Combustion Chemistry

Size: px
Start display at page:

Download "Combustion Chemistry"

Transcription

1 Cmbustin Chemistry Hai Wang Stanfrd University 2015 Princetn-CEFRC Summer Schl On Cmbustin Curse Length: 3 hrs June 22 26, 2015 Cpyright 2015 by Hai Wang This material is nt t be sld, reprduced r distributed withut prir written permissin f the wner, Hai Wang.

2 Lecture 1 1. THERMOCHEMISTRY Thermdynamics is the fundatin f a large range f physical science prblems. It prvides us with a basic understanding abut the driving frce f a physical prcess and the limits f such a prcess. The develpment f thermdynamic thery was intimately related t cmbustin. In particular, the secnd law f thermdynamics was cnceived largely t prve that it is impssible t perate a perpetual mtin machine. 1.1 Thermdynamic First Law The first law f thermdynamics states that the energy is cnserved when this energy is transfrmed frm ne frm t anther. In the cntext f cmbustin analysis, we state that fr a cntrl mass (r the wrking fluid), Q W = ΔU (1.1) where Q (kj r kcal) is the heat transferred frm the wrking fluid t the surrunding, W (kj r kcal) is the wrk dne by the wrking fluid t the surrunding, and U (kj r kcal) is the internal energy f the wrking fluid. It is imprtant t mentin that energy transfrmatin always invlves a prcess that has a initial state (1) and a final state (2). ΔU = U 2 U 1 is therefre the change f internal energy f the wrking fluid frm the initial state t the final state. Thermdynamic analysis fllw the cnventin that if the wrking fluid gives ff heat t the surrunding, Q < 0, and if the wrking fluid receives heat frm the surrunding, Q is psitive. Fr example, in a simple cling prcess, a fluid lses its internal energy t the surrunding (i.e., lwering its temperature) and thus ΔU < 0. Assuming that n wrk is dne (W = 0), then Q < 0. Likewise, if the wrking fluid des net wrk t the surrunding, W is psitive, and if the surrunding des net wrk t the wrking fluid, W < 0 (e.g., fr adiabatic cmpressin (Q = 0), the wrk dne by the surrunding t the wrking fluid serves t raise the internal energy f the wrking fluid, since W = ΔU > 0. The symbl U designates the internal energy f a given mass f a substance. Internal energy is a measure f the ttal energy f the cntrl mass. Fr example, excluding nuclear energy the internal energy f air is a sum f the kinetic energy f each atm. Therefre, the internal energy can be made a material prperty if it is defined as the internal energy per mass, an intensive prperty, dented here as u (kj/kg) r u (kj/kml). In this curse, we shall fllw the ntatin that intensive prperties are expressed in lwer cases. Assuming that we are running a thermdynamic prcess that the nly wrk dne during the prcess is that assciated with bundary wrk under a cnstant pressure P 1 = P 2 = P (e.g., a pistn wrk), the wrk dne may be calculated frm 1-1

3 Stanfrd University Versin 1.2 Putting Eq. (1.2) int Eq. (1.1), we have W = PdV = P( V V ). (1.2) Q = ( U 2 + P 2 V 2 ) ( U 1 + P 1 V 1 ), (1.3) where V is the vlume f the wrking fluid. In this case, the heat transferred during the prcess crrespnds t a net change f the cntrlled mass in the quantity U + PV between the initial and final states. We find it cnvenient t define a new thermdynamic prperty, the enthalpy H = U + PV, (1.4a) h = u + Pv, (1.4b) h = u + Pv. (1.4c) Here v and v are the specific vlumes, having the units (m 3 /kg) and (m 3 /kml) respectively. Clearly, these specific vlumes are related t the mass density ρ and mlar density (r cncentratin) c, respectively, i.e., v = 1/ρ and v = 1/c. In general, the internal energy u and enthalpy h depend n nly tw independent prperties that specifying the thermdynamic state, e.g., (T, P), (T, v), r (P, v). Fr a lw-density gas like air r cmbustin gases, T, P, and v are related by the ideal gas law r the equatin f state, Pv = R ' T, (1.5a) Pv = R u T, (1.5b) where R u is the universal gas cnstant (8.314 kj/kml-k), R is the specific gas cnstant and equal t R u/mw, and MW is the mlecular weight f the substance. Fr a lw-density gas, the internal energy is primarily a functin f T, i.e., u u( T ). This relatinship may be expressed by defining a cnstant-vlume specific heat c v (kj/kml-k) c v u = T v. (1.6) Fr an ideal gas we have du = c v dt. Likewise, the relatinship between enthalpy and temperature may be established by defining a cnstant-pressure specific heat c p (kj/kml-k) " c p = h % $ ' # T & p, (1.7) 1-2

4 Stanfrd University Versin 1.2 and dh = c p dt. In ther wrds, the tw specific heats defined abve characterize the heat required t raise the temperature f a substance by 1 K. Since fr an ideal gas dh = du + d( pv ) = du + RudT and u u( T ), we see that h and c p are als functin f temperature nly. The relatin between dh and du als yields cp = cv + R u. Here it is imprtant t nte that the enthalpy discussed thus far invlves nly the heating r cling a substance. This type f enthalpy is knwn as the sensible enthalpy r sensible heat. Later, we will intrduce tw ther types f enthalpy, ne f which is critical t cmbustin prblems. The first law f thermdynamics is quite insufficient t describe energy cnversin. Equatin (1.1) states that it is pssible t cl a substance f a given mass spntaneusly (i.e., lwering its internal energy U), and transfer this energy t the surrunding. In ther wrds, within the first law f thermdynamics, it is pssible t transfrm heat frm a lwtemperature bdy t a high temperature bdy. We knw that this cannt be true. The secnd law f thermdynamics, t be discussed belw, will address this prblem. 1.2 Thermdynamic Secnd Law and Entrpy In cntrast t the first law f thermdynamics, the secnd law is mre difficult t understand. The Kelvin-Planck statement f this law is It is impssible t cnstruct a device that will perate in a cycle and prduce n effect ther than the raising f a weight and the exchange f heat with a single reservir. In ther wrds, it is impssible t cnstruct a heat engine that (a) receives heat cntinuusly frm a heat reservir, (b) turns the heat transferred entirely t wrk, (c) withut having t leave any marks n the surrunding. Withut diverging int a lengthy discussin f the secnd law f thermdynamics, let us define entrpy S (kj/k) as δ Q S = T int rev. (1.8) where (δq) int rev is the heat a cntrl mass received during an infinitesimal, internally reversible prcess. Based n an analysis f thermdynamic cycles, it may be shwn that fr a spntaneus prcess t ccur, the entrpy f the cntrl mass must be equal t r greater than zer, ΔS = S 2 S 1 0. (1.9) Neither the Kelvin-Planck statement nr Eq. (1.8) really tells us what entrpy is. An understanding f entrpy will have t cme smetime later when we intrduce statistical thermdynamics. Here let us place sme discussin abut entrpy in a nn-rigrus fashin. Entrpy is a measure f mlecular randmness. This randmness may be measured by the predictability f the psitins f atms in a substance. A crystal material wuld have a small entrpy because atms are mre r less lcked int the crystal lattice. In fact, the third law f thermdynamics states that the entrpy f a pure crystalline substance at abslute 1-3

