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  • The exact amount of the internal energy can not be determined. However, we can measure the change in internal energy of the system by measuring changes in temperature.
    If U is positive that means the U final is greater than U initial. If U is negative it is the reverse.
  • a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – Δu is negative; second condition – Δu is positive
    b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – Δu is positive; second condition – Δu is negative
  • Lecture2

    1. 1. Quicker question If you carried out an exothermic reaction in the laboratory, the Container would feel __________. (a) cool (b) warm Answer: b
    2. 2. Measurement of thermal energy gained or lost during a chemical reaction
    3. 3. A Coffee- Cup Calorimeter Made of Two Styrofoam Cups
    4. 4. A Bomb Calorimeter.
    5. 5. A Bomb Calorimeter
    6. 6. Assuming an exothermic reaction: qrxn = - qsolution qsolution = masssolution× specific heatsolution × T⊿ How can we know the end of the reaction? Measure the temperature change of the sample of solution experiences. In accurate work, the amount of thermal energy lost to the surrounding air and the amount of thermal energy absorbed by the coffee-cup calorimeter can’t be negligible.
    7. 7. When 1.095 g of NaOH are dissolved in 150.00 g of water in a coffee cup calorimeter, the temperatue changes from 23.50 to 25.32 , the heat℃ ℃ capacity of the calorimeter is 32.9 J· ℃-1 , determine the heat of solution for 1 mol of NaOH solid. SAMPLE PROBLEM 1
    8. 8. SAMPLE PROBLEM 2 Under constant-volume conditions, the heat of combustion of benzoic acid, HC7H5O2, is 26.38kJ/g. A 1.200 g sample of benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increases from 22.45 to 26.10 . What is the total℃ ℃ heat capacity of the calorimeter?
    9. 9. SOLUTIO N:
    10. 10. A 1.00M aqueous solution of NaOH, a 0.50M aqueous solution of H2SO4, and a coffee-cup calorimeter were allowed to stand at a room temperature of 25.4 until the temperature of all℃ three reached 25.4 .A 50.0 mL sample of the℃ 1.00M NaOH was then placed in the calorimeter, 50.0mL of H2SO4 was added as rapidly as possible, and the two solutions were mixed thoroughly. The temperature rose to 31.9 .℃ From this experiment, what is the heat of neutralization of one mole of H2SO4? For simplicity, assume that the densities of the NaOH and H2SO4 solutions were 1.00g/mL and that the specific heat of the solution after PRACTICE PROBLEM (PAGE30 2.3)
    11. 11. 2.3 Enthalpy Internal energy (U) : sum of all the kinetic and potential energies of the systems component parts includes the motion of the atoms, or molecules, electrons and nuclei. 2H2(g) + O2(g) ---> 2H2O(l) + heat ⊿U ⊿U ⊿U=Ufinal – Uinital =Uproducts-Ureactants
    12. 12. • Gas A2 reacts with gas B2 to form gas A2B2. The internal energy of A2B2 is much greater than that of either A2 or B2. Is the reaction for the formation of A2B2 exothermic or endothermic? – Exothermic – Endothermic Concept Check
    13. 13. Energy is transferred by adding or removing heat, or by doing or having work done on it. ⊿U= q + w •Endothermic Process: q is positive •Exothermic Process: q is negative •System does work on surroundings: w is negative •Surroundings do work on the system: w is positive Heat involves a transfer of energy Work – force acting over a distance
    14. 14. Work vs. Energy Flow 6.1
    15. 15. Concept Check Determine the sign of ∆U for each of the following with the listed conditions: a) An endothermic process that performs work. – work > heat – work < heat b) Work is done on a gas and the process is exothermic. – work > heat – work < heat 6.1
    16. 16. Question • For a particular process, q = –10 kJ and w = 25 kJ. Which of the following statements is true? – Heat flows from the surroundings to the system. – The surroundings do work on the system. ∆U = –35 kJ.
