(A) The Arrhenius equation is given by $k = A e^{-E^*/RT}$.Taking the logarithm on both sides,we get $\log \, k = \log \, A - \frac{E^*}{2.303 R} \cdo

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(A) The Arrhenius equation is given by $k = A e^{-E^*/RT}$.Taking the logarithm on both sides,we get $\log \, k = \log \, A - \frac{E^*}{2.303 R} \cdot \frac{1}{T}$.This equation follows the linear form $y = mx + c$,where $y = \log \, k$ and $x = \frac{1}{T}$.Plotting $\log \, k$ versus $\frac{1}{T}$ yields a straight line with a slope equal to $-\frac{E^*}{2.303 R}$.Thus,the activation energy $E^*$ can be determined from the slope of this plot.

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

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The temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^/RT}$. The activation energy $(E^)$ of the reaction can be calculated by plotting:

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A
$\log \, k$ vs $\frac{1}{T}$
B
$k$ vs $T$
C
$k$ vs $\frac{1}{\log \, T}$
D
$\log \, k$ vs $T$

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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 temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^*/RT}$. The activation energy $(E^*)$ of the reaction can be calculated by plotting:

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In the given graph,$E_{a}$ for the reverse reaction will be (in $kJ$)

The addition of a catalyst during a chemical reaction alters which of the following quantities?

For $N_2 + 3H_2 \rightarrow 2NH_3$,$\Delta H = -22 \ kcal$,and $E_a = 70 \ kcal$. Hence $E_a$ for $2NH_3 \rightarrow N_2 + 3H_2$ is $.....$ $kcal$.

The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

If for two reactions $E_{a_1} > E_{a_2}$ and $\mu_1$ and $\mu_2$ are their respective temperature coefficients,then which of the following is correct?

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The temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^/RT}$. The activation energy $(E^)$ of the reaction can be calculated by plotting: