Question

Jul 16 UI 23 Consider these reactions, where M represents a generic metal. 1. AH = -875.0 kJ 2M(s) + 6 HCl(aq) → 2 MCI, (aq) + 3H2(g) HCI(g) HCl(aq) 2. AH = -74.8 kJ 3. H2(g) + Cl2(g) → 2 HCl(g) AH; = -1845.0 kJ 4. MCI,(8) MCI, (aq) AH = -158.0 kJ Use the given information to determine the enthalpy of the reaction 2 M(s) + 3 C12(g) — 2 MCI,(8) AH = -6128.8

Answer 1

Hess's law is applied - Enthalky change for a reaction is same whether reaction completes in a step or multiple step. Enthalky is state function ů do is to depends only on intral & final Valne not in path later to get that final value » 2 Mt 6HCl (ag) 2 me (aq) +3428 sh=-875.0k5 Helly Hlapp ex Jl₂ = -74.845 Helpt ebay 2HCl (qj , 243 = -18450 6 M3 (3) MC cap al = -158.8kJ Desired - 2015) + 34118) 201₃ (5) reaction Cl, is on reactant side & 3 inmole So maithting multiply Ey (3 by 3 V 341gs + 3Cl y 6HCl (g) Al 3 = 34-1845 = -5535kJ I

Mullakty Ex to comel GH by 6 in Helysbriel (ov) Loth = 6X-74.0 :-448.8K5 * her is not in desired reacion) Multiply Ex 4 by 2 & revestila as melz (s) in on se product side & 2 in mole. 4. RMClz (og) - 20ebiss Alle + 158.0 kJ Add , 4 3 W 2 m/s) + Gretag mellan ol, = -87545 Grill (g) - 64 allen -448.8 kJ 3 + 34 g) – billigs, DH; : -5535 kJ 2 metz as - 2MU), Ally'= +158 ok] 2 M(S) + 3 Clig 20 (3) on- (-875)+(-448.8)+(-5535) + (+158 2 - 6,700.8 kJ 2-6700.8K5 . 4)