    17. 17. Answer •b)The surroundings do work on the system. •Because heat is negative, it flows from the system to the surroundings. The internal energy change ∆U = q + w = 15 kJ. Because work is positive, the surroundings do work on the system.
    18. 18. w is P V for expanding gases.⊿ w = -P V⊿ If in a chemical reaction the work done is accomplished by expansion ⊿U= q -P V⊿
    19. 19. qp = ∆U + p·∆V = ( Uf - Ui ) + p( Vf - Vi ) = ( Uf - Ui ) + (pfVf - piVi ) = ( U + p V ) - ( U + p V ) (1) at constant volume: ∆V = 0 ∆U = qV w = 0 w = -P V⊿ ∆U = q + w (2) at constant pressure: ⊿U= q -P V⊿ qp = U + P V⊿ ⊿ ∆H = qp = U+P V⊿ ⊿ H ≡ U + pV =Hf -Hi
    20. 20. enthalpy change ( H△ ) : The thermal energy gained or lost when a change takes place under constant pressure. △H =Hfinal - Hinitial △Hfus----the enthalpy change of fusion △Hvap----the enthalpy change of vaporization △H----the enthalpy change of chemical reaction
    21. 21. eg: H2O (s) = H2O(l) melting or fusion △Hfus----the enthalpy change of fusion
    22. 22. △Hvap----the enthalpy change of vaporization eg: H2O (l) = H2O(g) vaporization
    23. 23. reaction △H----the enthalpy change of chemical reaction
    24. 24. Thermochemical equation : The thermal energy can be shown as a production in a thermochemical equation. CH4(g)+2O2(g) CO2(g) +2H2O(l)+890.32kJ eg 1: One mole of methane gas reacts with two moles of oxygen gas. The products are one mole of carbon dioxide gas, two moles of liquid water, and 890.32 kJ of thermal energy
    25. 25. eg 2: CH4(g)+2O2(g) CO2(g) +2H2O(l) H = -⊿ 890.32kJ “-” : the reaction is exothermic. “+”: the reaction is endothermic. NOICE: When H is given for a reaction, the equation⊿ for the reaction must be interpreted in terms of moles!!!
    26. 26. Attentions! 1. Enthalpy is an extensive property . 2 H2 (g) + O2 (g) --->2 H2 O (l) H2 (g) + 1/2O2 (g) ---> H2 O (l) H = -286 kJ⊿ 2. The enthalpy change for a reaction is equal in magnitude and opposite in sign to H for the⊿ reverse reaction ⊿H = -572 kJ H2 O (l) ---> H2 (g) + 1/2O2 (g) ⊿H = +286 kJ
    27. 27. 3.The enthalpy of the reaction depends on the state of the reactants and products H2 (g) + 1/2O2 (g) ---> H2 O (l) H = -286 kJ⊿ H2 (g) + 1/2O2 (g) ---> H2 O (g) ⊿H = -47 kJ ⊿H = -239 kJ
    28. 28. SAMPLE PROBLEMFor the burning of methane gas. CH4(g)+2O2(g) CO2(g) +2H2O(l) H = -⊿ 890.32kJ How many kilojoules will be given off by the burning of 451 g of methane? (The formula mass of CH4 is 16.043u ) SOLUTIO N: Converting the quantity of methane to moles. mol2.28 mol/g0.16 g451 n )CH( 4 == kJ1051.2mol/kJ890mol2.28 4 ×=×
    29. 29. PRACTICE PROBLEMS For the burning of isooctane. 2(CH3)2CHCH2C(CH3)3(l)+25O2(g) 16CO2(g) +18H2O(l) H⊿ = -10930.9kJ How many kilojoules will be given off by the burning of 369 g of isooctane? (The formula mass of isooctaneis 114u ) SOLUTIO N: Converting the quantity of isooctane to moles. mol molg g n 24.3 /114 369 )isooctane( == kJ molkJ mol 4 1077.1) 2 /9.10930 (24.3 ×= − ×
    30. 30. ASSIGNMENT: Page 31 2.3 Page 33 2.5
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