(D) Given: $T_1 = 300 \ K$,$T_2 = 310 \ K$,$k_2 = 2k_1$,$R = 8.3 \ J \cdot mol^{-1} \cdot K^{-1}$,$\log 2 = 0.3$. Using the Arrhenius equation: $\log

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(D) Given: $T_1 = 300 \ K$,$T_2 = 310 \ K$,$k_2 = 2k_1$,$R = 8.3 \ J \cdot mol^{-1} \cdot K^{-1}$,$\log 2 = 0.3$. Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$ $\log 2 = \frac{E_a}{2.303 \times 8.3} \left( \frac{310 - 300}{300 \times 310} \right)$ $0.3 = \frac{E_a}{19.1149} \times \frac{10}{93000}$ $E_a = \frac{0.3 \times 19.1149 \times 93000}{10} \ J \cdot mol^{-1}$ $E_a = 53330.57 \ J \cdot mol^{-1} \approx 53.33 \ kJ \cdot mol^{-1}$.

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

Easy

The rate constant of a first order reaction was doubled when the temperature was increased from $300 \ K$ to $310 \ K$. What is its approximate activation energy (in $kJ \cdot mol^{-1}$)? ($R = 8.3 \ J \cdot mol^{-1} \cdot K^{-1}$; $\log 2 = 0.3$)

AP EAMCET 2018 All AP EAMCET

A
$5.33$
B
$533.3$
C
$53333$
D
$53.33$

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Class 12 Questions

Class 12

Chemistry Questions

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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AP EAMCET 2018 Paper All AP EAMCET years →

For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

Medium

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The rate of reaction is tripled for a $10^\circ C$ rise in temperature. The increase in the reaction rate if the temperature is increased by $60^\circ C$ is $............$ times.

Medium

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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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Which of the following graphs between the rate constant $(k)$ and temperature $(T)$ represents the Arrhenius equation?

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Explain: How is the value of activation energy determined based on the Arrhenius equation?

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

The rate constant of a first order reaction was doubled when the temperature was increased from $300 \ K$ to $310 \ K$. What is its approximate activation energy (in $kJ \cdot mol^{-1}$)? ($R = 8.3 \ J \cdot mol^{-1} \cdot K^{-1}$; $\log 2 = 0.3$)

Similar Questions

For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

The rate of reaction is tripled for a $10^\circ C$ rise in temperature. The increase in the reaction rate if the temperature is increased by $60^\circ C$ is $............$ times.

Which of the following graphs between the rate constant $(k)$ and temperature $(T)$ represents the Arrhenius equation?

Explain: How is the value of activation energy determined based on the Arrhenius equation?

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