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Heat of Reaction - Calculations Help

  1. Jan 2, 2005 #1

    I would really appreciate help on the calculations, because I am just not quiet getting it.

    I had 3 different reactions I had to do.

    First, some things to note:

    Since all the solutions are dilute, then the density of each soltuion can be assumed to be 1.00g/mL. So, 100mL has a mass of 100g.

    Reaction 1 was the dissolution of NaOH. So it was 200mL of water. The initial temp was taken, then 5.5g of NaOH was placed in, and a second temp was taken.

    Reaction 2 was a heat of reaction between aqueous NaOH and aqueous HCL. 100mL of 1.0mol/L of HCl in a vessel, and the ame of NaOH. Initial temps are taken of each substance. Then, they are mixed together, and the new temp. is recorded.

    Finally, Reaction 3 is the heat of reaction between solid NaOH and aqueous HCl. We have 200mL of 1.0mol/L of HCl in a vessel, we record the temp. Then we take 5.5g of NaOH and we pour it into the acid. We then take that new temp.

    Of course the equation that is used for Heat or Rection is deltaH = m * deltaT * C.

    The calculations are difficult because there is more then that calculation...

    Not, for Reaction 1, dont think, this is how I did it:

    deltaH = m * c * deltaT
    deltaH = 200g * 1cal/moldegcel * 0.00418kj/cal * (13degcel)
    deltaH = 10.868kJ

    since its 5.5g though:

    -10.868kj/0.1375 = -79.04kj/mol

    However, Reaction 2 is abit different, and I know its not the same way..and I dont know how to do it. Same with 3.

    Any help would be appreciated.
  2. jcsd
  3. Jan 3, 2005 #2
    anybody have any idea?
  4. Jan 3, 2005 #3
    For your second reaction, just do exactly what you're doing to figure out the delta H. When you get to the number of moles, just use the .1 L * 1 mol/L to figure out the number of moles for either HCl or NaOH. After that again do what you were doing by dividing the deltaH by the number of moles to figure out the enthalpy change per mole of HCl or NaOH.

    For your third reaction, again do exactly what you were doing in question 1 for the deltaH. After that take the deltaH you determined and divide it by the number of moles you determined for each reactant to figure out the enthalpy change per mole for HCl and NaOH.

    I'm pretty sure this is how you're suppose to figure out the third reaction, but I've only started this unit recently... I hope some one else can help out later on, to see if it was correct or not.
    Last edited: Jan 3, 2005
  5. Jan 3, 2005 #4
    Thanks apchemstudent!

    Here is reaction 1 and 2. I think using Hess's law you can figure out what the third reaction answer should be as well. (a way to check if its right) Can someone who knows show me that.


    deltaH = m * c * deltaT
    deltaH = 200g * 1cal/moldegcel * 0.00418kj/cal * (13degcel)
    deltaH = 10.868kJ

    since its 5.5g though:

    -10.868kj/0.1375 = -79.04kj/mol


    deltaH = m * c * deltaT
    deltaH = 200g * 1cal/moldegcel * 0.00418kJ/cal * (10degcel)
    deltaH = -8.36kJ


    BTW, this is what I did before for Reaction 2. I think you are right, but I figured a neutralization reaction should be lower than the dissolution of NaOH, but guess not.
  6. Jan 3, 2005 #5
    I have somewhat an understanding for Hess's Law:

    HCl + Na(+)(aq) + OH(-)(aq) -> NaCl + H2O delta H = ?

    NaOH(s) - > Na(+)(aq) + OH(-)(aq) delta H = ?


    HCl + NaOH(s) -> NaCl + H2O delta H = ?

    You have the molar enthalpy for the both of the reactions and you are given the number of moles already. So multiply the molar enthalpy changes respectively by their correct moles and simply add them, to obtain the correct reaction formula and enthalpy change. I guess for the third reaction you don't have to do figure out the molar enthalpy change... If so you already know how to do that.
  7. Jan 4, 2005 #6
    ok, but the way I am thinking, wouldn't calculation three look the same as one then?
    I have the 5.5g in Reaction 3, so I would be using the moles of NaOH (0.1375)?

    Hess's law would be deltaH1 + deltaH2 = deltaH3
  8. Jan 4, 2005 #7
    Calculation three does not look the same as 1 because it involves the heat released from the dissolution of NaOH(s) and as well it also involves the heat released from the neuatralization of HCl and NaOH(aq).

    I made a mistake in my explanation of reaction 3...

    Since NaOH is the limiting reagent, then that means .2 - .1375 moles of HCl remains unreacted. I think we're assuming the unreacted HCl does not release any heat.

    If we trace each reaction individually:
    Step 1: Solid NaOH is turned into ions which also releases heat.

    NaOH(s) -> Na(+)(aq) + OH(-)(aq) <--- you've determined the enthalpy change per mole of NaOH(s) so you can figure out the amount of heat released by this reaction.

    Step 2: The ions from the NaOH are reacted with HCl(aq) to release even more energy.

    Na(+)(aq) + OH(-)(aq) + HCl(aq) -> NaCl + H2O <---- again you've determined the enthalpy change per mole of the NaOH(aq), which was the same as the HCl(aq), but since the NaOH is the limiting reagent, we should use it instead.

    Now we have delta H1 + delta H2 = delta H3 to see if it was close to your experimental values.
  9. Jan 5, 2005 #8


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    If both reactants (NaOH and water) were at the same initial temperature, the heat of dissolution is simply

    [tex]Q_{diss} = m_{tot}C \Delta T [/tex]

    Divide this by the number of moles of NaOH (n=5.5/40) to get [itex] \Delta H_{diss} [/itex]

    Write the equation for the reaction between NaOH and HCl, and balance it. You'll find that 1 mole of NaOH reacts with 1 mole of HCl completely. Now since your experiment uses 0.1 moles (1 mol/L * 0.1 L) of both, they will both react fully. So, find Q as above, and divide by the number of moles (n=0.1) again, to find [itex] \Delta H_{reac} [/itex]

    Start with the balanced equation : it takes one mole of NaOH for 1 mole of HCl. Now clearly, there's 0.2 moles of HCl and 0.1375 (=5.5/40) moles of NaOH. So, only 0.1375 moles of HCl will react with it, the rest is unreacted.

    As always, calculate Q from the temperatures, and divide by the number of moles that have contributed to the reaction (ie: 0.1375). This gives you a number representing the sum [itex] \Delta H_{diss} + \Delta H_{reac} [/itex], since both processes are involved here. Within reasonable experimental errors, this number should be the sum of the above two.

    This is correct. Just note the small correction I've made (in bold) to the units.

    PS : I believe apchem has said the same thing.
    Last edited: Jan 5, 2005
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