Given the thermochemical equation: H2(g) + Cl2(g) → 2 HCl(g); ΔH = -184 kJ, what is the value of the enthalpy change per mole of HCl(g) produced?

• A. -92 kJ
• B. -46 kJ
• C. -184 kJ
• D. +92 kJ

Answer: (A)

Verified step by step guidance

1. Identify the given thermochemical equation: \(\mathrm{H_2(g) + Cl_2(g) \rightarrow 2\ HCl(g)}\) with an enthalpy change \(\Delta H = -184\ \mathrm{kJ}\). This means that when 2 moles of HCl are formed, the enthalpy change is -184 kJ.
2. Understand that the enthalpy change given corresponds to the formation of 2 moles of HCl. To find the enthalpy change per mole of HCl, you need to divide the total enthalpy change by the number of moles of HCl produced.
3. Set up the calculation for the enthalpy change per mole of HCl as: \(\Delta H_{\text{per mole}} = \frac{-184\ \mathrm{kJ}}{2\ \mathrm{mol}}\).
4. Perform the division to find the enthalpy change associated with the formation of 1 mole of HCl. Note the sign of \(\Delta H\) remains the same because the reaction is exothermic.
5. Interpret the result as the enthalpy change per mole of HCl produced, which helps understand the energy released or absorbed when one mole of HCl forms under the given reaction conditions.