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Heat Of Reaction

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known dataGiven that for the reaction KOH(aq) + HCL(aq) ----> KCL(aq) + H2O(l) H = - 54kjmol^-1.
    What is the quantity of heat involved in the reaction? 2NaOH(aq) + H2SO4(aq) ----> Na2SO4(aq) + 2H2O(l)?



    2. Relevant equations
    Heat of reaction Hrn =heat content of product Hp - heat content of reactant Hr. It can be written as Hrn= Hp-Hr
    3. The attempt at a solution
    Since Hrn=Hp-Hr. I don't know what to do again in other to get the value of Hp and Hr so that I can take the difference and Hrn will appear from no where.
     
    Last edited: Sep 12, 2012
  2. jcsd
  3. Sep 12, 2012 #2

    Borek

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    You don't need "heat content" of reactants/products.

    First things first - write net ionic reaction.
     
  4. Sep 12, 2012 #3

    AGNuke

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    This is a neutralization reaction between a strong acid and a strong base. Any neutralization reaction between a strong base and a strong acid yields the same heat of neutralization, -54 KJ/mol.

    To better understand this, as Borek said, try to write the net ionic equation for both the reaction. You will see the magic then.
     
  5. Sep 13, 2012 #4
    For the reaction KOH(aq) + HCL(aq) ----> KCL(aq) + H2O(l)
    Net ionic equation
    H++OH---->H2O
    For the reaction:
    2NaOH(aq) + H2SO4(aq) ----> Na2SO4(aq) + 2H2O(l)
    Net ionic equation
    2OH-+2H+
    ---->2H2O
    So what is the heat of reaction?
     
    Last edited: Sep 13, 2012
  6. Sep 13, 2012 #5

    Borek

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    If burning 1 candle produces x kJ, how many kJ will be produced when you burn 2 candles?
     
  7. Sep 13, 2012 #6
    If 1 mole of water produces -54kJ, then 2 moles of water will produce -108kJ. Could that be what you are tying to ask?
     
  8. Sep 13, 2012 #7

    Borek

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    That's not what I am trying to ask, that's what you are trying to understand :tongue2:

    It is not "moles of water produce 108kJ", it is "108kJ evolved while producing 1 mole of water in the neutralization reaction". But otherwise you are right. Note the sign!
     
  9. Sep 13, 2012 #8
    What does the sign have to the with the computation?
     
  10. Sep 13, 2012 #9

    Borek

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    Earlier you wrote:

    That's incorrect. If anything, when 1 mole of water is made in neutralization reaction, 54 kJ are produced. That means standard enthalpy of reaction is -54 kJ - because by convention enthalpy change for exothermic reaction is negative (lost by the system, not gained by the surroundings).

    When you write "-54 kJ is produced" it suggests system gained 54 kJ, and the reaction was endothermic.
     
  11. Sep 13, 2012 #10
    Why do I have to write the net ionic equation before solving?
     
  12. Sep 14, 2012 #11

    AGNuke

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    You have to understand first what is actually going on before trying to intercept the situation. You don't eat food without seeing what is served to you, do you?
     
  13. Sep 14, 2012 #12
     
  14. Sep 14, 2012 #13

    AGNuke

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    Same here. Try to actually see what reaction is going on. This case, just simple neutralization, formation of water by strong acid and strong base.

    Always try to write down the reaction when solving the problem. This makes the problem visually easy and enables you to solve it rather easily, now that you are able to see the reaction.
     
  15. Sep 14, 2012 #14
     
  16. Sep 15, 2012 #15
    Here are the possible answers which the authors of the question set for the problem A.-108kj B. -54kj C. -27kj D. +27kj E. +54kj.
    They went and choose C. = -27kj instead of A. = -108kj. I don't blame them as such, maybe that mistake is due to typo error. Don't you think the same?
     
  17. Sep 15, 2012 #16

    Borek

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    27 kJ (regardless of the sign) doesn't make any sense.
     
  18. Sep 15, 2012 #17
    I don't understand what you mean.
     
  19. Sep 15, 2012 #18

    Borek

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    Neither 27 kJ nor -27 kJ is a correct answer.
     
  20. Sep 15, 2012 #19
    Thanks for the asistance. I do appreciate!
     
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