Answer:
The ΔH for the reaction is -2486 KJ
Explanation:
C2H4 (g) + 6 F2 (g) → 2 CF4 (g) + 4 HF (g) ΔH = ?
The enthalpy of formation of C2H4, CF4 and HF were given from the question which were +52.3 kJ, -680 kJ and -537 kJ respectively.
Using Hess's law, which states that 'the total enthalpy change in a chemical reaction is independent of the route provided the initial and final condition are the same'.
( Please find the attached document for the diagram of the enthalpy cycle)
52.3 + ΔH =(2 × -680) + (2× -537)
52.3 + ΔH = -2434
ΔH = -2434 - 52.3
ΔH = -2486.3 KJ
The enthalpy of formation of CF4 and HF were multiplied by 2 because they both had 2 moles from the balanced equation.
