Calculate the enthalpy of reaction for the combustion of ethene.

Question

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

Answer 1

Write and balance the equation. Then
dHrxn = (n*dHf products) - (n*dHf reactants)
Post your work if you get stuck.

Answer 2

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?

Answer 3

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

Answer 4

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.

Answer 5

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