What is the change in enthalpy (ΔH) when 1 mol of H2(g) is completely reacted according to the following thermochemical equation? 2 H2(g) + O2(g) → 2 H2O(l)ΔH = -571.6 kJ

• A. -571.6 kJ
• B. +285.8 kJ
• C. -285.8 kJ
• D. 0 kJ

Answer: (C)

Verified step by step guidance

1. Identify the given thermochemical equation and its enthalpy change: 2 H\_2(g) + O\_2(g) → 2 H\_2O(l) with ΔH = -571.6 kJ. This means that when 2 moles of H\_2 react, the enthalpy change is -571.6 kJ.
2. Determine the enthalpy change for 1 mole of H\_2 by using the stoichiometric ratio from the balanced equation. Since the enthalpy change corresponds to 2 moles of H\_2, divide the total ΔH by 2 to find the change for 1 mole.
3. Set up the calculation: ΔH (per 1 mol H\_2) = \frac{-571.6 \text{ kJ}}{2}. This step involves simple division to scale the enthalpy change to the amount of substance in question.
4. Interpret the sign of ΔH: a negative value indicates the reaction is exothermic, meaning heat is released when 1 mole of H\_2 reacts.
5. Conclude that the enthalpy change for the reaction of 1 mole of H\_2 is half the given value, maintaining the negative sign to reflect the exothermic nature.