Question 12c) in 2000 section B about the question with Hess's Law

To get enthalpy change of the equation you first hav to manipulate the enthalpies of combustion of the C, H and the Hydrazine (equation given for that one).

For the equation for hydrogen I got: (the following is the original equation I keep mentioning)

2H2 + O2 --> 2H2O Enthalpy = -286

I need 3H2 instead of 2H2 so I multiply everything by 1.5 to get exactly the same equation shown in the answers and so I get an enthalpy change of (-286 x 1.5) -429. However, the anwers have it as -858. I understand that they have multiplied the original enthalpy by 3 to get -858, but this doesn't make sense to me as I thought you would have to multiply the original equation and enthalpy by 1.5 to get the desired equation.

I don't see where I have made the mistake but it could be that the original equation could be H2 + O2 --> H20 and that the 'answers' have multiplied everything by 3 but I don't think that is possible because the target equation has to be balanced (i.e it can't be 3H2 + 3O2 --> 3H20 and even the answers does not show this).

Could someone please help me. I understand if I have given a question that seems hard to follow but I just do not understand

YIC

## 2000 - Written Paper - Q12(c) If someone could help please :)

### Ally

Posted 27 May 2005 - 06:38 PM

QUOTE(YIC @ May 27 2005, 07:32 PM)

So I shouldn't balance equations at the start (before I start manipulating them?)

I think I understand now, thanks

I think I understand now, thanks

Nope.

For 1 mole of H

_{2}the balanced equation is:

H

_{2}+ 0.5O

_{2}---> H

_{2}O

For 2 moles of H

_{2}the balanced equation is:

2H

_{2}+ O

_{2}---> 2H

_{2}O

For 3 moles (the one in this question) the balanced equation is:

3H

_{2}+ 1.5O

_{2}---> 3H

_{2}O