Question

The activation energy for the reaction X \(\rightarrow\) Y is 150 kJ mol\(^{-1}\). The change in enthalpy for the above reaction is -135 kJ mol\(^{-1}\). Then the activation energy for Y \(\rightarrow\) X is

• A. 280 kJ mol\(^{-1}\)
• B. 285 kJ mol\(^{-1}\)
• C. 270 kJ mol\(^{-1}\)
• D. 15 kJ mol\(^{-1}\)

Hint:

Drawing a simple energy profile diagram can help you visualize the relationship and avoid sign errors. 1. Draw the energy level for reactants (X). 2. Since \(\Delta\)H is negative (exothermic), draw the energy level for products (Y) lower than X. The difference is 135. 3. Draw the transition state peak. The energy barrier from X to the peak is E\(_{a,f}\) = 150. 4. The activation energy for the reverse reaction, E\(_{a,b}\), is the energy barrier from Y to the peak. 5. From the diagram, you can see that E\(_{a,b}\) = E\(_{a,f}\) + | \(\Delta\)H | = 150 + 135 = 285.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The enthalpy change of a reaction ($\Delta H$) represents the overall energy difference between the products and reactants. It mathematically bridges the activation energy of the forward reaction and the backward reaction.
Step 2: Key Formula or Approach:
Use the fundamental thermodynamic-kinetic relationship: $\Delta H = E_{a(\text{forward})} - E_{a(\text{backward})}$.
Step 3: Detailed Explanation:
Given parameters:
Forward activation energy, $E_{af} = 150$ kJ mol$^{-1}$.
Enthalpy change, $\Delta H = -135$ kJ mol$^{-1}$ (The negative sign indicates an exothermic reaction where products are lower in energy than reactants).
We need to find the backward activation energy, $E_{ab}$ for the reverse reaction $Y \rightarrow X$.
Rearranging the formula:
\[ \Delta H = E_{af} - E_{ab} \]
\[ E_{ab} = E_{af} - \Delta H \]
Substitute the values carefully observing the signs:
\[ E_{ab} = 150 - (-135) \]
\[ E_{ab} = 150 + 135 = 285 \text{ kJ mol}^{-1}. \]
Step 4: Final Answer:
The activation energy for the backward reaction is 285 kJ mol$^{-1}$.