Chapter 6: Thermochemistry

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1 Chapter 6: Thermochemistry Chemical reactions obey 2 laws: conservation of mass (previous chapters) conservation of energy (this chapter) 6.1 Energy and Types of Energy A. Definitions Energy - capacity to do work Work - (physicists) force x distance (chemists definition) directed energy change resulting from a process B. Types of Energy 1. Kinetic Energy - energy produced by moving object K.E = ½ mv 2 where m = mass and v = velocity 2. Radiant energy - (solar energy) energy from the sun; from chemical reactions of the sun; affect weather and plant growth 3. Thermal energy - associated with random motion of atoms and molecules. may be calculated from the temperature and amount of molecules Example: cup of coffee at 70 C has less thermal energy than a bathtub of water at 40 C 4. Potential Energy - energy available by virtue of an objects position, e.g., rock at the top of a cliff 5. Chemical Energy - energy stored in bonds; potential energy associated with chemical bonds:

2 bond breaking: bond making: energy is required energy is liberated All forms of energy may theoretically be converted to other forms of energy. Law of Conservation of Energy - the total quantity of energy in the Universe is constant. 6.2 Energy Changes in Chemical Reactions These are as important as the mass changes, e.g., combustion of hydrocarbons is done to make use of the energy liberated in the reaction. 1. Heat - (generally the form of energy in chemical reactions) transfer of thermal energy between two bodies at different temperatures: hotter colder 2. Thermochemistry - study of heat changes in chemical reactions 3. System - part of the Universe of interest to us (beaker, automobile...) 4. Surroundings - rest of the Universe 5. Types of Systems: open system - exchanges mass and energy (usually heat) with surroundings; e.g., open beaker of H 2 O (water can evaporate) closed system - allows heat transfer, not mass transfer; e.g. beaker with lid isolated system - allows no heat or mass transfer; e.g., insulated beaker with a lid

3 Exothermic process - gives off heat transfers heat to surroundings heat is essentially a product of the reaction system warms up 6. Endothermic process - heat must be supplied by the surroundings heat is essentially a reactant system cools off

6.3 Introduction to Thermodynamics Thermodynamics (broader than thermochemistry) scientific study of the interconversion of heat and other kinds of energy 1.

4 6.3 Introduction to Thermodynamics Thermodynamics (broader than thermochemistry) scientific study of the interconversion of heat and other kinds of energy 1. State Functions - state of a system defined by macroscopic properties, i.e. composition, energy, temperature, pressure, volume our concern change in the system from the point of initial and final states so if the volume changes we want to know V = Vf - Vi Energy is a state function Figure 6.4 Gain in potential energy is the same her (gravitational pull) regardless of pathway.

5 First Law of Thermodynamics Energy can be converted from one form to another, but cannot be created or destroyed. Potential energy to kinetic energy Example is a chemical reaction, e.g., burning sulfur S(s) + O2(g) SO2(g) cannot know exactly the total energy in the S, O2, and SO2, but can measure the change in energy (in this case heat evolved) E = Eprod - Ereac Esystem = Esurroundings Esystem = q + w where q = heat exchanged; w = work q and w are negative if heat moves out of system or work is done by the system and w are positive if heat is added to system or work is done on the system

Work involving gases Figure 6.5 piston moves upward as gas expands w = force x distance w = - P V units of work = L. atm conversion factor: 1 L. atm = 101.

6 Work involving gases Figure 6.5 piston moves upward as gas expands w = force x distance w = - P V units of work = L. atm conversion factor: 1 L. atm = J Example: A gas expands form 264 ml to 971 ml at constant temperature. Calculate the work done in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 4.00 atm.

