Animated Solution for Chemistry - Chemical Kinetics: The rate of a reaction double when its temperature changes from $300\text{ K}$ to $310\text{ K}$. Activation energy of such a reaction will be ($R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$ and $\log 2=0.301$)

Have you ever wondered why we keep milk in the refrigerator? Or why food cooks faster in a pressure cooker? It all comes down to the fascinating relationship between temperature and the rate of a chemical reaction. In the world of Chemical Kinetics, this relationship is beautifully captured by the Arrhenius Equation.

Imagine a chemical reaction as a group of hikers trying to cross a mountain pass. The height of this mountain is what we call the Activation Energy ($E_a$). It is the minimum extra energy the reactant molecules must acquire to successfully transform into products. At higher temperatures, more molecules have enough kinetic energy to scale this mountain, leading to a faster reaction rate.

When we want to compare the rate constants ($k_1$ and $k_2$) of a reaction at two different temperatures ($T_1$ and $T_2$), we use the logarithmic form of the Arrhenius equation:

This equation is a powerful tool. It tells us exactly how sensitive a reaction is to temperature changes. A higher activation energy means the reaction rate will spike dramatically even with a small increase in temperature.

In our specific problem, we are given a scenario where a mere $10\text{ K}$ rise in temperature (from $300\text{ K}$ to $310\text{ K}$) causes the reaction rate to double.

Since the rate of a reaction is directly proportional to its rate constant (assuming concentrations remain unchanged), saying the "rate doubles" is mathematically equivalent to saying:

We are also equipped with the universal gas constant, $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$, and the value of $\log 2=0.301$.

Let's carefully substitute our known values into the Arrhenius equation. Watch out for the units here; since $R$ is in Joules, our resulting $E_a$ will also be in Joules.

Now, we execute the atomic computations step-by-step:

Rearranging the equation to isolate our target variable, $E_a$:

Our calculated activation energy is approximately $53604{\text{ J mol}}^{-1}$. However, standard conventions and our multiple-choice options require the answer in kiloJoules per mole (${\text{kJ mol}}^{-1}$). By dividing by $1000$, we arrive at our final destination:

This elegant calculation shows how a macroscopic observation (the rate doubling) allows us to peek into the microscopic energy barriers governing the molecules. It's a perfect example of why the Arrhenius equation is a cornerstone of physical chemistry!