1 / 17

Heat of Solution

Heat of Solution.

juancarlos-rodriguez
Télécharger la présentation

Heat of Solution

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Heat of Solution When a 3.88 g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 23.0°C to 18.4°C. Calculate ΔH in kJ per mole of NH4NO3 for the solution process. Assume the specific heat of the solution is the same as that of pure water. NH4NO3 (s)  NH4+(aq) + NO3-(aq) ΔH = ?

  2. Heat of Solution qliquid is the heat gained by the solution (water + NH4NO3) in the calorimeter: qliquid = C ΔT qliquid = (specific heat of solution)(mass of solution)(ΔT) = (4.18 J )(63.88 g solution)(-4.6 K) gK = -1228 J This is NOT the answer. The sp. heat of the solution is assumed to be the same as that of water.

  3. Heat of Solution -qrxn = qliquid qrxn = -(-1228) J = 1228 J qrxn = 1228 J = ΔH ΔH for a mol of NH4NO3 : 1228 J x 80.05 g = 25335 J 3.88g mol NH4NO3 mol NH4NO3 …still not the answer … for 3.88 g of NH4NO3 ΔH = 25 kJ/mol

  4. Heat of Solution ΔH = 25 kJ/mol thermochemical equation: NH4NO3(s)  NH4+(aq) + NO3-(aq) ΔH = 25 kJ Note that, in a thermochemical equation, the units are kJ and not kJ/mol. CRC Handbook value: 25.77 kJ. What is the % difference in our result? % dif = 100 x experimental value – accepted value accepted value

  5. Bomb Calorimetry for Heats of Combustion At constant volume, the heat flow for a reaction measured in a bomb calorimeter equals the change in internal energy for the reacting system: qrxn = ΔE The design of the calorimeter is such that no heat escapes the system. The heat generated by the reaction qrxn being studied will be absorbed solely by the calorimeter, the temperature of which we can measure. constant-volume bomb calorimeter

  6. Bomb Calorimetry qrxn = ΔE ΔE = q + w, but w = 0 because the bomb is constant volume. How is the heat flow related to the enthalpy, ΔH? ΔH = ΔE + Δ(PV) = qrxn + PΔV + VΔP but ΔV = 0 = qrxn + VΔP = qrxn + RTΔn (≈2.5 kJ/mol) qrxn is generally within 1 or 2% of ΔH For our purposes, in a bomb calorimeter, qrxn = ΔE ≈ ΔH constant-volume bomb calorimeter

  7. Bomb Calorimetry qrxn = -qbomb calorimeter = -Cbomb calorimeterΔT Example: A 0.5865 g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10°C to 24.95°C. Calculate the heat of combustion of lactic acid per gram and per mole. HC3H5O3(s) + 3O2(g)  3CO2(g) + 3H2O(l)

  8. Bomb Calorimetry qrxn = -qbomb calorimeter = -Cbomb calorimeterΔT The design of the bomb calorimeter is such that the components DO contribute to the heat capacity of the system. ΔrxnH≈ qrxn = -Cb.c.Δ T = - 4.812 kJ x 1.85 °C °C = - 8.902 kJ … for the combustion of 0.5865 g lactic acid

  9. Bomb Calorimetry qrxn = -qbomb calorimeter = -Cbomb calorimeterΔT ΔrxnH per gram lactic acid = -8.902 kJ 0.5865 g = -15.2 kJ/g ΔrxnH per mole lactic acid = -8.902 kJ x 90.08 g 0.5865 g mol = -1.37 x 103 kJ/mol

  10. Determining the Heat Capacity Ccal of a Bomb Calorimeter qrxn = -Cbomb calorimeterΔT The heat capacity Cbomb calorimeter must be determined in a separate experiment by combusting a known amount of a sample with a known heat of combustion. The measured temperature change will then give Cbomb calorimeter . The combustion of exactly 1 g of benzoic acid produces 26.38 kJ of heat. The combustion of 1.000 g benzoic acid in a calorimeter produces a ΔT of 4.857°C. Cbomb = 26.38 kJ/4.857°C = 5.431 kJ/°C.

  11. How to Write the Equation for a Combustion Reaction • One reactant is the chemical being combusted. You will either be given the formula for this chemical or you may be expected to know it (from, say, having used it in lab). • The second reactant is always oxygen, O2(g). • If the reactant contains C and H only (hydrocarbon) or C, H, and O, the products are CO2(g) and H2O(l). • Once you write the reactants and products, with their states, you may proceed to balance the equation.

  12. Bomb Calorimetry – Example 1 A 1.800 g sample of liquid phenol (C6H5OH) was combusted in a bomb calorimeter whose total heat capacity is 11.66 kJ/°C. The temperature of the calorimeter plus contents increased from 21.36°C to 26.37°C. a) Write the balanced chemical equation for the reaction. b) What is the heat of combustion per gram of phenol? Per mole of phenol? a) C6H5OH(l) + 7O2(g)  6CO2(g) + 3H2O(l)

  13. Bomb Calorimetry – Example 1 b) qrxn = -CbombΔT = -11.66 kJ x (26.37-21.36)°C °C = -11.66 x 5.01 = -58.417 kJ (per 1.800 g) ΔrxnH≈ qrxn = -58.417 kJ/1.800g = -32.4 kJ/g ΔrxnH ≈ qrxn = -32.45 kJ x 94.11 g g mol = -3.05 x 103 kJ/mol

  14. Bomb Calorimetry – Example 2 Under constant-volume conditions, the heat of combustion of glucose (C6H12O6) is -15.57 kJ/g. A 2.500 g sample of glucose is burned in a bomb calorimeter. The temperature of the calorimeter increased from 20.55°C to 23.25°C. a) What is the total heat capacity of the calorimeter? b) If the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been? a) qrxn = -15.57 kJ x 2.500 g = -38.925 kJ (keep extra SFs for now) g qrxn = -Cbomb calorimeterΔT or Cbomb calorimeter = -qrxn/ΔT Cbomb calorimeter= +38.925 kJ/2.70°C = 14.4 kJ/°C b) qrxn = -15.57 kJ x 5.000 g = -77.85 kJ g ΔT = -qrxn/Cbomb calorimeter= +77.85 kJ/(14.42 kJ/°C) = 5.40 °C

  15. Summary 1. Work and heat are ways to transfer energy. 2. First Law of Thermodynamics: ΔE = q + w 3. For chemical reactions, an often-encountered form of work is P-V work, from the production or consumption of gases. P-V work: w = -PΔV 4. Enthalpy: H = E + PV Change in enthalpy: ΔH = ΔE + Δ(PV) For constant-pressure processes, ΔH = ΔE + PΔV. For constant-pressure processes when only P-V work is performed, ΔH = qP .

  16. Summary 5. Calorimetry lets us measure temperatures, we then calculate ΔT, then heat flow, q (which is ΔH). a) constant pressure calorimetry (coffee-cup calorimetry) qrxn or process = ΔH We measure T in the calorimeter and calculate ΔT and qliquid: qrxn or process = -qliquid= -Cliquid ΔT Cliquid = heat capacity of solution in calorimeter b) constant-volume calorimetry (bomb calorimetry) qrxn = ΔE ≈ΔH We measure T in the calorimeter and calculate ΔT andqrxn: qrxn = -qbomb calorimeter = -Cbomb calorimeterΔT

  17. Summary 6. Because of its relationship to qrxn, knowing ΔH lets us determine whether a reaction is exothermic or endothermic and will eventually lead us to determine whether a reaction is spontaneous.

More Related