(C) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.Taking natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{RT}$.Converting t

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

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

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The Arrhenius equation is represented as $k = A e^{-E_a/RT}$. The activation energy $E_a$ of the reaction can be calculated by plotting:

A
$k$ against $T$
B
$k$ against $\frac{1}{\log T}$
C
$\log k$ against $1/T$
D
$\log k$ against $\frac{1}{\log T}$

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Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

Collision theory, Energy of activation and Arrhenius equation

For the reaction $A \to B$,$K_1 = 10^8 \, e^{-6000/8.34T}$ and for the reaction $P \to Q$,$K_2 = 10^{10} \, e^{-8000/8.34T}$. At what temperature $T$ will $K_1 = K_2$ (in $K$)?

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For a particular reaction,the rate constant becomes double on increasing temperature from $27^{\circ} C$ to $37^{\circ} C$. Calculate the approximate activation energy (in $kcal \ mol^{-1}$,$R=2 \ cal \ mol^{-1} \ K^{-1}$).

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How does the rate of reaction change with an increase in temperature?

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Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

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Select the correct statement.

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The Arrhenius equation is represented as $k = A e^{-E_a/RT}$. The activation energy $E_a$ of the reaction can be calculated by plotting:

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For the reaction $A \to B$,$K_1 = 10^8 \, e^{-6000/8.34T}$ and for the reaction $P \to Q$,$K_2 = 10^{10} \, e^{-8000/8.34T}$. At what temperature $T$ will $K_1 = K_2$ (in $K$)?

For a particular reaction,the rate constant becomes double on increasing temperature from $27^{\circ} C$ to $37^{\circ} C$. Calculate the approximate activation energy (in $kcal \ mol^{-1}$,$R=2 \ cal \ mol^{-1} \ K^{-1}$).

How does the rate of reaction change with an increase in temperature?

Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

Select the correct statement.

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