(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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According to Arrhenius theory,the activation energy is

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The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.

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If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

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Consider the following statements:
$(i)$ Increase in concentration of reactant increases the rate of a zero order reaction
$(ii)$ Rate constant $k$ is equal to $A$ if $E_a = 0$
$(iii)$ Rate constant $k$ is equal to $A$ if $E_a = \infty$
$(iv)$ $\log_e k$ vs $T$ is a straight line
$(v)$ $\log_e k$ vs $1/T$ is a straight line
Which of the following statements are correct?

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The reaction rate at a given temperature becomes slower,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

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According to Arrhenius theory,the activation energy is

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