The decomposition of ethane,$\frac{d[C_2H_6]}{dt} = k[C_2H_6]$,proceeds through a complex mechanism,which includes $5$ steps. The overall rate constant $(k)$ is expressed as $k = \frac{k_1 k_2 k_3}{k_2 k_5}$,where $k_1, k_2, k_3, k_4, k_5$ are the rate constants of the $5$ steps. If the activation energies of the steps are $E_1 = 1E, E_2 = 2E, E_3 = 3E, E_4 = 4E, E_5 = 5E$,where $E = 20 \ kJ/mol$,find the overall activation energy of the decomposition.
- A$6.67 \ kJ/mol$
- B$3.33 \ kJ/mol$
- C$20 \ kJ/mol$
- D$10 \ kJ/mol$
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For the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$
Difficult
View SolutionGiven below is an expression for the rate constant of a first order reaction occurring at a certain temperature, $T (\text{K})$.
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k$ is $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k$ is $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)
DifficultNEET 2026
View SolutionFor reaction $A \to B$,the rate constant $k_1 = A_1 e^{-E_{a_1} / (RT)}$ and for the reaction $X \to Y$ the rate constant $k_2 = A_2 e^{-E_{a_2} / (RT)}$. If $A_1 = 10^8$,$A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal/mol$,$E_{a_2} = 1800 \ cal/mol$,then the temperature at which $k_1 = k_2$ is (Given : $R = 2 \ cal/K \cdot mol$)
Difficult
View SolutionThe rate of reaction increases with a rise in temperature because of:
Which of the following graphs between the rate constant $(k)$ and temperature $(T)$ represents the Arrhenius equation?
Easy
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