(D) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Converting to

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$\log k$ versus $1/T$ will give a straight line

(D) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Converting to base $10$ logarithm: $\log k = \log A - \frac{E_a}{2.303 R} \left( \frac{1}{T} \right)$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log k$ and $x = 1/T$,we get a straight line with a slope of $-E_a / (2.303 R)$.Therefore,plotting $\log k$ versus $1/T$ yields a straight line.

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

Medium

$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} e^{-(25000/RT)} \, s^{-1}$. It means

A
$\log k$ versus $\log T$ will give a straight line with slope as $-25000$
B
$\log k$ versus $T$ will give a straight line with slope as $-25000$
C
$\log k$ versus $\log 1/T$ will give a straight line with slope as $-25000$
D
$\log k$ versus $1/T$ will give a straight line

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

For an effective collision to occur,the colliding molecules must possess:

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For an exothermic reaction $A \to B$,the activation energy is $65 \ kJ \ mol^{-1}$ and the heat of reaction is $-42 \ kJ \ mol^{-1}$. The activation energy for the reverse reaction $B \to A$ would be....... $kJ \ mol^{-1}$

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The $\Delta H$ value of the reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is $-44.12 \ kcal$. If $E_1$ is the activation energy of the backward reaction and $E_2$ is the activation energy of the forward reaction,then for the above reaction:

Medium

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With an increase in temperature,the rate constant $k$:

Medium

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The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} e^{-(25000/RT)} \, s^{-1}$. It means

Similar Questions

For an effective collision to occur,the colliding molecules must possess:

For an exothermic reaction $A \to B$,the activation energy is $65 \ kJ \ mol^{-1}$ and the heat of reaction is $-42 \ kJ \ mol^{-1}$. The activation energy for the reverse reaction $B \to A$ would be....... $kJ \ mol^{-1}$

The $\Delta H$ value of the reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is $-44.12 \ kcal$. If $E_1$ is the activation energy of the backward reaction and $E_2$ is the activation energy of the forward reaction,then for the above reaction:

With an increase in temperature,the rate constant $k$:

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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