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

  • Thread starter Thread starter Whalstib
  • Start date Start date
  • Tags Tags
    Law
Click For Summary

Discussion Overview

The discussion revolves around calculating the enthalpy of vaporization for water (H2O) using Hess's Law, based on given thermochemical equations. Participants explore the implications of balancing chemical equations and the relationship between enthalpy changes and the states of water.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Warren presents two thermochemical equations involving methane (CH4) and asks for help in finding the enthalpy of vaporization for H2O.
  • Some participants question whether all components of the equations should be halved, suggesting a potential misunderstanding of the balancing process.
  • Warren expresses confusion about the necessity of halving the coefficients for CH4 and CO2 in the equations provided by the textbook.
  • A participant suggests that the enthalpy of vaporization is defined for one mole of a substance and indicates that the states of H2O must be correctly identified as liquid and gas in the equations.
  • Warren speculates on the relationship between the enthalpy of vaporization and the number of moles, proposing a hypothetical scenario to illustrate his understanding.
  • Another participant asserts that the enthalpy of vaporization for water is a constant value of 44 kJ/mol, emphasizing that it cannot vary based on arbitrary numbers.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, with some expressing confusion about the balancing of equations and the implications for the enthalpy of vaporization. There are competing views on how to interpret the equations and the constants involved.

Contextual Notes

There are unresolved questions regarding the proper balancing of the equations and the interpretation of enthalpy changes, particularly in relation to the states of water. The discussion reflects varying levels of understanding of Hess's Law and thermochemical principles.

Whalstib
Messages
119
Reaction score
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
 
Physics news on Phys.org
Are you sure that everything wasn't suppose to be halved?
 
Nope! That's what the book says!

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

W
 
Whalstib said:

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
 

Attachments

  • scan0085.jpg
    scan0085.jpg
    12.7 KB · Views: 467
Last edited:
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
 
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
 
really...so despite any long winded problems 44kj would be the answer for the enthalpy of vaporization for H2O...Thanks...W
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
5K
Replies
1
Views
27K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
2
Views
20K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K