1 / 50

First exam

First exam. Wednesday October 16 th Exam is at the beginning of lecture Covers chapter 6 and chapter 7 sections 1-4, no lab material unless it is part of the chapters mentioned. Multiple choice, true or false and calculations questions.

tamira
Download Presentation

First exam

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. First exam • Wednesday October 16th • Exam is at the beginning of lecture • Covers chapter 6 and chapter 7 sections 1-4, no lab material unless it is part of the chapters mentioned. • Multiple choice, true or false and calculations questions. • Allowed is a cheat sheet, periodic table, calculator and the important tables.

  2. Determination of % NaHCO3 in Alka Seltzer Tablets Objectives: • To determine the amount of NaHCO3 in Alka Seltzer tablets by observing the amount of CO2 produced from the acid base reaction of HCO3- with acetic acid (in vinegar). • To practice stoichiometry. • To study the concept of limiting reactant.

  3. Determination of % NaHCO3 in Alka Seltzer Tablets • Alka Seltzer is an effervescent tablet that contains aspirin (acetylsalicylic acid), citric acid, and sodium bicarbonate (NaHCO3). As soon as the tablet dissolves in water, the NaHCO3 dissociates to form a bicarbonate ion (HCO3-) and a sodium ion (Na+) 1. NaHCO3(s)  Na+(aq) + HCO3-(aq) Acetic acid, which will ‘donate’ H+ to water to create H3O+(aq) is then added to the mixture. With the addition of acetic acid the following acid-base reaction then takes place: 2. HCO3-(aq) + H3O+(aq)  2H2O(l) + CO2 (g)

  4. Determination of % NaHCO3 in Alka Seltzer Tablets Weight of tablet is not • Alka Seltzer is an effervescent tablet that contains aspirin (acetylsalicylic acid), citric acid, and sodium bicarbonate (NaHCO3). • 1. NaHCO3(s)  Na+(aq) + HCO3-(aq) • 2. HCO3-(aq) + H3O+(aq)  2H2O(l) + CO2 (g) • Mass of CO2 (g) lost Use stoichiometry to go back and calculate %NaHCO3(s) in Alka Seltzer tablet

  5. Worksheet 2 • 1. Write the balanced equations for the combustion of the hydrocarbons (a) in the presences of excess oxygen (b) slightly limited supply of oxygen and (c) very limited supply of oxygen • C7H8. • a. C7H8(g) + 9O2(g)  7CO2(g) + 4H2O • b. 2C7H8(g) + 11O2(g)  14CO(g) + 8H2O • c. C7H8(g) + 2O2(g)  7C(s) + 4H2O •  C18H36. •  a. C18H36(g) + 27O2(g)  18CO2(g) + 18H2O •  b. C18H36(g) + 18O2(g)  18CO(g) + 18H2O • c. C18H36(g) + 9O2(g)  18C(s) + 18H2O

  6. Worksheet 2 2. Use the activity series to predict whether or not the following reaction occurs and if the reaction occurs write the product that forms. • 2Au3+(aq) + 3Ca(s)  • Au(s) + Ca2+(aq)  • Sn(s) + Na+(aq)  • Mn(s) + Co2+(aq)  • Cu(s) + H+(aq) 

  7. Learning check 2Au3+(aq) + 3Ca(s) Au(s) + Ca2+(aq) Sn(s) + Na+(aq) Mn(s) + Co2+(aq) Cu(s) + H+(aq) 2Au(s) + 3Ca2+(aq) rxn occurs NO reaction NO reaction Co(s) + Mn2+(aq) rxn occurs NO reaction

  8. Oxidizing & non-oxidizing acids

  9. Worksheet 2 • Write a balanced net ionic equation for the oxidation of metallic copper to copper (II) ion by hot, concentrated H2SO4 in which the sulfur is reduced to SO2 . • Hot concentrated sulfuric acid • 4H+ + SO42-(aq) + 2e- SO2 (g)+ 2H2O(l) • Cu(s) + 4H+ + SO42-(aq)Cu2+ (aq) +SO2 (g)+ 2H2O(l) • +4-2=+2 +2 • One copper, 4 hydrogen, one sulfur, 4 oxygen

  10. Worksheet 2 • Write a balanced net ionic equation for the oxidation of metallic copper to copper (II) ion by hot, concentrated H2SO4 in which the sulfur is reduced to SO2 . • Cu(s) + 2H2SO4(aq)CuSO4(aq) + SO2(g) + 2H2O(l) • Cu(s) + 4H+ + 2SO42-(aq)Cu2+ (aq) + SO42-(aq) + SO2 (g)+ 2H2O(l) • Cu(s) + 4H+ + SO42-(aq)Cu2+ (aq) +SO2 (g)+ 2H2O(l)

  11. Average class density You need the class data for the average density and for % calculations Part 1

  12. 7.5 Heat, work and the first law of thermodynamic • Heat of reaction is heat absorbed or released during chemical reaction • Determined by measuring temperature change they cause in surroundings • Measured using a calorimeter either under conditions of constant volume (using a rigid container that cannot change in volume) or at a constant pressure. Calorimeter • Instrument used to measure temperature changes

  13. heats of reaction • Calorimeter design depends on • Type of reaction • Precision desired • Usually measure heat of reaction under one of two sets of conditions • Constant volume, qV • Closed, rigid container • Constant pressure, qP • Open to atmosphere

  14. What is Pressure? Pressure is the amount of force acting on unit area Atmospheric Pressure • Pressure exerted by Earth’s atmosphere by virtue of its weight. • ~14.7 lb/in2 • Container open to atmosphere • Under constant P conditions • P ~ 14.7 lb/in2 ~ 1 atm ~ 1 bar

  15. Comparing qV and qP • Difference between qV and qP can be significant • Reactions involving large volume changes, • Consumption or production of gas • Consider gas phase reaction in cylinder immersed in bucket of water • Reaction vessel is cylinder topped by piston • Piston can be locked in place with pin • Cylinder immersed in insulated bucket containing weighed amount of water • Calorimeter consists of piston, cylinder, bucket, and water

  16. Comparing qV and qP • Heat capacity of calorimeter = 8.101 kJ/°C • Reaction run twice, identical amounts of reactants, once at constant volume and once at constant pressure Run 1: qV - Constant Volume • Pin locked; • ti = 24.00 C; tf = 28.91 C qCal= Ct = 8.101 J/C  (28.91 – 24.00)C = 39.8 kJ qV = – qCal = –39.8 kJ

  17. Comparing qV and qP Run 2: qP • Run at atmospheric pressure • Pin unlocked • ti = 27.32 C; tf = 31.54 C • Heat absorbed by calorimeter is qCal = Ct = 8.101 J/C  (31.54  27.32)C = 34.2 kJ qP = – qCal = –34.2 kJ

  18. Comparing qV and qP • qV = -39.8 kJ • qP = -34.2 kJ • System (reacting mixture) expands, pushes against atmosphere, does work • Uses up some energy that would otherwise be heat • Work = (–39.8 kJ) – (–34.2 kJ) = –5.6 kJ • Expansion work or pressure volume work • Minus sign means energy leaving system

  19. Work Convention Work = –P × V • P = opposing pressure against which piston pushes • V = change in volume of gas during expansion • V = Vfinal – Vinitial • For Expansion • Since Vfinal > Vinitial • V must be positive • So expansion work is negative • Work done by system

  20. Example Calculate the work associated with the expansion of a gas from 152.0 L to 189.0 L at a constant pressure of 17.0 atm. • 629 L atm • –629 L atm • –315 L atm • 171 L atm • 315 L atm Work = –P × V V = 189.0 L – 152.0 L w = –17.0 atm × 37.0 L

  21. Example A chemical reaction took place in a 6 liter cylindrical enclosure fitted with a piston. Over the course of the reaction, the system underwent a volume change from 0.400 liters to 3.20 liters. Which statement below is always true? • Work was performed on the system. • Work was performed by the system. • The internal energy of the system increased. • The internal energy of the system decreased. • The internal energy of the system remained unchanged.

  22. First law of thermodynamics • The first law of thermodynamics •  is a version of the law of conservations of energy, adapted for thermodynamic systems. • The law of conservation of energy states that energy can be transformed from one form to another, but cannot be created or destroyed. • First law of thermodynamic states that the change in the internal energy of a closed system is equal to the amount of heat (q) and the amount of work (w) that goes into the system.