5 Stanfrd University Versin 1.2 zer temperature is zer. In ther wrds, the atms in a pure crystal are frzen (n scillatin) at 0 K. Therefre, their spatial psitin is cmpletely predictive. In cntrast, a gas wuld have a large entrpy because mlecules that make up the gas cnstantly mve abut in the space, resulting in small predictability regarding their psitins. Mrever, an increase in temperature f the gas leads an increase in the speed f mlecular mtin and smaller predictability f the mlecular psitins. In ther wrds, entrpy increases with an increase in temperature. In cntrast, an increase in pressure leads t clser spacing amng mlecules. As a result, the mlecules becme mre cnfined spatially and the entrpy is smaller at higher pressures. The dissciatin f a chemical substance int gaseus fragments always leads t an increase in entrpy since it is harder t predict the spatial psitins f the fragments than the mlecules f their parent substance. The inequality expressed by Eq. (1.9) basically says that fr a spntaneus prcess t ccur, the entrpy f the cntrl vlume must increase, i.e., natural prcesses favr mre randmness than rderness. Cnceptually this makes sense since ur experience tells us that a building can spntaneusly cllapse int a pile f rubble, but a pile f rubble wuld nt spntaneusly transfrm int a building (nt withut ur interventin). Tw different gases, say, N 2 and O 2, wuld always mix and they never spntaneusly separate spatially, leading t better predictability f their psitins. The cncept f entrpy is als deeply rted in ur life. Take the life f a wrkahlic as an example, the first law f thermdynamics states that it is pssible fr him/her t receive heat Q in the frm f fd and hpefully withut gaining weight (ΔU = 0), t transfrm this heat entirely t wrk W. The secnd law says that he/she really cannt d this. That is, my ffice always gets messier ver time and I will need t clean it (i.e., nt all the heat ges t useful wrk) as time ges by. Because entrpy is a measure f randmness, which in turn, is determined by T and P, it is als a material prperty. It fllws that we can define and dente the entrpy f a substance by s (kj/kg-k) r s (kj/kml-k). Althugh we d nt knw fr the time being hw t directly measure entrpy, we may develp sme relatinships that can help us t determine the entrpy value. Here we apply the first law t a cnstant T and P, internally reversible prcess (e.g., cmpress a vlume immersed in a temperature bath by a pistn very slwly), but since δ Qint rev = TdS and δ Wint rev = r δq δw = du (1.10) int rev int rev, PdV, we have TdS = du + PdV (1.11) ds = + = c dt v + T T T T. (1.12) 1-4

6 Stanfrd University Versin 1.2 Replacing u by h Ts and rearranging, we btain ds = dh vdp = c dt p vdp T T T T. (1.13) Applying the ideal gas law, we may rewrite equatins (12) and (13) as One may integrate the abve equatins t shw that dt dv ds = cv + Ru, (1.14) T v dt dp ds = cp Ru T P. (1.15) dt Δ s = s2 s1 = cv + R 1 T 2 2 u ln v1 v, (1.16) Δ = = 2 dt P2 s s2 s1 cp Ru ln 1 T P. (1.17) Equatin (1.17) states that if c p is a cnstant, an increase f temperature by ΔT frm T causes the entrpy t increase by cp ln( 1+ΔT T ) and an increase f pressure by ΔP frm P leads t the entrpy t decrease by Ru ln( 1+ΔP P ). Given the third law f thermdynamics, which establish the abslute zer fr entrpy, the entrpy f an ideal gas at a given thermdynamic state (i.e., knwn T and P) can be easily determined if c p is knwn. Equatin (1.17) als states that unlike enthalpy and internal energy, the entrpy f an ideal gas is a functin f bth temperature and pressure. In applicatin, we define the standard entrpy s as 1 2 dt T2 dt T 1 dt Δ s = s2 s1 = cp = cp Ru ln( P ) cp Ru ln( P ) 1 T 0 0 T T, (1.18) r where T dt s = cp Ru ln P 0 T P is the standard pressure f 1 atm. Hence, ( ), (1.19a) Δ s = T 0 c p dt T, (1.19b) 1-5

7 Stanfrd University Versin 1.2 By tabulating this standard entrpy, we may easily determine the entrpy change f an ideal gas under an arbitrary cnditin by 1.3 Chemical Reactins P s ( T, P) = s ( T) Ru ln P. (1.20) Befre we apply the abve thermdynamic principles t cmbustin analysis, we need t take a mment t review a few aspects f chemical reactins. Frm a prcess pint f view, a chemical reactin may be viewed as the cnversin f reactants that enter int a reactr (the initial state) t prducts that leaves the reactr (the final state). Fr example, methane (CH 4) flws int a reactr with air (21%O 2 and 79% N 2). Suppse the mlar rati f xygen t methane is 2-t-1. We may write that t burn 1 mle f methane, CH O 2 + (2 79/21) N 2 Reactr CO 2 + 2H 2O + (2 79/21) N 2. Here the prducts include 1 mle CO 2, 2 mles f H 2O and (2 79/21) mles f N 2. Of curse, in writing the abve prcess reactin, we may neglect the bx and simply write CH O 2 + (2 79/21) N 2 CO 2 + 2H 2O + (2 79/21) N 2. (1.21) The abve reactin is knwn as the cmplete cmbustin reactin as all the carbn in the fuel is xidized t CO 2 and all the hydrgen is cnverted t H 2O. These cmpunds are called the cmplete cmbustin prducts. If there is n excess xygen (i.e., all xygen is cnsumed in the xidatin prcess), the characteristic fuel-t-xygen mlar rati is knwn as the stichimetric rati (equal t ½ fr methane). The stichimetric rati fr an arbitrary fuel C mh n may be readily determined by writing ut the cmplete, stichimetric reactin, C mh n + (m+ n 4 ) O 2 + (m+ n 4 ) (79/21) N2 m CO 2 + which gives the stichimetric rati equal t 1/(m+ n 4 ). n H 2 2O + (m+ n 4 ) (79/21) N2. (1.22) In a practical cmbustin prcess, hwever, the fuel-t-xygen mlar rati needs nt t be the stichimetric rati. Fr example, a gasline engine ften runs slightly abve the stichimetric rati at the cld start, fr reasns t be discussed later. T characterize fuelt-xygen rati in a practical cmbustin prcess, we intrduce the equivalence rati, defined as the mlar rati f fuel-t-xygen fr an actual cmbustin prcess by that f stichimetric cmbustin: φ = ( mles f fuel mles f xygen ) ( mles f fuel mles f xygen ) act. sti.. (1.23) 1-6

8 Stanfrd University Versin 1.2 Of curse, it may be shwn that the equivalence rati may be calculated using the mlar rati f fuel-t-air r the mass rati f fuel-t-xygen r fuel-t-air. By examining the equivalence rati, we can quickly tell the nature f the fuel/air mixture. That is, if φ = 1, we have stichimetric reactin; if φ < 1 we have excess xygen that is nt cmpletely used in a reactin prcess and the cmbustin is called fuel-lean cmbustin; if φ > 1 we have excess fuel and the cmbustin is called fuel-rich cmbustin. < 1 fuel lean φ = 1 stichimetric > 1 fuel rich Under the fuel rich cmbustin, the cmbustin reactin inherently yields incmplete cmbustin prducts, like CO, H 2 etc. 1.4 Enthalpy f Frmatin, Enthalpy f Cmbustin As we discussed in sectin 1.1, there are 3 types f enthalpy. The first type is assciated with heating r cling f a substance. The secnd type is latent enthalpy (r heat). This is the enthalpy assciated with the phase change f a substance. Fr example, the latent heat f evapratin f H 2O, h lg, is hlg = hg h l, (1.24) where h g and h l are the enthalpy f water in its vapr and liquid states, respectively. What is perhaps mre imprtant t cmbustin analysis is the reactin enthalpy. Fr example, reactin (21) releases an amunt f heat due t chemical bnd rearrangements. Cmbining Eqs (1.3) and (1.4a), we have Q = H2 H 1. Since state 1 crrespnds t the reactants, and state 2 crrespnds the prducts, the abve equatin states that (a) in a nn-adiabatic reactr, the heat released frm the reactr is equal t the ttal enthalpy f the cmbustin prducts subtracted by the ttal enthalpy f the reactant, and (b) since fr a cmbustin prcess Q < 0, H 2 < H 1, i.e., the ttal enthalpy f the prducts is lwer than that f the reactants. The nature f reactin enthalpy is very different frm the sensible enthalpy, as the frmer is due t re-arranging chemical bnds and the latter is simply due t heat and cling withut changing the chemical nature f the substance. T calculate the exact amunt f reactin enthalpy and therefre the amunt f heat release, we need t first understand the cncept and applicatin f enthalpy f frmatin. 1-7

