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 ....... $cal$.

  • A
    $2.303 \times 2$
  • B
    $2/2.303$
  • C
    $2$
  • D
    None of these

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The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

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The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

For an exothermic reaction,where the enthalpy of reaction is $\Delta H$ in $kJ/mol$,the minimum value for the activation energy will be:

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