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Chemistry Energy Problems

  1. Mar 3, 2005 #1
    Hello all, this question is really simple but I still seem to get it.

    A 50.0 mL sample of 0.500 mol/L [tex] MgSO_4 [/tex] solution at 24.4 degrees celsius is added to 50.0 mL of a 0.500 mol/L [tex] Ba(NO_3)_2 [/tex] solution at 24.4 degrees celsius. The temperature of the mixture rises to 26.3 degrees celsius. Calculate the heat of reaction.

    I tried using [tex] mc\triangle{t} + mc\triangle{t} [/tex] but there is no specific heat capacity for either of the compounds. I'm guessing the formula [tex] \triangle{H}=nH [/tex] fits in somewhere but I don't know where. Any hints or suggestions would be appreciated. Thanks.
  2. jcsd
  3. Mar 3, 2005 #2


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    The ΔH is determined from the WATER temperature change using {c = 4.184 J/goC} for water:
    {Water Mass} = (50 g) + (50 g) = (100 g)
    {Water ΔT} = (26.3) - (24.4) = (+1.9 oC)
    {Heat Energy Released} = (100)*(1.9)*(4.184) = (795 J) = (0.795 kJ)

    Heat of reaction is generally expressed relative to the primary reactant's mole number, so we use [tex] MgSO_4 [/tex]:
    {Moles [tex] MgSO_4 [/tex]} = (50.0 mL)*(0.500 mol/L) = (0.025 moles)

    {Heat of Reaction} = -(0.795 kJ)/(0.025 moles) = (-31.8 kJ/mole)

    (Note: Exothermic reactions are by convention expressed with (-)values.)

    Last edited: Mar 4, 2005
  4. Mar 4, 2005 #3
    Got it. Thanks for the help.
  5. Mar 4, 2005 #4


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    Homework Helper

    You may need to also employ [itex]q_{calorimter}[/itex]. If you were not required to do so, than just ignore.
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