given the following data:
2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O (delta)G=-6399 kJ
C + O2 --> CO2 (delta)G=-394 kJ
H2 + 1/2 O2 --> H2O (delta)G=-237 kJ
calculate (delta)G for the following reaction:
6C + 3H2 --> C6H6
so I assume this has to do with Hess's law, which is simple, but what I'm wondering is do we multiply delta G as well when we multiply the entire reaction? I know that you have to switch the sign when reversing the reaction, but I'm not sure about the multiplication, because I thought I heard that you leave that alone when calculating gibbs free energy? am I mistaken? With multiplication and whatnot I derived 124.5 kJ
2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O (delta)G=-6399 kJ
C + O2 --> CO2 (delta)G=-394 kJ
H2 + 1/2 O2 --> H2O (delta)G=-237 kJ
calculate (delta)G for the following reaction:
6C + 3H2 --> C6H6
so I assume this has to do with Hess's law, which is simple, but what I'm wondering is do we multiply delta G as well when we multiply the entire reaction? I know that you have to switch the sign when reversing the reaction, but I'm not sure about the multiplication, because I thought I heard that you leave that alone when calculating gibbs free energy? am I mistaken? With multiplication and whatnot I derived 124.5 kJ
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Yes, you multiply the delta G just like you would a delta H.