AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither created nr destryed. The ttal amunt f energy in the universe is cnstant. Ptential energy- energy due t psitin r cmpsitin Kinetic energy- energy due t the mtin f an bject -depends n mass and velcity f an bject KE = ½ mv 2 m = mass in kg v = velcity in m/s -units are J, since J = kg m 2 s 2 Heat- invlves a transfer f energy between tw bjects due t a temperature difference. Wrk- frce acting ver a distance -invlves a transfer f energy Temperature- a prperty that reflects randm mtins f the particles f a particular substance Exthermic- reactin which releases heat -energy flws ut f the system -ptential energy is changed t thermal energy -prducts have lwer ptential energy than reactants Endthermic- reactin which absrbs heat -energy flws int the system -thermal energy is changed int ptential energy -prducts have higher PE than reactants The system is ur reactin. The surrundings are everything else. Internal energy (E) f a system is the sum f the kinetic and ptential energies f all the particles in a system. E = q + w E is the change in the system s internal energy q represents heat w represents wrk (usually in J r kj) Thermdynamic quantities always cnsist f a number and a sign (+ r ). The sign represents the systems pint f view. (Engineers use the surrundings pint f view) Exthermic q (systems energy is decreasing) Endthermic +q (systems energy is increasing) Calculate E if q = 50 kj and w = +35kJ. AP CHEMISTRY Chapter 6 Therm ntes pg. 1 Kristen Jnes 05/31/14
Fr a gas that expands r is cmpressed, wrk can be calculated by: w = PV units: Latm = (atm)(l) Calculate the wrk if the vlume f a gas is increased frm 15 ml t 2.0 L at a cnstant pressure f 1.5 atm. There are several ways that the enthalpy change f a chemical reactin can be calculated, depending upn the infrmatin available. 1. Stichimetry 2. Calrimetry 3. Hess s Law 4. Heats f Frmatin Stichimetry example: Fr the reactin 2Na + 2H 2 O 2NaOH + H 2, H = 368 kj/ml rxn Calculate the heat change that ccurs when 3.5 g f Na reacts with excess water. Calrimetry - the science f measuring heat flw -based n bserving the temperature change when a bdy absrbs r discharges heat. -instrument is the calrimeter Heat Capacity C = Heat absrbed J r J Incr. in temp. C g C ml -heat capacity f H 2 O is 4.184 J/ g C Cnstant pressure calrimetry- -pressure remains cnstant during the prcess -simple calrimetry -used t determine heats f reactin Energy released as heat = (heat capacity) (mass f slutin) (increase in temp) J = (J/g C) ( g) (T) q = mct A cffee cup calrimeter cntains 150. g H 2 O at 24.6 C. A 110. g blck f mlybdenum is heated t 100 C and then placed in the water in the calrimeter. The cntents f the calrimeter cme t a temperature f 28.0 C. What is the heat capacity per g f mlybdenum? AP CHEMISTRY Chapter 6 Therm ntes pg. 2 Kristen Jnes 05/31/14
4.00g f ammnium nitrate are added t 100.0 ml f water in a plystyrene cup. The water in the cup is initially at a temperature f 22.5 C and decreases t a temperature f 19.3 C. Determine the heat f slutin f ammnium nitrate in kj/ml. Assume that the heat absrbed r released by the calrimeter is negligible. extensive prperty - depends n the amunt f substance intensive prperty - desn t depend n the amunt f substance heat f reactin is extensive temperature is intensive Hess s Law -the change in enthalpy (H) is the same whether the reactin ccurs in ne step r in several steps. -H is nt dependent n the reactin pathway. -The sum f the H fr each step equals the H fr the ttal reactin. 1. If a reactin is reversed, the sign f H is reversed. 2. If the cefficients in a reactin are multiplied by an integer, the value f H is multiplied by the same integer. Given the fllwing reactins and their respective enthalpy changes, calculate H fr the reactin: 2C + H 2 C 2 H 2. C 2 H 2 + 5/2 O 2 2CO 2 + H 2 O H = 1299.6 kj/ml C 2 H 2 C + O 2 CO 2 H = 393.5 kj/ml C H 2 + ½ O 2 H 2 O H = 285.9 kj/ml H 2 The heat f cmbustin f C t CO 2 is 393.5 kj/ml f CO 2, whereas that fr cmbustin f CO t CO 2 is 283.0 kj/ml f CO 2. Calculate the heat f cmbustin f C t CO. Standard enthalpy f frmatin ) -change in enthalpy that accmpanies the frmatin f ne mle f a cmpund frm its elements with all substances in their standard states at 25 C. Standard States- -fr gases, pressure is 1 atm -fr a substance in slutin, the cncentratin is 1 M -fr a pure substance in a cndensed state (liquid r slid), the standard state is the pure liquid r slid. -fr an element, the standard state is the frm in which the element exists under cnditins f 1 atm and 25 C The heat f frmatin f a free element in its standard state is zer. AP CHEMISTRY Chapter 6 Therm ntes pg. 3 Kristen Jnes 05/31/14