9 Stanfrd University Versin 1.2 The enthalpy f frmatin h f at a given temperature is defined as the heat released frm prducing 1 mle f a substance frm its elements at that temperature. By this definitin, the enthalpy f frmatin is zer fr the reference elements. These elements are, fr example, graphite [dented by C(S) hereafter), mlecular hydrgen H 2, mlecular xygen (O 2), mlecular nitrgen (N 2), and mlecular chlrine (Cl 2). The enthalpy f frmatin f CO 2, say at 298 K, may be cnceptually measured by reacting 1 mle f graphite and 1 mle f O 2 at 298 K, prducing 1 mle f CO 2 at the same temperature: C(S) + O 2 CO 2 + Q, where Q is the heat released frm the abve prcess (Q = kj). Using Eq. (1.24), we have ( CO ) ( ) ( ) ( ) Q = (kj)= H H = 1 hf K CO 1 hf K C(S) + hf K O = h f,298k 2 2 1,298 2,298, The enthalpy f frmatin f CO 2 is therefre h f = kj/ml at 298 K. Likewise the enthalpy f frmatin f CO is determined by measuring the heat release frm and C(S) + ½ O 2 CO + Q ( kj at 298 K), (1.25) h f (CO) = kj/ml at 298 K. The cnceptual definitin uses the same temperature fr the reactr, reactants, and prducts, and this cnditin is knwn as the standard cnditin. Fr this reasn, we use a superscript, i.e., h f, t designate this standard cnditin and the crrespnding enthalpy f frmatin is termed as the standard enthalpy f frmatin. Obviusly if reactin (25) is carried ut at a different temperature under the standard cnditin, we d nt necessarily get the same heat release. In ther wrds, the enthalpy f frmatin is dependent n temperature, yet this temperature dependence is related t sensible heat. T illustrate the relatinship f enthalpy f frmatin f a substance at tw different temperatures, we sketch the fllwing diagram: 1-8

10 Stanfrd University Versin 1.2 elements Δ H = H ( T ) H ( 298 ) elements = h ( T ) h ( 298 ) + h ( T ) h ( 298 ) C(S) O 2 1 h, 2 f T 1 2 h f,298 K Elements Substance (1 mle C(s) + 1 mle O 2 ) (1 mle CO 2 ) T Δ H = ( ) ( ) h T h CO CO K Recgnizing that enthalpy is a state functin, i.e., fr an ideal gas the enthalpy f a substance is fully defined if the temperature is knwn, and the change f enthalpy fr a prcess is independent f the path, we may write H 2' H 1 = H 2' H 2 + H 2 H 1 It fllws that ( ) = " h ( T ) h (298) = H 2' H 1' + ( H 1' H 1 ) = h f,t # ( CO 2 ) + " # h T ( ) $ % + h CO2 f,298 CO 2 ( ) h (298) $ % + " h T C(S) # ( ) h (298). $ % O2 h f,t ( CO 2 ) = h f,298 ( CO 2 ) + " # h ( T ) h (298) $ % CO2 " # h ( T ) h (298) $ % + " h T C(S) # ( ) h (298) $ { % O2 }, which may be generalized t ( ) ν ( ) h, = h,298 + h T h(298) h T h (298) substance, (1.25a) f T f i i elements where ν i is the stichimetric cefficients f the i th elements in the reactin that frms the substance. Therefre if the specific heat r sensible enthalpy f a substance is knwn, we nly need t knw the value f enthalpy f frmatin at a particular temperature, and in general this temperature is equal t 298 K. In cmbustin analysis, we ften grup the first and secnd term f Eq. (1.25a) by defining a ttal enthalpy as and h T h f,298 + # $ h T ( ) h(298) % &, substance (1.25b) h f,t ( ) h(298) = h T ν " i # h T $. (1.25c) elements % i Table 1.2 lists the enthalpy f frmatin f sme imprtant species fr cmbustin analysis. 1-9

11 Stanfrd University Versin 1.2 Table 1.2 Standard enthalpy f frmatin f key cmbustin species in the vapr state Species Name h (kj/ml) f,298 H 2O Water CO Carbn mnxide CO 2 Carbn dixide CH 4 Methane C 2H 6 Ethane C 3H 8 Prpane C 4H 10 Butane C 8H 18 Octane C 2H 4 Ethylene 52.5 C 2H 2 Acetylene CH 3OH Methanl C 6H 6 Benzene 82.9 H Hydrgen atm O Oxygen atm OH Hydrxyl radical 39.0 The standard enthalpy f cmbustin h c (kj/ml-fuel) is defined as the heat released frm the cmplete cmbustin f 1 mle f fuel at 298 K. Cnsider the cmplete cmbustin f methane (Eq. 1.21). We will again utilize the path independent prperty t illustrate that hc may be determined by the sum f enthalpy f frmatin f the prducts (multiplied by the mlar rati f the prduct-t-fuel) subtracted by the sum f enthalpy f frmatin f the reactants: h c CH 4 + 2O /21 N 2 CO 2 + 2H 2 O /21 N 2 Δ H 1 Δ H 2 C(S) + 2H 2 + 2O /21 N 2 ( ) ( ) ( ) h (kj/kml-fuel) =Δ H +Δ H = 1 h CH + 1 h CO + 2 h H O. c 1 2 f,298 4 f,298 2 f,298 2 Fr an arbitrary fuel (C mh n) underging cmbustin (1.22), we determine its enthalpy f cmbustin as 1-10

12 Stanfrd University Versin 1.2 h = ( ) + ( ) c(kj/kml-fuel) m h,298 CO n f 2 hf,298 H2 O h f,298 ( CmHn). 2 In additin, fr an arbitrary reactin given by ' νiai ν ' A ', (1.26) i i react. prd. ' where A and i A are the i i th reactants and prducts, respectively, and ν i are termed as the stichimetric cefficients, we determine the enthalpy f reactin at an arbitrary temperature T by ( ) ν ( ) Δ H = ν h A h A rt, i' T i' i T i prd. react. = νi' hf,298( Ai' ) νihf,298( Ai) + νi' h ( T) h(298) ν ( ) (298) ' i h T h i i prd. react. prd. react. =Δ Hr,298 + νi' h ( T) h(298) ν ( ) (298) i' i h T h i prd. react. Since the ttal numbers f the elements in the reactants and prducts are identical, the sensible enthalpy terms fr the elements in Eq. (1.25c) are canceled ut. If ΔH rt, is psitive, the reactin absrbs heat. This type f reactins is knwn t be endrthermic. If ΔH rt, < 0, the reactin releases heat as it prceeds t cmpletin. This type f reactins is knwn t be exthermic. Cnversely, If ΔH rt, > 0, the reactin requires heat t achieve cmpletin. This type f reactins is knwn t be endthermic. 1.5 Chemical Equilibrium The cmplete cmbustin reactins given by Eqs. (1.21) and (1.22) essentially crrespnd t maximum heat release. That is, if prducts ther than CO 2 and H 2O are frmed, the enthalpy f reactin will be decidedly lwer. In practical cmbustin prcesses, a cmbustin reactin can never reach cmpletin. Rather the prducts f cmbustin will acquire the state f chemical equilibrium. Althugh ften than nt the prducts will be dminated by the cmplete cmbustin prducts, incmplete cmbustin prducts (CO, H 2, st, NO etc) are inherent t a cmbustin prcess. Our experience tells us that a prcess r reactin wuld be spntaneus if it releases heat. Fr example, the cmbustin f methane spntaneusly prduces CO 2 and H 2O ( ΔH r,298 <0), but a mixture f CO 2 and H 2O wuld nt spntaneusly react and prduce methane and O 2. On the ther hand, the entrpy f 1 mle f CO 2 is decidedly smaller than the entrpy fr a mixture made f 1 mle f CO and 0.5 mle f O 2. Likewise the entrpy f

