Question

Calculate ΔH° for the process Sb(s) + 52Cl₂(g) → SbCl₅(s) from the following information:

Sb(s) + 32Cl₂(g) → SbCl₃(s) ΔH° = −314 kJ
SbCl3(s) + Cl₂(g) → SbCl₅(s) ΔH° = −80 kJ

Answer 1

To find the enthalpy change (ΔH°) for the reaction Sb(s) + 5/2 Cl₂(g) → SbCl₅(s), add the enthalpy changes of the two provided reactions. The result is ΔH° = − 394 kJ for the overall process.

Step-by-step explanation:

To calculate the enthalpy change (ΔH°) for the process Sb(s) + 5/2 Cl₂(g) → SbCl₅(s), we can use the enthalpy changes from the given reactions:

• Sb(s) + 3/2 Cl₂(g) → SbCl₃(s), ΔH° = − 314 kJ
• SbCl₃(s) + Cl₂(g) → SbCl₅(s), ΔH° = − 80 kJ

By adding these two reactions, we can derive the desired reaction:

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

SbCl₃(s) + Cl₂(g) → SbCl₅(s) ΔH° = − 80 kJ

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Sb(s) + 5/2 Cl₂(g) → SbCl₅(s) ΔH° = − 394 kJ

The sum of the enthalpy changes for the two given reactions gives the enthalpy change for the overall process. Therefore, ΔH° for the reaction Sb(s) + 5/2 Cl₂(g) → SbCl₅(s) is − 394 kJ.