How to Estimate the Standard Enthalpy of Formation of H2O at 100°C?

Click For Summary

Discussion Overview

The discussion revolves around estimating the standard enthalpy of formation of gaseous H2O at 100°C, starting from its known value at 298K. Participants explore the application of heat capacities and the necessary calculations involved in this estimation.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the known standard enthalpy of formation of H2O at 298K is -241.82 kJ mol-1 and seeks to estimate its value at 100°C using provided heat capacities.
  • Another participant suggests devising a pathway that involves transitioning from reactants at 100°C through the reaction at 25°C to the product at 100°C.
  • Several participants discuss the standard enthalpies of H2 and O2, questioning whether they are zero due to their definitions, and whether this affects their calculations.
  • There is a reiteration of the values for ΔH and heat capacities, with one participant expressing the need to ensure unit consistency in their calculations.
  • A participant mentions converting ΔH from kJ to J for consistency in units.
  • Another participant notes that if the value had been left in kJ, it would have led to a significantly different answer, implying that small temperature changes do not drastically affect heats of reaction.

Areas of Agreement / Disagreement

Participants generally agree on the values of standard enthalpies and heat capacities, but there is no consensus on the exact method for calculating the enthalpy at 100°C, as some steps and assumptions remain unclear.

Contextual Notes

Participants have not fully resolved the assumptions regarding the independence of heat capacities with temperature and the implications of using different units in calculations.

krootox217
Messages
51
Reaction score
2

Homework Statement



I have the following task:

The Standard enthalpy of formation of gaseous H2O at 298K is -241.82 kJ mol-1. Estimate its value at 100 °C given the following values of the molar heat capacities at constant pressure: H2O (g): 33.58 JK-1mol-1, H2 (g): 28.84 JK-1mol-1, O2 (g): 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature.

Homework Equations

The Attempt at a Solution



The only formula I can think of is this one:

HVSEQp6.png


But I don't know how to calculate this exactly. Can some help me getting started?

 
Physics news on Phys.org
First, you must devise a pathway to get from reactants at 100C, through the reaction at 25C, and finally to the product at 100C.
 
Ok, thx

is it correct that the standard enthalpie of H2 and O2 are 0 because of the definition?

Does this mean I that I can use my formula in this way?

e9af3459ba109af49490024c1ccdc164.gif
 
krootox217 said:
Ok, thx

is it correct that the standard enthalpie of H2 and O2 are 0 because of the definition?

Does this mean I that I can use my formula in this way?

e9af3459ba109af49490024c1ccdc164.gif
All correct.
 
  • Like
Likes   Reactions: krootox217
Therefore the ΔH(T1) is -241.82 kJ mol-1, the Cp(H2O) is 33.58 JK-1mol-1, the Cp(H2) is 28.84 JK-1mol-1, the Cp(O2) is 29.37 JK-1mol-1 and I just need to insert these values in the equation above and I get the result for ΔH at 373K and therefore my final result?
 
krootox217 said:
Therefore the ΔH(T1) is -241.82 kJ mol-1, the Cp(H2O) is 33.58 JK-1mol-1, the Cp(H2) is 28.84 JK-1mol-1, the Cp(O2) is 29.37 JK-1mol-1 and I just need to insert these values in the equation above and I get the result for ΔH at 373K and therefore my final result?
Check all your units for consistency.
 
I think everything is fine, I just need to transform ΔH(T1) from -241.82 kJ mol-1 to -241'820 Jmol-1 and then it should be consistent, I think
 
Go for it! (Note: If you had left it as kJ, you'd have gotten a much different answer. Experience will tell you that heats of reaction don't change drastically with moderate temperature change, like here, so a "wild" answer is a tipoff that something's wrong.)
 
  • Like
Likes   Reactions: krootox217
Thanks a lot!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K
Replies
6
Views
4K
  • · Replies 10 ·
Replies
10
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
3K
Replies
13
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K