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 relationship between activation energies and enthalpy change.
For a reversible reaction, the activation energy of the forward reaction (E\(_{a,f}\)), the activation energy of the reverse reaction (E\(_{a,b}\)), and the enthalpy change of the reaction (\(\Delta\)H) are related. This can be visualized with a reaction profile diagram. The relationship is: \[ \Delta H = E_{a,f} - E_{a,b} \] Where:

• \(\Delta\)H is the enthalpy change of the forward reaction.
• E\(_{a,f}\) is the activation energy of the forward reaction (X \(\rightarrow\) Y).
• E\(_{a,b}\) is the activation energy of the backward/reverse reaction (Y \(\rightarrow\) X).

Step 2: Identify the given values.

• Activation energy for the forward reaction, E\(_{a,f}\) = 150 kJ mol\(^{-1}\)
• Enthalpy change, \(\Delta\)H = -135 kJ mol\(^{-1}\) (The reaction is exothermic).
• We need to find the activation energy for the reverse reaction, E\(_{a,b}\).

Step 3: Rearrange the formula and calculate E\(_{a,b}\).
\[ E_{a,b} = E_{a,f} - \Delta H \] Substitute the given values: \[ E_{a,b} = (150 \text{ kJ mol}^{-1}) - (-135 \text{ kJ mol}^{-1}) \] \[ E_{a,b} = 150 + 135 = 285 \text{ kJ mol}^{-1} \] Step 4: Final Answer.
The activation energy for the reverse reaction (Y \(\rightarrow\) X) is 285 kJ mol\(^{-1}\).