Lecture 4. The First Law f Thermdynamics
THERMODYNAMICS: Basic Cncepts Thermdynamics: (frm the Greek therme, meaning "heat" and, dynamis, meaning "pwer") is the study f energy cnversin between heat and mechanical wrk, and subsequently the macrscpic variables such as temperature, vlume and pressure Energy: capacity t d wrk Wrk: mtin against an ppsing frce System: part f the wrld which we have interest r being investigated. Surrunding: is where we make ur bservatins
Definitin f Wrk: Wrk is mtin against ppsing frce. Wrk is defined as a frce acting thrugh a displacement x, the displacement being in the directin f the frce. w fx
Cnsider a Gas Expansin Wrk: Initial State Final State P 1, V 1, T P 2, V 2, T
Cnsider a Gas Expansin Wrk: w mg( h h ) 1 2 w mgh The external ppsing pressure,, P ex is : w P ex P ex mg A r Ah P w P ex ex V ( V V ) 1 2
Cnsider a Gas Expansin Wrk: Fr at a wrk every instant, P w dne against v v 1 2 P in in dv, w is nly P in infinitetissimal nrt and Pin V v2 dv w nrt v1 V V2 w nrt ln nrt ln V v v 1 1 2 is the P ex dv infinitesimally pressure f increase in vlume P1 P 2 greater the gas, than P ex
Cnsider a Gas Expansin Wrk: the external pressure is infinitetissimally smaller than the internal pressure at all stages f the expansin - reversible prcess. A reversible change in thermdynamics is a change that can be reversed by an infinitesimal mdificatin f a variable. in reversible prcesses, there is maximum amunt f wrk dne that culd pssible extracted frm a prcess
Gas Expansin Wrk: Sample Prblem A sample f 4.50 g f methane ccupies 12.7 L at 310 K. (A)Calculate the wrk dne when the gas expands isthermally against a cnstant external pressure f 200 Trr until its vlume has increased by 3.3 L. (B) Calculate the wrk that wuld be dne if the same expansin ccurred reversibly. A. since ppsing SOLUTION : pressure is cnstant we use the frmula (irreversible prcess) w (200 w P ex V 1 atm trr )(3.3 L) 760 trr w 0.87 L atm : w 0.87 L atm 101.3 J 1 L atm w 87.97 J
Cnsider a Gas Expansin Wrk: SOLUTION : B. fr a reversible gas expansin : w ( 16 4.5 g g ml w nrt ) (8.3145 J K -1-1 V ln V ml w -167.4 J f i -1 ) 310K ln (3.3 12.7) L 12.7L
Definitin f HEAT: Heat is the transfer f energy between tw bdies that are at different temperatures. Heat appears at the bundary f the system. 0th Law: heat is transferred frm the htter bject t the clder ne. Heat is path dependent. m mass, T q mst change in temperature s specific heat
Sample Prblem: Calculate the wrk needed fr a 65 kg persn t climb thrugh 4 m hill a) n the surface f the earth and b) n the mn (g = 1.6 m/s 2 ). If t d this wrk, the persn have t derive the energy frm drinking milk (720 cal/g) and nly 17% f the calric value is cnverted t mechanical energy. Hw much milk des the persn need t drink t d this task n the surface f the earth?
THE FIRST LAW OF THERMODYNAMICS: Energy can neither be created nr destryed but can be cnverted frm ne frm t anther. Law f Cnservatin f Energy. ENERGY OF THE UNIVERSE: and E E univ univ E E sys sys E E sys E surr the changes in energies fr therefre surr E a given system, 0 surr If a system underges an energy change, the surrundings must als underg a change equal in magnitude but ppsite in directin
Ttal ENERGY f a system; Internal Energy, U TOTAL ENERGY: E ttal KE PE U fr a system at rest, KE and E ttal U PE are 0 Internal Energy, U: energy assciated with the chemical system depends n the thermdynamic parameters such as T, P, V, cmpsitin, etc. cnsists f translatinal, rtatinal, vibratinal, electrnic energies, as well as intermlecular interactins
Internal Energy, U: Fr a given system, we d nt knw the exact nature f U, we will be nly interested in the changes in U fr a particular prcess that the system underges. U U 2 U 1 Mathematical expressin f the first law f thermdynamics: r fr an infinitesimal change: U q w du dq dw The change in the internal energy f a system in a given prcess is the sum f the heat exchange, q, between the system and its surrundings and the wrk dne n r by the system.