*****Values f H f are fund in Appendix 4**** H reactin = H f prducts - H f reactants Cnsider the reactin: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g) H = 1196 kj. Using the infrmatin in the table belw, calculate the H f fr ClF 3 (g). Cmpund NH 3 HF H f 46 kj/ml 271 kj/ml AP CHEMISTRY CHAPTER 17 THERMODYNAMICS Spntaneus prcess (thermdynamically favred) -ccurs withut utside interventin -may be fast r slw Entrpy, S - a measure f randmness r disrder -assciated with prbability (There are mre ways fr smething t be disrganized than rganized.) Entrpy increases ging frm a slid t a liquid t a gas. Entrpy usually increases when slutins are frmed. Entrpy increases in a reactins when mre atms r mlecules are frmed. Entrpy increases with increasing temperature 2nd law f thermdynamics In any spntaneus prcess there is always an increase in the entrpy f the universe. The energy f the universe is cnstant but the entrpy f the universe is increasing. Free energy, G G = H TS Free energy change is a measure f the spntaneity(thermdynamic favrability) f a reactin. It is the maximum wrk available frm the system. A spntaneus(thermdynamically favred) reactin carried ut as cnstant temperature and pressure has a negative G. Fr example, when ice melts H is psitive (endthermic), S is psitive and G = 0 at 0 C. If: S psitive and H negative = spntaneus at all temps S psitive and H psitive = spntaneus at high temp S negative and H negative = spntaneus at lw temp S negative and H psitive = nnspntaneus at all temps. Third Law f Thermdynamics The entrpy f a perfect crystal at 0 K is zer. AP CHEMISTRY Chapter 6 Therm ntes pg. 4 Kristen Jnes 05/31/14
S reactin = S prd S react Ex. Given the fllwing standard mlar entrpies, calculate S fr the reactin: 2Al(s) + 3MgO(s) 3Mg(s) + Al 2 O 3 (s) Al(s) = 28.0 J/K MgO(s) = 27.0 J/K Mg(s) = 33.0 J/K Al 2 O 3 (s) = 51.0 J/K G = standard free energy change -change in free energy that ccurs if the reactants in their standard states are cnverted t prducts in their standard states G =G f prd - G f react at standard cnditins G f fr a free element in its standard state is zer. Ex. Given the equatin N 2 O 4 (g) 2NO 2 (g) and the fllwing data, calculate G. G f fr N 2 O 4 (g) = 97.82 kj/ml G f fr NO 2 (g) = 51.30 kj/ml G =H T S (***When wrking this, change S t kj***) This is called the Gibbs-Helmhltz equatin. Ex. Fr the given reactin and the fllwing infrmatin, calculate G at 25 C. 2PbO(s) + 2SO 2 (g) 2PbS(s) + 3O 2 (g) H (kj/ml) S (J/mlK) PbO(s) 218.0 70.0 SO 2 (g) 297.0 248.0 PbS(s) 100.0 91.0 O 2 (g) ----- 205.0 Remember: H = H f prd H f react Yu can "add" equatins similar t Hess' law methd f H. See example 17.10, pg. 807 The equilibrium pint ccurs at the lwest value f free energy available t the reactin system. G = G prd G react = 0 AP CHEMISTRY Chapter 6 Therm ntes pg. 5 Kristen Jnes 05/31/14
Ex. Given fr the reactin Hg(l) Hg(g) that H = 61.3 kj/ml and S = 100.0 J/mlK, calculate the nrmal biling pint f Hg. G = RT ln K When G = 0, free energy f reactants and prducts are equal when all cmpnents are in their standard states. During a phase change, G = 0. Ex. Calculate the apprximate standard free energy fr the inizatin f hydrfluric acid, HF (K a = 1.0 10 3 ), at 25 C. External surces f energy can be used t drive reactins with psitive G values. 1. Electricity may be used t cause a reactin t ccur thrugh electrlysis. We will lk at this in ur electrchemistry unit (Ch 18). 2. Light may be used (phtinizatin r absrptin f phtns (as in phtsynthesis)) 3. The reactin may be cuple with anther that has a very negative G value, as in the cnversin f ADP t ATP. AP CHEMISTRY Chapter 6 Therm ntes pg. 6 Kristen Jnes 05/31/14