Question

For the chemical reaction H₂ (g) + F₂ (g) → 2 HF (g) ΔH° = -79.2 kJ/mol. What is the molar enthalpy (in kJ/mol) for the reaction 3 H₂ (g) + 3 F₂ (g) → 6 HF (g)?

A) -237.6
B) 237.6
C) -79.2
D) 79.2

Answer 1

Molar enthalpy for the scaled reaction of hydrogen with fluorine to form hydrogen fluoride is found by tripling the original value, resulting in -237.6 kJ/mol.

Step-by-step explanation:

The molar enthalpy for the reaction of hydrogen gas with fluorine gas to form hydrogen fluoride, H₂ (g) + F₂ (g) → 2 HF (g), with a ΔH° of -79.2 kJ/mol, can be scaled up directly for the reaction involving three moles of each reactant and six moles of product. To find the molar enthalpy for the reaction 3 H₂ (g) + 3 F₂ (g) → 6 HF (g), you simply multiply the original ΔH° by three: -79.2 kJ/mol times 3 equals -237.6 kJ/mol. Therefore, the correct answer is Option A) -237.6 kJ/mol.