13 Stanfrd University Versin 1.2 mle f H 2O is smaller than the entrpy fr a mixture made f 1 mle f H 2 and 0.5 mle f O 2. Therefre it may be said that reactin (1.21) CH O 2 + (2 79/21) N 2 CO 2 + 2H 2O + (2 79/21) N 2. (1.21) prduces the largest heat but with a smaller entrpy change, whereas reactin (1.27) prduces less heat but with a larger change f entrpy upn reactin: CH O 2 + (2 79/21) N 2 CO + 2H 2 + O 2 + (2 79/21) N 2. (1.27) We learned frm the secnd law f thermdynamics that withut heat release r absrptin, a spntaneus prcess will prceed in the directin t increase entrpy. Therefre a cmprmise between enthalpy and entrpy Figure 1.1. releases Variatin must f be the made. enthalpy, entrpy, and Gibbs This cmprmise is respnsible fr functin chemical fr equilibrium. an exthermic It reactin. may be quantified by the Gibbs functin r Gibbs free energy, G(T,P) = H(T) - TS(T,P) G(T,P) H(T) -TS(T,P) 0 Equilibrium 1 100% Reactants 100% Prducts Reactin Prgress, ε G H TS, (1.28a) g h Ts, (1.28b) g h Ts, (1.28c) Figure 1.1 shws a pssible scenari fr variatin f the enthalpy, entrpy times temperature, Gibbs functin fr an exthermic reactin as it prgresses t cmpletin at a given T and P. We define here a reactin prgress variable ε, such that ε = 0 fr pure reactants and ε = 1 fr pure prducts. If the reactin is exthermic, the ttal enthalpy f the reacting gases (reactants and prducts) decreases as ε increases. Here the spntaneus release f chemical energy is driving the reactin twards cmpletin. If the reactin is accmpanied by a decrease in the entrpy (e.g., H 2 + 1/2O 2 H 2O), the TS(T,P) functin wuld mntnically increase as reactin prgresses. This entrpy reductin gives resistance twards the cmpletin f reactin. Overall the Gibbs functin must decrease initially, but because f the rise f the TS term, it eventually will have t g 1-12

14 Stanfrd University Versin 1.2 up as ε increases. In ther wrds, the Gibbs functin must reach a minimum at sme pint. The definitin f chemical equilibrium is therefre r simply (, ) dg T P dε = 0, (1.29a) dg( T, P ) = 0. (1.29b) Again, the abve equilibrium criterin represents a cmprmise f H and ST, since bth f them prefer t minimize themselves. Therefre, the driving frce f chemical reactin lies in the minimizatin f the Gibbs functin. Nw let us cnsider an arbitrary reactin given by Eq. (1.26). The Gibbs functin f the reacting gas may be written as i i i' i' (, ) = (, ) + (, ) GTP ng TP ng TP, (1.30) react. where n i is the mlar number f the i th species. Putting Eq. (1.30) int (1.29a), we btain, fr cnstant T and P, (, ) Cnservatin f mass requires that prd. dg T P dni dni' = gi ( T, P) + gi' ( T, P) = 0, (1.31) dε dε dε react. prd. 1 dn 1 dn 1 dn 1 = 2 N 1 dn 1 dn 1 dn = = 1' = 2' N... = ' = γ ( ε), (1.32) ν dε ν dε ν dε ν dε ν dε ν dε 1 2 n 1' 2' n' where N and N are the ttal numbers f reactants and prducts, respectively, and γ( ε ) is a functin that depends nly n ε. Cmbining equatins (1.31) and (1.32), we btain γ( ε) νigi ( T, P) + νi' gi' ( T, P ) = 0. (1.33) react. prd. Since γ( ε) 0, we see that equilibrium state is given by i i i' i' ( ) ν ( ) ν g T, P + g T, P = 0. (1.34) react. The functin g i is the Gibbs functin f species i, which may be expressed by prd. 1-13

15 Stanfrd University Versin 1.2 P gi( T, P) = hf ( T) Tsi ( T) = hf ( T) T si ( T) Ru ln. (1.35) P We nw define a standard Gibbs functin (, = 1 atm) and re-write Eq. (1.35) as g T P as ( ) = ( ) ( ) g T h T Ts T, (1.36) f i Putting Eq. (1.37) int (1.34) and rearranging, we have P gi ( T, P) = g ( T) + RuT ln. (1.37) P P P i νi i νi i u νi νi ln, (1.38) P P i' ' g '( T) g ( T) = R T ' ln prd. react. prd. react. where P i is the partial pressure f species i, and f curse, P = 1 atm. The left-hand side f the abve equatin may be defined as the standard Gibbs functin change f reactin, ( ) ν ( ) ν ( ) ΔG T g T g T (1.39) r i' i' i i prd. react. The right-hand side f Eq. (1.38) may be re-arranged t yield r K p ν i ' i ' prd. Δ Gr ( T) = RuT ln ν i Pi react. ( T) P νi ' Pi ' prd. ΔGr = exp ν i Pi RuT react. ( T). (1.40). (1.41) where K p(t) is the equilibrium cnstant f the reactin. Nte that by neglecting (1.40) and (1.41), we have frced P i t take the unit f atm. P in Eqs. The equilibrium cnstant may als be defined by the cncentratins f the reactants and prducts, 1-14

16 Stanfrd University Versin 1.2 c ν i' i' P ν i' i' prd. prd. Δν ( ) = ( ) = ( )( ) K T R T K T R T c νi ν u p u i ci Pi react. react. Δν, (1.42) where Δ ν = v v. i' prd. react. There are several imprtant facts abut the equilibrium cnstant. (a) While K p is defined as the pressure rati f the prducts and reactants (Eq. 1.41), this equilibrium cnstant is a functin f temperature nly. i (b) Cnsider the reactin H 2O = H 2 + ½ O 2. (1.43f) The equilibrium cnstant fr the frward directin f the reactin is K p, f P P 12 H2 O2 ( T) = P HO 2. We may als write the reactin in the back directin, H 2 + ½ O 2 = H 2O, (1.43b) and its equilibrium cnstant K P 2 ( T) = HO pb, 12 PH P 2 O2. Obviusly, K p, f ( T) (c) Reactin (43f) may be written alternatively as 1 = K T. pb, ( ) 2H 2O = 2H 2 + O 2, (1.43f ) with its equilibrium cnstant K P P 2 ' p, f 2 PHO 2 H2 O2 ( T) =. 1-15

