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.