Given the following information calculate the heat of formation of C2H4.

C2H4 + 3 O2 ¡æ 2 CO2 + 2 H2O ¥ÄH¡Æ = -414 kJ/mol
C + O2 ¡æ CO2 ¥ÄH¡Æ = -393.5 kJ/mol
H2 + ¨ö O2 ¡æ H2O ¥ÄH¡Æ = -241.8 kJ/mol

Refer to your other problem that Jai worked for you.

To calculate the heat of formation of C2H4 (ethylene), we need to use the given thermochemical equations and their enthalpy values.

The heat of formation (ΔHf) is the enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure.

We can start by writing the balanced equation for the formation of C2H4:

C2H4 + 3 O2 → 2 CO2 + 2 H2O

Now, we need to use the given thermochemical equations to determine the enthalpy change for this reaction.

First, we'll double the equation for the formation of CO2 because we need two moles of CO2 in the balanced equation:

2(C + O2 → CO2) ΔH = 2*(-393.5 kJ/mol) = -787 kJ/mol

Next, we'll multiply the equation for the formation of H2O by 2 because we need two moles of H2O in the balanced equation:

2(H2 + 0.5 O2 → H2O) ΔH = 2*(-241.8 kJ/mol) = -483.6 kJ/mol

Now, we can add up all the enthalpy changes:

2 CO2 + 2 H2O = -787 kJ/mol + (-483.6 kJ/mol) = -1270.6 kJ/mol

Finally, we can calculate the heat of formation of C2H4 by subtracting the enthalpy change for the balanced equation from the given enthalpy change:

ΔHf(C2H4) = ΔH(C2H4 + 3 O2 → 2 CO2 + 2 H2O) - ΔH(2 CO2 + 2 H2O)
= -414 kJ/mol - (-1270.6 kJ/mol)
= -414 kJ/mol + 1270.6 kJ/mol
= 856.6 kJ/mol

Therefore, the heat of formation of C2H4 is 856.6 kJ/mol.