# Molar Heat of Neutralization

1. Dec 31, 2013

### PVnRT81

1. The problem statement, all variables and given/known data

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.

2. Dec 31, 2013

### Staff: Mentor

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. Dec 31, 2013

### PVnRT81

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. Jan 1, 2014

### Staff: Mentor

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?

5. Jan 6, 2014

### PVnRT81

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

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

6. Jan 6, 2014

### Staff: Mentor

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.

7. Jan 6, 2014

### PVnRT81

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. Jan 7, 2014

### Staff: Mentor

Assume solutions were heated separately to the final temperature and mixed afterwards.

9. Jan 7, 2014

### PVnRT81

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. Jan 8, 2014

### Staff: Mentor

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.

11. Jan 9, 2014

### PVnRT81

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