Final answer:
By manipulating the given reactions and adding their enthalpies, we can determine the enthalpy change of the desired reaction: 2M(s) + 3Cl₂(g) ⟶ 2MCl₃(s). The enthalpy change for this reaction is -1922.6 kJ.
Step-by-step explanation:
The enthalpy change for the reaction 2M(s) + 3Cl₂(g) ⟶ 2MCl₃(s) can be calculated using Hess's Law and the given information. We need to use the enthalpies of the given reactions to determine the enthalpy change of the desired reaction. The enthalpy change of the desired reaction can be determined by adding the enthalpies of the given reactions according to their stoichiometric coefficients. From the given reactions, we have:
Reaction 1: 2M(s) + 6HCl(aq) ⟶ 2MCl₃(aq) + 3H₂(g) (ΔH1 = -575.0 kJ)
Reaction 2: HCl(g) ⟶ HCl(aq) (ΔH2 = -74.8 kJ)
Reaction 3: H₂(g) + Cl₂(g) ⟶ 2HCl(g) (ΔH3 = -1845.0 kJ)
Reaction 4: MCl₃(s) ⟶ MCl₃(aq) (ΔH4 = -222.0 kJ)
Step 1: Reverse Reaction 4:
MCl₃(aq) ⟶ MCl₃(s) (ΔH4 = 222.0 kJ)
Step 2: Multiply Reaction 1 by 2:
4M(s) + 12HCl(aq) ⟶ 4MCl₃(aq) + 6H₂(g) (ΔH' = -1150.0 kJ)
Step 3: Multiply Reaction 2 by 2:
2HCl(g) ⟶ 2HCl(aq) (ΔH'' = -149.6 kJ)
Step 4: Add the enthalpies of all the reactions:
ΔH desired reaction = ΔH' + ΔH'' + ΔH3 + ΔH4
ΔH desired reaction = -1150.0 kJ + (-149.6 kJ) + (-1845.0 kJ) + 222.0 kJ
ΔH desired reaction = -1922.6 kJ
Therefore, the enthalpy change for the reaction 2M(s) + 3Cl₂(g) ⟶ 2MCl₃(s) is -1922.6 kJ.