7 6.4 Enthalpy, H, of Chemical Reactions the total energy of a chemical system at constant pressure (conditions of most reactions) depends on amount of substance cannot measure enthalpy directly, but can measure changes in enthalpy H = Hproducts - Hreactants endothermic reaction: exothermic reaction: H > 0 (positive) -- heat is absorbed (deposit) H < 0 (negative) -- heat is released (withdrawal) This semester we'll be concerned primarily with H and not E Enthalpy and the First Law of Thermodynamics Enthalpy (H) of system: H = E + PV Change in enthalpy (at constant pressure) (change is what we can easily measure) H = E + PV or rearranged E = H - PV remember that Esystem = q + w so q = H at constant pressure and PV = nrt, so H = E + RT n E = H - RT n

8 Thermochemical Equations Example: One mole of ice melts to liquid at 0 C. (1 atmosphere, a standard condition) 18 grams (1 mole) of ice 6.01 kj of energy absorbed H = 6.01 kj If the process is reversed, i.e., the liquid water is frozen, then H = kj Some energy conversion factors 1 Joule = 1 J = 1 Kg m2/s2 = 1 N. m where N = Newton, m = meters, s = seconds 1 kj = 1000 J J = 1 cal 1000 cal = 1 Kcal = 1 Calorie (Food Calorie) For example: The reaction for the combustion of ethylene: C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O (l) H = - 1,411 kj (very exothermic) i.e., 1,411 kj of heat energy are released in the reaction of 1 mole of C 2 H 4 with 3 moles of O 2 Rules for thermochemical equations: work in moles reversing an equation reverses the sign of H multiplying the equation by a factor applies to H as well as to moles must specify state of reaction and products Example: If 10.0 g of C 2 H 4 are burned, how much heat is produced? 10.0 g x 1 mole C2 H kj 28.0 g x mole C 2 H 4 = 504 kj

9 6.5 Calorimetry measurement of heat change A. Specific Heat (s) - amount of heat needed to raise the temperature of 1 gram of a substance by 1 C. (units are J/g. C) (intensive property) B. Heat Capacity (C)- amount of heat required to raise the temperature of a given quantity substance by 1 C. (extensive property) Example: specific heat of water is J/g. C heat capacity of 60.0g of water is 60.0 g x J = 251 J g. C C C. Calculating amount of heat from s and change in temperature ( t) q = ms t q = C t where m = mass t = change in temperature q is positive for endothermic process q is negative for exothermic process Example: An iron bar of mass 869 g cools from 94 C to 5 C. Calculate the heat released in kj. q = ms t t = 94 C - 5 C = 89 C from Table 6.1, p. 210: s = J/g. C q = 869g x J x 89 C = 34,339 J g. C q = kj (heat released)

ignition wire O2 inlet calorimeter bucket bomb sample cup Procedure: measure

10 D. Constant Volume Calorimetry see Figure 6.8 (isolated system - no heat leaves or enters) thermometer insulated jacket ignition wire O2 inlet calorimeter bucket bomb sample cup Procedure: measure starting temperature measure mass of sample ignite sample measure temperature after reaction q out of rexn = q into the bomb and water

11 qsystem = qwater + qbomb + qrxn = 0 but we can also say qrxn = (qwater + qbomb) qwater = ms t = (m water )(4.184 J/g C) t qbomb = Cbomb t C bomb = (m bomb )(s bomb ) q out of rexn = q into the bomb and water Example A quantity of g of methanol (CH 3 OH) was burned in a bomb calorimeter. Consequently, the temperature of the H 2 O rose by 4.20 C. If the quantity of water surrounding the calorimeter was exactly 2000 g and the heat capacity of the calorimeter was 2.02 kj/ C, calculate the molar heat of combustion of methanol. q rxn = (q water + q bomb ) q out of rexn = q into the bomb and water q rxn = 2.02 x 10 3 J x 4.20 o C g x J o C g o C x 4.20 o C qrxn = 8484 J J = 43,629.6 J = 43.6 kj BUT we know that the rexn was EXOTHERMIC so the qrxn is 43.6 kj!!!!!! Calculate the molar heat of combustion: g x 1 mole CH 3 OH g = moles kj kj = moles mole

6.6 Standard Enthalpy of Formation and Reaction cannot measure absolute enthalpy, only the change in enthalpy thus use a standard (like sea level)- accepted convention H f = H for the formation of

4 and in Appendix 3 H f for elements in most stable form is 0 e.g., O2, H f = 0 O 3, H f = 142 kj/mole C graphite, H f = 0 C diamond, H f = 1.

