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: CH 4 + O 2 CO 2 + H 2 O Bonds stronger in CO 2 and H 2 O molecules than in CH 4 + O 2 Heat goes from system to surroundings Endothermic Reaction: System gains heat (Entering) Ex: Ice Melting: H 2 O (s) H 2 O (l) Energy needed to break attractions between H 2 O 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 U system =U final -U initial Internal Energy of System: U system = U kinetic + U potential 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: U reaction = U products -U reactants 7
Work and Heat Energy transfer is caused by changes in heat & work in a system U system = q + w Work (w) = force x distance = -P V Negative work (-w): System loses energy to surroundings Positive work (+w): System gains energy from surroundings Work (w) = -P V V = V f -V i 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. U sys = U final -U initial =350J = q + w V = 2L P = 1atm T= 273K a. Calculate work done against a vacuum & a pressure of 1atm. w = -P V = -(0atm x 2L) = 0atm L = 0J w = -P V = -(1atm x 2L) = -2atm L = -202.6J (1Latm= 101.32J) b. Calculate the heat required for each system. q = U sys -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 = q p -P V Some work used to expand system against surroundings. Enthalpy: (H) Experimentally measured and tabulated At constant pressure: Assume negligible volume change: H= U + P V H= U = q p There will be slight differences between H and U 11
Real Differences between U and H H measured experimentally as heat 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) H=-367.5kJ/mol Volume change of H 2 gas V=nRT/P T= 298K, P=1.0atm, n=1.0mol H 2 V= 24.5L U = H- P V H = -367.5kJ P V = 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 = q rxn ) Amount of heat exchanged between system and surroundings during a chemical reaction H reaction = H products -H reactants Exothermic Reactions Temperature increases in an isolated system. Heat released to surroundings is treated as a product 2SO 2 (g) + O 2 (g) 2SO 3 (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 + 2SO 3 2SO 2 (g) + O 2 (g) H = +197.8kJ State function: Independent of path 13
Rules for Working with Heat and Enthalpy 1. Use same multiplier as used in chemical reaction 2SO 2 (g) + O 2 (g) 2SO 3 (g) H = -197.8kJ Decrease or increase reactants and products, do the same for H 1SO 2 (g) + 1/2O 2 (g) 1SO 3 (g) H = -197.8kJ/2= -98.9kJ Note: Write H with units that match mole ratio in equation H = -98.9 kj /mol SO 3 (g) 2. Reverse the reaction, reverse sign of H 2SO 3 2SO 2 (g) + O 2 (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 NO 2 (g) according to this reaction: 2NO(g) + O 2 (g) 2NO 2 (g) H = -114 kj Heat required for following reaction: 2NO 2 (g) 2NO(g) + O 2 (g) H = +114 kj H = +114 kj is needed to decompose of 2 moles NO 2 Calculate molar mass of NO 2 to convert mass to moles of NO 2. moles NO 15.0g 1mol 2 2 x 0. 326 2 1 NO 46.0g Find the amount of heat for the amount of NO 2 actually present. H NO NO 114kJ 0.326molNO 2 x 2mol 1 18. 6 rxn NO 2 2 mol kj NO 2 15
Calorimetry 16
Calorimetry Calorimetry: Measurement of heat changes Calorimeter: Device to measure heat ( T) produced by a chemical reaction T= T final T initial Units: C Specific Heat (s): Heat needed to raise T of 1 gram by 1 C. s = q/(m x T) Units: J/g C s water = 4.184 J/g C Heat capacity (C): Quantity of heat needed to raise T by 1 C C = m x s Units: J/ C or J/K Heat of Reaction (q rxn ) = m x s x T Units: J 17
Commonly Used Specific Heats 18
Calorimetry Calculations Reaction (system) heat is transferred to water (surroundings) q sys = -q H2O 1. Measure T of water q H2O = m H2O x s H2O x T H2O Use q H2O to find system variables Crossover: q H2O = -q sys q rxn = m sys x s sys x T sys Always have 2 sets of variables! 1 for water (surroundings) 1 for reaction (system) Make 2 columns of variables System: s, m, T initial, T final Water: s, m, T initial, T final 19
