I finished my working out for a question (question not needed):
1) 6C + 6O2 → 6CO2 ΔH = -2360.4
2) 3H2 + 3 ½ O2 → 3H2O ΔH = -857.4
3) C6H6 + 7 ½ O2 → 6CO2 + 3H2O ΔH = -3278
6C (s) + 3H2 (g) → C6H6 (l)
ΔH = (-2360.4) + (-857.4) + (-3278) = -6495.8 kJ
I finished and got the answer.
My question is:
Looking at my equations (1, 2, 3) why is oxygen not equal on both sides? Hydrogen and carbon are equal. I still left my answer as the starting equation: 6C (s) + 3H2 (g) → C6H6 (l).
I just don't understand why the oxygen in total is not equal on both sides.
Can I submit my answer like this? What is happening with the oxygen?
eqn 1 has 12 O atoms on each side
eqn 2 is strange. It evidently has 3 * .5O2 or in other words 3 O atoms on each side. Nutty
eqn 3 would be balanced if it meant 7.5 * 2 = 15 on the left
for eqn 2 try 6 H2 + 3 O2 ---> 6 H2O
for eqn 3
2C6H6 + 15 O2 → 12CO2 + 6H2O
What about:
1) 6C + 6O2 → 6CO2
2) 3H2 + 3 ½ O2 → 3H2O
3) 6CO2 + 3H2O → C6H6 + 7 ½ O2
Looking at my equations (1, 2, 3) why is oxygen not equal on both sides? Hydrogen and carbon are equal. I still left my answer as the starting equation: 6C (s) + 3H2 (g) → C6H6 (l).
I just don't understand why the oxygen in total is not equal on both sides.
I think you DO need the question to know how to answer it. If you're trying to use Hess' law to get 6C + 3H2 ==> C6H6 by adding the three equations as listed you're in for a long night. I didn't work it through but I don't think it will work without reversing at least one of the equations. One problem is that equation 2 is not balanced. 2) 3H2 + 3 ½ O2 → 3H2O ΔH = -857.4 You have 7 O atoms on the left and 3 on the right. It should be written as 3H2 + 3/2 O2 → 3H2O ΔH = -857.4
That equation comes from the original of 2H2 + O2 ==> 2H2O and you multiply that equation by 3/2 all the way through to arrive at 3H2 + 3/2 O2 ==> 3H2O.
Do you not need to multiply the whole equation by the same thing? With 3H2 + 3/2 O2 ==> 3H2O this was not done.
The question was:
Use the information provided in the below equation and Table, to show how the enthalpy of formation for benzene was determined using Hess’ Law. You should include an enthalpy cycle diagram and show your workings clearly.
6C(s) + 3H2(g) → C6H6(l) deltaHf = +60.2kJ mol-1
1) C + O2 → CO2 ΔHc = -393.4
2) H2 + ½ O2 → H2O ΔHc = -285.8
3) C6H6 + 7 ½ O2 → 6CO2 + 3H2O ΔHc = -3278
3H2 + 3/2 O2 ==> 3H2O
What was not done? I probably should not have meddled half way through the problem but if your talkig about the 2H2O + O2 ==> 2H2O I started with and I multiply thorugh by 3/2 I did multiply the whole thing by 3/2.
2H2O x 3/2 = 3H2O
O2 x 3/2 = 3/2 O2
2H2O x 3/2 = 3H2O
Or have I misunderstod you?
I wrote that I got the same equation and the same deltaHf as the question. My values did add up to this.
I understand the answer. You did do everything.
How would you get to that same answer from the question given?
The question was:
Use the information provided in the below equation and Table, to show how the enthalpy of formation for benzene was determined using Hess’ Law. You should include an enthalpy cycle diagram and show your workings clearly.
6C(s) + 3H2(g) → C6H6(l) deltaHf = +60.2kJ mol-1
1) C + O2 → CO2 ΔHc = -393.4
2) H2 + ½ O2 → H2O ΔHc = -285.8
3) C6H6 + 7 ½ O2 → 6CO2 + 3H2O ΔHc = -3278