Calculate the standard enthalpy change for the formation of ammonium chloride (NH4Cl) from its constituent elements, nitrogen (N2), hydrogen (H2), and chlorine (Cl2), given the following information:Reaction equation: N2(g) + 4H2(g) + 2Cl2(g) → 2NH4Cl(s)Standard enthalpy of formation for NH4Cl: -314.4 kJ/molStandard enthalpy of formation for N2(g): 0 kJ/molStandard enthalpy of formation for H2(g): 0 kJ/molStandard enthalpy of formation for Cl2(g): 0 kJ/mol Express your answer in kilojoules per mole (kJ/mol).

Question

Calculate the standard enthalpy change for the formation of ammonium chloride (NH4Cl) from its constituent elements, nitrogen (N2), hydrogen (H2), and chlorine (Cl2), given the following information:Reaction equation: N2(g) + 4H2(g) + 2Cl2(g) → 2NH4Cl(s)Standard enthalpy of formation for NH4Cl: -314.4

Answer 1

To calculate the standard enthalpy change for the formation of ammonium chloride  NH4Cl  from its constituent elements, we can use the following equation:H reaction  =  Hf products  -  Hf reactants In this case, the reactants are N2 g , H2 g , and Cl2 g , and the product is NH4Cl s . The standard enthalpy of formation for the reactants is 0 kJ/mol, and for NH4Cl, it is -314.4 kJ/mol.Since the reaction equation is: N2 g  + 4H2 g  + 2Cl2 g   2NH4Cl s , we need to multiply the standard enthalpy of formation for NH4Cl by 2 to account for the stoichiometry of the reaction.H reaction  = [2   -314.4 kJ/mol ] - [0 kJ/mol + 0 kJ/mol + 0 kJ/mol]H reaction  = -628.8 kJ/molThe standard enthalpy change for the formation of ammonium chloride  NH4Cl  from its constituent elements is -628.8 kJ/mol.