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Enthalpy of neutralisation

  1. Feb 11, 2004 #1
    Hello all,

    I don't quite understand how to calculate the enthalpy of neutralisation. Example, if I add [itex]50cm^3[/itex] of a [itex]1.0 mol dm^-3[/itex] solution of HCL to a polystyrene cup. And then add [itex]50cm^3[/itex] of a [itex]1.1 mol dm^-3[/itex] of NaOH to the same cup, we'll get a reaction with a temperature rise. The temperature rise can be noted and a calculation can be made to find the enthalpy of neutralisation.

    So, assuming no heat is lost to the suroundings and the specific heat capacity of the solution is the same as that of water, then

    Heat absorbed by solution = m * c * t
    = 100 * 4.2 * 6.5
    = 2730J

    This is the part i don't understand, do I devide this figure with the moles of HCL in the solution or the moles of NaOH?

    Any help would be much appreciated. Thank you
  2. jcsd
  3. Feb 11, 2004 #2


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    Write out a balanced equation and find the limiting reagent from the info provided. Use this mol value.

    It does not matter the way you state it, just kJ/mol will do as long as you associated with the particular reaction.

    Hope this answers your question.
  4. Feb 18, 2004 #3
    calculate number of moles for HCL through given data .... ( u know molarity....find moles) and same for NaOH ....

    Since the first one is strong acid ..the other is strong base...so their Enthalpy Of Nuetralization will be 57.1 KJ/mol which is constant for strong acids and bases.

    calculate ....RISE IN TEMP ....cuz u got the net heat released...
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