(D) The Arrhenius equation is given by: $\log K = \log A - \frac{E_a}{2.303 RT}$ Given: $A = 1.45 \times 10^{11} s^{-1}$,$E_a = 35 \times 10^3 \, cal\

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(D) The Arrhenius equation is given by: $\log K = \log A - \frac{E_a}{2.303 RT}$ Given: $A = 1.45 \times 10^{11} s^{-1}$,$E_a = 35 \times 10^3 \, cal\, mol^{-1}$,$T = 300 + 273 = 573 \, K$,$R = 1.987 \approx 2 \, cal\, K^{-1} mol^{-1}$ Substituting the values: $\log K = \log (1.45 \times 10^{11}) - \frac{35000}{2.303 \times 2 \times 573}$ $\log K = 11.161 - \frac{35000}{2640.038} \approx 11.161 - 13.257 = -2.096$ $K = \text{antilog}(-2.096) = 7.94 \times 10^{-3} s^{-1}$

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

Medium

The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

A
$10 \times 10^{-2} s^{-1}$
B
$5.37 \times 10^{10} s^{-1}$
C
$5 \times 10^{-4} s^{-1}$
D
$7.94 \times 10^{-3} s^{-1}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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Consider the following plots of log of rate constant $k$ $(\log k)$ vs $\frac{1}{T}$ for three different reactions. The correct order of activation energies of these reactions is

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$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:

$Step$ $Rate\ Constant\ /\ Activation\ energy$

$Step\ 1$ $k_1, E_{a_1} = 180\ kJ/mol$

$Step\ 2$ $k_2, E_{a_2} = 80\ kJ/mol$

$Step\ 3$ $k_3, E_{a_3} = 50\ kJ/mol$

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Which of the following plots is in accordance with the Arrhenius equation?

Easy

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If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

MediumTS EAMCET 2018

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$Step$	$Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$	$k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$	$k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$	$k_3, E_{a_3} = 50\ kJ/mol$

The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

Similar Questions

For a chemical reaction,the rate of reaction doubles when the temperature increases by $10\,^oC$. If the temperature is increased from $10\,^oC$ to $100\,^oC$,the rate of reaction will increase by a factor of .....

Consider the following plots of log of rate constant $k$ $(\log k)$ vs $\frac{1}{T}$ for three different reactions. The correct order of activation energies of these reactions is

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If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

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