[jaysup_2006]

Consider the following data.
2 C6H6(l) + 15 O2(g) ---> 12 CO2(g) + 6 H2O(l)  G° = -6399 kJ

C(s) + O2(g) -->  CO2(g)  G° = -394 kJ

H2(g) + 1/2 O2(g) ---> H2O(l)  G° = -237 kJ

Calculate G° for the following reaction.
6 C(s) + 3 H2(g) --> C6H6(l)

Im confused, I leave the first reaction alone, I then reverse the second reaction and multipy it by twelve to get:12CO2 --->12C +
12O2...therefore delta g reverses sign and 12 times before..4728

Then I combine the first and the second, to receive a simplified chemical equation; 2C6H6 + 302--->12C+6H2O

Then I reverse the third to get H2O(l) --->H2(g) + 1/2 O2(g), and I multipy it by 6, the delta g now becomes: 1422

So the two reactions I am left with are:  2C6H6 + 302--->12C+6H2O
                                                        6H2O(l) --->6H2(g) + 3O2(g)..canceling out the common things leave me with..

2C6H6 ---> 12C + 6H2 which has a delta g of -249...I then divided this by two, to get the actual final equation, so now the delta G becomes 124.5, I then again reverse the equation, to my final delta g is 124.5.

...I dont know if this is exactly right, so if someone could help me, thanks.

[jaysup_2006]

Yeah, I actually did everything right, and the answer is correct!!!!!!!!!!!!

[reyrey389]

yeah i just did it for a hw problem, its 124.5 kJ, the trick to these kind of problems is to first balance the substances that you see appearing in the final equation first and if that substance appears more than once in the set of equations then balance another one