Question

From the enthalpies of reactionH2 (g) + F2 (g) → 2 HF (g) ΔH = -537 kJC (s) + 2 F2 (g) → CF4 (g) ΔH = -680 kJ2 C (s) + 2 H2 (g) → C2H4 (g) ΔH = +52.3 kJcalculate ΔH for the reaction of ethylene with F 2:C2H4 (g) + 6 F2 (g) → 2 CF4 (g) + 4 HF (g)

Answer 1

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.