Calculate the enthalpy of reaction for the combustion of ethene.

Express the enthalpy of reaction calculated in question above as a molar enthalpy of reaction per mole of carbon dioxide.

To calculate the enthalpy of reaction for the combustion of ethene, we need to know the enthalpy values for the reactants and products involved in the reaction.

The balanced chemical equation for the combustion of ethene (C2H4) is:

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

In this reaction, one mole of ethene reacts with three moles of oxygen gas to produce two moles of carbon dioxide and two moles of water.

Now, to calculate the enthalpy of reaction, we need to use the enthalpy values for the reactants and products. The enthalpy change for the combustion reaction can be calculated using the bond enthalpies or the standard enthalpies of formation.

Using bond enthalpies:
The bond enthalpy is the energy required to break a particular bond. We can calculate the enthalpy change by subtracting the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products.

For the combustion of ethene, the bond enthalpy values are as follows:
C-C = 348 kJ/mol
C-H = 413 kJ/mol
O=O = 495 kJ/mol
C=O = 803 kJ/mol
O-H = 463 kJ/mol

Now, calculate the enthalpy change by subtracting the sum of bond enthalpies of reactants from the sum of bond enthalpies of products:
Enthalpy change = (2 × C=O bond enthalpy + 2 × O-H bond enthalpy) - (1 × C-C bond enthalpy + 4 × C-H bond enthalpy + 3 × O=O bond enthalpy)

Enthalpy change = (2 × 803 kJ/mol + 2 × 463 kJ/mol) - (1 × 348 kJ/mol + 4 × 413 kJ/mol + 3 × 495 kJ/mol)

Simplifying the equation gives:
Enthalpy change = 3826 kJ/mol - 3919 kJ/mol
Enthalpy change = -93 kJ/mol

Therefore, the enthalpy change for the combustion of ethene is -93 kJ/mol.

To express the enthalpy of reaction per mole of carbon dioxide, we know that for every 2 moles of CO2 produced, there is 1 mole of ethene burned. Thus, we need to divide the enthalpy change of -93 kJ/mol by 2.

Enthalpy of reaction per mole of CO2 = -93 kJ/mol ÷ 2
Enthalpy of reaction per mole of CO2 = -46.5 kJ/mol CO2

Therefore, the enthalpy of reaction for the combustion of ethene expressed as a molar enthalpy of reaction per mole of carbon dioxide is -46.5 kJ/mol CO2.

Write and balance the equation. Then

dHrxn = (n*dHf products) - (n*dHf reactants)
Post your work if you get stuck.

ok i got to

(2mol CO2 x -393.5kJ/i mol CO2 + 2mol H2O x -241.8kJ/ 1 mol H2O) - (1mol C2H6 x -94.0/ 1 mol C2H6 + 3mol O2 x 0kJ/ 1 mol O2)
= (-1270.6kJ0 - (-84.0)
= -1186.6kJ
is this right? and how do i do the second part about expressing the enthalpy of reaction calculated in question above as a molar enthalpy of reaction per mole of carbon dioxide?

how do i express the enthalpy of reaction calculated in question above as a molar enthalpy of reaction per mole of carbon dioxide?

Thanks for showing your work. You show C2H4 as 94 in the first step and 84.0 in the final calculation; however, you addition doesn't add up either so I don't know exactly what is missing. My text shows 84.86 for C2H4. The number you have calculated (when you correct it) is for the reaction. Since there are two mols CO2 produced in the reaction, that number divided by 2 will be per mol CO2.

I should point out that delta H per reaction is one question and per mol CO2 is a second question. You have run them together in the post.

im doing ethene so it is -84.0. Howe do i spilt the two up? the second question confuses me and im completely frazled about what to do.

C2H6(g) + 3O2(g) = 2CO2(g) + 2H2O(g)
=(2mol CO2 x -393.5kJ/1 mol CO2 + 2mol H2O x -241.8kJ/ 1 mol H2O) - (1mol C2H6 x -84.0/ 1 mol C2H6 + 3mol O2 x 0kJ/ 1 mol O2)
= (-1270.6kJ) - (-84.0kJ)
= -1186.6kJ