Answer :

Snor1ax

Answer: -1323 kJ mol^-1

Explanation:

[tex]\begin{aligned}&\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+3 \mathrm{O}_{2(g)} \longrightarrow 2 \mathrm{CO}_{2(g)}+2 \mathrm{H}_{2} \mathrm{O}_{(1)} \quad \Delta \mathrm{H}=? \\&\Delta \mathrm{H}=\left(\sum \Delta \mathrm{Hf} \text { products }\right)-\left(\sum \Delta \mathrm{Hf} \text { Reac tan ts }\right) \\&\Rightarrow[(2 \times-393.7)+(2 \times-241.8)]-[(52.3+0)] \\&\Rightarrow[-787.4 - 483.6]-52 \\&\Rightarrow-1323 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}[/tex]

pstnonsonjoku

The enthalpy of combustion for the burning of ethylene is -1323.33 kJ/mol

What is combustion?

The term combustion has to do with the burning of a particular substance in oxygen. In this case, the equation of the reaction is;

C2H4 + 3O2 ----->2CO2 + 2H2O

The enthalpy of combustion can be obtained from the relation;

ΔH rxn = [(2(-393.7) + 2(-241.8)] - [(52.3 + 3(0)]

ΔH rxn = (-787.4) + (-483.6) - 52.3

ΔH rxn = -1323.33 kJ/mol

Learn more about heat of combustion: https://brainly.com/question/10103458

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