What is the value of the enthalpy change per mole of HCl(g) produced? — Thermodynamics Chemistry Question

STEPS:

1. Apply Hess's Law to find the overall enthalpy of the reaction ($\Delta H^\circ_{\text{overall}}$):
Enthalpy is a state function, meaning that the overall enthalpy change of a multi-step reaction is equal to the sum of the enthalpy changes of its individual elementary steps:
* Step 1: $\Delta H^\circ_1 = 242\text{ kJ/mol}_{\text{rxn}}$
* Step 2: $\Delta H^\circ_2 = 4\text{ kJ/mol}_{\text{rxn}}$
* Step 3: $\Delta H^\circ_3 = -432\text{ kJ/mol}_{\text{rxn}}$

Summing these values yields:
$$\Delta H^\circ_{\text{overall}} = 242\text{ kJ} + 4\text{ kJ} + (-432\text{ kJ}) = \mathbf{-186\text{ kJ/mol}_{\text{rxn}}}$$

2. Analyze the stoichiometry of the overall reaction:
By combining and canceling the intermediates ($\text{Cl}$ and $\text{H}$) in the three steps, we get the overall balanced chemical equation:
$$\text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons \mathbf{2\ \text{HCl}(g)}$$
This balanced equation demonstrates that 2 moles of $\text{HCl}(g)$ are produced per mole of overall reaction.

3. Scale the enthalpy change per mole of product:
The calculated overall enthalpy change of $-186\text{ kJ}$ is associated with the reaction as written, which generates 2 moles of $\text{HCl}$. To find the energy change per single mole of $\text{HCl}$ produced, we must divide the overall enthalpy change by 2:
$$\Delta H^\circ_{\text{per mole of HCl}} = \frac{-186\text{ kJ}}{2\text{ mol HCl}} = \mathbf{-93\text{ kJ/mol HCl}}$$

This identifies Option A as the correct answer.

*

WHY_OTHERS_WRONG:

• Option B is incorrect (-121 kJ): A student might arrive at this value by taking the enthalpy of Step 1 ($242\text{ kJ}$), dividing it by 2 to get $121\text{ kJ}$, and arbitrarily applying a negative sign. This error occurs if a student incorrectly assumes that only the slow, rate-determining step determines the enthalpy change of the reaction.
• Option C is incorrect (-186 kJ): This represents the overall enthalpy change ($\Delta H^\circ_{\text{overall}}$) for the reaction as written. Choosing this option means the student correctly used Hess's Law but failed to read the question carefully and forgot to divide by the 2 moles of $\text{HCl}$ produced.
• Option D is incorrect (-242 kJ): This is simply the negative of the enthalpy change of Step 1 ($+242\text{ kJ}$). It ignores the other steps in the mechanism and fails to calculate the enthalpy of the final product formation.