The standard enthalpy change for the following reaction is -542 kJ: H₂(g) + F₂(g) $\rightarrow$ 2HF(g)
Calculate the standard enthalpy of formation of hydrogen fluoride.

Question

The standard enthalpy change for the following reaction is -542 kJ: H₂(g) + F₂(g) $\rightarrow$2HF(g) Calculate the standard enthalpy of formation of hydrogen fluoride.

Quizplus · Accepted Answer

By comparing the given reaction with the standard enthalpy of formation equations for HF and F2, we can write: H₂(g) + F₂(g) → ½F2(g) ∆H°f = -542 kJ/mol F2(g) → 2F(g) ∆H°f = +79 kJ/mol H2(g) + F2(g) → 2HF(g) ∆H°f = -272 kJ/mol From the above equations, we can calculate the ∆H°f of HF using Hess's law: ∆H°f of HF = [2(∆H°f of F) + ∆H°f of H2] - ∆H°f of F2 = [2(79 kJ/mol) + 0] - (-542 kJ/mol) = -271 kJ/mol Therefore, the answer is D) -271 kJ/mol.

The Standard Enthalpy Change for the Following Reaction Is -542 →\rightarrow→

The Standard Enthalpy Change for the Following Reaction Is -542 $\rightarrow$