  23. First law of thermodynamics • In an isolated system, the change in internal energy (E) is constant: E = Ef – Ei = 0 • Can’t measure internal energy of anything • Can measure changes in energy E is state function E = heat+ work E = q + w E = heat input + work input

  24. First law of thermodynamics • Energy of system may be transferred as heat or work, but not lost or gained. • If we monitor heat transfers (q) of all materials involved and all work processes, can predict that their sum will be zero • Some energy transfers will be positive, gain in energy • Some energy transfers will be negative, a loss in energy • By monitoring surroundings, we can predict what is happening to system

  25. Example Which of the following is not an expression for the First Law of Thermodynamics? • Energy is conserved • Energy is neither created nor destroyed • The energy of the universe is constant • Energy can be converted from work to heat • The energy of the universe is increasing

  26. First law of thermodynamics • E = q + w • Endothermic reaction • E = + • Exothermic reaction • E = –

  27. E is independent of path q and w • NOT path independent • NOT state functions • Depend on how change takes place

  28. Discharge of Car Battery Path a • Short out with wrench • All energy converted to heat, no work • E = q (w = 0) Path b • Run motor • Energy converted to work and little heat • E = w + q (w >> q) • E is same for each path • Partitioning between two paths differs

  29. Example A gas releases 3.0 J of heat and then performs 12.2 J of work. What is the change in internal energy of the gas? • –15.2 J • 15.2 J • –9.2 J • 9.2 J • 3.0 J E = q + w E = – 3.0 J + (–12.2 J)

  30. 7.6 Heats of reactions • Apparatus for measuring E in reactions at constant volume • Vessel in center with rigid walls • Heavily insulated vat • Water bath • No heat escapes • E = qv Bomb Calorimeter (Constant V)

  31. Heats of reactions • Heat of reaction at a constant volume (qv) equals E when the system does no pressure-volume work as in a bomb calorimeter. • E = qv pressure-volume work

  32. Calorimeter Problem When 1.000 g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from 22.000 ˚C to 26.049 ˚C. a) How many Calories are in olive oil, per gram? The heat capacity of the calorimeter is 9.032 kJ/˚C. t = 26.049 ˚ C – 22.000 ˚C = 4.049˚C qabsorbed by calorimeter = Ct = 9.032 kJ/°C × 4.049˚C = 36.57 kJ qreleased by oil = – qcalorimeter = – 36.57 kJ –8.740 Cal/g oil

  33. Calorimeter Problem (cont) Olive oil is almost pure glyceryl trioleate, C57H104O6. The equation for its combustion is C57H104O6(l) + 80O2(g)  57CO2(g) + 52H2O What is E for the combustion of one mole of glyceryl trioleate (MM = 885.4 g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate. E= qV = –3.238 × 104 kJ/mol oil

  34. Example A bomb calorimeter has a heat capacity of 2.47 kJ/K. When a 3.74×10–3 mol sample of ethylene was burned in this calorimeter, the temperature increased by 2.14 K. Calculate the energy of combustion for one mole of ethylene. • –5.29 kJ/mol • 5.29 kJ/mol • –148 kJ/mol • –1410 kJ/mol • 1410 kJ/mol qcal = Ct = 2.47 kJ/K × 2.14 K= 5.286 kJ qethylene = – qcal = – 5.286 kJ

  35. Enthalpy (H) • The internal energy of a system at a constant pressure is called the system’s enthalpy. Like internal energy, enthalpy is a state function. • By definition H = E + PV • So when P is constant the change in H can • H =E + PV • H = state function

  36. Enthalpy (H) • So when the only work a system can do is expansion work ( pressure-volume) against the apposing atmospheric pressure then H =E + PV = (qP+ w) + PV If only work is P–V work,w = – P V H = (qP+ w) – w = qP

  37. Enthalpy Change (H) H is a state function • H = Hfinal – Hinitial • H = Hproducts – Hreactants • Significance of sign of H Endothermic reaction • System absorbs energy from surroundings • H positive Exothermic reaction • System loses energy to surroundings • H negative

  38. Coffee Cup Calorimeter • Simple • Measures qP • Open to atmosphere • Constant P • Let heat be exchanged between reaction and water, and measure change in temperature • Very little heat lost • Calculate heat of reaction • qP= Ct

  39. Coffee Cup Calorimetry NaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH. The initial t = 25.5 °C and final t = 32.2 °C. What is H in kJ/mole of HCl? Assume for these solutions s = 4.184 J g–1°C–1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(aq) qabsorbed by solution= mass st massHCl = 50.0 mL  1.02 g/mL = 51.0 g massNaOH = 50.0 mL  1.04 g/mL = 52.0 g massfinal solution = 51.0 g + 52.0 g = 103.0 g t = (32.2 – 25.5) °C = 6.7 °C

