Question

Calculate the reaction enthalpy for the formation 2 al(s) 3 cl2(g) −→ 2 alcl3(s), of anhydrous aluminum chloride using the data 2 al(s) 6 hcl(aq) −→ 2 alcl3(aq) 3 h2(g) ∆h◦ = −1049 kj hcl(g) −→ hcl(aq) ∆h◦ = −74.8 kj h2(g) cl2(g) −→ 2 hcl(g) ∆h◦ = −185 kj alcl3(s) −→ alcl3(aq) ∆h◦ = −323 kj

Answer 1

To determine the reaction enthalpy for the formation of anhydrous aluminum chloride, apply Hess's Law using the provided reactions and manipulate them to represent the formation of AlCl3(s) from Al(s) and Cl2(g). The provided reactions are combined with their enthalpy changes, paying attention to reverse and multiplication when necessary.

Step-by-step explanation:

To calculate the reaction enthalpy for the formation of anhydrous aluminum chloride (2Al(s) + 3Cl2(g) → 2AlCl3(s)), we use Hess's Law, which allows us to add or subtract given reactions to obtain the enthalpy for the target reaction.

The reactions provided are:

1. 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) ΔH° = −1049 kJ
2. HCl(g) → HCl(aq) ΔH° = −74.8 kJ
3. H2(g) + Cl2(g) → 2HCl(g) ΔH° = −185 kJ
4. AlCl3(s) → AlCl3(aq) ΔH° = −323 kJ

To find the enthalpy change of the desired reaction, we need to manipulate and combine these reactions.

• From Reaction (1), we know the enthalpy change for 2Al(s) + 6HCl(aq). We'll reverse Reaction (4) to make AlCl3(s) into AlCl3(aq) and add these two reactions, but we must change the sign of ΔH° for the reversed reaction.
• Next, we'll use Reaction (3) twice to account for the formation of HCl(g) from H2(g) and Cl2(g).
• Finally, we will subtract the ΔH° of Reaction (2) twice because we'll be going from HCl(aq) to HCl(g), instead of the provided direction.

By adding the enthalpies for these steps, we will find the ΔH° for the target reaction, which is the formation of AlCl3(s) from Al(s) and Cl2(g).