Question

Given the following information:

	Li(s) → Li(g)	enthalpy of sublimation of Li(s) = 166 kJ/mol
	HF(g) → H(g) + F(g)	bond energy of HF = 565 kJ/mol
	Li(g) → Li+(g) + e–	ionization energy of Li(g) = 520. kJ/mol
	F(g) + e– → F–(g)	electron affinity of F(g) = -328 kJ/mol
	Li+(g) + F–(g) → LiF(s)	lattice energy of LiF(s) = -1047 kJ/mol
	H2(g) → 2H(g)	bond energy of H2 = 432 kJ/mol

Calculate the change in enthalpy for:

2Li(s) + 2HF(g) → H₂(g) + 2LiF(s)

-1064 kJ

-556 kJ

-680 kJ

184 kJ

None of these choices are correct.

Answer 1

2Li(s) + 2HF(g) -> 2LiF(s) + H₂(g)

Li(s) -> Li(g) sublimation = 2 * 166 kJ/mol = 332
Li(g) -> Li⁺(g) + e^- IELi = 2 * 520 kJ/mol == 1040

HF(g) -> H(g) + F(g) BEHF = 2 * 565 kJ/mol = 1130
H₂(g) -> 2H(g) BEH₂ = -1 * 432 kJ/mol == -432

F(g) +e^- -> F^-1 EAF = 2 * -328 kJ/mol = -656
Li⁺(g) + F^-(g) -> LiF(s) LE = 2 * -1047kJ/mol = -2094


Calculate the net change in energy for the reaction:

2Li(s) + 2HF(g) → H₂(g) + 2LiF(s)

= 332 + 1040 + 1130 +(-432) + (-656) + (-2094)

= - 680 kJ