Calculating Molar Heat of Neutralization

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Discussion Overview

The discussion revolves around calculating the molar heat of neutralization for nitric acid (HNO3) using experimental data. Participants explore the methodology for determining the heat released during the neutralization reaction, addressing the calculations involved and the assumptions made regarding temperature and reactant amounts.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the total heat released by nitric acid using the formula q=mcΔT, arriving at a value of -75 kJ/mol for the molar heat of neutralization.
  • Another participant questions the validity of the initial temperature readings, noting that the temperatures of the two solutions were different before mixing.
  • Some participants suggest that the amount of lithium hydroxide should also be considered in the calculations, raising the question of why only the amount of nitric acid was calculated.
  • There is a discussion about whether the equilibrium temperature of the solutions needs to be calculated, with some participants asserting that the final temperature is already provided.
  • Concerns are raised about the possibility of one reactant being limiting and whether all acid was neutralized, prompting a need to check this condition.
  • Participants discuss the assumption that the solutions were heated separately to the final temperature before mixing, and how this affects the calculations.
  • One participant expresses confusion about whether the solutions would heat up again during the neutralization reaction and seeks clarification on how to proceed with the calculations.
  • Another participant suggests calculating the heat absorbed by each solution separately and then combining these values to determine the total heat produced by the reaction.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions regarding the temperatures and amounts of reactants. There is no consensus on the correct approach to take, and multiple competing perspectives remain throughout the discussion.

Contextual Notes

Participants highlight the importance of considering the different initial temperatures of the solutions and the potential for limiting reactants, which may affect the calculations. The discussion also reflects uncertainty regarding the correct methodology for integrating these factors into the heat calculations.

PVnRT81
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Homework Statement



Calculate the molar neutralization heat of nitric acid (HNO3) using the following data:

Before Neutralization:

HNO3 (0.5 mol/L)
V = 200 ml
T= 23 Celsius

LiOH (1 mol/L)
V = 200 mL
T = 25 Celsius

After Neutralization:

V = 400 mL
T = 27.5 Celsius

2. The attempt at a solution

Calculating the total heat released by Nitric Acid:

q=mcDT
= (400)(4.184)(27.5-23)
= 7531.2 Joules

Calculating moles of Nitric Acid:

C= n/v
(0.5 mol/L) x (0.2) = 0.1 moles Nitric Acid

Calculating Molar Heat of Neutralization:

= (-7532.2 Joules / 1000) / (0.1 moles)

= -75 KJ / mol

Can someone please verify if this is done correctly? Thank you.
 
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PVnRT81 said:
q=mcDT
= (400)(4.184)(27.5-23)
= 7531.2 Joules

Doesn't look OK to me, as you wrote initial temperatures of both mixed solutions were different.

Why have you calculated amount of nitric acid and not of lithium hydroxide? (I am not saying it is wrong, just asking why this approach).
 
Borek said:
Doesn't look OK to me, as you wrote initial temperatures of both mixed solutions were different.

Why have you calculated amount of nitric acid and not of lithium hydroxide? (I am not saying it is wrong, just asking why this approach).

So, I would need to first calculate the equilibrium temp of both acids?


The question asks for the nitric acid, so that's what I tried doing.
 
PVnRT81 said:
So, I would need to first calculate the equilibrium temp of both acids?

I don't see two acids here. I see two separate solutions.

The question asks for the nitric acid, so that's what I tried doing.

So you missed the fact that one of the reactants can be limiting and you should check if all acid was neutralized?
 
Borek said:
I don't see two acids here. I see two separate solutions.



So you missed the fact that one of the reactants can be limiting and you should check if all acid was neutralized?

Yes. Sorry about that, I meant to say solutions...

Also, the exact problem statement added that the acid had been completely neutralized.
 
PVnRT81 said:
Also, the exact problem statement added that the acid had been completely neutralized.

OK, but I don't know exact problem statement, I know only what you posted.

So, I would need to first calculate the equilibrium temp of both acids?

Final temperature is given, so you don't need to calculate it. However, you need to take into account fact that for each part of the solution ΔT was different.
 
Borek said:
OK, but I don't know exact problem statement, I know only what you posted.



Final temperature is given, so you don't need to calculate it. However, you need to take into account fact that for each part of the solution ΔT was different.

Calculating the total heat released by Nitric Acid:

q=mcDT
= (400)(4.184)(27.5-23)
= 7531.2 Joules


From here on, I don't now exactly how I can integrate the fact that the two initial temperatures were different.
 
Assume solutions were heated separately to the final temperature and mixed afterwards.
 
Borek said:
Assume solutions were heated separately to the final temperature and mixed afterwards.

At the same temperature of 27.5 Celsius, they are mixed; ok, so what I fail to understand is that even after they are heated and mixed, wouldn't they heat up again when the neutralization reaction commences?

How would I proceed from here?
 
  • #10
No, they are not mixed at the final temp.

Calculate separately amount of heat absorbed by each solution - you know mass and ΔT of each, so you can do it easily.

If you add these two numbers, you will get amount of heat that was absorbed by both solutions together - and it equals amount of heat produced by the reaction.
 
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  • #11
Borek said:
No, they are not mixed at the final temp.

Calculate separately amount of heat absorbed by each solution - you know mass and ΔT of each, so you can do it easily.

If you add these two numbers, you will get amount of heat that was absorbed by both solutions together - and it equals amount of heat produced by the reaction.

Aha! It actually makes sense. Thank you very much.
 

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