Q.Which of the following is affected by catalyst ? (A) $\Delta H$ (B) $\Delta G$ (C) $E_a$ (D) $\Delta S$

The Arrhenius equation is the starting point here. It tells us that the rate constant $k$ depends on temperature as:

where $A$ is the pre-exponential factor (related to collision frequency and orientation), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature.

The exponential term $e^{-E_a/RT}$ is the key. It represents the fraction of molecules that have enough energy to overcome the activation barrier — that is, molecules with kinetic energy equal to or greater than $E_a$. This comes from the Maxwell–Boltzmann distribution of molecular energies: the fraction of molecules with energy $\geq E_a$ is proportional to $e^{-E_a/RT}$.

So the question is simply asking: which of the given expressions matches this Boltzmann factor?

Let’s check each option:

1. Option (A): $+\dfrac{E_a}{RT}$

   This is a positive number (since $E_a$, $R$, $T$ are all positive). It grows as temperature decreases or activation energy increases. But a fraction must lie between 0 and 1 — this expression can be much larger than 1, so it cannot represent a fraction of molecules. Discard.

2. Option (B): $e^{-E_a/RT}$

   This is the Boltzmann factor. For typical values ($E_a \sim 50$ kJ/mol, $T \sim 300$ K, $R = 8.314$ J/mol·K), $E_a/RT \approx 20$, so $e^{-20} \approx 2 \times 10^{-9}$ — a tiny fraction, which makes sense: only a very small proportion of molecules have enough energy to react at room temperature. This matches the physical meaning exactly.

3. Option (C): $-\dfrac{E_a}{RT}$

   This is negative. A fraction cannot be negative. Discard. …

The Arrhenius equation is one of the most elegant relationships in chemical kinetics — it connects the rate constant $k$ to temperature $T$ through two parameters: the activation energy $E_a$ and the pre-exponential factor $A$. The equation is:

If you plot $k$ directly against $T$, you get a curve. But the trick is to take logarithms — that turns the exponential into a straight line. Why does that help? Because a straight line is easy to interpret: its slope and intercept give you $E_a$ and $A$ directly.

Let’s see how.

1. Take the natural logarithm of both sides

   Starting from $k = A e^{-E_a/(RT)}$, we get:

This is already in the form $y = mx + c$, where:

• $y = \ln k$
• $x = 1/T$
• slope $m = -E_a/R$
• intercept $c = \ln A$

So a plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ and intercept $\ln A$.

2. But the question uses $\log k$ — that’s base 10

   In many Indian exam contexts, $\log$ means $\log_{10}$. To convert from natural log to base 10, use:

Substitute into the equation above:

Divide through by 2.303:

3. Identify slope and intercept

   Now compare with $y = mx + c$:

   • $y = \log k$
   • $x = 1/T$
   • slope $m = -\dfrac{E_a}{2.303 \, R}$
   • intercept $c = \dfrac{\ln A}{2.303} = \log A$

   That’s exactly what option (D) says. …

The key to this question lies in distinguishing between kinetics (how fast a reaction occurs) and thermodynamics (whether a reaction is spontaneous and how much energy is exchanged). A catalyst is a kinetic tool — it speeds up a reaction without being consumed, but it never alters the starting point or the destination of the reaction.

Think of a mountain pass. The reactants are at the base of one side, the products at the base of the other. The height difference between them is $\Delta H$ (enthalpy change). The overall steepness and direction of the slope is $\Delta G$ (free energy change). The disorder along the path is $\Delta S$ (entropy change). A catalyst is like a tunnel through the mountain — it lowers the peak you have to climb over (the activation energy $E_a$), but the heights of the two bases and the distance between them remain exactly the same.

Let’s examine each option carefully.

1. $\Delta H$ (Enthalpy change)

   $\Delta H$ is the difference in enthalpy between products and reactants. It depends only on the initial and final states of the system. A catalyst does not appear in the overall balanced equation and is regenerated at the end. Since the reactants and products are identical with or without the catalyst, $\Delta H$ is unchanged.

   $\Delta H_{\text{catalysed}} = \Delta H_{\text{uncatalysed}}$

2. $\Delta G$ (Gibbs free energy change)

   $\Delta G = \Delta H - T\Delta S$ is the thermodynamic driving force for a reaction. It determines spontaneity. A catalyst cannot make a non-spontaneous reaction spontaneous — it only accelerates a reaction that is already thermodynamically favourable. The equilibrium constant $K$ is related to $\Delta G$ by $\Delta G^\circ = -RT\ln K$, and a catalyst does not shift equilibrium. So $\Delta G$ remains the same.

3. $E_a$ (Activation energy) …