Heat of rxn per mol H2O formed in acid-base reaction

Click For Summary

Discussion Overview

The discussion revolves around calculating the heat of reaction per mole of water formed in an acid-base reaction between NaOH and HCl. Participants explore the implications of mass and temperature changes in the solution, as well as the assumptions made regarding specific heat and density.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the heat of reaction using the formula ΔHrxn = -mcΔT, arriving at a value of -7.8 kJ based on the mass of the solution and temperature change.
  • Another participant suggests that the entire mass of the solution should be considered, not just the water, and notes that the density of the solutions is not exactly 1 g/mL.
  • A later reply questions the relevance of calculating heat per mole of water formed, indicating that while it is important, it does not affect the heat calculation directly in this context.
  • One participant confirms the heat calculation and suggests dividing the total heat by the moles of water produced to find the heat per mole, leading to a proposed value of -55 kJ/mol.
  • Another participant acknowledges that they have not verified the calculations but agrees with the approach taken by the previous participant.

Areas of Agreement / Disagreement

Participants express differing views on the importance of considering the total mass of the solution versus just the mass of water. There is no consensus on the best approach to calculating the heat of reaction per mole of water formed, as participants explore various assumptions and methods.

Contextual Notes

Participants note limitations regarding the assumptions of density and specific heat capacity, as well as the implications of using the total mass of the solution in calculations.

Nuchem
Messages
3
Reaction score
0

Homework Statement


Calculate the heat of reaction per mole of water formed in the acid-base reaction. Assume that the total mass of the solution is 150g.

75mL of 2.0M NaOH added to 75mL of 2.0M HCl in a styrofoam cup. Average initial temp = 295.65K. final temp = 308.8K

Homework Equations


ΔHrxn = qrxn = -mcΔT


The Attempt at a Solution


qrxn = -(141.234g)(4.186J/gK)(308.8K-295.65K) = -7.8kJ

based on: .15 mol of HCl (5.5g) and .15 mol NaOH (6.0g) reacted to form .15 mol H2O (2.7g) and .15 mol NaCl (8.8g) therefore since the entire reaction is aqueous and the total mass of the solution is 150g: 150g -8.8g (the amount of the product that is not water) = 141.2g

in a simpler train of thought i decided 2.70g of water were actually formed and therefore:
qrxn = -(2.70g)(4.186J/gK)(308.8K-295.65K) = -1.49*102J

I feel that i am grossly over thinking this, however, and I need a bit of advice
 
Physics news on Phys.org
Everything gets heated - so you can't ignore product and assume only water mass is important.

Simplest approach is to assume you have 150 mL of solution before and after the reaction and assume density and specific heat capacity of water.

That's not exactly true, but close enough. Note that density of neither solution was exactly 1 g/mL, so mass of the solution is not 150 g. But you had the same problem with your approach.
 
Alright so "per mole of water formed" isn't really important in the calculations? Assuming the specific heat and density of the products is equal to water (I am actually instructed to do this at a later point) and assuming the mass of the solution is 150g (also instructed to do so) you're saying qrxn = -(150g)(4.186J/gK)(308.8K-295.65K) = -8.3kJ per mole of water formed?
 
Nuchem said:
Alright so "per mole of water formed" isn't really important in the calculations??

It is important, just not when you calculate amount of heat that evolved in this particular experimental setup. And you calculated this amount of heat correctly - now just check how many moles of water were produced to be able to calculate heat per mole.
 
Oh! Okay that's a lot more clear now, thank you. So I would take the -8.3kJ and divide by moles of water (.15 seeing that HCl + NaOH → H2O + NaCl is 1:1:1:1) to get
-8.3kJ/.15mol = -55kJ/mol
 
I have not checked the numbers, just skimmed - and what you did looks OK now.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
24K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 7 ·
Replies
7
Views
24K
  • · Replies 6 ·
Replies
6
Views
21K