(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Given: $k_2 = 2k_1$,$T_1 = 300 \

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(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Given: $k_2 = 2k_1$,$T_1 = 300 \ K$,$T_2 = 310 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.$\log 2 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{310 - 300}{300 \times 310} \right]$$0.301 = \frac{E_a}{19.147} \left[ \frac{10}{93000} \right]$$E_a = \frac{0.301 \times 19.147 \times 9300}{10} \approx 53598 \ J \ mol^{-1} \approx 53.6 \ kJ \ mol^{-1}$.

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

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When the temperature changes from $300 \ K$ to $310 \ K$,the rate of the reaction doubles. The activation energy of the reaction is ..... $kJ \ mol^{-1}$. $(R = 8.314 \ J \ K^{-1} \ mol^{-1} \text{ and } \log 2 = 0.301)$

A
$53.6$
B
$48.6$
C
$58.5$
D
$60.5$

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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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Reactant $A$ shows two reactions:
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When the temperature changes from $300 \ K$ to $310 \ K$,the rate of the reaction doubles. The activation energy of the reaction is ..... $kJ \ mol^{-1}$. $(R = 8.314 \ J \ K^{-1} \ mol^{-1} \text{ and } \log 2 = 0.301)$

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The reaction: $Cr_{2}O_{3} + 2 Al \rightarrow Al_{2}O_{3} + 2 Cr$ $\quad (\Delta_{r}G^{\Theta} = -421 \ kJ)$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

What does $P \cdot Z_{AB} \cdot e^{-\frac{E_a}{RT}}$ indicate in the rate equation?

Reactant $A$ shows two reactions:$A \xrightarrow{K_1} B$,activation energy $= Ea_1$$A \xrightarrow{K_2} C$,activation energy $= Ea_2$If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:

The Arrhenius plots of two reactions,$I$ and $II$ are shown graphically. The graph suggests that

It is found that for every $10^{\circ}C$ rise in temperature,the rate of the reaction doubles itself. This is mainly because

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Reactant $A$ shows two reactions:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is: