# Simple Hess's Law Problem

1. Apr 6, 2010

### Whalstib

1. The problem statement, all variables and given/known data

Here's the formula I was given
CH4 (g) + 2O2 (g) >CO2(g) + 2H20 (l) : -803Kj
CH4 (g) + 2O2 (g) . CO2 (g) + 2H2O (g) : -891 Kj

Find enthalpy for vaporization of H2O

2. Relevant equations

3. The attempt at a solution
My soln
CO2(g) + 2H20 (l)>CH4 (g) + 2O2 (g) > : +803Kj
CH4 (g) + 2O2 (g)>CO2 (g) + 2H2O (g) : -891 Kj

BUT Book says:

1/2CO2(g) + 2H20 (l) > 1/2CH4 (g) + 2O2 (g) > : 1/2(+803Kj)
1/2CH4 (g) + 2O2 (g) > 1/2CO2 (g) + 2H2O (g) : 1/2(-891 Kj)

Why would the balanced eqn have to be be balanced by halving CO2's and CH4's? I'm stumped

Any help would be great!

Thanks
Warren
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 6, 2010

### sdoyle1

Are you sure that everything wasn't suppose to be halved?

3. Apr 6, 2010

### Whalstib

Nope! That's what the book says!

Makes no sense to me....but I'm new at this....

W

4. Apr 6, 2010

### sameh1

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5. Apr 6, 2010

### Whalstib

Oh!
So base it on an "empirical" formula and since we balanced the CH4 and all we deal with is H2O, 2H2O>2H2O becomes H2O>H2O

Would it then follow if say (number totally made up) 10H2O(l)>10H20(g) 1000KJ
The enthalpy of vaporization be: 100KJ? IE 1000/10? Since the "empirical" formula would be H2O>H2O? Regardless of the other reactants/products?

I think that's it as I scan the problem! Thanks!
Warren

6. Apr 7, 2010

### sameh1

let me tell you something The enthalpy of vaporization, (symbol ΔHvap), also known as the heat of vaporization or heat of evaporation, is the energy required to transform a given quantity of a substance into a gas it is constant for one substance for example water and it is 44 KJ/mol you cant have any number like 1000

7. Apr 7, 2010

### Whalstib

really...so despite any long winded problems 44kj would be the answer for the enthalpy of vaporization for H2O.....Thanks...W