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Transcript
Chapter Six Energy Relationships in Chemical Reactions 1 Energy (U): Capacity to Do Work Radiant energy Energy from the sun Nuclear energy Energy stored in the nucleus of an atom Thermal energy Energy associated with temperature Kinetic energy: molecular movement Chemical energy Energy stored in chemical bonds Potential energy: Object postion 2 Thermochemistry Study of heat change in chemical reactions System: The part of the universe being studied Open: Energy & matter exchange with surrounding Closed: Only energy exchange with surroundings Isolated: No energy or matter exchange- rare Surroundings: Part of the universe not being studied 3 Heat Transfer in a System Heat, q: Transfer of energy due to temperature difference Exothermic Reaction: System gives off heat (Exiting) Ex: Methane Burning: CH4 + O2 → CO2 + H2O Bonds stronger in CO2 and H2O molecules than in CH4 + O2 Heat goes from system to surroundings Endothermic Reaction: System gains heat (Entering) Ex: Ice Melting: H2O (s) → H2O (l) Energy needed to break attractions between H2O molecules Heat goes from surroundings into the system 4 Thermodynamics Study of the conversions between heat & energy 5 Using Thermodynamics State of a system: Describes specific conditions defined by state functions Measurement of these properties Potential energy of ball at the top of the mountain all State functions: Properties defined by final - initial values only Composition, temperature, pressure, energy, volume Height of mountain Path functions: final - initial values vary by process Heat, work: interdependent variables # steps taken, Time to top energy per step, etc. 6 First Law of Thermodynamics Energy can be converted from 1 form to another but cannot be created or destroyed Usystem= Ufinal-Uinitial Internal Energy of System: Usystem = Ukinetic+ Upotential Kinetic energy: Amount of molecular motion Associated with temperature Potential energy: Energy stored in bonds Strength of chemical bonds Cannot separate two types of energy For chemical reactions: Ureaction = Uproducts- Ureactants 7 Work and Heat Energy transfer is caused by changes in heat & work in a system Usystem = q + w Work (w) = force x distance = -PV Negative work (-w): System loses energy to surroundings Positive work (+w): System gains energy from surroundings Work (w) = -PV V = Vf -Vi For this situation: +V -w 8 Work and Heat as Path Functions A gas expands in volume from 2L to 4L at STP. 350J of energy are needed to maintain the new conditions. Usys = Ufinal- Uinitial =350J = q + w V = 2L P = 1atm T= 273K a. Calculate work done against a vacuum & a pressure of 1atm. w = -PV = -(0atm x 2L) = 0atm L = 0J w = -PV = -(1atm x 2L) = -2atm L = -202.6J (1Latm= 101.32J) b. Calculate the heat required for each system. q = Usys - w Under a vacuum: q = 350J-0J = 350J (All energy is heat) Under 2.0atm: q = 350J-(-202.6J) = 552.6J (need extra heat!) 9 Enthalpy of Chemical Reactions 10 Pressure/Volume Relationships with U For systems at constant pressure (usually ~1atm): U = q + w = qp - PV Some work used to expand system against surroundings. Enthalpy: (H) Experimentally measured and tabulated At constant pressure: Assume negligible volume change: H= U + PV H= U = qp There will be slight differences between H and U 11 Real Differences between U and H H measured experimentally as heat 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) H=-367.5kJ/mol Volume change of H2 gas V=nRT/P T= 298K, P=1.0atm, n=1.0mol H2 V= 24.5L U = H- PV H = -367.5kJ PV = 1.0atm x 24.5L = 24.5Latm x 0.101kJ/Latm = 2.5kJ U = -367.5kJ -2.5kJ= 370kJ Difference between H and U is only 2.5kJ ~0.7% difference considered negligible 12 Enthalpy of Reaction (H = qrxn) Amount of heat exchanged between system and surroundings during a chemical reaction Hreaction = Hproducts- Hreactants Exothermic Reactions Temperature increases in an isolated system. Heat released to surroundings is treated as a product 2SO2 (g) + O2(g) 2SO3 (g) + 197.8kJ H = -197.8kJ Endothermic Reactions Temperature decreases in an isolated system. Heat absorbed by system is treated as a reactant 197.8kJ + 2SO32SO2 (g) + O2(g) State function: Independent of path H = +197.8kJ 13 Rules for Working with Heat and Enthalpy 1. Use same multiplier as used in chemical reaction 2SO2 (g) + O2(g) 2SO3 (g) H = -197.8kJ Decrease or increase reactants and products, do the same for H 1SO2 (g) + 1/2O2(g) 1SO3 (g) H = -197.8kJ/2= -98.9kJ Note: Write H with units that match mole ratio in equation H = -98.9 kJ /mol SO3(g) 2. Reverse the reaction, reverse sign of H 2SO3 2SO2 (g) + O2(g) H = +197.8kJ Use rules to calculate the H for any amount of product or reactant 14 Calculate how much heat is required to decompose15.0g NO2(g) according to this reaction: 2NO(g) + O2(g) 2NO2(g) H = -114 kJ Heat required for following reaction: H = +114 kJ 2NO2(g) 2NO(g) + O2(g) H = +114 kJ is needed to decompose of 2 moles NO2 Calculate molar mass of NO2 to convert mass to moles of NO2. moles NO2 15.0 g NO2 1 x 1mol NO2 46.0 g NO2 0.326mol NO2 Find the amount of heat for the amount of NO2 actually present. H rxn 114kJ 0.326mol NO2 x 18.6kJ 2mol NO2 1 15 Calorimetry 16 Calorimetry Calorimetry: Measurement of heat changes Calorimeter: Device to measure heat (T) produced by a chemical reaction T= Tfinal –Tinitial Units: C Specific Heat (s): Heat needed to raise T of 1 gram by 1C. s = q/(m x T) Units: J/gC swater = 4.184 J/gC Heat capacity (C): Quantity of heat needed to raise T by 1C C=mxs Units: J/C or J/K Heat of Reaction (qrxn) = m x s x T Units: J 17 Commonly Used Specific Heats 18 Calorimetry Calculations Reaction (system) heat is transferred to water (surroundings) qsys = -qH2O 1. Measure T of water qH2O = mH2O x sH2O x TH2O Use q H2O to find system variables Crossover: qH2O = -qsys qrxn= msys x ssys x Tsys Always have 2 sets of variables! 1 for water (surroundings) 1 for reaction (system) Make 2 columns of variables System: s, m, Tinitial, Tfinal Water: s, m, Tinitial, Tfinal 19 A 30.0g sample of metal is heated to 100.0C by placing it in a boiling water bath. The metal is removed from the boiling water, and placed in 25.0g of water (4.18 J/gC) at 22.0C. The final temp was 27.6C, what is the specific heat of the metal? What is your equation? Smetal = qmetal/mT qmetal = - qwater qwater = masswater x Swater x Twater Twater = 27.6C – 22.0C = 5.6C qwater = 25.0g x 4.18 J/gC x 5.6C = 585 J qmetal = - qwater = -585 J What is the metal mass? 30.0g What is qmetal? What is Tmetal ? What is S. H. metal? (27.6C -100.0C) = -72.4C SH metal 585 J 1 1 0.27 J x x g C 1 30.0 g 72.4 C 20 Heat Evolved During a Chemical Reaction Reaction conducted in a Styrofoam cup calorimeter The calorimeter does not absorb or give off heat. Heat absorbed by the solution in the calorimeter, qcal is the heat that has been given off by the chemical reaction, -qrxn Two solutions are mixed in the calorimeter 40.0 mL of 1.00 M KOH(aq) 40.0mL of 0.500M H2S04(aq) Ti of both solutions = 21.00C Data on final solution Density: 1.02 g/mL Volume: 80.0 mL S.H.: 4.00 J/gC Temperature: 27.85C 21 Calculate the enthalpy change, H, of this reaction per mole of H2O formed. What is the chemical reaction? What is the equation for H? Calculate Heat of Reaction (q). 