Why does the rate of a reaction increase with a rise in temperature?

  • A
    The number of collisions increases.
  • B
    The activation energy decreases.
  • C
    $A$ larger fraction of molecules possess energy greater than the activation energy.
  • D
    The frequency of effective collisions increases due to higher kinetic energy.

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Similar Questions

The rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$

The rate constant of a reaction increases by five times on increase in temperature from $27^{\circ} C$ to $52^{\circ} C$. The value of activation energy in $kJ \, mol^{-1}$ is $....$ . (Rounded-off to the nearest integer) $[R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

The Arrhenius equation is represented as $k = A e^{-E_a/RT}$. The activation energy $E_a$ of the reaction can be calculated by plotting:

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For an exothermic reaction $A \rightarrow B$,the activation energy of $A$ is $17 \, kJ/mol$. The heat of reaction is $40 \, kJ$. Calculate the activation energy for the reverse reaction $B \rightarrow A$.

The specific rate constant of a $1^{st}$ order reaction depends on the

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