Question

Calculate the enthalpy of formation of the anhydrous salt of aluminum bromide (AlBr 3) using the following data:

2Al (s) + 6HBr (aq) → 2AlBr3 (aq) + 3H2 (g) Δ? = -1061 kJ mol-1

HBr (g) → HBr (aq)∆? = -81.15 kJ mol-1

H 2 (g) + Br2 (l) → 2HBr (g) Δ? = -72.80 kJ mol-1

AlBr 3 (s) → AlBr 3 (aq)∆? = -368 kJ mol-1

Answer 1

ches) Enthalhy is state function so it e depends only on final and initial. state, it does not depends upon path. 30 2A1C5) + 6 HBO (aq) -> 2 AlBt; (aq) + 37241) AH = = 1061 kJ/mol here this eq. and an mutiply by 1/2 Alboka dl (s) + 3 HB3(19) — Ml Btz(ar) + 3 Hy(3) eq H= -1061 -530.5 kJ/mel. HBr (8) HBr(aq) DH= -81015 kJ/mol. multiply this eq. by 3 ( because we need to eleminate 3 HBr (aq) in eq (i)] 3 HBr (9) 34Br(99) Geacii) DU= 341-81015) =-243.45 kJ/mol H₂(g) + B&z (1) - 2 HB8 (9) AN=- 72.8 kJ/mol. nulltiply by this eq by 3 ( because we need eleminate 3 12 in eq ] 3.1518) + 3. Bryll) - 34B8(9) - erlin DH 2 34-72.b) = -109.2 KIT mal

Alboz (aq) AH = -368 kJ/mol Al B&3(5) Reverse this eq. dl Boz (29) — AL BOZ (5) SH=-(-368) = 368 kJ/mol. AH Step II find neededes, eq. de (s) + 3hBo/29) > AlBogust 3,12() -530.5 34 B8 (9) 34 B8 (69) 31469) + 3B02 (l) - 3 HBO(3) -109.2 - AlBoz (aq) ALBoz (5) 368 -243.45 -515015 du (5) + 3. Bd2(l) —— AU BOZ(3) anhy.formation eq Ml (5) + 3B02 (1) > Boz (3) hon = - 515015 KJ tad Thank you