Animated Solution for Chemistry - Chemical Kinetics: The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from $27^{\circ }\text{C}$ to $42^{\circ }\text{C}$. Its energy of activation in $\text{J/mol}$ is ......... . (Take $\ln 5=1.6094$; $R=8.314{\text{ J mol}}^{-1}{\text{ K}}^{-1}$)

Unlocking Activation Energy

Imagine a sealed container filled with gas molecules zipping around chaotically. At any given temperature, these molecules don't all travel at the same speed; instead, they possess a wide distribution of kinetic energies, beautifully described by the Maxwell-Boltzmann distribution.

For a chemical reaction to occur, a collision between molecules isn't enough. They must collide with a minimum amount of energy, known as the activation energy ($E_a$). On our distribution graph, this is represented by a threshold line. Only the molecules in the shaded area to the right of this line have enough energy to react.

When we increase the temperature from $27^{\circ }\text{C}$ ($300\text{ K}$) to $42^{\circ }\text{C}$ ($315\text{ K}$), something remarkable happens. The entire distribution curve flattens out and shifts to the right. While the average energy increases only slightly, the area under the curve beyond the activation energy threshold expands dramatically.

The problem states that this specific area—the number of activated molecules—increases five-fold. Because the rate constant ($k$) of a reaction is directly proportional to the fraction of molecules possessing energy greater than $E_a$, we can confidently state that the new rate constant is five times the original:

Arrhenius

To connect these rate constants, temperatures, and the elusive activation energy, we invoke the Arrhenius equation. By taking the natural logarithm of the ratio of the rate constants at two different temperatures, we get a powerful linear relationship:

Let's substitute our known values into this equation. We know the ratio $\frac{k_{T_2}}{k_{T_1}}$ is $5$, $T_1$ is $300\text{ K}$, $T_2$ is $315\text{ K}$, and the universal gas constant $R$ is $8.314{\text{ J mol}}^{-1}{\text{ K}}^{-1}$.

Now, it's time for some careful algebra. Let's simplify the temperature term inside the bracket by finding a common denominator:

We are given the value of $\ln (5)$ as $1.6094$. Substituting this in and rearranging the equation to isolate $E_a$ gives us:

Executing this final multiplication and division yields our activation energy:

This result perfectly illustrates how even a modest $15^{\circ }\text{C}$ increase in temperature can drastically accelerate a reaction by exponentially increasing the population of highly energetic molecules.