Final answer:
The standard molar enthalpy of formation (ΔH°f) for HCl(g) is -93 kJ/mol, calculated by dividing the total enthalpy change for the reaction (-186 kJ) by two since two moles of HCl are formed.
Step-by-step explanation:
The value of ΔH° for the reaction H2(g) + Cl2(g) → 2HCl(g) is given as -186 kJ. Since one mole of hydrogen gas reacts with one mole of chlorine gas to produce two moles of hydrogen chloride (HCl), the standard molar enthalpy of formation of HCl(g), or ΔH°f for HCl(g) can be calculated per mole. The equation for the formation of HCl(g) clearly shows that two moles of product are formed, which means that the entire reaction enthalpy change should be divided by two in order to obtain the enthalpy of formation per mole of HCl(g).
Thus, if the total enthalpy change for the formation of two moles of HCl is -186 kJ, then the enthalpy of formation of HCl(g) per mole is:
ΔH°f[HCl(g)] = ΔH° / 2 = -186 kJ / 2 = -93 kJ/mol.