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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The rate of a reaction $A$ doubles on increasing the temperature from $300 \, K$ to $310 \, K$. By how much should the temperature of reaction $B$ be increased from $300 \, K$ so that its rate doubles,if the activation energy of reaction $B$ is twice that of reaction $A$ (in $, K$)?
From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$
Difficult
View SolutionThe decomposition of formic acid on a gold surface follows first-order kinetics. If the rate constant at $300 \ K$ is $1.0 \times 10^{-3} \ s^{-1}$ and the activation energy $E_a = 11.488 \ kJ \ mol^{-1}$,the rate constant at $200 \ K$ is ............ $\times 10^{-5} \ s^{-1}$.
(Round off to the Nearest Integer).
(Given: $R = 8.314 \ J \ mol^{-1} K^{-1}$)
(Round off to the Nearest Integer).
(Given: $R = 8.314 \ J \ mol^{-1} K^{-1}$)
DifficultJEE MAIN 2021
View SolutionWhich graph shows zero activation energy?
Medium
View SolutionThe rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T},$ respectively. The temperature at which $k_1 = k_2$ is
DifficultAIPMT 2008
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