Question

According to the following figure, the magnitude of the enthalpy change of the reaction \(A + B \to M + N\) in kJ mol\(^{-1}\) is equal to \(\dots\dots\dots\). (Integer answer) 
[Given: \(x = 20\text{ kJ mol}^{-1}\), \(y = 45\text{ kJ mol}^{-1}\), \(z = 15\text{ kJ mol}^{-1}\)] 

Hint:

If the backward activation energy is greater than the forward, the reaction is exothermic (\(\Delta H<0\)), as seen in this diagram.

Answer 1

Correct Answer: 45

Step 1: Understanding the Concept:
In an energy profile diagram, the enthalpy change (\(\Delta H\)) of a reaction is the difference between the potential energy of the products and that of the reactants.
Step 2: Key Formula or Approach:
\[ \Delta H = \text{Activation Energy (forward)} - \text{Activation Energy (backward)} \]
Or, \(\Delta H = PE_{\text{products}} - PE_{\text{reactants}}\).
Step 3: Detailed Explanation:
From the provided graph:
- \(x\) is the height from the reactant level (\(A+B\)) to the peak of the barrier. This is the forward activation energy, \(E_{a(f)} = 20\text{ kJ mol}^{-1}\).
- \(y\) is the height from the product level (\(M+N\)) to the peak of the barrier. This is the backward activation energy, \(E_{a(b)} = 45\text{ kJ mol}^{-1}\).
Calculation of Enthalpy Change:
\[ \Delta H = E_{a(f)} - E_{a(b)} \]
\[ \Delta H = 20 - 45 = -25\text{ kJ mol}^{-1} \]
The question asks for the magnitude of the enthalpy change.
\[ |\Delta H| = |-25| = 25\text{ kJ mol}^{-1} \]
Step 4: Final Answer:
The magnitude of the enthalpy change is 25.