Animated Solution for Chemistry - Chemical Kinetics: For the reaction of ${\text{H}}_2$ with ${\text{I}}_2$, the rate constant is $2.5\times 10^{-4}{\text{ dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$ at $327^{\circ }\text{C}$ and $1.0{\text{ dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$ at $527^{\circ }\text{C}$. The activation energy for the reaction, in ${\text{kJ mol}}^{-1}$ is ($R=8.314{\text{ J K}}^{-1}{\text{mol}}^{-1}$)

Temperatures and Rate Constants

Let's start by writing down what we know. We are given the rate constants for a chemical reaction at two different temperatures.

Notice that the temperatures are provided in Celsius. In thermodynamics and kinetics, we must always work with absolute temperatures. So, our very first job is to convert them to Kelvin by adding $273$.

For the first state: $T_1=327^{\circ }\text{C}+273=600\text{ K}$ $k_1=2.5\times 10^{-4}{\text{ dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$

For the second state: $T_2=527^{\circ }\text{C}+273=800\text{ K}$ $k_2=1.0{\text{ dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$

Arrhenius Two-Temperature Form

To connect rate constants, temperatures, and activation energy, we use the two-temperature form of the Arrhenius equation. This equation is derived from the linear form $\ln k=\ln A-\frac{E_a}{RT}$. When we evaluate this at two different temperatures and subtract, the pre-exponential factor $A$ cancels out, leaving us with:

This equation is incredibly powerful because it allows us to find the activation barrier $E_a$ just by observing how the reaction speed changes with temperature.

Navigating the Math

Now, let's carefully substitute our values into the formula. We put in our $k$ values on the left, and our temperatures and the universal gas constant $R$ on the right. Don't rush the calculation here.

Let's simplify the left-hand side first.

Taking the log of $4000$ gives us:

Next, we simplify the right-hand side. The temperature difference is $200$, and the product is $480,000$. This simplifies to $\frac{1}{2400}$. Multiplying $2.303$ and $8.314$ gives about $19.147$.

Activation Energy

Finally, we isolate the activation energy by multiplying the terms across the equals sign.

Because the gas constant $R$ was in Joules, our answer is in Joules. Converting this to kilojoules gives us $165.5$, which rounds off nicely to $166{\text{ kJ mol}}^{-1}$.

Food for thought: What if we added a catalyst? The activation energy $E_a$ would drop, making the reaction faster at both temperatures. Interestingly, the ratio of the rate constants $\frac{k_2}{k_1}$ would actually decrease. A great concept to ponder upon!