(B) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 \ RT}$.Comparing this with the given equation $\log k = -\frac{20}{T} + 4$:

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$92.12 \ cal \ mol^{-1}$ and $10^4 \ s^{-1}$

(B) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 \ RT}$.Comparing this with the given equation $\log k = -\frac{20}{T} + 4$:We get $\log A = 4$,which implies $A = 10^4 \ s^{-1}$.Also,$\frac{E_a}{2.303 \ R} = 20$.Using $R = 2 \ cal \ K^{-1} \ mol^{-1}$,we get $E_a = 20 \times 2.303 \times 2 = 92.12 \ cal \ mol^{-1}$.Thus,the activation energy is $92.12 \ cal \ mol^{-1}$ and the pre-exponential factor is $10^4 \ s^{-1}$.

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

Difficult

For a first order reaction $(A \rightarrow B)$,the temperature $(T)$ dependent rate constant $(k)$ in $s^{-1}$ was found to follow the equation: $\log k = \left(-\frac{20}{T}\right)+4$. The activation energy $(E_a)$ and pre-exponential factor $(A)$ respectively,are

TS EAMCET 2020 All TS EAMCET

A
$46.06 \ cal \ mol^{-1}$ and $10^{-4} \ s^{-1}$
B
$92.12 \ cal \ mol^{-1}$ and $10^4 \ s^{-1}$
C
$46.06 \ cal \ mol^{-1}$ and $10^4 \ s^{-1}$
D
$9.212 \ cal \ mol^{-1}$ and $10^{-4} \ s^{-1}$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

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The activation energy for a simple chemical reaction $A \to B$ is ${E_a}$ in the forward direction. The activation energy for the reverse reaction:

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The activation energy for the forward reaction is $150 \ kJ \ mol^{-1}$ and that of the reverse reaction is $260 \ kJ \ mol^{-1}$. The enthalpy change for the reaction is:

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For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

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$A$ chemical reaction was carried out at $300 \, K$ and $280 \, K$. The rate constants were found to be $K_1$ and $K_2$ respectively. The energy of activation is $1.157 \times 10^4 \, cal \, mol^{-1}$ and $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$. Then:

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For a first order reaction $(A \rightarrow B)$,the temperature $(T)$ dependent rate constant $(k)$ in $s^{-1}$ was found to follow the equation: $\log k = \left(-\frac{20}{T}\right)+4$. The activation energy $(E_a)$ and pre-exponential factor $(A)$ respectively,are

Similar Questions

The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

The activation energy for a simple chemical reaction $A \to B$ is ${E_a}$ in the forward direction. The activation energy for the reverse reaction:

The activation energy for the forward reaction is $150 \ kJ \ mol^{-1}$ and that of the reverse reaction is $260 \ kJ \ mol^{-1}$. The enthalpy change for the reaction is:

For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

$A$ chemical reaction was carried out at $300 \, K$ and $280 \, K$. The rate constants were found to be $K_1$ and $K_2$ respectively. The energy of activation is $1.157 \times 10^4 \, cal \, mol^{-1}$ and $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$. Then:

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