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Molar Heat of Neutralization

  1. Dec 31, 2013 #1
    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. jcsd
  3. Dec 31, 2013 #2

    Borek

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    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).
     
  4. Dec 31, 2013 #3
    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.
     
  5. Jan 1, 2014 #4

    Borek

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    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?
     
  6. Jan 6, 2014 #5
    Yes. Sorry about that, I meant to say solutions...

    Also, the exact problem statement added that the acid had been completely neutralized.
     
  7. Jan 6, 2014 #6

    Borek

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    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.
     
  8. Jan 6, 2014 #7
    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.
     
  9. Jan 7, 2014 #8

    Borek

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    Staff: Mentor

    Assume solutions were heated separately to the final temperature and mixed afterwards.
     
  10. Jan 7, 2014 #9
    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?
     
  11. Jan 8, 2014 #10

    Borek

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    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.
     
  12. Jan 9, 2014 #11
    Aha! It actually makes sense. Thank you very much.
     
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