Which equation is true to calculate the energy of activation,if the rate of reaction is doubled by increasing temperature from $T_1 \ K$ to $T_2 \ K$?

  • A
    $\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
  • B
    $\log_{10} 2 = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
  • C
    $\log_{10} \frac{k_1}{k_2} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$
  • D
    $\log_{10} \frac{1}{2} = \frac{E_a}{2.303} \left[ \frac{1}{T_2} - \frac{1}{T_1} \right]$

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Similar Questions

$A \rightarrow B$
The rate constants of the above reaction at $200 \, K$ and $300 \, K$ are $0.03 \, min^{-1}$ and $0.05 \, min^{-1}$ respectively. The activation energy for the reaction is $.... \, J$ (Nearest integer).
(Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 5 = 0.70$,$\log 3 = 0.48$,$\log 2 = 0.30$)

The activation energy for a reaction at the temperature $T \ K$ was found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of the rate constant to the Arrhenius factor is

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.
[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]

The rate constant of a first order reaction is $3.46 \times 10^{-2} \ s^{-1}$ at $298 \ K$. What is the rate constant of the reaction at $350 \ K$ if its activation energy is $50.1 \ kJ \ mol^{-1}$ (in $s^{-1}$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$(\log 2 = 0.3010)$

Which one of the following does not represent the Arrhenius equation?

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