The heat of formation of ethene, C 2 H 4 is 50 kJmol ^ -1 , and that of ethane, C 2 H 6 is -82kJmol ^ -1…

Analyze the question: You are given the heats of formation of ethene (+50 kJ/mol) and ethane (-82 kJ/mol). You need the enthalpy change for hydrogenating ethene to ethane.

Concept: The enthalpy change of a reaction is

ΔH = ΣΔHf(products) - ΣΔHf(reactants)

The heat of formation of an element in its standard state, such as H₂, is zero.

Solution:

Reaction: C₂H₄ + H₂ → C₂H₆

• Products: ΔHf(C₂H₆) = -82 kJ/mol
• Reactants: ΔHf(C₂H₄) + ΔHf(H₂) = 50 + 0 = 50 kJ/mol

ΔH = (-82) - (50) = -132 kJ/mol

The negative sign means the reaction is exothermic. Heat is evolved, which matches the question.

Answer: B (-132 kJ)

Tip: Always use "products minus reactants". Remember that elements in their standard states (H₂, O₂, N₂, C graphite) have ΔHf = 0. A common mistake is to add 82 and 50 and get 132 without thinking about the sign, or to subtract in the wrong order and get +132. Check the sign against the wording: "heat evolved" means exothermic, so ΔH is negative.