1. C2H4(g)+3O2(g)-->
2CO2(g)+2H2O(l) deltaH= -1411kJ
2. C(graphite)+3H2(g)+(1/2)O2(g)-->
C2H5O2 deltaH= -278 kJ
3. C2H4(g)+H2O(l)-->
C2H5OH(l) deltaH= -44 kJ
final: C2H5OH(l)+3O2(g)-->
3H2O(l)+2CO2(g)
I can't figure out how to cancel the second equation because there is a single C and H2 there.Can anyone explains please?step by step solution please.Thank you.
2006-10-14
16:14:48
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5 answers
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asked by
ichigokun
1
in
Science & Mathematics
➔ Chemistry
I know it is little weird ..but the product of the equation is under the reactant..and i do that because it seems like some parts of the equation is not seen if i type the whole thing on the same line or something..please help.thank you
2006-10-14
16:16:55 ·
update #1
Recorrect equation #2
C(graphite)+3H2(g)+(1/2)O2(g)
-->C2H5OH(l) deltaH= -278kJ
this is the correct one
2006-10-14
16:22:20 ·
update #2
Hey..so is that mean I can just ignore equation #2?I don't have to use all the equation in this case?can anyone explains more please?=)
2006-10-14
16:26:06 ·
update #3