Activation energy of any chemical reaction can be calculated if one knows the value of:
The Arrhenius equation relates the rate constant (k) of a chemical reaction to the temperature (T) and the activation energy (Ea):
$$ k = A e^{-E_a/RT} $$
Where:
If you have the rate constant at two different temperatures (\( k_1 \) at \( T_1 \) and \( k_2 \) at \( T_2 \)), you can use a modified form of the Arrhenius equation to calculate the activation energy:
$$ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$
By rearranging this equation, you can solve for \( E_a \):
$$ E_a = -R \frac{\ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)} $$
Therefore, knowing the rate constant at two different temperatures is necessary to calculate the activation energy.
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :