Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: jaysup_2006 on April 07, 2007, 11:45:16 PM
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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.
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Yeah, I actually did everything right, and the answer is correct!!!!!!!!!!!! ;D
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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