An exothermic reaction X - Y has an activation energy 30 kJ mol^-1. If energy change ΔE during the

Q: Why is $\Delta H$ given as a negative value?

In an exothermic reaction, energy is released into the surroundings. This means the products have less intrinsic potential energy than the reactants, resulting in a negative change in enthalpy.

Q: Can activation energy ever be negative?

No, activation energy is the minimum extra energy required to initiate a reaction. It represents a physical barrier that must be overcome, so it is always a positive quantity.

Q: Why do we subtract $\Delta H$ from $E_a^f$ to get $E_a^b$?

Look at the energy profile. The backward activation energy ($E_a^b$) spans from the product's energy level all the way to the peak. This total height is the sum of the forward barrier ($E_a^f$) and the energy drop (magnitude of $\Delta H$).

Q: What would happen if this was an endothermic reaction?

For an endothermic reaction, the products would be at a higher energy level than the reactants. $\Delta H$ would be positive, and the backward activation energy ($E_a^b$) would be smaller than the forward activation energy ($E_a^f$).

Q: Does adding a catalyst change the value of $\Delta H$?

Absolutely not! A catalyst provides an alternative pathway with a lower activation energy peak. It decreases both $E_a^f$ and $E_a^b$ by the exact same amount, leaving the energy difference between reactants and products ($\Delta H$) completely unchanged.

Question

An exothermic reaction $X ightarrow Y$ has an activation energy $30 	ext{ kJ mol}^{-1}$. If energy change $\Delta E$ during the reaction is $-20 	ext{ kJ mol}^{-1}$, then the activation energy for the reverse reaction in kJ is ............ .

Sigma Percentile · Accepted Answer

### The Anatomy of an Exothermic Reaction

Imagine you are standing at the edge of a high plateau, looking down into a deep valley. To get to the valley, you first have to hike over a small hill that blocks your path. This physical landscape is the perfect analogy for the energy profile of an exothermic chemical reaction.

In our reaction, $X ightarrow Y$, the reactant $X$ is resting on the high plateau. It possesses a certain amount of intrinsic potential energy. However, it cannot simply fall into the valley to become product $Y$. It must first absorb enough energy to break its existing chemical bonds and reach a highly unstable, high-energy state known as the transition state. This "hill" it must climb is called the **forward activation energy**, denoted as $E_a^f$.

Once the molecules reach the peak, they cascade down into the valley, releasing energy as new, more stable bonds are formed to create product $Y$. Because the valley is lower than the initial plateau, the overall process releases energy into the surroundings. This net difference in altitude between the reactant plateau and the product valley is the **enthalpy change**, $\Delta H$. For an exothermic reaction, $\Delta H$ is always negative, signifying a loss of system energy.

### Decoding the Energy Profile

Now, let's look at the journey in reverse. What if product $Y$ wanted to climb back up and become reactant $X$?

Standing in the deep valley, product $Y$ looks up at the transition state peak. To reach it, $Y$ must climb the entire height of the mountain. This massive climb is the **backward activation energy**, $E_a^b$.

If you visualize this geometrically on an energy profile graph, a beautiful and simple mathematical relationship emerges. The total height of the mountain from the valley floor ($E_a^b$) is exactly equal to the height of the hill from the plateau ($E_a^f$) plus the depth of the valley relative to the plateau (the magnitude of $\Delta H$).

### The Master Equation of Energy Conservation

We can express this geometric reality with a fundamental thermodynamic equation:

$$\Delta H = E_a^f - E_a^b$$

This equation is the master key to solving our problem. We are given two crucial pieces of information:
1. The forward activation energy, $E_a^f = 30 	ext{ kJ mol}^{-1}$.
2. The energy change (enthalpy change), $\Delta H = -20 	ext{ kJ mol}^{-1}$.

Notice the negative sign on $\Delta H$. It is absolutely critical. It mathematically enforces the fact that the products are at a lower energy state than the reactants.

### The Final Calculation

Our goal is to find the backward activation energy, $E_a^b$. Let's rearrange our master equation to isolate the unknown variable:

$$E_a^b = E_a^f - \Delta H$$

Now, we carefully substitute our known values into the rearranged equation. This is where many students make a fatal sign error, so we must proceed with caution:

$$E_a^b = 30 - (-20)$$

Subtracting a negative number is mathematically equivalent to adding a positive number. The two negative signs cancel each other out:

$$E_a^b = 30 + 20$$

$$E_a^b = 50 	ext{ kJ mol}^{-1}$$

The backward activation energy is $50 	ext{ kJ mol}^{-1}$.

Does this make physical sense? Absolutely. Since the reaction is exothermic, the products are trapped in a deeper energy well than the reactants. Therefore, the energy required to climb out of that well and reach the transition state ($E_a^b$) must be significantly larger than the energy required for the reactants to reach the same state ($E_a^f$).

### The Catalyst Thought Experiment

Before we conclude, let's run a quick thought experiment. What would happen to these values if we introduced a positive catalyst into the reaction vessel?

A catalyst works by providing an entirely new reaction pathway with a lower transition state. Imagine a tunnel being bored through the top of our mountain.

Because the peak is lowered, the forward activation energy ($E_a^f$) decreases. Consequently, the backward activation energy ($E_a^b$) also decreases by the exact same amount. However, the altitudes of the reactant plateau and the product valley remain completely untouched. Therefore, the enthalpy change ($\Delta H$) is a thermodynamic state function that is entirely independent of the reaction pathway or the presence of a catalyst.

Mastering this visual intuition will make you unstoppable in Chemical Kinetics!

Animated Solution for Chemistry - Chemical Kinetics: An exothermic reaction $X \to Y$ has an activation energy $30 kJ mol^{- 1}$ . If energy change $Δ E$ during the reaction is $- 20 kJ mol^{- 1}$ , then the activation energy for the reverse reaction in kJ is ............ .

Enter Numerical Value:

Visualized Solution

Energy Profile Diagram

Activation Energies \& \Delta H

The Master Equation

Rearranging for $E_{a}^{b}$

Final Calculation

Effect of Catalyst

The Sigma Insight: Theories of Chemical Reaction

The Anatomy of an Exothermic Reaction

Decoding the Energy Profile

The Master Equation of Energy Conservation

The Final Calculation

The Catalyst Thought Experiment

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