17 Stanfrd University Versin 1.2 Cmparing the equilibrium cnstants fr the tw frward reactins, we see that (d) Cnsider the fllwing tw reactins ( ) ( ) ' 2 K T = K T. p, f p, f H 2 = 2 H. H 2O = H + OH (1.44f) (1.45f) (where the dentes that the species is a free radical). We have K p,44 f P P 2 H ( T) = H2 K P P = P H OH and ( ) p,45 f T HO 2. A linear cmbinatin f reactins (43f-45f) yield H 2O = ½ H 2 + OH. (1.46f) A little algebra tells us that (e) While K p is nt a functin f pressure, K c generally is dependent n pressure s lng as Δν 0. On the ther hand, if Δ ν =0, K c(t) = K p(t). (f) The equilibrium cnstant f a given reactin may be determined if the enthalpy f frmatin and the entrpy f reactants and prducts are knwn thrugh Eqs. (1.36), (1.39) and (1.41). (g) The definitin f K p tells us that the reactin wuld be mre cmplete if K p is larger. A larger K p may be accmplished with a larger, negative ΔG r ( T ). Cmbining Eqs. (1.36) and (39), we see that ΔGr ( T) νi' hf, i' ( T) νihf, i ( T) T νi' si' ( T) νisi ( T) prd. react. prd. react., (1.47) where Δ r ( ) ΔH r ( ) S ( ) ( ) ( ) =ΔH T TΔS T r r S T is termed as the entrpy f reactin. Therefre a large, negative T (reactin being highly exthermic) favrs a large K p, whereas a large, psitive Δ r T (reactin creating a large amunt f entrpy) als favrs a large K p r prmtes the cmpletin f the reactin. 1-16

18 Stanfrd University Versin Adiabatic Flame Temperature With the cncepts f chemical equilibrium understd, we may nw try t calculate the equilibrium cmpsitin f a cmbustin reactin. In ding s, we wish t define the adiabatic flame temperature. Cnsider an adiabatic cmbustin prcess whereby the reactants enters int a cmbustr at temperature T 0, and prducts exit the cmbustr at the adiabatic flame temperature T ad. Since the prcess is adiabatic (Q = 0 ), we have ( ) ( ) Hprd. Tad Hreact. T 0 = 0. (1.48) We nw expand Eq. (1.48) using the ttal enthalpy equatin fr each species (Eq. 1.25b), ν h ν h = ν h ν h + ν h ( T ) h ( 298) ad 0.(1.49) i' T, i' i T, i i' f,298, i' i f,298, i i' ad i ' prd. react. prd. react. prd. ( ) h ( ) ν i h T0 298 = 0 react. Obviusly the first term n the right-hand side f Eq. (1.49) is the standard enthalpy f reactin ΔH r,298. The secnd term determines the sensible heat needed t heat the prducts frm 298 K t the adiabatic flame temperature T ad. T simplify ur analysis, we shall assume that the reactants enter int the reactr at T 0 = 298 K s the third term becmes 0. Rearranging Eq. (1.49), we see that ( ) ( 298) i Δ H = ν h T h. (1.50) r,298 i' ad i ' prd. In ther wrds, the adiabatic flame temperature is btained when all the heat released frm a cmbustin reactin is used t raise the prduct temperature frm 298 t T ad. The existence f chemical equilibrium makes the calculatin f this adiabatic flame temperature a bit mre invlved. Specifically, while the values f ν i are always well defined, is nt since it is dependent n the equilibrium cmpsitin f the prducts. ν ' i Cnsider the cmbustin f 1 mle f carbn (graphite) in 1 mle f xygen at a pressure f 1 atm. 1 mle C(S) + 1 mle O 2 x CO 2 + yco + zo 2. (1.51) The reactant temperature is 298 K. The principle f chemical equilibrium states that the prducts cannt be entirely CO 2. Rather, a small amunt f CO (y mles) must be prduced alng with z mles f O 2 unused. These prducts are in equilibrium at the adiabatic flame temperature amng themselves thrugh 1-17

19 Stanfrd University Versin 1.2 with its equilibrium cnstant given by CO 2 = CO + ½ O 2, (1.52) 1 2 K p ( T ad ) = P P CO O 2 = yz 1 2 P P CO2 x x + y + z = exp( ΔG r R u T ad ). (1.53) (We need t recgnize that the prducts f a cmbustin prcess cannt be in equilibrium with the reactants f the prcess. Rather it is the prducts that are in equilibrium amng themselves.) Since there are fur unknwns in Eq. (1.53) (i.e., x, y, z and T ad), we need t prvide three mre equatins t slve this prblem. Tw f these equatins cme frm mass cnservatin: Carbn: x+ y =1 ml, (1.54) Oxygen: 2x+ y+ 2z = 2 ml. (1.55) The last equatin is given by Eq. (1.50), which may be expanded t give ( CO ) ( CO) ( ) ( 298) xh + f yh f = x h Tad h,298 2,298 CO2 ( ) ( 298) ( ) ( 298) + y h Tad h + z h Tad h CO O2.(1.56) Clearly the cupled Eqs ( ) cannt be slved analytically. We shall defer t sectin 1.8 and use Excel t slve the equatin. Any mre realistic cmbustin equilibrium prblems will have t be slved numerically by a cmputer a tpic we will discuss als in sectin 1.8. Figure 1.2 shws the variatin f the adiabatic flame temperature as a functin f equivalence rati fr the cmbustin f methane, prpane, ethylene and benzene in air at 1 atm pressure. As expected, the flame temperature peaks at an equivalence rati arund unity, slightly n the fuel rich side. The decrease f T ad twards smaller φ is caused by dilutin f unused xygen as well as the greater amunt f nitrgen brught int the cmbustr with xygen. The decrease f the flame temperature twards larger φ is because f xygen deficiency, which des nt allw the fuel t be fully xidized. Figure 1.3 shws the effect f pressure n the adiabatic flame temperature. It is seen that as pressure increases, (a) the adiabatic flame temperature becmes higher and (b) the peak shifts twards φ = 1. Here it may be nted that the pressure serves t reduce the extent f dissciatin f CO 2 and H 2O, and in ding s, frce the reactin twards better cmpletin. T explain the variatin f Tad as a functin f pressure, we plt in Figure 1.4 the equilibrium cmpsitin at φ = 1. It is seen that as pressure decreases, the extent f CO 2 and H 2O dissciatin int H 2, O 2, CO, and even highly unstable free radical species, including H, O, and OH becmes mre and mre significant. 1-18

20 Stanfrd University Versin Adiabatic Flame Temperature, T ad (K) Figure 1.3. Adiabatic Figure flame temperature f Figure 1.4. prpane 1.2. Mle cmbustin Adiabatic fractins in flame f air equilibrium temperature prducts cmputed fr at several pressures. f methane, prpane prpane, cmbustin ethylene in and air (φ=1). benzene at 1 atm pressure. ethylene benzene prpane methane Equivalence Rati, φ Adiabatic Flame Temperature, T ad (K) 100 atm 10 atm atm atm Equivalence Rati, φ H 2 O 10-1 CO 2 Mle Fractin CO O 2 H 2 H OH O Pressure, P (atm) 1-19

