(C) The activation energy $(E_a)$ can be calculated using the Arrhenius equation.The equation is given by:$\log \left( \frac{k_2}{k_1} \right) = \frac

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rate constant at two different temperatures

(C) The activation energy $(E_a)$ can be calculated using the Arrhenius equation.The equation is given by:$\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$From this equation,it is clear that $E_a$ can be determined if the rate constants ($k_1$ and $k_2$) are known at two different temperatures ($T_1$ and $T_2$).

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

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Activation energy of any chemical reaction can be calculated if one knows the value of

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A
probability of collision
B
orientation of reactant molecules during collision
C
rate constant at two different temperatures
D
rate constant at standard temperature

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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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For reactions $M, N, O$ and $P$,the activation energy order is $E_M < E_N < E_O < E_P$. For which reaction will the ratio $K_{310}/K_{300}$ be maximum?

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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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If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

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For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

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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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For reactions $M, N, O$ and $P$,the activation energy order is $E_M < E_N < E_O < E_P$. For which reaction will the ratio $K_{310}/K_{300}$ be maximum?

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

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

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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