can anyone tell me how I calculate the enthalpy change of formation of ethene given that the enthalpies of combustion of ethane, carbon and hydrogen are -1411, 393 -286 kj/mol respectively
2007-04-28 08:19:29 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
1. write the equation 2C + 2H2 -----> C2H4 (formation of ethene)
2. add 3O2 to both sides: 3O2 + 2C + 2H2 ----> 3O2 + C2H4
3. what you now need to calculate is (2 x enthalpy of combustion of carbon) + (2 x enthalpy of combustion of hydrogen) - enthalpy of combustion of ethene. Take care with the signs.
2007-04-28 10:02:59 · answer #1 · answered by Gervald F 7 · 0⤊ 0⤋
Sorry, it's been a while since I've done this, so I forget the details, but the principle is as follows.
Start with the chemical equation, and work out how many carbons, oxygens and hydrogens you need to make the equation balance. Then it's a question of working out the difference in enthalpies between the forms. I seem to remember drawing the equations in a triangle formation with ethane and oxygen in one corner, ethene, oxygen and hydrogen in the second, hydrogen, oxygen and carbon in the third.
Sorry I can't be more useful than that.
2007-04-28 10:05:07 · answer #2 · answered by Anonymous · 0⤊ 0⤋