Unit 14 Thermochemistry Notes

Transcription

1 Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website:

2 Page 2 f 12 Unit 14 Ntes 14.1 WHAT IS THERMOCHEMISTRY? THERMOCHEMISTRY is the study f the transfers f energy as heat that accmpany chemical reactins and physical changes (changes in state). In Thermchemistry, we will lk at hw heat flws between a system and its surrundings. The system is the bject r bjects yu are investigating. The surrundings are everything else. In thermchemical equatins, heat flw is given frm the pint f view f the system. EXAMPLE 1: Ice melts System: water Surrundings: everything else Heat + H2O (s) H2O (l) ENDOTHERMIC Heat is a reactant Reverse the prcess Water freezes System: water Surrundings: everything else H2O (l) Heat + H2O (s) EXOTHERMIC Heat is a prduct EXAMPLE 2: Anther example is a burning lg that is releasing heat, but that energy was cntained in the chemical bnds as ptential energy f the lg befre it began t burn.

3 Unit 14 Ntes Page 3 f 12 Energy, Heat, and Temperature Energy the capacity fr ding WORK r prducing HEAT. Energy has n MASS Law f the Cnservatin f Energy - Energy is neither created nr destryed, which is the basis fr the study f thermchemistry. Chemical Bnd Energy Ptential Energy stred in the bnds f chemical cmpunds. Heat (H) ENERGY that transfers frm ne bject t anther because f a TEMPERATURE difference between them. The standard internatinal unit (SI unit) f heat is the Jule ( J ). Heat as energy can als be measured in calries (remember, Calries n fd label are kilcalries!) Heat is NOT temperature! *Heat flws frm a warmer bject _t_ a cler bject. Temperature (T) measure f the average kinetic energy f the particles in a sample f matter. Units f Temperature are C r K A change temperature f 1K is the same magnitude as a change in temp f 1 C Heat Energy transferred during a reactin Every reactin invlves a change in chemical ptential energy. Reactants transfrm int prducts which either have GREATER r LESS ptential energy than the reactants. S, in any reactin there is always either a ABSORBTION r RELEASE f energy. This change is called the Change in Enthalpy f Reactin. (In the past it was als called heat f reactin.)

4 Page 4 f 12 Unit 14 Ntes Change in Enthalpy f Reactin (H) = the CHANGE f heat fr a reactin under cnstant pressure. Energy released r Energy absrbed H = H prducts - H reactants Heat lst t surrundings EXOTHERMIC Heat gained by system ENDOTHERMIC Heat flws ut f system Heat flws int the system Heat expressed n the Prduct side f rxn Heat expressed n the Reactant side f rxn - ΔH + ΔH ΔH +ΔH Ptential Energy f system Thermal energy f surrundings FEELS HOT Ptential Energy f system Thermal energy f surrundings FEELS COLD Practice: The graph shws the energy change during the reactin A + B C.

5 1) The prducts have ( MORE / LESS ) ptential energy than the reactants. Unit 14 Ntes Page 5 f 12 2) The reactin is ( ENDOTHERMIC / EXOTHERMIC). S, Heat is _GAINED BY THE SYSTEM. Practice: On the graph belw, draw the reactin prgress f an exthermic reactin. Practice: On the graph belw, draw the reactin prgress f an endthermic reactin.

6 Page 6 f 12 Unit 14 Ntes 14.2 ENTHALPIES OF REACTION Calrimetry is the accurate and precise measurement f HEAT_ change fr chemical prcesses. Calrimetry is based n the fllwing principle. Heat LOST by a system = Heat GAINED by its surrundings q lst by system = q gained by calrimeter In rder fr the prcess t be accurate and precise it must be carried ut in a device called a CALORIMETER. Cnstant vlume (bmb) calrimetry: 1) Allws reactin t take place inside an enclsed cntainer (bmb) which is surrunded by water 2) Measures the temperature change f the water. The bmb absrbs sme energy s its heat capacity must be cnsidered in calculatins. 3) Uses the specific heat equatin, q=mcprt, t calculate the heat gained r lst by the calrimeter. A thermchemical equatin is a chemical equatin fr a reactin that includes HEAT. *Thermchemical equatins must be BALANCED.. *Treat the cefficients as number f mles, never as number f mlecules *Thermchemical equatins treat enthalpy change (_H ) just like any ther REACTANT r PRODUCT Practice 1. 4Fe (s) + 3O2 (g) 2Fe2O3 (s) kj Des this reactin release heat r absrb heat? RELEASES Hw much? 1625 kj What des kj mean? 1000 JOULES (measurement f ENERGY ) Endthermic r exthermic? EXOTHERMIC

