The energies of activation for forward and reverse reactions for A2 + B2 2AB are 180 kJ mol^-1 a

Q: Why is the enthalpy change negative, but the correct option is positive 20?

Great observation! The negative sign indicates that the reaction is exothermic (energy is released). However, the options provided are magnitudes. In multiple-choice questions, if only positive values are given for an energy change, you should select the magnitude.

Q: Does a catalyst ever change the enthalpy of a reaction?

Never! Enthalpy is a state function, meaning it only depends on the initial state (reactants) and final state (products). A catalyst changes the path taken (kinetics) but not the start and end points (thermodynamics).

Q: Why does the catalyst lower both activation energies by the exact same amount?

A catalyst lowers the peak of the energy barrier. Since both the forward and reverse reactions must pass over this exact same peak, lowering the peak by 100 kJ/mol reduces the climb for both sides equally.

Q: What if the catalyst only lowered the forward activation energy?

That is physically impossible! If a catalyst only lowered the forward activation energy, it would change the equilibrium constant and violate the First Law of Thermodynamics by creating a perpetual motion machine.

Q: How do we know this reaction is exothermic?

Because the forward activation energy (180 kJ/mol) is less than the backward activation energy (200 kJ/mol). This means the products sit at a lower energy level than the reactants, releasing the difference as heat.

Question

The energies of activation for forward and reverse reactions for $A_2 + B_2 ightleftharpoons 2AB$ are $180 	ext{ kJ mol}^{-1}$ and $200 	ext{ kJ mol}^{-1}$, respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 	ext{ kJ mol}^{-1}$. The enthalpy change of the reaction ($A_2 + B_2 ightarrow 2AB$) in the presence of catalyst will be (in $	ext{kJ mol}^{-1}$)

Options:

(A) 300

(B) 120

(C) 280

(D) 20

Sigma Percentile · Accepted Answer

### The Energy Landscape

Imagine you are standing at the foot of a massive mountain, representing the energy barrier of a chemical reaction.

To get to the other side, you must climb to the peak. This climb is the **activation energy**.

For our reaction, $A_2 + B_2 ightleftharpoons 2AB$, the forward climb is $E_{a(f)} = 180 	ext{ kJ mol}^{-1}$.

If you were to make the return journey from the products back to the reactants, the climb would be even steeper: $E_{a(b)} = 200 	ext{ kJ mol}^{-1}$.

### The Master Equation for Enthalpy

Now, what exactly is the enthalpy change ($\Delta H$) of this reaction?

Enthalpy change is simply the difference in altitude between your starting point and your final destination.

Mathematically, it is the difference between the forward and backward activation energies:

$$ \Delta H = E_{a(f)} - E_{a(b)} $$

Let's calculate this for our uncatalyzed reaction.

Substituting the values, we get:

$$ \Delta H = 180 - 200 = -20 	ext{ kJ mol}^{-1} $$

The negative sign is crucial here. It tells us that the final destination is lower than the starting point, meaning the reaction is **exothermic** and releases energy.

### Enter the Catalyst

Next, we introduce a catalyst into the system.

Think of a catalyst as a secret tunnel through the mountain. It provides an alternative pathway with a much lower peak.

In our problem, the catalyst lowers the activation energy barrier by $100 	ext{ kJ mol}^{-1}$.

Because the peak itself is lowered, the climb is reduced equally for both the forward and reverse journeys!

The new forward activation energy becomes:

$$ E'_{a(f)} = 180 - 100 = 80 	ext{ kJ mol}^{-1} $$

And the new reverse activation energy becomes:

$$ E'_{a(b)} = 200 - 100 = 100 	ext{ kJ mol}^{-1} $$

### The Grand Conclusion

Let's find the new enthalpy change using our catalyzed pathway.

$$ \Delta H' = E'_{a(f)} - E'_{a(b)} $$

$$ \Delta H' = 80 - 100 = -20 	ext{ kJ mol}^{-1} $$

Notice something beautiful? The enthalpy change remains exactly the same!

This highlights a fundamental law of nature: **a catalyst only affects the kinetics (speed) of a reaction, never its thermodynamics (initial and final states).**

Since the question asks for the magnitude of the enthalpy change, our final answer is **$20 	ext{ kJ mol}^{-1}$**.

Select Answer:

Visualized Solution

Energy Profile Diagram

$Δ H = E_{a (f)} - E_{a (b)}$

$Δ H$ without Catalyst

Effect of Catalyst

$Δ H$ with Catalyst

Thermodynamics vs Kinetics

The Sigma Insight: Theories of Chemical Reaction

The Energy Landscape

The Master Equation for Enthalpy

Enter the Catalyst

The Grand Conclusion

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