AP Chemistry Lab Report help

Discussion in 'Community Discussion' started by fireshot91, Nov 27, 2010.

  1. macrumors 601


    Jul 31, 2008
    Northern VA
    So, I could really use some help in here. It's for my lab report, and I am utterly clueless.

    We had 3 reactions:
    NaOH + HCl -> H2O
    NaOH + NH4Cl -> NH3 + H2O
    HCl + NH3 -> NH4Cl

    We calculated maximum and minimum temperatures for each of the reactions in a calirometer.
    Reaction 1) Max = 34.24, Initial = 22.41
    Reaction 2) Max = 22.17 ,Initial = 21.89
    Reaction 3) Max= 34.32, Initial = 22.57


    Question 1:Calculate the amount of heat energy, q, produced in each reaction. Use 1.03g/mL for each density, use 4.18 J/(g*C) for specific heat of all solutions.

    Reaction 1) 1.03g/mL * 50mL = 51.5 g (11.83)(4.18) = 2550 J
    Reaction 2) 51.5g (-0.28)(4.18) = -60 J
    Reaction 3) 51.5g(11.75)(4.18) = 2530 J

    *Correct? Thoughts?*

    Question 2: Calculate the enthalpy change, ∆H, for each reaction in temrs of kJ/mol of each reactant
    Reaction 1) 2.55 kJ/.10 mol = 25.5 kJ/mol
    Reaction 2)-.060 kJ/.10 mol = .6 kJ/ mol
    Reaction 3) 2.53 kJ/.10 mol = 25.3 kJ/mol

    *Is that right?*

    Question 3: Use your answers from 2 above and Hess's law to determine the experimental molar enthalpy for Reaction 3.
    *No clue*

    Question 4: Use Hess's Law, and the accepted values of ∆H in the Pre-Lab exercise to calculate the ∆H for Reaction 3. How does the accepted value compare to your experimental value?
    * NO clue*
  2. macrumors 68040

    Designer Dale

    Mar 25, 2009
    Folding space
    You are in an Advanced Placement Chemistry class. Trust your instincts and see if they are right.

    Good luck

  3. macrumors 6502a


    May 7, 2006
    I won't spell out step-by-step how to do your homework for you, but I will point you in the right direction.

    You're on the right track with the first part; I'm not going to do the math but it should be fairly obvious if your answers look plausible.

    As for dH: all Hess' Law is: algebra. Given known enthalpy changes for given reactions, you can sum the dH for multiple steps of your net reaction to get your net dH. Be sure to flip the sign if the process is going in the opposite direction as your known reaction(eg, 1+2 --> 3 vs 3 --> 1+2 ); when changing the coefficients in your known reactions, be sure to multiply the dH by the same value you're multiplying the coefficiants. It should all cancel pretty easily/neatly; it's just a simple system of linear equations.

    This should get you on the right track; good luck.
  4. macrumors regular


    Oct 16, 2007
    Los Angeles
    Q= mass(g)*specific heat(J/(g*C))*(Tfinal-Tinital)

    That is the equation for the heat in the reaction. You used Tmax, I'm not quite sure what that is.

    You have a negative value for the 2nd reaction. Which indicates that heat leaves the system, but your values don't support that.

    Enthalpy change seems right as deltaH=Q

    Here's a tutorial for Hess's Law that just happens to have one of your reactions.
  5. thread starter macrumors 601


    Jul 31, 2008
    Northern VA
    Oh..I get number 3, I think.

    You flip reaction 2, and you add it to Reaction 1 to get Reaction 3....

    Sorry, I thought that you were supposed to add all three reactions (1,2,3) and then find the enthalpy change of a reaction that they give you. (Which equals to all three of the reactions being added somehow).
  6. thread starter macrumors 601


    Jul 31, 2008
    Northern VA
    We used a Vernier Temperature kit for our lab. Put Solution 1 into calirometer, put Temperature probe in, then added the other solution in. At the end, when the reaction was complete, we were told to go into the graph info in the program and record the Max and Min temperatures....so that's what I did .
  7. macrumors newbie

    Dec 2, 2012
    I am also in this class, and one thing you need to realize for determining heat evolved in each reaction is that there is 100ml used, not 50ml (unless you combined 25ml solutions. But based on your kj/mol, you got x/.1 thus .050L•2.0M=.1 mol
    Your steps look correct nonetheless, but your numbers are off.

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