{k_5}\right )^{0.5}.

If the activation energies are 40, 60, 50 and 10 kJ/mol, respectively, what is the overall energy of activation (in kJ/mol)?

Note: The above image is not the described reaction.

Image credit: Wikipedia Wikimuzg
This problem is a part of My picks for JEE 2


The answer is 25.

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Pranjal Jain by Pranjal Jain , 23, India

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2 solutions

According to Arrhenius Equation, note that k = A × e − E a R T k = A \times e^{-\frac{E_{a}}{RT}} Since A A , R R and T T are constants, k ∝ e E a k \propto e^{E_{a}} k ∝ e 60 − 50 × ( e 40 − 10 ) 0.5 k \propto e^{60-50} \times (e^{40-10})^{0.5} k ∝ e 25 ⇒ E a = 25 k \propto e^{25} \Rightarrow E_{a} = 25

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Great Solution Thanks Keep up the Good Work

Somyaneel Sinha - 5 years, 5 months ago
Prince Loomba
Jan 29, 2016

Convert / into - , * into + , ^ into *.
(60-50)+((40-10))0.5=25. Although it is similar to vishnu's solution. But I have used this to solve the problem.

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Do you mean taking logarithms? Well, I call it that.

Pranjal Jain - 5 years, 4 months ago

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yes like log exactly.

Prince Loomba - 5 years, 3 months ago

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