Verified answer

C+O2-->CO2 ΔH=-393 kj => x2

2C+2O2 --> 2CO2 ΔH1=-786 kj

2CO+O2-->2CO2 ΔH=-566 kj => reverse it

2CO2 --> 2CO+O2 ΔH2= +566 kj

add them =>

2C(g)+O2(g)--->CO(g) ΔH= ΔH1 + ΔH2 = -220 kj

have a good day ;-)

OK. Your equation is not balanced it should be: 2C + O2 -----> 2CO

With that as a start, here is what I do.

Write down the equation you want, but leave 4 or 5 lines of blank space above it, then draw a straight line over the top of the equation like this:

________________________

2C + O2 -----> 2CO

Now of the two equations you are given, which one has C on the left of the arrow? Yes, the 1st one, BUT it only has ONE C doesn't it and your given equation has 2C. Well to fix this just multiply the C + O2 ----> CO2 by 2 IINCLUDING THE -393!!

So you get 2C + 2O2 -----> 2CO2 -786. Now write that just above the line like this.

2C + 2O2 -----> 2CO2 -786

___________________

2C + O2 -----> 2CO

OK, that takes care of the C's. No the CO's. Notice that in your equation the CO is on the right and there are 2COs. Use the second equation to fix that problem. However, the second equation has the 2CO on the left of the arrow, and you want it on the RIGHT of the arrow. How to fix this, easy, simpley FLIP the equation arround to get 2CO2 -----> 2CO + O2, also CHANGE THE SIGN ON THE ENERGY FROM - TO +

Right that equation above the others

2CO2 -----> 2CO + O2, +566

2C + 2O2 -----> 2CO2 -786

___________________

2C + O2 -----> 2CO

Now you can cross off anything on the left side of the arrows from anything on the right side of the arrows. So you can cross off the 2CO2s, and the ONE O2 from the right on the top equation, but leave ONE O2 on the second equation instead of 2.

What you should have left is 2C and an O2 on the left and just 2CO on the right side.

just add +566 - 786 to find the energy.

Hope you can follow that.