Calculate the enthalpy of reaction for the combustion of ethene?

C2H4(g) + 3O2(g) => 2CO2(g) + 2H2O(g)

On the right hand side of the above expression we know the enthalpy of formation of the CO2 and the H2O from their atoms. It is
2(-393.5) + 2(-285.8) = - (787 + 571.6) = -1358.6 kJ. This means that in forming 2 moles of CO2(g) and two moles of H2O(l) from their atoms, 1358.6 kJ are released.

To get there from the left hand side of the equation, we must first form one mole of C2H4 from its atoms and then burn it with 3 moles of O2. ΔHc is the heat of combustion.

ΔHf[C2H4(g)] +ΔHc[C2H4(g)] = 2ΔH°f[CO2(g)]+2ΔH°f[H2O(l)]

ΔHf[C2H4(g)] -1411kJ = -1358.6 kJ.

ΔHf[C2H4(g)] = 52.4 kJ.....and since we were talking about one mole of ethylene, that's 52.4 kJ/mole, i.e., to form ethylene from its atoms we have to put energy INTO it

i saw this answer at yahoo. is it correct though??

Post Merge: 10 years ago

i saw this answer at yahoo. is it correct though??