To determine the standard enthalpy of formation for ammonium chloride NH4Cl , we need to manipulate the given reactions to obtain the desired reaction: NH3 g + HCl g NH4Cl s . First, we need to find the enthalpy change for the formation of HCl g . We can do this by manipulating the third reaction:3 N2 g + 2H2 g + 2Cl2 g 2NH4Cl s H = -634.2 kJ/molDivide the entire reaction by 2:1/2 N2 g + H2 g + Cl2 g NH4Cl s H = -317.1 kJ/molNow, subtract the first reaction from the new reaction: 1/2 N2 g + H2 g + Cl2 g NH4Cl s - N2 g + 3H2 g 2NH3 g This simplifies to:2H2 g + Cl2 g 2HCl g H = -271.1 kJ/molNow, divide the entire reaction by 2:H2 g + 1/2 Cl2 g HCl g H = -135.55 kJ/molNow we have the enthalpy change for the formation of HCl g . Next, we need to add the enthalpy change for the formation of NH3 g and HCl g to find the enthalpy change for the formation of NH4Cl s :1 N2 g + 3H2 g 2NH3 g H = -46.0 kJ/mol divide by 2 1/2 N2 g + 3/2 H2 g NH3 g H = -23.0 kJ/molAdd the two reactions:1/2 N2 g + 3/2 H2 g NH3 g H = -23.0 kJ/molH2 g + 1/2 Cl2 g HCl g H = -135.55 kJ/mol--------------------------------------------1/2 N2 g + 2H2 g + 1/2 Cl2 g NH3 g + HCl g H = -158.55 kJ/molNow, we can compare this reaction to the second given reaction:2 NH3 g + HCl g NH4Cl s H = -176.2 kJ/molSince the reactions are the same, the standard enthalpy of formation for ammonium chloride NH4Cl is:H NH4Cl = -176.2 kJ/mol