HEAT and WORK sign cnventins: Prcess Sign Wrk dne by the system n the surrundings - Wrk dne n the system by the surrundings + Heat absrbed by the system frm the surrundings + Heat absrbed by the surrundings frm the system -
Cnstant vlume adiabatic bmb calrimeter Sample inside an xygen filled cntainer is ignited by an electrical discharge. Heat released is measured by the increase in the temperature f the water. U q v U qv w PV, at cnstant V, PV 0 q v Adiabatic means n heat exchange with the surrundings
ENTHALPHY In prcesses carried ut under cnstant P, U q Fr a cnstant P prcess; p 2 U U 1 ( U 2 q v q PV Enthalpy, w PV, r p P( V 2 2 ) ( U V ), 1 1 PV H U PV H U PV dh du PV r fr infinitetisimal change, 1 ) Ttal enthalpy f a system can nt be measured directly, s the change in enthalpy, H, is a mre useful value than H itself.
Entalphy, H vs. Internal Energy, U Cnsider the fllwing reactin: 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) Heat evlved by the reactin is 367.5 kj, reactin takes place at cnstant pressure, Internal Energy U: q p = H = -367.5KJ. U H PV The vlume f the H 2 (1 mle) generated by the reactin ccupies 24.5 L, therefre PV = -24.5 L at r -2.5 kj Finally, - Difference is due t expansin wrk. U = -367.5 kj 2.5 kj U = -370 kj.
Sample Prblem: A 0.4089 g sample f benzic acid was burned in a cnstant vlume bmb calrimeter. Cnsequently, the temperature f the water in the inner jacket rse frm 20.17 C t 22.22 C. If the effective heat capacity f the bmb calrimeter plus water is 5267.8 J/K. calculate the values f U and H fr the cmbustin f Benzic acid in KJ/ml.
SAMPLE PROBLEM: Cmpare the difference between H and U fr the fllwing physical changes: A) 1 mle f ice t 1 mle f water at 273 K and 1 atm. and b) 1 ml water t 1 mle steam at 373 K and 1 atm. The mlar vlumes f water at 273 K are 0.0196 L/ml and 0.0180 L/ml, respectively, and the mlar vlumes f water and steam at 373 K are 0.0188 L/ml and 30.61 l/ml respectively.
Cnstant Pressure Calrimeter: A cnstant-pressure calrimeter made f tw plastic cups. The uter cup helps insulate the reacting mixture frm the surrundings. Tw slutins f knwn vlume cntaining the reactants at the same temperature are carefully mixed in the calrimeter. The heat prduced r absrbed by the reactin can be determined by the temperature change, the quantities and specific heats f the slutins used, and the heat capacity f the calrimeter.
HEAT CAPACITIES: Calrimetry: measurement f heat changes in chemical and physical prcesses. When heat is added t the system, the crrespnding temperature rise will depend n a)the amunt f heat delivered, b)the amunt f the substance, c)the chemical nature and the physical state f the substance, d)the cnditins at which the energy is added t the system. Thus fr a given amunt f a substance, change in temperature is related t the heat added: q CT, r C q T J K -1, C is heat capacity C n C q nt J K -1 ml -1, C is mlar heat capacity
HEAT CAPACITY at cnstant vlume at cnstant vlume, U qv U Cv T T du C dt U T 2 CvdT Cv( T T1 ) 2 T v 1 r q v nc v T
HEAT CAPACITY at cnstant pressure prcesses at q p C p T dh C dt H cnstant T T p H T 2 C pdt C p ( T T1 ) 2 1 pressure r prcess, nc H p T fr gases, C p > C v because f the wrk t be dne t the surrundings in cnstant pressure prcesses fr cndensed phases, C p and C v are identical fr mst purpses q p fr ideal gases, C p C v R
SAMPLE PROBLEM Calculate the values f U and H fr heating f 55.40 g f Xenn fr 300 K t 400 K. Assume ideal gas behavir and the heat capacities are independent f temperature. Xenn is a mnatmic gas. Its C v is equal t 12.47 J K -1 ml -1.