7 Unit 14 Ntes Page 7 f C (s) + 2S (s) kj CS2 (l) Is heat released r absrbed in this thermchemical reactin? ADSORBED Hw much? 89.3 kj Endthermic r exthermic? ENDOTHERMIC Thermchemical reactins are nrmally written by designating the value f H, rather than by writing the energy as a reactant r prduct. 2H 2(g) = O 2(g) -> 2H 2O (g) H = kj A negative H indicates a exthermic reactin. Chemical equatins invlving H are similar t _STOICHIOMETRY prblems; they depend n the number f MOLES f reactants and prducts invlved, fr example: CaO(s) + H2O(l) Ca(OH)2 (s) kj and 2 CaO(s) + 2 H2O(l) 2 Ca(OH)2 (s) kj Practice H2 (g) + F2 (g) 2HF(g) H = -536 kj Calculate the heat change (in kj) fr the cnversin f 10.1 g f H 2 gas t HF gas at cnstant pressure. 1 mle H2 563 kj 10. 1g x x = 2680 kj released g H2 1 ml H2 Practice 2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe (s) H = -851 kj Calculate the heat change (in kj) fr the thermite reactin f 320 g f Fe 2O 3 int Al 2O 3 at cnstant pressure 1 mle Fe2O3 320 g Fe2O3 x g Fe2O3 x 852 kj = 1643 kj r 1600 kj absrbed 1 ml Fe2O3

8 Page 8 f 12 Unit 14 Ntes Practice K 2O (s) + H 2O (l) 2KOH (aq) H = 215 kj What is the heat change fr the abve reactin, at cnstant temperature pressure if yu begin with g f K 2O? 1 mle K2O 282 g K2O x g K2O x 215 kj = 853 kj absrbed 1 ml K2O Standard Heat f Frmatin (H f ) f a cmpund is the H that results frm the frmatin f ne mle f cmpund frm its elements at 25 C and 1atm. N2 (g) + O2 (g) NO2 Hf = kj/ml NOTE: the Hf f a ELEMENT OR DIATOMIC GAS in its standard state _IS ZERO. S... Hf f Fe(s) = 0 Hf f O2 (g) = 0 Hf f H2 (g) = 0 Hf f Na (s) = 0 Questin: - Hw d yu determine the Enthalpy f Reactin that yu d nt knw? Answer: Use Hess s Law! Hess s Law - The verall enthalpy change in a reactin is equal t the sum f the enthalpy changes fr the individual steps in the prcess. Hess s Law als states the enthalpy difference between reactants and prducts is INDEPENDENT f pathway. Imprtance f Hess s Law Any ENTHALPY OF REACTION may be calculated using enthalpies f frmatin fr all the substances in the reactin withut knwing anything else abut hw the reactin ccurs. Mathematically, the verall equatin fr enthalpy change is H = sum f [(H f f prducts) x (ml f prducts)] sum f [(H f f reactants) x (ml f reactants)]

