Simple Hess's Law Problem

  • Thread starter Whalstib
  • Start date
  • #1
119
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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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
23
0
Are you sure that everything wasn't suppose to be halved?
 
  • #3
119
0
Nope! That's what the book says!

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

W
 
  • #4
9
0

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
 

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  • #5
119
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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
9
0
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
119
0
really...so despite any long winded problems 44kj would be the answer for the enthalpy of vaporization for H2O.....Thanks...W
 

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