Question Detail
Question
The standard enthalpy of formation of NH₃(g) at 298 K is -46 kJ mol⁻¹. Which of the following values is correct for the standard enthalpy change in kJ mol⁻¹ for the following reaction? 2NH₃(g) → N₂(g) + 3H₂(g)
Correct Answer
Option A is the correct answer.
Detailed Explanation
ΔH°f(NH₃) = -46 kJ/mol. For 2NH₃, ΔH°f = -92 kJ. The reaction is the reverse of formation of 2NH₃ from elements. ΔH°rxn = -ΔH°f(products as elements?) Actually: ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants) = [0 + 0] - [2 × (-46)] = 0 - (-92) = +92 kJ.
Hint
Reverse reaction changes the sign of ΔH. Also, elements in standard state have ΔHf° = 0.