Question

the value of δh° for the reaction below is -186 kj. h 2 (g) cl 2 (g) → 2hcl (g) the value of δh° f for hcl (g) is ________ kj/mol.

Answer 1

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.

Answer 2

The value of ∆H°f for HCl(g) based on the given reaction is -92.3 kJ/mol, which is the amount of energy released when one mole of HCl(g) is formed at standard state conditions.

Step-by-step explanation:

The student has asked for the standard molar enthalpy of formation (∆H°f) for HCl(g) based on a given reaction enthalpy. The reaction H₂(g) + Cl₂(g) → 2HCl(g) has a given standard enthalpy change (∆H°) of -186 kJ. For this reaction, which produces 2 moles of HCl, the ∆H°f per mole of HCl(g) can be calculated by dividing the total enthalpy change by the number of moles of HCl formed, resulting in -186 kJ ÷ 2 = -93 kJ/mol. This is close to the value provided in Appendix G of -92.307 kJ/mol for the standard molar enthalpy of formation of HCl(g).

Therefore, the value of ∆H°f for HCl(g) is -92.3 kJ/mol.