The enthalpy of formation of ammonia gas is -46.0 kJ/mol. The enthalpy change for the reaction: 2(g) — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The standard enthalpy of formation of $NH_3$(g) refers to the reaction: 1/2 $N_2$(g) + 3/2 $H_2$(g) -> $NH_3$(g) with δ H = -46.0 kJ/mol. The given reaction: 2$NH_3$(g) -> $N_2$(g) + 3$H_2$(g) is the reverse of the formation of 2 moles of $NH_3$. Therefore, δ H_rxn = -2 * (-46.0 kJ) = +92.0 kJ.