Energy, Heat and Chemical Change

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1 Energy, Heat and Chemical Change Chemistry 35 Fall 2000 Thermochemistry A part of Thermodynamics dealing with energy changes associated with physical and chemical reactions Why do we care? -will a reaction proceed spontaneously? -if so, to what extent? It won t tell us: -how fast the reaction will occur -the mechanism by which the reaction will occur 2 1

2 What is Energy? Energy is the capacity to do work or to transfer heat -Kinetic Energy: energy associated with mass in motion (recall: E k = ½mv 2 ) -Potential Energy: energy associated with the position of an object relative to other objects (energy that is stored - can be converted to kinetic energy) 3 The System We must define what we are studying: System Surroundings System: portion of the universe under study Surroundings: everything else If exchange is possible, system is open 4 2

3 Energy Transfer Energy can be transferred in two different ways: 1. By doing Work (applying a force over a distance) W = F x d 2. Heat - q (results in a change in temperature) Note: W, q, and E all have the same units (Joule), but: - W & q depend on path (path function) - E is independent of path (state function) 5 First Law of Thermodynamics The total energy of the universe is constant. Energy is neither created or destroyed in a process, only converted to another form. -Conservation of Energy You can t win... you can only break even. Change in energy of the system E = q + w Work: + is done on system Heat Flow: - is done by system + is into system - is out of system 6 3

4 Example: Gas in a cylinder (fixed volume) Suppose we have a gas in a cylinder with a movable piston: What if: The piston is locked in a fixed position -Volume is constant, so: w = 0 -> w = f x d OR w = P V = 0 Thus: E = q v Now, if we add heat to the system: q > 0, so E > 0 (E incr.) -temp incr -> E k = 3/2 RT -> E k = 3/2 R T 7 Example: Gas in a cylinder (variable volume) Ok, now let s allow the piston to move, but let s disallow heat flow (q = 0) - adiabatic process So: E = w = -P V Movie! Gas Compression: V < 0 -> E is + -work is done on the system -temperature of the gas increases Gas Expansion: V > 0 -> E is - -work is done by the system -temperature of the gas decreases 8 4

5 Measuring E We can measure q by the change in temperature: If q > 0: heat is added to the system -endothermic (system absorbs heat) -temperature (of the surroundings) decreases If q < 0: heat is given off by the system -exothermic (system loses heat) -temperature (of the surroundings) increases 9 Calorimetry: Measuring E We can quantify heat flow (q) by measuring the change in temperature of a system: Calorimetry Heat Capacity (C): amount of heat (q) required to raise the temperature of a substance by 1 K -if we calculate C for a specific amount of a particular substance, we call it the Molar Heat Capacity (c v at constant volume) or the Specific Heat Capacity (c s, per gram of the substance) Example: for g H 2 O -> specific heat = J/g - K C = mass x specific heat = m x c s = g x J/g - K = J/K 10 5

6 Heat flow and temperature Now we can relate q and temperature: Example: q = C (T final - T initial ) How much heat is required to raise the temperature of g H 2 O from 25.0 o C to 78.0 o C? q = C T = (m x specific heat)( T) = (232.0 g x J/g -K)( ) = ( J/K)(53.0 K) = J = 5.14 x 10 4 J = 51.4 kj 11 Constant-Volume Calorimetry Use a bomb calorimeter: -typically used with combustion reactions -heat (q) from rxn is transferred to the water and the calorimeter -knowing the heat capacity of the calorimeter: q rxn = -C cal x T So: q v = E constant volume) 12 6

7 Enthalpy Chemistry is commonly performed at constant pressure, so: -it is easy to measure heat flow (q p ) -work (P V) is small (but finite) and hard to measure Define a new term: Enthalpy (H) H = E + PV 13 Relating Enthalpy and Heat Recall: E = q + w At constant pressure: E = q p - P V Rearranging: q p = E + PV q p = (E + PV) Substituting: q p = H So, if we measure q p, then we can obtain the enthalpy change ( H) directly 14 7

8 Constant-Pressure Calorimetry So-called coffee-cup calorimetry: Add reactants to cup Measure resulting temperature increase (or decrease) q soln = -q rxn So: H = -q soln = -m soln (sp heat) T You will do this this week in lab! 15 How are E and H related? From the definition of Enthalpy: H = E + P V -for an ideal gas: P V = RT n So: H = E + RT n q p q v PV work 16 8

9 Enthalpies of Reactions Enthalpy (H) is a state function, so all we need to know are the initial and final values for a reaction: H rxn = H (products) - H (reactants) Together with a balanced reaction expression, we have a thermochemical equation: 2 H 2 (g) + O 2 (g) 2 H 2 0 (g) H = kj Balanced, with physical states Note sign: exothermic 17 More on Enthalpy Enthalpy is an extensive property -the magnitude of H depends on the quantity of reactant H forward = - H reverse - reverse reaction will have same H, but with the opposite sign Reactions with H << 0 often occur spontaneously -but not always; there are exceptions (wait until Thermodynamics to resolve this fully!) 18 9

10 Enthalpy of Reaction Calculation Example Calculate the H for the reaction that occurs when 1.00 g H 2 O 2 decomposes according to the following reaction: H 2 O 2 (l) H 2 O (l) + ½ O 2 (g) H = kj convert: g H 2 O 2 kj 1.00 g H 2 O 2 x 1 mol H 2 O 2 x kj = g H 2 O 2 1 mol H 2 O 2 = kj 19 Hess s Law H is a state function so we can use any sequence of reactions (that sum to the desired reaction) to calculate a value of H rxn : IF: Rxn(1) + Rxn(2) = Rxn(3) Then: H 1 + H 2 = H 3 An example: Rxn(1): Sn(s) + Cl 2 (g) SnCl 2 (s) kj Rxn(2): SnCl 2 (s) + Cl 2 (g) SnCl 4 (l) kj Rxn(3): Sn(s) + 2Cl 2 (g) SnCl 4 (l) kj H 20 10

