Dentista08

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I dont get it, if we are doing products-reactants shouldn't it be
(2(+283.3)- 2(-393.7)
Product CO produces so we reverse the sign?
Totally confused by this particular problem...
 

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Tommymkea

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You want to rearrange the two given equations so that they add up looking like the target one.

For the first one: 2*(C + O2 --> CO2) = 2C + 2(O2) --> 2(CO2)

For the second one (reverse and multiply by 2) you get:

2*(CO2 --> CO + 1/2(O2) = 2(CO2) --> 2CO + O2

Adding up the two equations gives you the target one. Same idea with the delta Hs. For the first one you multiplied everything by 2 (the delta H as well). For the second delta H you change its sign and multiply it by two just like you did in the second equation. Then you add the two up (just like to get to the taraget equation):

2(-393.7KJ) + (-(2(-283.3KJ))) = -220.8KJ

Hope it helps. Buona fortuna :luck:
 

Dentista08

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You want to rearrange the two given equations so that they add up looking like the target one.

For the first one: 2*(C + O2 --> CO2) = 2C + 2(O2) --> 2(CO2)

For the second one (reverse and multiply by 2) you get:

2*(CO2 --> CO + 1/2(O2) = 2(CO2) --> 2CO + O2

Adding up the two equations gives you the target one. Same idea with the delta Hs. For the first one you multiplied everything by 2 (the delta H as well). For the second delta H you change its sign and multiply it by two just like you did in the second equation. Then you add the two up (just like to get to the taraget equation):

2(-393.7KJ) + (-(2(-283.3KJ))) = -220.8KJ

Hope it helps. Buona fortuna :luck:
Grazie Mille!
I get that part now, but I guess I still don't understand why do you add the two together if its supposed to be products MINUS reactants. I'm tempted to subtract (2(+283.3)) from 2(-393.7)
 
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gentile1225

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damn you dentista! lol i was working it out while you answered, haha. good explanation.
 

zuma35

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You use products minus reactants when you are given the heats of formation for each individual product and reactant.
Here, you are given the delta H for two individual whole reactions. Hess's law is that if a reaction is done in steps, the total delta H is equal to the sum of the individual steps. That's why you add.
 

Dentista08

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You use products minus reactants when you are given the heats of formation for each individual product and reactant.
Here, you are given the delta H for two individual whole reactions. Hess's law is that if a reaction is done in steps, the total delta H is equal to the sum of the individual steps. That's why you add.
u made that very clear. thank you! I was confusing the 2
 
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