SAMPLE PROBLEM: When 229 J f energy is supplied as heat at cnstant pressure t 3.00 mle CO 2 (g), the temperature f the sample increases by 2.06 K. Calculate the mlar heat capacities at cnstant vlume (C v ) and cnstant pressure (C p ) f the gas.
SAMPLE PROBLEM: Use the C p and C v values frm the previus prblem t calculate the change in a) mlar enthalpy b) mlar internal energy when carbn dixide is heated frm 15 C (temperature when air is inhaled) t 37 C (bld temperature, temperature in ur lungs).
GAS EXPANSION: ISOTHERMAL EXPANSION Isthermal The heat transfer a slw rate that prcess : int r ut the thermal temperature is held f cnstant. the system typically equilibrium is maintained. must happen at such Fr ideal gases, the internal energy is dependent n T, Fr isthermal then, U 0 therefre it fllws that w - q expansin by an ideal gas, the heat absrbed is equal t the wrk dne by the H U PV ideal gas n its since U 0, and PV is cnstant at cnstant T, surrundings. then H 0
GAS EXPANSION: ADIABATIC EXPANSION Adiabatic prcess : n heat exchange between system and surrundings q 0, V V 1 2 P P 2 1 where C C p v r P1V 1 P2V 2 fr mnatmic gases,c s that 5 3 fr diatmic gases, s that 7 5 C v v 3 2 R and C 5 2 R and C p p 5 2 R, 7 5 R
SAMPLE PROBLEM: Calculate the values f U, and H fr the reversible adiabatic expansin f 1.0 mle f a mnatmic ideal gas frm 2 L t 6 L. The temperature f the gas is initially 298 K.
THERMOCHEMISTRY: Heat f reactin is the heat change in the transfrmatin f reactants at a given temperature and pressure t prducts at the same cnditins. Fr cnstant pressure prcesses, the heat f reactin q p is equal t enthalpy change f the reactin, r H. Exthermic reactins : surrundings. Endthermic reactins : frm the surrundings. r system gives ff H is negative. system absrbs heat r H is psitive. heat t
THERMOCHEMISTRY: 2H 2 (g) + O 2 (g) 2HO 2 (g) 241.8 kj f heat is given ff. The enthalpy change fr this prcess is called standard enthalphy f reactin, r H In general, standard enthalpy change f a chemical reactin is the ttal enthalpy f the prducts minus the ttal enthalpy f the reactants H r H vh ( prducts ) is the standard mlar enthalpy; v is stichimetric cefficient vh ( reactants )
THERMOCHEMISTRY: aa bb cc dd the standard enthalpy f the reactin is : r H ch r in general; (C ) dh (D) (ah ( A) bh (B)) r H v f H ( prducts ) v f H (reac tan ts ) H f is the standard mlar enthalpy f frmatin. - is the enthalpy change when 1 mle f a cmpund is frmed frm its cnstituent elements at 1 bar and 298 K.
THERMOCHEMISTRY: fr elements in their mst stable alltrpic at a particular temperature : f f H H : (H 2 (g )) 0; f H (O 2 f H (g )) 0 0 frms 2H 2 (g) + O 2 (g) H 2 O(g) H f 241.8 kj/mle
MEASUREMENTS OF f H DIRECT METHOD: measure f H f direct synthesis frm their elements: 2H 2 (g) + O 2 (g) H 2 O(g) r H r H 2 f H v f H ( prducts ) (H O(g )) 2 r H 2 f H (H 241.8 v 2 f H (g )) (reac tan ts ) f H (O 2 (g ))
SAMPLE PROBLEM: Calculate the reactin enthalpy f the fllwing reactin: 3HN 3 (l) + 2NO(g) H 2 O 2 (l) + 4N 2 (g) Using their standard enthalpies f frmatin. f H values: kj/ml HN 3 (l) 264.0 NO(l) 90.25 H 2 O 2 (l) -187.78
MEASUREMENTS OF f H INDIRECT METHOD: HESS S LAW: When reactants are cnverted t prducts, the change in enthalpy is the same whether the reactin takes place in ne step r in a series f steps. fh fr carbn mnxide: C(graphite) + 1/2O 2 (g) CO(g) C(graphite) + O 2 (g) CO 2 (g) H r 393.5 kj/mle CO + 1/2O 2 (g) CO 2 (g) H r 283.0 kj/mle
MEASUREMENTS OF f H fh fr carbn mnxide: C(graphite) + O 2 (g) CO 2 (g) H r 393.5 kj/mle CO 2 (g) 1/2O 2 (g) + CO(g) H r 283.0 kj/mle C(graphite) + 1/2O 2 (g) CO(g) H r 110.5 kj/mle
MEASUREMENTS OF f H C(graphite) + O 2 (g) CO 2 (g) rh 393.5 kj/mle C(graphite) + O 2 (g) CO 2 (g) + 393.5kJ Enthalpy changes are additive. the H f the reverse reactin will have the ppsite sign r multiplying a reactin by a factr, yu als multiply the r H by the same factr. Hess s Law can be rephrased as: The standard enthalpy f an verall reactin is the sum f the standard enthalpies f the individual reactins int which the reactin might be divided.
