Hess's Law Problem: Finding Enthalpy of Vaporization for H2O

[Whalstib]

Homework Statement

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

Homework Equations

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

Answer 88 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)

Answer 44Kj

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

 

[sdoyle1]

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

 

[Whalstib]

Nope! That's what the book says!

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

W

 

[sameh1]

Homework Statement

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

Homework Equations

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

Answer 88 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)

Answer 44Kj

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

I attached you an explnation to your problem every thing need to be halved because the
enthalpy for vaporization of H2O defined for 1 mol I believe that there is a proplem with your equations H2O in the first equation must be gas And H2O in the second equation must be liquid check out your proplem in http://en.wikipedia.org/wiki/Hess's_law

 

[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

 

[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 can't have any number like 1000

 

[Whalstib]

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