The standard molar enthalpy of formation of NH₃(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N₂(g) and 1.96 g H₂(g) react to produce NH₃(g)?

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The enthalpy change for the reaction can be calculated using the standard molar enthalpies of formation of the reactants and products. The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol, the standard molar enthalpy of formation of N2(g) is 0 kJ/mol, and the standard molar enthalpy of formation of H2(g) is 0 kJ/mol. Therefore, the enthalpy change for the reaction is:ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants)ΔH°rxn = [1 mol NH3(-45.9 kJ/mol)] - [1 mol N2(0 kJ/mol) + 3 mol H2(0 kJ/mol)]ΔH°rxn = -45.9 kJ/molThe enthalpy change for the reaction is -45.9 kJ/mol. To find the enthalpy change for the given masses of reactants, we need to convert the masses to moles and then multiply by the enthalpy change per mole.9.51 g N2 * (1 mol N2 / 28.01 g N2) = 0.339 mol N21.96 g H2 * (1 mol H2 / 2.016 g H2) = 0.973 mol H2ΔH°rxn = (0.339 mol N2 + 0.973 mol H2) * (-45.9 kJ/mol)ΔH°rxn = -29.8 kJ/mol-rxnTherefore, the enthalpy change for the reaction is -29.8 kJ/mol-rxn.

The Standard Molar Enthalpy of Formation of NH3(g) Is -45

The Standard Molar Enthalpy of Formation of NH₃(g) Is -45