11 Hess s Law: Example We can use Hess s Law to calculate H for reactions without ever actually having to perform the reaction: Find H for the following reaction: 2NH 3 (g) + 3N 2 O(g) 4N 2 (g) + 3H 2 O(l) What we know: H Rxn(1): 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) kj Rxn(2): N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(l) kj Rxn(3): H 2 (g) + ½O 2 H 2 O(l) kj 21 Hess s Law Example: Cont d We need to react only TWO mol NH 3, so divide rxn(1) by 2: 2NH 3 (g) + 3/2 O 2 (g) N 2 (g) + 3H 2 O(l) We need to react THREE mol N 2 O, so multiply rxn(2) by 3: 3N 2 O(g) + 3H 2 (g) 3N 2 (g) + 3H 2 O(l) We need to get rid of the stuff not in the final rxn (H 2, O 2, and excess H 2 O), so reverse rxn(3) and multiply by 3: Add em up! 3H 2 O(l) 3H 2 (g) + 3/2 O 2 (g) 2NH 3 (g) + 3/2 O 2 (g) + 3N 2 O(g) + 3H 2 (g) + 3H 2 O(l) N 2 (g) + 3H 2 O(l) + 3N 2 (g) + 3H 2 O(l) + 3H 2 (g) + 3/2 O 2 (g) 22 11

12 Hess s Law Example: The Thrilling Conclusion! Whatever was done to the reaction equations, we need to do to the H values: H 1 2 = = kj H 2 x 3 = x 3 = kj H 3 x (-3) = x (-3) = kj H rxn = kj 23 Enthalpies of Formation A consistent way to tabulate H values: -for a defined reaction type (formation from elements) -for a fixed amount of compound (1 mol) -under standard conditions (25 o C, 1 atm) Called: The Standard Molar Enthalpy of Formation or H o f Example: Ag(s) + ½Cl 2 (g) AgCl(s) H = kj Then: H o f (AgCl(s)) = kj/mol 24 12

13 Using Enthalpies of Formation Hess s Law and tablulated H o f values are a powerful tool for predicting enthalpy changes for reactions: H o = Σ H o f (products) - Σ Ho f (reactants) Example: We know: So: The Thermite Reaction 8 Al(s) + 3 Fe 3 O 4 (s) 4 Al 2 O 3 (s) + 9 Fe(s) H o f (Fe 3 O 4 ) = kj H o f (Al 2 O 3 ) = kj H o = [4( kj/mol) + 9(0)] - [3( kj/mol) + 8(0)] H o = ( kj) - ( ) = kj 25 Demo Example: Burning Mg The overall reaction is: 2 Mg(s) + CO 2 (s) 2 MgO(s) + C(s) H o f : 0?? kj/mol 0 How do we find H o f for SOLID CO 2? Use Hess s Law! We know: H o f for CO 2 (g) = kj/mol But how do we deal with phase changes? 26 13

14 Enthalpies of Phase Changes There are enthalpy changes associated with physical processes: So: CO 2 (s) CO 2 (l) H fusion = 6.01 kj/mol CO 2 (l) CO 2 (g) H vap = kj/mol CO 2 (s) CO 2 (g) H sublimation = H fusion + H vap H sublimation = kj/mol We want: CO 2 (g) CO 2 (s) H = - H sub = kj/mol 27 Putting it all together So, to get H o f for CO 2 (g): C(s) + O 2 (g) CO 2 (g) CO 2 (g) CO 2 (s) Adding gives: C(s) + O 2 (g) CO 2 (s) H o f = kj/mol H = kj/mol H o f = kj/mol And, finally: 2 Mg(s) + CO 2 (s) 2 MgO(s) + C(s) H o f H o rxn : kj/mol kj/mol 0 = 2( kj/mol) - ( kj/mol) = kj 28 14

15 Bond Enthalpies Armed with Hess s law and H o f values, let s dissect the reaction process (i.e., the energetics of bond making and breaking): For CH 4 (methane): We know: C(s) + 2 H 2 (g) CH 4 (g) H o f = kj/mol - We want to know: the bond enthalpy for the C-H bond So, look at the reverse reaction: CH 4 (g) C(g) + 4 H (g) H = 4 x C-H bond enthalpy 29 Hess s Law to the rescue We want to do this: Enthalpy Cycle - H o f CH 4 (g) C(g) + 4 H (g) H o vap 4 x H o f (H(g)) C(s) + 2 H 2 (g) CH 4 (g) C(s) + 2 H 2 (g) - H o f = - (-74.9 kj) C(s) C(g) H o vap = kj 2 H 2 (g) 4 H(g) 4 x H o f = 4(217.9 kj) CH 4 (g) C(g) + 4 H (g) kj So, C-H bond enthalpy = kj/4 = kj 30 15

16 Bond Enthalpies: What good are they? They indicate how strong a bond is: N N H-H In general: > = > kj/mol 436 kj/mol If we assume that bond enthalpies for a particular bond are constant in all compounds, then we can use them to estimate H rxn : H rxn = Reactant Bond Enthalpies - Product Bond Enthalpies (energy added to break bonds) (energy released by making bonds) -This works well if there are no intermolecular forces (i.e., in the gas phase!) 31 16

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