(B) Given the rate constants: $k_1 = 10^{16} e^{-2000/T}$ and $k_2 = 10^{15} e^{-1000/T}$.At the temperature where $k_1 = k_2$,we have:$10^{16} e^{-20

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(B) Given the rate constants: $k_1 = 10^{16} e^{-2000/T}$ and $k_2 = 10^{15} e^{-1000/T}$.At the temperature where $k_1 = k_2$,we have:$10^{16} e^{-2000/T} = 10^{15} e^{-1000/T}$Dividing both sides by $10^{15} e^{-2000/T}$:$\frac{10^{16}}{10^{15}} = \frac{e^{-1000/T}}{e^{-2000/T}}$$10 = e^{1000/T}$Taking the natural logarithm $(\ln)$ on both sides:$\ln(10) = \frac{1000}{T}$Since $\ln(10) = 2.303 \log_{10}(10) = 2.303$,we get:$2.303 = \frac{1000}{T}$$T = \frac{1000}{2.303} \ K$

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

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The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T},$ respectively. The temperature at which $k_1 = k_2$ is

AIPMT 2008 All AIPMT

A
$2000 \ K$
B
$\frac{1000}{2.303} \ K$
C
$1000 \ K$
D
$\frac{2000}{2.303} \ K$

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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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AIPMT 2008 Paper All AIPMT years →

The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

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Medium

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For a reaction $A \rightarrow B$,the enthalpy of reaction is $-4.2 \ kJ \ mol^{-1}$ and the enthalpy of activation is $9.6 \ kJ \ mol^{-1}$. The correct potential energy profile for the reaction is shown in which option?

MediumNEET 2021

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For a first-order reaction,if the rate constant is $k_1$ at temperature $T_1$ and $k_2$ at temperature $T_2$,which of the following relations is correct? ($E_a$ = activation energy)

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Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is

MediumAP EAMCET 2007

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The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T},$ respectively. The temperature at which $k_1 = k_2$ is

Similar Questions

The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$

For a reaction $A \rightarrow B$,the enthalpy of reaction is $-4.2 \ kJ \ mol^{-1}$ and the enthalpy of activation is $9.6 \ kJ \ mol^{-1}$. The correct potential energy profile for the reaction is shown in which option?

For a first-order reaction,if the rate constant is $k_1$ at temperature $T_1$ and $k_2$ at temperature $T_2$,which of the following relations is correct? ($E_a$ = activation energy)

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.The correct answer is

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Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is