![]() ![]() ![]() Let's do that (the flip only, the division by 2 happens below) and keep everything else the same:ĢNH 4Cl(s) -> N 2(g) + 4H 2(g) + Cl 2(g) ΔH rxn = +630.78 kJģ) I know that I have two (and only two) missing products. This is because I know only NH 4Cl is on the reactant side in the target equation. N 2(g) + 3H 2(g) -> 2NH 3(g) ΔH rxn = −296.4kJĢ) I know that the second data equation will have to be flipped and divided by 2. N 2(g) + 3H 2(g) -> 2NH 3(g) ΔH rxn = −296.4 kJĭetermine the identity of the two missing products and calculate the ΔH rxn for this reaction:ġ⁄ 2H 2(g) + 1⁄ 2Cl 2(g) -> HCl(g) ΔH rxn = −92.3 kJ/mol Problem #11: The standard enthalpy of formation for hydrogen chloride is −92.3 kJ/mol. Problems 11 - 25 Hess' Law - three equations and their enthalpies - Probs 1-10 Hess' Law - two equations and their enthalpies Hess' Law - standard enthalpies of formation Hess' Law - three equations and their enthalpies Hess' Law - four or more equations and their enthalpies Hess' Law - bond enthalpies Thermochemistry menu Using three equations and their enthalpies ChemTeam: Hess' Law - three equations and their enthalpies - Probs 11-25 ![]()
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