Lecture 11/9/98, Chapter 6

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Calorimetry

  1. Dissolving 4.058 g of a certain salt in 50.00 g of water at 25.72 oC decreases the temperature to 23.58 oC in a calorimeter. The specific heat of the solution of the salt in the calorimeter is 33.910 J g-1 oC-1. (Question from Kask, U.; Rawn, D. General Chemistry (W.C. Brown, Dubuque, IA, 1993)
    1. Is the process exothermic or endothermic? Endothermic, temperature decreased.
    2. Assuming that no heat is gained or lost by the calorimeter walls, calculate the heat of solution of the salt in joules per gram of the salt.
      1. [delta] T = 2.14 oC
      2. Mass = 54.058 g
      3. Specific heat is
      4. Erxn = (33.91 J g-1 C-1) * (2.14 oC) * (54.058 g)
      5. E = 3923 J
      6. 4.058 g salt sample so 966.7 J/gram
  2. (a) Burning a 15.00 g sample of methane, CH4, in a bomb calorimeter increases the temperature in the calorimeter by 2.190 oC. The heat of combustion of methane is 890.3 kJ mol-1. (Question from Kask, U.; Rawn, D. General Chemistry (W.C. Brown, Dubuque, IA, 1993)
    1. Calculate the heat capacity of the calorimeter in kJ oC-1.
      1. 15.00 g methane = 0.937 moles
      2. [delta] Hcombustion = 890.3 kJ/mol
      3. [delta] H = 834.6 kJ
      4. [delta] T = 2.190 C
      5. heat capacity = 381.1 kJ/oC
    2. When 2.266 g of unknown compound is burned in the same calorimeter , the temperature of the calorimeter increases by 1.059 oC. Calculate the heat of combustion of the unknown compound in kJ g-1.
      1. [delta] T = 1.059 C so
      2. 403.6 kJ
      3. 2.266 g sample so:
      4. 178.1 kJ/gram of unknown
  3. Calculate the heat released (or absorbed) when 25.0 grams of propane (C3H8) reacts with 25 grams of oxygen gas. If this heat is used to change the temperature of 500 grams of water at 73.15 K, what is the final temperature of the water? Lecture Problem
  4. A 2.27 g sample of the explosive, nitroglycerin, C3H5(NO3)3, is allowed to explode in a bomb calorimeter at standard conditions. The amount of heat evolved is 15.1 kJ. The explosion of nitroglycerin occurs according to the equation. (Question from Kask, U.; Rawn, D. General Chemistry (W.C. Brown, Dubuque, IA, 1993)

    2 C3H5(NO3)3(l) -> 6 CO2(g) + 5 H2O(l) + 3 N2(g) + 1/2 O2(g)

    Using the necessary data from your textbook, calculate Hf for nitroglycerin.

    1. First explaination
      1. 15.1 kJ for 2.27 grams
      2. MW = 227 (so 2.27 grams is 0.0100 mole)
      3. For 2 moles of nitro, 3020 kJ would be evolved
      4. Make 6 moles of CO2 -2361 kJ
      5. Make 5 moles H2O -1429 kJ
      6. Make 3 moles N2 0
      7. Make 1/2 mole O2 0
      8. Total energy of formation -3790 kJ
      9. Since energy of reaction is -3020 kJ
      10. H formation for 2 moles nitro is -770 kJ
      11. So Hf = -385 kJ/mol for nitroglycerin
    2. Alternate Explination for problem: There are many different ways to solve this problem, and approaches to take. Let's start with a long and involved way that clearly shows how the steps fit together.

      First let's balance this equation. The stoichiometry here is a bit tricky. Here is one balanced equation: 4 C3H5(NO3)3(l) -> 12 CO2(g) + 10 H2O(l) + 6 N2(g) + 1 O2(g)

      The reaction of 2.27 g of nitroglycerin produced 15.1 kJ of energy, so:

      1. MW = 227 g/mole
      2. mass = 2.27 grams
      3. mole = 0.01 mole of nitroglycerin
      4. The energy released by the explosion of 1 mole of nitroglycerin would be 15.1 kJ/0.01 mole = 1510 kJ/mole.
      5. The energy released by the explosion of 4 moles of nitroglycerin (in the reaction as it is written here would be 1510 kJ/mole * 4 mole = 6040 kJ for the reaction as written.
      6. The energy for the reaction is released, so /\Hrxn for the explosion of 4 moles of nitroglycerin is -6040 kJ for the reaction as written.

      Now the question askes for Hf for nitroglycerin, the equation for that would be:
      3 C + 5/2 H2 + 3/2 N2 + 9/2 O2 -> C3H5(NO3)3(l)

      The fractions in the equation may be a bit bothersome, but remember Hf is the energy required to form 1 mole, so we need fractions for some of the elements in their natural states.

      This problem will require using Hess's law, that we can add up different steps to obtain the energy for a reaction. The pieces that we know are;

      1. The energy released by the explosion of 2.27 grams of nitroglycerin. From which we can calcualte the energy released when 1 mole (or 2 moles in the equation above) react.
      2. The heat of formation for many the products of the above equation (CO2(g), H2O(l), N2(g), and O2(g)) are all easily found in tables.
      3. The heat of formation for the products minus the heat of formation for the reactants is equal to the heat of the reaction. ie: Hrxn = Hproducts - Hreactants

      So we can now write out all the relevant equations:

      Reaction /\Hrxn Source
      2 C3H5(NO3)3(l) -> 6 CO2(g) + 5 H2O(l) + 3 N2(g) + 1/2 O2(g) -3020kJ Calorimetry experiment above
      C + O2 -> CO2 -393.509 kJ /\Hf table in textbook
      H2 + 1/2 O2 -> H2O(l) -285.83 kJ /\Hf table in textbook

      Now for the last step, we need to fit all these equations together. The "short" way of doing this is to recall that:
      /\Hrxn = /\Hproducts - /\Hreactants

      Since we know

      1. /\Hrxn = -6040 kJ
      2. /\Hproducts = 12*/\Hf CO2 + 10 * /\Hf H2O + 6 * /\Hf N2 + 1 * /\Hf O2
        /\Hproducts = 12 * (-393.509 kJ) + 10 * (-285.83 kJ) + 6 * 0 kJ + 1 * 0 kJ
        /\Hproducts = -7580.4 kJ
      3. /\Hreactants = /\Hproducts - /\Hrxn
        /\Hreactants = -7580.4 kJ - (-6040 kJ)
        /\Hreactants = -1540.4 kJ
      4. /\Hf nitroglycerin = 1/4 /\Hreactants
        /\Hf nitroglycerin = 1/4 * -1540 kJ
        /\Hf nitroglycerin = -385.1 kJ

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