2 KOH + H2SO4 K2SO4(aq) + 2H2O Hrxn= qrxn = -qH2O q = S x m x T s = 4.00 J/gC (assume water) m = 81.6g (mass of both solns) T = 6.85C (+ temp went up) m = 1.02 g/mLx 80.0 mL = 81.6g T = 27.85C - 21.00C = 6.85C qH2O= 4.00 J/gC x 81.6g x 6.85C = 2236 J qrxn = -qH2O= -2236 J Note: this is heat that was transferred into the water IN CALORIMETER, not water in chemical reaction!!! Amount of heat generated for 81.6g of reactants, not per mole H2O! 22 Calculating the H per mole H2O How many moles of water are produced in the reaction? 2 KOH + H2SO4 K2SO4(aq) + 2H2O 40.0 mL of 1.00 M KOH (aq) = 0.040 mol KOH = 40.0 mL of 0.500 M H2S04(aq) = 0.020 mol H2S04 = 0.040 mol H2O 0.040 mol H2O Calculate H per 1 mole of water produced. Hrxn H per mole water H per mole water = qrxn = -2236 J = -2236 J / 0.040 mol H2O = -55900 J/mol = -55.9kJ/mol 23 Heats of Reactions 24 Standard Enthalpy of Formation and Reaction 25 The Direct Method Standard Enthalpies of Formation, Hf The standard state of an element: Hf=0 Pure element in its most stable form at a pressure of 1 atm and 20C For solvents in aqueous solution, they are at a concentration of 1 M H2(g) N2(g) O2(g) Cl2(g) Br2(l) Hg(l) Na(s) The standard molar enthalpy of formation, Hf Hf for 1 mol of substance produced from elements in standard states Tabulated for compounds at 1atm and 20C C(s) + O2(g) CO2(g) Hf = -393.5kJ/mol Calculating H of a chemical reaction Subtract Hf values of reactants from Hf values of products Multiply by the stoichiometric coefficient for that species. C(s) and O2(g) Hf = 1 x 0kJ/mol = 0kJ/mol each CO2(g) Hf = 1 x -393.5kJ/mol = -393.5kJ/mol Hf = H products - H reactants= -393.5 - (0+0) = -393.5kJ/mol 26 Calculate H for the combustion of C2H5OH(l) Hfvalues are given below C2H5OH(l)+3 O2(g)2CO2(g) + 3H2O(l) Hf Reactants C2H5OH(l) = -277.7 kJ/mol x 1 mol O2(g) = 0 kJ/mol x 3 mol Total Reactants: = -277.7kJ = 0.0kJ = -277.7kJ Hf Products CO2(g) = -393.5 kJ/mol x 2 mol H2O (l) = -285.8 kJ/mol x 3 mol Total Products: = -787.0kJ = -857.4kJ =-1644.4kJ H=products - reactants = (-1644.4kJ)-(-277.7kJ) = -1366.7kJ 27 The Indirect Method: Hess’s Law Of Constant Heat Summation The heat of a reaction, H, is constant, whether the reaction is carried out directly in one step or through a number of steps. The equation for the overall reaction is the sum of two or more other equations (reaction steps). Hrxn = H1 + H2 + H3 +... Reverse a chemical equation: Reverse the sign of H. Hright = - Hleft 28 An Enthalpy Diagram 1. Find each reactant in 1 equation only Change equation if necessary 2. Find each product in 1 equation only Change equation if necessary 3. Add reactions Cross out product/reactant pairs Add multiples 4. Check final equation Find: C(graphite) + O2(g) CO(g) + 1/2O2(g) a. CO(g) + 1/2O2(g) CO2 b. C(graphite) + O2(g) CO2 H= ? H= -283.0kJ H= -393.5kJ 1. C(graphite) + O2(g) CO2 CO(g) + 1/2O2(g) 2. CO2 3. C(graphite) + O2(g) CO(g) + 1/2O2(g) H= -393.5kJ H= +283.0kJ H= -110.5kJ 29 A Styrofoam cup calorimeter was used to determine Hsol, of KOH(s). Hsol = -58.4 kJ for the reaction: KOH(s)KOH(aq) What is Hsol,when KOH(s) reacts with H2S04 (aq)? Use the H from the earlier calorimetry example 2 KOH(aq) + H2SO4(aq)K2SO4(aq) + 2H2O(l) H = 2 moles x –55.9 kJ/mol = -111.8 kJ Look up H of KOH(s) to KOH(aq) and convert for 2 moles KOH KOH(s)KOH(aq) H = -58.4 kJ/mol H = 2 moles x -58.4 kJ/mol = -116.8 kJ Add the Hs for all steps of the reactions 2 KOH(s) 2 KOH(aq) 2 KOH(aq) + H2SO4(aq)K2SO4(aq) + 2H20(l) 2 KOH (s) + H2SO4(aq)K2SO4(aq) + 2H20(l) H = -116.8 kJ H = -111.8 kJ H = -228.6 kJ 30