Calculating Molar Heat of Neutralization

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In summary: So, I need to calculate the amount of heat absorbed by the lithium hydroxide solution and then subtract it from the total heat released by the nitric acid solution in order to get the total heat released by the reaction?Yes, exactly.Aha! It actually makes sense. Thank you very much. So, I need to calculate the amount of heat absorbed by the lithium hydroxide solution and then subtract it from the total heat released by the nitric acid solution in order to get the total heat released by the reaction?Yes, exactly.In summary, using the provided data, the molar neutralization heat of nitric acid (HNO3) was calculated to be -75 kJ/mol. The initial temperatures of the
  • #1
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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  • #2
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).
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
Assume solutions were heated separately to the final temperature and mixed afterwards.
 
  • #9
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.
 

FAQ: Calculating Molar Heat of Neutralization

1. What is the molar heat of neutralization?

The molar heat of neutralization is the amount of heat released or absorbed when one mole of an acid and one mole of a base react to form water and a salt. It is a measure of the energy change during a neutralization reaction.

2. How is the molar heat of neutralization calculated?

The molar heat of neutralization is calculated by dividing the heat released or absorbed during a neutralization reaction by the moles of acid or base used. This gives the amount of heat released or absorbed per mole of reactant.

3. What factors affect the molar heat of neutralization?

The molar heat of neutralization is affected by the strength of the acid and base used, the concentration of the reactants, and the temperature at which the reaction takes place. Stronger acids and bases tend to have higher molar heats of neutralization compared to weaker ones.

4. Why is the molar heat of neutralization important in chemistry?

The molar heat of neutralization is important in chemistry because it helps determine the extent of a neutralization reaction and the energy changes that occur during the reaction. It also provides information about the strength of acids and bases and their ability to neutralize each other.

5. How is the molar heat of neutralization used in practical applications?

The molar heat of neutralization is used in practical applications such as in the design of chemical reactions, determining the concentration of acids and bases, and in the production of heat in exothermic reactions. It is also used in industries such as the pharmaceutical and food industries to control the pH of products and to ensure the safety and effectiveness of products.

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