# Problem: Calculate ΔH°298 for the process Sb(s) + 5/2Cl2(g) ⟶ SbCl5(s)from the following information:Sb(s) + 3/2Cl2(g) ⟶ SbCl3(s)             ΔH°298 = −314 kJSbCl3(s) + Cl2(g) ⟶ SbCl5(g)            ΔH°298 = −80 kJ

🤓 Based on our data, we think this question is relevant for Professor Gowda's class at WVU.

###### FREE Expert Solution

We’re being asked to determine the enthalpy change (ΔH°298) for the chemical reaction:

Sb(s) + 5/2Cl2(g) ⟶ SbCl5(s)

We can use Hess’s Law to determine the enthalpy change of the overall reaction from the given reactions:

1. Sb(s) + 3/2Cl2(g) ⟶ SbCl3(s)             ΔH°1 = −314 kJ

2. SbCl3(s) + Cl2(g) ⟶ SbCl5(g)             ΔH°2 = −80 kJ

We now need to find a combination of reactions that when added up, gives us the overall reaction. ###### Problem Details

Calculate ΔH°298 for the process Sb(s) + 5/2Cl2(g) ⟶ SbCl5(s)
from the following information:
Sb(s) + 3/2Cl2(g) ⟶ SbCl3(s)             ΔH°298 = −314 kJ
SbCl3(s) + Cl2(g) ⟶ SbCl5(g)            ΔH°298 = −80 kJ