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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.
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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.
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.
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)
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
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
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)
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
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
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)
Average class density You need the class data for the average density and for % calculations Part 1
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
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
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
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
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= Ct = 8.101 J/C (28.91 – 24.00)C = 39.8 kJ qV = – qCal = –39.8 kJ
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 = Ct = 8.101 J/C (31.54 27.32)C = 34.2 kJ qP = – qCal = –34.2 kJ
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
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
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
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.
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.
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
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
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
First law of thermodynamics • E = q + w • Endothermic reaction • E = + • Exothermic reaction • E = –
E is independent of path q and w • NOT path independent • NOT state functions • Depend on how change takes place
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
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)
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)
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
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 = Ct = 9.032 kJ/°C × 4.049˚C = 36.57 kJ qreleased by oil = – qcalorimeter = – 36.57 kJ –8.740 Cal/g oil
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
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 = Ct = 2.47 kJ/K × 2.14 K= 5.286 kJ qethylene = – qcal = – 5.286 kJ
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 + PV • H = state function
Enthalpy (H) • So when the only work a system can do is expansion work ( pressure-volume) against the apposing atmospheric pressure then H =E + PV = (qP+ w) + PV If only work is P–V work,w = – P V H = (qP+ w) – w = qP
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
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= Ct
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 st 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
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
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
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)
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)
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
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
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
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
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
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
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