12 6.6 Standard Enthalpy of Formation and Reaction cannot measure absolute enthalpy, only the change in enthalpy thus use a standard (like sea level)- accepted convention H f = H for the formation of one mole of substance from its elements in their standard states at a pressure of one atmosphere Standard State = 1 atmosphere, 25 C H f for some substances found in Table 6.4 and in Appendix 3 H f for elements in most stable form is 0 e.g., O2, H f = 0 O 3, H f = 142 kj/mole C graphite, H f = 0 C diamond, H f = 1.90 kj/mole Example of a "formation" reaction (remember: H f applies to 1 mole of a compound formed from its elements): H 2 (g) + 1/2 O 2 (g) H 2 O (l) H f (liq water) = kj/mole practice writing formation reactions -- e.g., Na 2 SO 4 2 Na (s) + 2 O 2 (g) + S (s) Na 2 SO 4 (s) H f = kj/mole

13 B. Standard Enthalpy of Reaction aa + bb cc + dd H rexn = H f (products) - H f (reactants) H rexn = [c. H f (C) + d. H f (D)] - [a. H f (A) + b. H f (B)] 1. Direct Method - applies when reactants are elements in most stable state S (rhombic) + 3 F2 (g) SF 6 (g) In these cases may directly measure by doing the reaction 2. Indirect method (Hess's Law) (Law of Heat Summation) When reactants are converted to products, the change in H is the same whether the reaction takes place in one or several steps. This applies when compounds cannot be made directly from the elements Example Calculate the H f of acetylene, C 2 H 2 gas from its elements. 2 C (graphite) + H 2 (g) C 2 H 2 (g) The equations for each step and the corresponding enthalpy changes are: C (graphite) + O2(g) CO2 (g) H2 (g) + ½ O2 (g) H2O (l) H f = kj H f = kj 2 C 2 H 2 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O (l) H rexn = kj First: reverse last reaction to put the product (C 2 H 2 ) on the correct side of the equation 4 CO 2 (g) + 2 H 2 O (l) 2 C 2 H 2 (g) + 5 O 2 (g) H = kj

14 Second: There is no H2O or CO2 in the formation equation, so place these reagents on opposite sides. The goal is to get these to cancel when the 3 reactions are summed. C (graphite) + O 2 (g) CO 2 (g) H2 (g) + ½ O2 (g) H2O (l) H f = kj H f = kj Third: Multiply by appropriate factors to make sure the same number of each reagent that must be canceled occurs on right and left sides. 4 [ C (graphite) + O2(g) CO2 (g)] 4 H f = kj 2 [ H 2 (g) + ½ O 2 (g) H 2 O (l)] 2 H f = kj 4 CO2 (g) + 2 H 2 O (l) 2 C 2 H 2 (g) + 5 O 2 (g) H = kj 4 C (graphite) + 2 H2 (g) a) 2 C2H2 (g) H = kj 2 C (graphite) + H 2 (g) a) C 2 H 2 (g) H f = kj 6.7 Heat of Solution and Dilution examples: dissolving salts, hot and cold packs in first aid kits H soln - heat generated or absorbed when a solute dissolves Lattice Energy (U) - energy required to completely separate one mole of a solid ionic compound into gaseous ions: NaCl (s) + energy Na + (g) + Cl - (g) H hyd - heat of hydration: enthalpy change associate with hydration H soln = H hyd + U see Figure 6.11 NaCl (s) Na + (g) + Cl - (g) U = 788 kj Na + (g) + Cl - (g) H 2 O Na + (aq) + Cl - (aq) H hyd = -784 kj H 2 O NaCl (s) Na + (aq) + Cl - (aq) H soln = 4 kj Thus the mixture cools slightly.

16 1 calorie (cal) = J 1000 cal = 1 Calorie (Cal) = 1 food Calorie These numbers can be determined in a bomb calorimeter.

17 IA Periodic Table of the Elements VIIIA (1) (18) H IIA IIIA IVA VA VIA VIIA He (2) (13) (14) (15) (16) (17) Li Be B C N O F Ne Na Mg IIIB IVB VB VIB VIIB VIIIB IB IIB Al Si P S Cl Ar (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Uuu Uub Uug (261) (262) (266) (264) (270) (268) (281) Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Solutions and Ions. Pure Substances

Solutions and Ions. Pure Substances Class #4 Solutions and Ions CHEM 107 L.S. Brown Texas A&M University Pure Substances Pure substance: described completely by a single chemical formula Fixed composition 1 Mixtures Combination of 2 or more