(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

Chemistry Questions

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

The time taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

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Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

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The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:
$T / ^{\circ}C$ $0$ $20$ $40$ $60$ $80$
$10^{5} \times k / s^{-1}$ $0.0787$ $1.70$ $25.7$ $178$ $2140$

Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.

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What will be the effect of temperature on the rate constant?

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At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

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$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

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

Similar Questions

The time taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

What will be the effect of temperature on the rate constant?

At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

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