the standard entalpy of combustion of ethene gas C2h4 IS -1411.1 KJ/mol at SATP. Calculate Hf for ethene gas.

my work so far:
C2H4 +302--> 2C02 +2H20
H=[(2mol)(-241.8kj/mol)+(2)(-393.5)]-[x]
-1411.1kj/mol=-1270.6-x
x=-1270.6+1411.1

Ans: 52.5 kj which I can't get to

Hf.

To calculate the enthalpy of formation (Hf) for ethene gas (C2H4), you need to use the following equation:

C2H4 + 3O2 -> 2CO2 + 2H2O

Given that the standard enthalpy of combustion for ethene gas is -1411.1 kJ/mol, you can use it to solve for the enthalpy of formation.

First, calculate the enthalpy change for the combustion reaction:

ΔHcombustion = [(2 mol CO2)(-393.5 kJ/mol) + (2 mol H2O)(-241.8 kJ/mol)] - [-1411.1 kJ/mol]
ΔHcombustion = -787.0 kJ/mol - (-1411.1 kJ/mol)
ΔHcombustion = -787.0 kJ/mol + 1411.1 kJ/mol
ΔHcombustion = 624.1 kJ/mol

Now, since the formation reaction is the reverse of the combustion reaction, the enthalpy change for the formation of ethene gas will be the negative of the enthalpy change for combustion:

ΔHformation = -ΔHcombustion
ΔHformation = -624.1 kJ/mol

Therefore, the enthalpy of formation (Hf) for ethene gas is -624.1 kJ/mol.

To find the standard enthalpy of formation (ΔHf) for ethene gas (C2H4), you need to use the given information about the standard enthalpy of combustion and apply the concept of Hess's law.

Hess's law states that the change in enthalpy for a chemical reaction is the same, regardless of the number of steps taken to convert the reactants to products.

First, write the balanced chemical equation for the combustion of ethene:

C2H4 + 3O2 → 2CO2 + 2H2O

Based on the equation, you correctly determined that the standard enthalpy change of combustion is -1411.1 kJ/mol.

Now, you need to consider the reverse reaction of combustion, which is the formation of ethene. The reverse reaction will have the same magnitude of enthalpy change but with the opposite sign. Therefore, the standard enthalpy change of the formation of ethene (ΔHf) can be expressed as:

ΔHf (C2H4) = -ΔHc (combustion of C2H4)

Substituting the value you have:

ΔHf (C2H4) = -(-1411.1 kJ/mol)
ΔHf (C2H4) = 1411.1 kJ/mol

Therefore, the standard enthalpy of formation for ethene gas (C2H4) is 1411.1 kJ/mol.

It seems like in your attempt, you made an error in the calculation. Make sure to double-check your math and signs when subtracting the enthalpy values.