21 Stanfrd University Versin Tabulatin and Mathematical Parameterizatin f Thermchemical Prperties Key thermdynamic r thermchemical prperties discussed s far include the cnstantpressure specific heat, sensible enthalpy, enthalpy f frmatin, and entrpy under the standard cnditin. It is imprtant t recgnize that under the ideal gas cnditin, c p is nt a functin f pressure, but it is a functin f temperature. A cmmn methd is t tabulate, amng thers, ( ) p c T, s ( T ), h(t) h(298) and h ( ) f T as a functin f temperature. Such table is knwn as the JANAF table. 1 Appendix 1.A gives a truncated versin f these tables fr species listed in Table 1.2. In research publicatins, these tables are usually cndensed t smething that lks like Table 1.3. Table 1.3. Thermchemical prperties f selected species in the vapr state. h f ( 298K) s ( 298K) cp ( ) T (J/ml-K) Species (kj/ml) (J/ml-K) C(S) H O H 2O CO CO CH C 2H C 3H C 4H C 8H C 2H C 2H CH 3OH C 6H H O OH The thermchemical data f a large number f species has been cmpiled by the Natinal Institute f Standards and Technlgy. These data may be fund at Anther web-based surce f data is the Active Tables (ATcT): 1 Chase, M. W., Jr., J. Phys. Chem. Ref. Data, 4 th Editin, Mn. 9, Suppl. 1 (1998a). 1-20

22 Stanfrd University Versin 1.2 Fr cmbustin calculatins, a very gd surce f thermchemical data is: Alexander Burcat and Brank Ruscic Third Millennium Thermdynamic Database fr Cmbustin and Air-Pllutin Use, 2005 ( r The database is the result f extensive wrk by Prfessr Alexander Burcat f Technin University, Israel ver the last 20 years. In this database, the thermchemical data are expressed in the frm f plynmial functin and are thus mre cmpact than the JANAF table. A typical recrd f thermchemical data may lk like the fllwing: Species name Reference surce Cmpsitin T Lw, T high, T mid Phase CO2 L 7/88C 1O 2 0 0G E E E E E E E E E E E E E E E+05 4 a i (i = 1,7) fr T mid < T < T high a i (i = 1,7) fr T lw < T < T mid The plynmial fits are made fr tw separate temperature ranges (T lw T T mid and T mid T T high). There are 7 plynmial cefficients, a i (i,=1,..7), fr each temperature range. The thermchemical data are calculated frm these fits as c R = a + a T + a T 2 + a T 3 + a T 4, (1.57) p u s R = a lnt + a T + a T a T 3 3+ a T a. (1.58) u h R = at + a T + a T + a T + a T + a (1.59) T u where h T is the ttal enthalpy (Eq. 1.25b), s is the standard entrpy (1 atm). Nte that these equatins are in strict agreement with knwn relatinships amng the thermchemical prperties, i.e., T ht ( T) = cpdt + h f,298.15, (1.60) T s ( T) = cpd lnt + s ( ) (1.61) T calculate and tabulate the thermchemical data prperties in the JANAF type frm, c p and s are directly calculated with Eqs. (1.57) and (1.58). The sensible enthalpy is determined as ( ) ( ) = ( ) ( ) h T h h T h, (1.62) where h T is calculated using Eq. (1.59). Fr enthalpy f frmatin, we use T ( ) = ( ) ν ( ) f T i T, i elements T h T h T h T (1.63) 1-21

23 Stanfrd University Versin 1.2 where ν i represents the mlecular cmpsitin f the substance. Fr example, a C xh yo z species has ν C = x, ν H = y 2 and ν O = z An EXCEL spreadsheet has been prepared fr the JANAF like tabulatin. The file may be dwnladed frm the curse web site t JANAF.xls. Burcat s database may als be dwnladed in text frm Therm.txt, and in Excel frm Therm.xls. 1.8 Slutin f an Equilibrium and Adiabatic Flame Temperature Prblem We shall nw return t the prblem f carbn (graphite) xidatin in sectin 1.6. We wish t calculate the adiabatic flame temperature fr cmbustin f 1 mle f carbn (graphite) in 1 mle f xygen at a pressure f 1 atm. The initial temperature is 298 K. The fur equatins are x + y = 1 2x + y+ 2z = yz P 0 = exp( ΔGr RuTd) x x + y+ z ( ) + xh f CO yh f ( CO) = x h ( Tad) h ( 298) + y h ( Tad ) h ( 298) + z h ( T ) ( 298) CO ad h O2,298 2,298 CO2 Slutin f the abve prblem is prvided in an Excel sheet dwnladable frm xidatin.xls. Nte that t run the Excel slver requires the user t dwnlad the thermchemical prperty tables frm t JANAF.xls. This file shuld be placed in the same directry as the carbn xidatin.xls file. The slutin f this set f nnlinear algebraic equatins gives x = ml y = ml z = ml Tad = 3537 K. 1-22

24 Stanfrd University Versin 1.2 Nw suppse that the carbn is burned in air, instead f pure xygen, the set f nnlinear algebraic equatins may be revised by including the mlar number f N 2 (=1 mle O 2 79/21) and the sensible enthalpy required t heat up the nitrgen, x + y = 1 2x + y+ 2z = yz P 0 = exp( ΔGr RuTd) x x + y+ z ( ) + xh f,298 CO2 yh f,298 ( CO) = x h ( Ta d ) h ( 298) CO2 + y h ( Tad ) h ( 298) + z h ( T ) ( 298) ( ) ( 298) CO ad h h T O ad h The slutin is x = ml y = ml z = ml Tad = 2312 K 2 N2 Cmparing the tw sets f slutin, we find that (a) the adiabatic flame temperature is ntably lwer when air is used, and (b) the reactin is less cmplete when pure xygen is used because the higher adiabatic flame temperature frces a greater extent f CO 2 dissciatin int CO and O 2.. Tw cmmnly used equilibrium slvers are Stanjan (r the equilibrium slver EQUIL - f the ChemKin suite f package) and the NASA Equilibrium cde (cec86). We will use the equilibrium slver f the ChemKin suite f package fr the current class. Instructins abut the cmputer cde can be fund n p

25 Stanfrd University Versin 1.2 Appendix A1. Truncated JANAF tables Graphite (C(S)) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

26 Stanfrd University Versin 1.2 Hydrgen (H2) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

27 Stanfrd University Versin 1.2 Oxygen (O2) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

28 Stanfrd University Versin 1.2 Water vapr (H2O(v)) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

29 Stanfrd University Versin 1.2 Carbn mnxide (CO) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

30 Stanfrd University Versin 1.2 Carbn dixide (CO2) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

31 Stanfrd University Versin 1.2 Methane (CH4) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

32 Stanfrd University Versin 1.2 Ethane (C2H6) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

33 Stanfrd University Versin 1.2 Prpane (C3H8) T cp ( T ) s ( T ) h(t)-h(298) hf ( T ) g f ( T ) (K) (J/ml-K) (J/ml-K) (kj/ml) (kj/ml) (kj/ml)

Thermodynamics and Equilibrium

Thermodynamics and Equilibrium Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,

More information

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY 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

More information

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change? Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S

More information

Thermodynamics Partial Outline of Topics

Thermodynamics Partial Outline of Topics Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)

More information

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany

More information

Unit 14 Thermochemistry Notes

Unit 14 Thermochemistry Notes 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: https://cincchem.pbwrks.cm

More information

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review) CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal

More information

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3

More information

Thermochemistry. Thermochemistry

Thermochemistry. Thermochemistry Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk

More information

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY CHEMICAL REACTIONS INVOLVE ENERGY The study energy and its transrmatins is knwn as thermdynamics. The discussin thermdynamics invlve the cncepts energy, wrk, and heat. Types Energy Ptential energy is stred

More information

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system

More information

Chapter 17 Free Energy and Thermodynamics

Chapter 17 Free Energy and Thermodynamics Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics

More information

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C? NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic

More information

Lecture 12: Chemical reaction equilibria

Lecture 12: Chemical reaction equilibria 3.012 Fundamentals f Materials Science Fall 2005 Lecture 12: 10.19.05 Chemical reactin equilibria Tday: LAST TIME...2 EQUATING CHEMICAL POTENTIALS DURING REACTIONS...3 The extent f reactin...3 The simplest

More information

Chemistry 1A Fall 2000

Chemistry 1A Fall 2000 Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur

More information

Spontaneous Processes, Entropy and the Second Law of Thermodynamics

Spontaneous Processes, Entropy and the Second Law of Thermodynamics Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer

More information

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics CHEM 116 Electrchemistry at Nn-Standard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam

More information

Lecture 4. The First Law of Thermodynamics

Lecture 4. The First Law of Thermodynamics 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

More information

ChE 471: LECTURE 4 Fall 2003

ChE 471: LECTURE 4 Fall 2003 ChE 47: LECTURE 4 Fall 003 IDEL RECTORS One f the key gals f chemical reactin engineering is t quantify the relatinship between prductin rate, reactr size, reactin kinetics and selected perating cnditins.

