(C) Given $k_1 = 10^{16} \times e^{-2000/T}$ and $k_2 = 10^{15} \times e^{-1000/T}$.For $k_1 = k_2$:$10^{16} \times e^{-2000/T} = 10^{15} \times e^{-1

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(C) Given $k_1 = 10^{16} \times e^{-2000/T}$ and $k_2 = 10^{15} \times e^{-1000/T}$.For $k_1 = k_2$:$10^{16} \times e^{-2000/T} = 10^{15} \times e^{-1000/T}$$10 \times e^{-2000/T} = e^{-1000/T}$$10 = \frac{e^{-1000/T}}{e^{-2000/T}}$$10 = e^{1000/T}$Taking the natural logarithm on both sides:$\ln(10) = \frac{1000}{T}$$2.303 \times \log_{10}(10) = \frac{1000}{T}$$2.303 = \frac{1000}{T}$$T = \frac{1000}{2.303} \text{ K}$

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

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The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is

KCET 2023 All KCET

A
$\frac{2000}{2.303} \text{ K}$
B
$2000 \text{ K}$
C
$\frac{1000}{2.303} \text{ K}$
D
$1000 \text{ K}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

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Define the following terms:
$(1)$ Fraction of molecules
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At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
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The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is

Similar Questions

The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

For a certain reaction,consider the plot of $\ln k$ versus $1/T$ given in the figure. If the rate constant of this reaction at $400 \ K$ is $10^{-5} \ s^{-1}$,then the rate constant at $500 \ K$ is:

Define the following terms:$(1)$ Fraction of molecules$(2)$ Frequency factor

At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?$2N_2O_5 \to 4NO_2 + O_2$

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Define the following terms:
$(1)$ Fraction of molecules
$(2)$ Frequency factor

At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
$2N_2O_5 \to 4NO_2 + O_2$