  40. Coffee cup calorimetry qcal = 103.0 g  4.184 J g–1 °C–1  6.7 °C = 2890 J Rounds to qcal = 2.9  103 J = 2.9 kJ qrxn = –qcalorimeter = –2.9 kJ = 0.0500 mol HCl Heat evolved per mol HCl = = -58 kJ/mol

  41. 7.7 Thermochemical Equations • Focus on System • Endothermic • Reactants + heat  products • Exothermic • Reactants  products + heat • Want convenient way to use enthalpies to calculate reaction enthalpies • Need way to tabulate enthalpies of reactions

  42. The standard state • A standard state specifies all the necessary parameters to describe a system. Generally this includes the pressure, temperature, and amount and state of the substances involved. • Standard state in thermochemistry • Pressure = 1 atmosphere • Temperature = 25 °C = 298 K • Amount of substance = 1 mol (for formation reactions and phase transitions) • Amount of substance = moles in an equation (balanced with the smallest whole number coefficients)

  43. thermodynamic quantities • E and H are state functions and are also extensive properties • E and H are measurable changes but still extensive properties. • Often used where n is not standard, or specified • E ° and H ° are standard changes and intensive properties • Units of kJ /mol for formation reactions and phase changes (e.g. H°f or H°vap) • Units of kJ for balanced chemical equations(H°reaction)

  44. H in chemical reactions Standard Conditions for H 's • 25 °C and 1 atm and 1 mole Standard Heat of Reaction (H ° ) • Enthalpy change for reaction at 1 atm and 25 °C Example: N2(g) + 3H2(g)  2 NH3(g) 1.000 mol 3.000 mol 2.000 mol • When N2 and H2 react to form NH3 at 25 °C and 1 atm 92.38 kJ released • H= –92.38 kJ

  45. Thermochemical Equation • Write H immediately after equation N2(g) + 3H2(g) 2NH3(g)H = –92.38 kJ • Must give physical states of products and reactants • Hdifferent for different states CH4(g) + 2O2(g)  CO2(g) + 2H2O(l )H ° rxn = –890.5 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)H ° rxn= –802.3 kJ • Difference is equal to the energy to vaporize water

  46. Thermochemical Equation • Write H immediately after equation N2(g) + 3H2(g) 2NH3(g)H= –92.38 kJ • Assumes coefficients is the number of moles • 92.38 kJ released when 2 moles of NH3 formed • If 10 mole of NH3 formed 5N2(g) + 15H2(g) 10NH3(g)H= –461.9 kJ • H° = (5 × –92.38 kJ) = – 461.9 kJ • Can have fractional coefficients • Fraction of mole, NOT fraction of molecule ½N2(g) + 3/2H2(g) NH3(g)H°rxn= –46.19 kJ

  47. State Matters! C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g) ΔH °rxn= –2043 kJ C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l ) ΔH °rxn = –2219 kJ Note: there is difference in energy because states do not match If H2O(l )→ H2O(g) ΔH °vap = 44 kJ/mol 4H2O(l )→ 4H2O(g)ΔH °vap = 176 kJ/mol Or –2219 kJ + 176 kJ = –2043 kJ

  48. Example! Consider the following reaction: 2C2H2(g) + 5O2(g)→ 4CO2(g) + 2H2O(g) ΔE° = –2511 kJ The reactants (acetylene and oxygen) have 2511 kJ more energy than products. How many kJ are released for 1 mol C2H2? –1,256 kJ

  49. Given the equation below, how many kJ are required for 44 g CO2 (MM = 44.01 g/mol)? 6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g) ΔH˚reaction = 2816 kJ If 100. kJ are provided, what mass of CO2 can be converted to glucose? Example! 470 kJ 9.38 g

  50. Example! Based on the reaction CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g) H˚reaction = – 434 kJ/mol CH4 What energy change occurs when 1.2 moles of methane reacts? • – 3.6 × 102 kJ • +5.2 × 102 kJ • – 4.3 × 102 kJ • +3.6 × 102 kJ • – 5.2 × 102 kJ H = –434 kJ/mol × 1.2 mol H = –520.8 kJ = 5.2 × 102 kJ

More Related