SAMPLE PROBLEM: Calculate the standard mlar enthalpy f frmatin f acetylene (C 2 H 2 ) frm its elements. 2C(graphite) + H 2 (g) C 2 H 2 (g) The equatins fr cmbustin and the crrespnding enthalpy changes are: (1) C(graphite) + 1/2O 2 (g) CO 2 (g) (2) H 2 (g) + 1/2O 2 (g) H 2 O(l) H r H r 393.5 kj/mle 285.8 kj/mle (3) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(l) H r 2598.8 kj/mle
SAMPLE PROBLEM: fh fr carbn mnxide: C(graphite) + O 2 (g) CO 2 (g) H r 393.5 kj/mle CO 2 (g) 1/2O 2 (g) + CO(g) H r 283.0 kj/mle C(graphite) + 1/2O 2 (g) CO(g) H r 110.5 kj/mle
SAMPLE PROBLEM:
SAMPLE PROBLEM: Use the thermchemical equatin shwn belw t determine the enthalpy f the reactin: H 2 (g) + 1/2O 2 (g) H 2 O(g) CO 2 (g) C(s) + O 2 (g) f H = +590.2 kj/ml C 2 H 6 (g) 2C(s) + 3H 2 (g) = 127.0 kj/ml C 2 H 6 (g) + 7/2O 2 (g) 2CO 2 (g) + 3H 2 O (l) =-2340.7 kj/ml
TEMPERATURE DEPENDENCE OF ENTHALPY CHANGE KIRCHHOFFS s LAW: The difference in enthalpies f a reactin at tw different temperatures, T 1 and T 2, is just the difference in the enthalpies f heating the prducts and reactants frm T 1 t T 2. r H 2 rh1 C p 2 T ( T ) 1 Where C p is the difference in mlar heat capacities between the prducts and reactants
SAMPLE PROBLEM: The standard enthalpy change fr the reactin: 3O 2 (g) 2O 3 (g) Is given by r H = 285.4 kj/ml at 298 K and 1 bar. Calculate the value f r H at 380 K. Assume that the C p values are all independent f temperature.
Bnd Dissciatin Enthalpy: N 2 (g) 2N(g) Bnd Enthalpy: change in enthalpy when bnds frm r break in diatmic mlecule Bnd Dissciatin Enthalpy: average change in enthalpy in plyatmic mlecules when individual bnds frm r break. H 2 O(g) H(g) + OH(g) OH(g) H(g) + O(g) r H = 502 kj/ml r H = 427 kj/ml
SAMPLE PROBLEM:
ENTHALPY CHANGE and BOND ENERGIES: The Enthalpy f a Reactin can be estimated by the enthalpies f the ttal number f bnds brken and frmed in the reactin. H BE(reactants) BE(prducts) r ttal energy input - ttal energy released
Bnd Enthalpy Changes
SAMPLE PROBLEM: Estimate the enthalpy f cmbustin f methane: CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Using bnd enthalpies in Table 4.4. Cmpare yur result with that calculated frm the enthalpies f frmatin f prducts and reactants.
End f Chapter