To calculate the standard enthalpy of formation of hydrogen fluoride HF , we need to manipulate the given reactions to obtain the target reaction: 1/2 H2 g + 1/2 F2 g HF g . First, let's divide the first reaction by 2 to get 1 mole of H2 g on the reactant side:1/2 2H2 g + O2 g 2H2O g H2 g + 1/2 O2 g H2O g H = -242 kJNow, let's divide the second reaction by 2 to get 1 mole of H2 g on the reactant side:1/2 H2 g + Cl2 g 2HCl g H2 g + 1/2 Cl2 g HCl g H = -92 kJWe can see that the third reaction is already in the desired form:1/2 F2 g + H2 g HF g H = -273 kJNow, we need to add the reactions in such a way that we obtain the target reaction. We can do this by adding the first and second reactions and then subtracting the third reaction: H2 g + 1/2 O2 g H2O g + H2 g + 1/2 Cl2 g HCl g - 1/2 F2 g + H2 g HF g This simplifies to:1/2 H2 g + 1/2 F2 g HF g Now, we need to add the enthalpy changes for these reactions:H = -242 kJ + -92 kJ - -273 kJ = -334 kJSo, the standard enthalpy of formation of hydrogen fluoride HF is -334 kJ/mol.