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

Imagine you are trying to push a heavy boulder over a hill. The height of that hill represents the activation energy ($E_a$) of a chemical reaction. It is the minimum extra energy required by the reacting molecules to successfully collide and form products. When you increase the temperature, you are essentially giving the molecules more kinetic energy, making it easier for them to overcome this energy barrier.

In this problem, we are given a fascinating scenario: raising the temperature from $27^{\circ }\text{C}$ to $52^{\circ }\text{C}$ causes the rate constant to skyrocket, becoming five times its original value. Our mission is to calculate the exact height of that energy barrier, the activation energy.

Arrhenius to the Rescue

To connect the rate constants at two different temperatures with the activation energy, we rely on the logarithmic form of the Arrhenius equation. This equation is a cornerstone of chemical kinetics:

Before we plug anything in, we must remember a golden rule of physical chemistry: always use absolute temperature (Kelvin).

Let's convert our given temperatures: $T_1=27^{\circ }\text{C}+273=300\text{ K}$ $T_2=52^{\circ }\text{C}+273=325\text{ K}$

We are also given that the ratio of the new rate constant to the old one is 5, meaning $\frac{k_2}{k_1}=5$.

Now, let's substitute our known values into the Arrhenius equation. The universal gas constant $R$ is given as $8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

Let's simplify the terms step-by-step to avoid any silly mistakes. The value of $\log (5)$ is approximately $0.699$. The product of $2.303$ and $8.314$ is roughly $19.147$. The temperature difference in the numerator is $25$, and the product in the denominator is $97500$.

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

After carefully multiplying and dividing, we find:

We have our answer, but there is a catch! The question specifically asks for the activation energy in ${\text{kJ mol}}^{-1}$. To convert Joules to kilojoules, we simply divide by $1000$:

Finally, rounding off to the nearest integer, we arrive at our final answer:

This result tells us a lot about the reaction. A higher activation energy means the reaction rate is highly sensitive to temperature changes. Just a $25^{\circ }\text{C}$ jump caused a massive five-fold increase in the rate! Understanding this sensitivity is crucial for controlling reactions in industrial processes.