Q.2. The activation energy for two reactions are $E_a$ and $E'_a$ with $E_a$ > $E'_a$. If the temperature of reacting system increases from $T_1$ to $T_2$, predict which of the following is correct? a. $\dfrac{k'_1}{k_1} = \dfrac{k'_2}{k_2}$
b. $\dfrac{k'_1}{k_1} > \dfrac{k'_2}{k_2}$
c. $\dfrac{k'_1}{k_1} < \dfrac{k'_2}{k_2}$
d. $\dfrac{k'_1}{k_1} < 2\ \dfrac{k'_2}{k_2}$ k values are rate constants at lower temperature and k values at higher temperature.

[!NOTE]
The final line is the book's own footnote, printed exactly as shown -- the second "k" is missing its prime. As the options' algebra requires, $k_1$, $k_2$ are the rate constants of the two reactions at the lower temperature $T_1$ and $k'_1$, $k'_2$ those at the higher temperature $T_2$.

The rate constant $k$ is not just a number we measure—it encodes how the molecular-scale energy barrier controls reaction speed. To see why a catalyst must increase $k$, we need the Arrhenius equation.

This equation tells us that $k$ depends exponentially on the activation energy. A catalyst works by providing an alternative reaction pathway with a lower $E_a$—it doesn't change the thermodynamics (reactants and products stay the same), but it reduces the energy hill molecules must climb to react.

Step-by-step reasoning

1. What a catalyst does at the molecular level

   A catalyst participates in the reaction mechanism but is regenerated at the end. It creates intermediate steps with lower energy barriers than the uncatalyzed path. The net effect: $E_a$ (catalyst) $< E_a$ (no catalyst).

2. Impact on the exponential term

   When $E_a$ decreases, the exponent $-E_a/RT$ becomes less negative (closer to zero). Since $e^x$ is an increasing function, $e^{-E_a/RT}$ becomes larger.

3. The pre-exponential factor $A$

   This factor relates to collision frequency and orientation. A catalyst typically doesn't change $A$ significantly—the main effect is on $E_a$.

4. Independence from reaction order

   The rate constant $k$ appears in the rate law (e.g., $\text{rate} = k[A]^n$), but its value is determined by the Arrhenius equation, not by the order $n$. The order tells us how concentration affects rate; the activation energy tells us the intrinsic speed at given concentrations. A catalyst lowers $E_a$ regardless of whether the reaction is zeroth, first, second, or any other order.

5. Quantitative example

   Suppose $E_a = 100 \, \text{kJ/mol}$ without catalyst and $E_a = 50 \, \text{kJ/mol}$ with catalyst at $T = 300 \, \text{K}$ (with $R = 8.314 \, \text{J/(mol·K)}$): …

Why activation energy matters

The Arrhenius equation connects temperature to the rate constant through the activation energy—the minimum energy barrier reactants must overcome to transform into products. The equation captures a fundamental idea: higher temperatures give molecules more kinetic energy, increasing the fraction that can surmount the barrier.

The exponential term $e^{-E_a/RT}$ embodies the temperature dependence. When $E_a$ is large, even small temperature changes dramatically alter $k$. But what happens when there's no barrier at all?

The special case: $E_a = 0$

1. Substitute zero activation energy into the Arrhenius equation:

2. Interpret the result:

   The rate constant collapses to just the pre-exponential factor $A$. This factor represents the frequency of collisions with proper orientation—it depends on molecular properties and collision geometry, but crucially, it has no temperature dependence built into the exponential term.

3. Physical meaning:

   A zero activation energy means every collision between properly oriented molecules leads to reaction, regardless of their kinetic energy. There's no energy threshold to cross. Temperature might still affect collision frequency slightly through changes in molecular speed, but the dominant exponential temperature dependence vanishes. …