9 Unit 14 Ntes Page 9 f 12 Example 1 - Calculate the Enthalpy f Reactin (H ) (Remember that H f f elements and diatmic gasses are zer.) NO(g) + ½ O 2 (g) -> NO 2 (g) Given: ½ N 2 (g) + ½ O 2 (g) -> NO(g) H f = kj ½ N 2 (g) + O 2 (g) -> NO 2(g) H f = kj Answer: Methd 1 Cmbine Thermchemical Equatins NO(g) ½ N 2 (g) + ½ O 2 ½ N 2 (g) + O 2 (g) NO 2(g) NO(g) + ½ O 2 (g) -> NO 2 (g) H f = kj H f = kj H f = kj Methd 2, use equatin fr calculating H : H = sum f [(Hf f prducts) x (ml f prducts)] sum f [(Hf f reactants) x (ml f reactants)] H = [33.2 kj x 1] [90.29kJ x 1] = kj Practice - Calculate the Enthalpy f Reactin (H ) (Hint: Balance the reactin first!) 2_ CO(g) + 1_ O2 (g) 2_ CO2 (g) Given: C(s) + ½ O2 (g) -> CO(g) Hf = -110 kj C(s) + O2(g) -> CO2(g) Hf = -393 kj Answer: Methd 1 Cmbine Thermchemical Equatins 2 CO(g) 2 C(s) + 1 O2 (g) Hf = +220 kj (Nte that yu need 2X f Hf ) 2 C(s) + 2 O2(g) 2 CO2(g) Hf = -786 kj (Nte that yu need 2X f Hf ) 2 CO(g) + 1 O2 (g) 2 CO2 (g) Hf = -566 kj Methd 2, use equatin fr calculating H : H = [-393 kj x 2] [-110 kj x 2] = kj

10 Page 10 f 12 Unit 14 Ntes 13.3 DRIVING FORCES OF REACTIONS The majrity f chemical reactins are EXOTHERMIC, where prducts are mre stable. In nature, reactins tend t prceed in a directin that leads t lwer energy state. Hwever, sme endthermic reactins d ccur SPONTANEOUSLY! Questin: Why d sme reactins ccur spntaneusly that are endthermic? Answer: Tw factrs determine whether a reactin will ccur spntaneusly. 1. Change in ENTHALPY and 2. Change in RANDOMNESS f the particles in the system, which is knwn as ENTROPY Example f Endthermic Spntaneus Change: Melting Ice Ice will melt SPONTANEOUSLY at rm temperature by absrbing heat frm the surrundings even thugh this is endthermic prcess. The well-rdered arrangement f water mlecule in ice crystals is lst, and the less-rdered liquid phase f higher energy is frmed. There is tendency in nature t prceed in directin that INCREASES the randmness f a system. Entrpy (S) is the measure f the RANDOMNESS f the particles, such as mlecules, in a system. In general S(gases) > S(liquids) > S(slids) Entrpy Change (S) can be measured fr reactins and is the difference between the entrpy f the prducts and the reactants.

11 Unit 14 Ntes Page 11 f 12 Practice: Predict whether the value f S fr each f the fllwing reactins will greater than, less than, r equal t zer. 3H2 (g) + N2 (g) -> 2 NH3 (g) 2Mg (s) + O2 (g) -> 2 MgO (s) C6C12O6(s) + O2(g) -> 6 CO2(g) + 6 H2O(g) KNO3(s) - > K + 1(aq) + NO3-1 (aq) S is less than zer S is less than zer S is greater than zer S is greater than zer Free Energy Prcesses in nature are driven in tw directins: tward least ENTHALPY and tward greatest ENTROPY A functin called Gibbs Free Energy (G ) has been defined that simultaneusly assesses the tendency fr enthalpy and entrpy t change. Natural prcess prceed in the directin that lwers the Gibbs Free Energy. G = H - TS The fllwing table shws the fur pssible scenaris fr Gibbs Free Energy. Remember that fr a reactin t be spntaneus, the Gibbs Free energy shuld be NEGATIVE fr the reactin. H S G - value (exthermic) + value (mre randm) always negative - value (exthermic) - value (less randm) negative at lwer temperatures + value (endthermic) + value (mre randm) negative at higher temperature + value (endthermic -value (less ransm) never negative Practice: Predict whether the reactins will be spntaneus at 298K by calculating G frm the given enthalpies and entrpies. 1. C 2H 2(g) + H 2O (g) -> C2H6 (g) H = kj/ml and -S = kj/(ml*k) G = H - TS = kj/ml (298K x kj/(ml*k) = kj/ml (Spntaneus) 2. CH 4(g) + H 2O(g) -> CO(g) + 3 H 2(g) H = kj/ml and -S = kj/(ml*k) G = H - TS = kj/ml (298K x kj/(ml*k) = +270 kj/ml (Nt Spntaneus)

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