A 30.0g sample of metal is heated to 100.0 C 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/g C) at 22.0 C. The final temp was 27.6 C, what is the specific heat of the metal? What is your equation? What is q metal? What is the metal mass? What is T metal? S metal = q metal /m T q metal = - q water q water = mass water x S water x T water T water = 27.6 C 22.0 C = 5.6 C q water = 25.0g x 4.18 J/g C x 5.6 C = 585 J q metal = - q water = -585 J 30.0g (27.6 C -100.0 C) = -72.4 C What is S. H. metal? SH metal 585J 1 x 1 30.0 g x 1 72.4 C 0.27J g 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, q cal is the heat that has been given off by the chemical reaction, -q rxn Two solutions are mixed in the calorimeter 40.0 ml of 1.00 M KOH(aq) 40.0mL of 0.500M H 2 S0 4 (aq) T i of both solutions = 21.00 C Data on final solution Density: 1.02 g/ml Volume: 80.0 ml S.H.: 4.00 J/g C Temperature: 27.85 C 21
Calculate the enthalpy change, H, of this reaction per mole of H 2 O formed. What is the chemical reaction? 2 KOH + H 2 SO 4 K 2 SO 4 (aq) + 2H 2 O What is the equation for H? H rxn = q rxn = -q H2O Calculate Heat of Reaction (q). q = S x m x T s = 4.00 J/g C (assume water) m = 81.6g (mass of both solns) m = 1.02 g/mlx 80.0 ml = 81.6g T = 6.85 C (+ temp went up) T = 27.85 C - 21.00 C = 6.85 C q H2O = 4.00 J/g C x 81.6g x 6.85 C = 2236 J q rxn = -q H2O = -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 H 2 O! 22
Calculating the H per mole H 2 O How many moles of water are produced in the reaction? 2 KOH + H 2 SO 4 K 2 SO 4 (aq) + 2H 2 O 40.0 ml of 1.00 M KOH (aq) = 0.040 mol KOH = 0.040 mol H 2 O 40.0 ml of 0.500 M H 2 S0 4 (aq) = 0.020 mol H 2 S0 4 = 0.040 mol H 2 O Calculate H per 1 mole of water produced. H rxn H per mole water H per mole water = q rxn = -2236 J = -2236 J / 0.040 mol H 2 O = -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, H f The standard state of an element: H f =0 Pure element in its most stable form at a pressure of 1 atm and 20 C For solvents in aqueous solution, they are at a concentration of 1 M H 2 (g) N 2 (g) O 2 (g) Cl 2 (g) Br 2 (l) Hg(l) Na(s) The standard molar enthalpy of formation, H f H f for 1 mol of substance produced from elements in standard states Tabulated for compounds at 1atm and 20 C C(s) + O 2 (g) CO 2 (g) H f = -393.5kJ/mol Calculating H of a chemical reaction Subtract H f values of reactants from H f values of products Multiply by the stoichiometric coefficient for that species. C(s) and O 2 (g) H f = 1 x 0kJ/mol = 0kJ/mol each CO 2 (g) H f = 1 x -393.5kJ/mol = -393.5kJ/mol H f = H products - H reactants = -393.5 - (0+0) = -393.5kJ/mol 26
Calculate H for the combustion of C 2 H 5 OH(l) H f values are given below C 2 H 5 OH(l)+3 O 2 (g) 2CO 2 (g) + 3H 2 O(l) H f Reactants C 2 H 5 OH(l) = -277.7 kj/mol x 1 mol = -277.7kJ O 2 (g) = 0 kj/mol x 3 mol = 0.0kJ Total Reactants: = -277.7kJ H f Products CO 2 (g) = -393.5 kj/mol x 2 mol = -787.0kJ H 2 O (l) = -285.8 kj/mol x 3 mol = -857.4kJ Total Products: =-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). H rxn = H 1 + H 2 + H 3 +... Reverse a chemical equation: Reverse the sign of H. H right = - H left 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) + O 2 (g) CO(g) + 1/2O 2 (g) H=? a. CO(g) + 1/2O 2 (g) CO 2 H= -283.0kJ b. C(graphite) + O 2 (g) CO 2 H= -393.5kJ 1. C(graphite) + O 2 (g) CO 2 H= -393.5kJ 2. CO 2 CO(g) + 1/2O 2 (g) H= +283.0kJ 3. C(graphite) + O 2 (g) CO(g) + 1/2O 2 (g) H= -110.5kJ 29
A Styrofoam cup calorimeter was used to determine H sol, of KOH(s). H sol = -58.4 kj for the reaction: KOH(s) KOH(aq) What is H sol,when KOH(s) reacts with H 2 S0 4 (aq)? Use the H from the earlier calorimetry example 2 KOH(aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 O(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) H = -116.8 kj 2 KOH(aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 0(l) H = -111.8 kj 2 KOH (s) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 0(l) H = -228.6 kj 30