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium Lecture 17: 11.07.05 Free Energy f Multi-phase Slutins at Equilibrium Tday: LAST TIME...2 FREE ENERGY DIAGRAMS OF MULTI-PHASE SOLUTIONS 1...3 The cmmn tangent cnstructin and the lever rule...3 Practical

More information

General Chemistry II, Unit II: Study Guide (part 1)

General Chemistry II, Unit II: Study Guide (part 1) General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reactin Equilibrium in the Gas Phase General Chemistry II Unit II Part 1 1 Intrductin Sme chemical reactins have a significant amunt

More information

Matter Content from State Frameworks and Other State Documents

Matter Content from State Frameworks and Other State Documents Atms and Mlecules Mlecules are made f smaller entities (atms) which are bnded tgether. Therefre mlecules are divisible. Miscnceptin: Element and atm are synnyms. Prper cnceptin: Elements are atms with

More information

More Tutorial at

More Tutorial at Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,

More information

Chem 75 February 16, 2017 Exam 2 Solutions

Chem 75 February 16, 2017 Exam 2 Solutions 1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml

More information

CHEM 103 Calorimetry and Hess s Law

CHEM 103 Calorimetry and Hess s Law CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the

More information

Chemical Thermodynamics

Chemical Thermodynamics Chemical Thermdynamics Objectives 1. Be capable f stating the First, Secnd, and Third Laws f Thermdynamics and als be capable f applying them t slve prblems. 2. Understand what the parameter entrpy means.

More information

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25 CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing

More information

Lecture 16 Thermodynamics II

Lecture 16 Thermodynamics II Lecture 16 Thermdynamics II Calrimetry Hess s Law Enthalpy r Frmatin Cpyright 2013, 2011, 2009, 2008 AP Chem Slutins. All rights reserved. Fur Methds fr Finding H 1) Calculate it using average bnd enthalpies

More information

GLOBAL CLIMATE AND ENERGY PROJECT STANFORD UNIVERSITY. Energy Tutorial: Exergy 101 GCEP RESEARCH SYMPOSIUM 2012 STANFORD, CA.

GLOBAL CLIMATE AND ENERGY PROJECT STANFORD UNIVERSITY. Energy Tutorial: Exergy 101 GCEP RESEARCH SYMPOSIUM 2012 STANFORD, CA. GLOBAL LIMATE AND ENERGY PROJET STANFORD UNIVERSITY Energy Tutrial: Exergy 101 GEP RESEARH SYMPOSIUM 2012 STANFORD, A hris Edwards Prfessr Department f Mechanical Engineering Stanfrd University GLOBAL

More information

Chapter 4 Thermodynamics and Equilibrium

Chapter 4 Thermodynamics and Equilibrium Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume

More information

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013 CHEM-443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal

More information

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis

More information

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes. Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs

More information

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin

More information

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra Chem 115 POGIL Wrksheet - Week 8 Thermchemistry (Cntinued), Electrmagnetic Radiatin, and Line Spectra Why? As we saw last week, enthalpy and internal energy are state functins, which means that the sum

More information

General Chemistry II, Unit I: Study Guide (part I)

General Chemistry II, Unit I: Study Guide (part I) 1 General Chemistry II, Unit I: Study Guide (part I) CDS Chapter 14: Physical Prperties f Gases Observatin 1: Pressure- Vlume Measurements n Gases The spring f air is measured as pressure, defined as the

More information

GOAL... ability to predict

GOAL... ability to predict THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict

More information

1 The limitations of Hartree Fock approximation

1 The limitations of Hartree Fock approximation Chapter: Pst-Hartree Fck Methds - I The limitatins f Hartree Fck apprximatin The n electrn single determinant Hartree Fck wave functin is the variatinal best amng all pssible n electrn single determinants

More information

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia: University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,

More information

Computational modeling techniques

Computational modeling techniques Cmputatinal mdeling techniques Lecture 4: Mdel checing fr ODE mdels In Petre Department f IT, Åb Aademi http://www.users.ab.fi/ipetre/cmpmd/ Cntent Stichimetric matrix Calculating the mass cnservatin relatins

More information

AP Chemistry Assessment 2

AP Chemistry Assessment 2 AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND

More information

Semester 2 AP Chemistry Unit 12

Semester 2 AP Chemistry Unit 12 Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved

More information

CHEM 1001 Problem Set #3: Entropy and Free Energy

CHEM 1001 Problem Set #3: Entropy and Free Energy CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre.

More information

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals f Diffusin Diffusin: Transprt in a slid, liquid, r gas driven by a cncentratin gradient (r, in the case f mass transprt, a chemical ptential

More information

CHAPTER PRACTICE PROBLEMS CHEMISTRY

CHAPTER PRACTICE PROBLEMS CHEMISTRY Chemical Kinetics Name: Batch: Date: Rate f reactin. 4NH 3 (g) + 5O (g) à 4NO (g) + 6 H O (g) If the rate f frmatin f NO is 3.6 0 3 ml L s, calculate (i) the rate f disappearance f NH 3 (ii) rate f frmatin

More information

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition)

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition) Name Chem 163 Sectin: Team Number: ALE 24. Vltaic Cells and Standard Cell Ptentials (Reference: 21.2 and 21.3 Silberberg 5 th editin) What des a vltmeter reading tell us? The Mdel: Standard Reductin and

More information

Chemistry 114 First Hour Exam

Chemistry 114 First Hour Exam Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)

More information

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is

More information

A Chemical Reaction occurs when the of a substance changes.

A Chemical Reaction occurs when the of a substance changes. Perid: Unit 8 Chemical Reactin- Guided Ntes Chemical Reactins A Chemical Reactin ccurs when the f a substance changes. Chemical Reactin: ne r mre substances are changed int ne r mre new substances by the

More information

17.1 Ideal Gas Equilibrium Constant Method. + H2O CO + 3 H2 ν i ν i is stoichiometric number is stoichiometric coefficient

17.1 Ideal Gas Equilibrium Constant Method. + H2O CO + 3 H2 ν i ν i is stoichiometric number is stoichiometric coefficient 17.1 Ideal Gas Equilibrium Cnstant Methd CH 4 + H2O CO + 3 H2 ν i -1-1 1 3 ν i is stichimetric number is stichimetric cefficient ν i (1) dn1 dn = ν1 ν 2 2 Prcess 1. # phases? Methd? (K a methd). 2. Find

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk

More information

Chapter 8 Reduction and oxidation

Chapter 8 Reduction and oxidation Chapter 8 Reductin and xidatin Redx reactins and xidatin states Reductin ptentials and Gibbs energy Nernst equatin Disprprtinatin Ptential diagrams Frst-Ebswrth diagrams Ellingham diagrams Oxidatin refers

