Question

Given the following data

2 CIF(g) + O₂ (g) → Cl₂O(g) + F₂O(g) ΔH= 167 kJ
2 ClF₃ (g) + 202 (g) → Cl₂O(g) +3F₂O(g) ΔH= 341.4kJ
2 F₂ (g) + O₂ (g) → 2F₂O(g).4 AH-34 ΔH=-43.4 kJ
calculate ΔH for the reaction
CIF(g)+F₂(g) →CIF₃ (9)
ΔΗ=

Answer 1

The enthalpy change ΔH for the reaction ClF(g) + F₂(g) → ClF₃(g) is calculated to be -104.5 kJ by using Hess's law to sum the enthalpy changes of other relevant reactions.

Step-by-step explanation:

To calculate the enthalpy change ΔH for the reaction ClF(g) + F₂(g) → ClF₃(g) using Hess's Law, we manipulate the given reactions so that their sum corresponds to the reaction of interest. We apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps into which the reaction can be divided.

The given reactions are:

• 2 ClF(g) + O₂ (g) → Cl₂O(g) + F₂O(g), ΔH= 167 kJ
• 2 ClF₃ (g) + 2O₂ (g) → Cl₂O(g) +3F₂O(g), ΔH= 341.4kJ
• 2 F₂ (g) + O₂ (g) → 2F₂O(g), ΔH= -43.4 kJ

By cancelling out identical species on opposite sides of the equations and altering coefficients appropriately, the reactions yield the desired equation:

• ClF(g) + F₂(g) → ClF₃ (g)

The sum of the modified enthalpy changes is:

• ΔH° = (+107.0 kJ) + (24.7 kJ) + ( – 236.2 kJ) = -104.5 kJ

Thus, the enthalpy change for the reaction ClF(g) + F₂(g) → ClF₃(g) is -104.5 kJ.