(C) The overall rate constant is given by $K = \frac{K_1 K_2}{K_3}$.Substituting the Arrhenius equation $K = A \cdot e^{-E_a / RT}$ for each step:$A \

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(C) The overall rate constant is given by $K = \frac{K_1 K_2}{K_3}$.Substituting the Arrhenius equation $K = A \cdot e^{-E_a / RT}$ for each step:$A \cdot e^{-E_a / RT} = \frac{(A_1 e^{-E_{a1} / RT}) \cdot (A_2 e^{-E_{a2} / RT})}{(A_3 e^{-E_{a3} / RT})}$$A \cdot e^{-E_a / RT} = \frac{A_1 A_2}{A_3} \cdot e^{-(E_{a1} + E_{a2} - E_{a3}) / RT}$Comparing the exponents,we get $E_a = E_{a1} + E_{a2} - E_{a3}$.Substituting the given values: $E_a = 40 + 50 - 60 = 30 \ kJ/mol$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

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For a reaction taking place in three steps at the same temperature,the overall rate constant is $K = \frac{K_1 K_2}{K_3}$. If $E_{a1}$,$E_{a2}$,and $E_{a3}$ are $40 \ kJ/mol$,$50 \ kJ/mol$,and $60 \ kJ/mol$ respectively,the overall activation energy $E_a$ is . . . . . . $kJ/mol$.

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$20$
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$10$
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$30$
D
$45$

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Collision theory, Energy of activation and Arrhenius equation

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For a reaction taking place in three steps at the same temperature,the overall rate constant is $K = \frac{K_1 K_2}{K_3}$. If $E_{a1}$,$E_{a2}$,and $E_{a3}$ are $40 \ kJ/mol$,$50 \ kJ/mol$,and $60 \ kJ/mol$ respectively,the overall activation energy $E_a$ is . . . . . . $kJ/mol$.

Similar Questions

Regarding the velocity in a reversible reaction,which is the correct explanation of the effect of a catalyst?

For an endothermic reaction,the energy of activation is $E_a$ and the enthalpy of reaction is $\Delta H$ (both in $kJ/mol$). The minimum value of $E_a$ will be:

The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value ($R=$ gas constant,$E_a=$ Activation energy)

The relation between the rate constant and temperature according to the Arrhenius equation is:

The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

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The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$