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)

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.

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