More information

Differentiation Applications 1: Related Rates

Differentiation Applications 1: Related Rates Differentiatin Applicatins 1: Related Rates 151 Differentiatin Applicatins 1: Related Rates Mdel 1: Sliding Ladder 10 ladder y 10 ladder 10 ladder A 10 ft ladder is leaning against a wall when the bttm

More information

Heat Effects of Chemical Reactions

Heat Effects of Chemical Reactions eat Effects f hemical Reactins Enthalpy change fr reactins invlving cmpunds Enthalpy f frmatin f a cmpund at standard cnditins is btained frm the literature as standard enthalpy f frmatin Δ (O (g = -9690

More information

3. Review on Energy Balances

3. Review on Energy Balances 3. Review n Energy Balances Objectives After cmpleting this chapter, students shuld be able t recall the law f cnservatin f energy recall hw t calculate specific enthalpy recall the meaning f heat f frmatin

More information

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b . REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but

More information

, which yields. where z1. and z2

, which yields. where z1. and z2 The Gaussian r Nrmal PDF, Page 1 The Gaussian r Nrmal Prbability Density Functin Authr: Jhn M Cimbala, Penn State University Latest revisin: 11 September 13 The Gaussian r Nrmal Prbability Density Functin

More information

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry

More information

Advanced Chemistry Practice Problems

Advanced Chemistry Practice Problems Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with

More information

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s)

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s) Chapter 9 - Stichimetry Sectin 9.1 Intrductin t Stichimetry Types f Stichimetry Prblems Given is in mles and unknwn is in mles. Given is in mles and unknwn is in mass (grams). Given is in mass and unknwn

More information

Lecture 13: Electrochemical Equilibria

Lecture 13: Electrochemical Equilibria 3.012 Fundamentals f Materials Science Fall 2005 Lecture 13: 10.21.05 Electrchemical Equilibria Tday: LAST TIME...2 An example calculatin...3 THE ELECTROCHEMICAL POTENTIAL...4 Electrstatic energy cntributins

More information

lecture 5: Nucleophilic Substitution Reactions

lecture 5: Nucleophilic Substitution Reactions lecture 5: Nuclephilic Substitutin Reactins Substitutin unimlecular (SN1): substitutin nuclephilic, unimlecular. It is first rder. The rate is dependent upn ne mlecule, that is the substrate, t frm the

More information

" 1 = # $H vap. Chapter 3 Problems

 1 = # $H vap. Chapter 3 Problems Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius

More information

A Few Basic Facts About Isothermal Mass Transfer in a Binary Mixture

A Few Basic Facts About Isothermal Mass Transfer in a Binary Mixture Few asic Facts but Isthermal Mass Transfer in a inary Miture David Keffer Department f Chemical Engineering University f Tennessee first begun: pril 22, 2004 last updated: January 13, 2006 dkeffer@utk.edu

More information

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry Chapter 19 lectrchemistry Part I Dr. Al Saadi 1 lectrchemistry What is electrchemistry? It is a branch f chemistry that studies chemical reactins called redx reactins which invlve electrn transfer. 19.1

More information

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command NUPOC SUDY GUIDE ANSWER KEY Navy Recruiting Cmmand CHEMISRY. ph represents the cncentratin f H ins in a slutin, [H ]. ph is a lg scale base and equal t lg[h ]. A ph f 7 is a neutral slutin. PH < 7 is acidic

More information

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature

More information

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s .9 Kinetic Mlecular Thery Calculate the effective (rms) speeds f the He and Ne atms in the He-Ne gas laser tube at rm temperature (300 K). Slutin T find the rt mean square velcity (v rms ) f He atms at

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Chapter 11: Atmosphere

Chapter 11: Atmosphere Chapter 11: Atmsphere Sectin 1: Atmspheric Basics Objectives 1. Describe the cmpsitin f the atmsphere. 2. Cmpare and cntrast the varius layers f the atmsphere. 3. Identify three methds f transferring energy

More information

188 CHAPTER 6 THERMOCHEMISTRY

188 CHAPTER 6 THERMOCHEMISTRY 188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101.

More information

Chem 111 Summer 2013 Key III Whelan

Chem 111 Summer 2013 Key III Whelan Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general

More information

NUMBERS, MATHEMATICS AND EQUATIONS

NUMBERS, MATHEMATICS AND EQUATIONS AUSTRALIAN CURRICULUM PHYSICS GETTING STARTED WITH PHYSICS NUMBERS, MATHEMATICS AND EQUATIONS An integral part t the understanding f ur physical wrld is the use f mathematical mdels which can be used t

More information

Compressibility Effects

Compressibility Effects Definitin f Cmpressibility All real substances are cmpressible t sme greater r lesser extent; that is, when yu squeeze r press n them, their density will change The amunt by which a substance can be cmpressed

More information

4 Fe + 3 O 2 2 Fe 2 O 3

4 Fe + 3 O 2 2 Fe 2 O 3 UNIT 7: STOICHIOMETRY NOTES (chapter 9) INTRO TO STOICHIOMETRY Reactin Stichimetry: Stichimetry is simply a way t shw f smething this is. Relatinship between a given and an unknwn: GIVEN UNKNOWN Type 1

More information

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces

More information

Pressure And Entropy Variations Across The Weak Shock Wave Due To Viscosity Effects

Pressure And Entropy Variations Across The Weak Shock Wave Due To Viscosity Effects Pressure And Entrpy Variatins Acrss The Weak Shck Wave Due T Viscsity Effects OSTAFA A. A. AHOUD Department f athematics Faculty f Science Benha University 13518 Benha EGYPT Abstract:-The nnlinear differential

More information

CHM112 Lab Graphing with Excel Grading Rubric

CHM112 Lab Graphing with Excel Grading Rubric Name CHM112 Lab Graphing with Excel Grading Rubric Criteria Pints pssible Pints earned Graphs crrectly pltted and adhere t all guidelines (including descriptive title, prperly frmatted axes, trendline

More information

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate.

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate. CEM1405 2007-J-2 June 2007 In the spaces prvided, explain the meanings f the fllwing terms. Yu may use an equatin r diagram where apprpriate. 5 (a) hydrgen bnding An unusually strng diple-diple interactin

More information

Experiment #3. Graphing with Excel

Experiment #3. Graphing with Excel Experiment #3. Graphing with Excel Study the "Graphing with Excel" instructins that have been prvided. Additinal help with learning t use Excel can be fund n several web sites, including http://www.ncsu.edu/labwrite/res/gt/gt-

More information

Thermodynamics EAS 204 Spring 2004 Class Month Day Chapter Topic Reading Due 1 January 12 M Introduction 2 14 W Chapter 1 Concepts Chapter 1 19 M MLK

Thermodynamics EAS 204 Spring 2004 Class Month Day Chapter Topic Reading Due 1 January 12 M Introduction 2 14 W Chapter 1 Concepts Chapter 1 19 M MLK Thermdynamics EAS 204 Spring 2004 Class Mnth Day Chapter Tpic Reading Due 1 January 12 M Intrductin 2 14 W Chapter 1 Cncepts Chapter 1 19 M MLK Hliday n class 3 21 W Chapter 2 Prperties Chapter 2 PS1 4

More information

Chemistry 20 Lesson 11 Electronegativity, Polarity and Shapes

Chemistry 20 Lesson 11 Electronegativity, Polarity and Shapes Chemistry 20 Lessn 11 Electrnegativity, Plarity and Shapes In ur previus wrk we learned why atms frm cvalent bnds and hw t draw the resulting rganizatin f atms. In this lessn we will learn (a) hw the cmbinatin

More information

BIT Chapters = =

BIT Chapters = = BIT Chapters 17-0 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride

More information

A) 0.77 N B) 0.24 N C) 0.63 N D) 0.31 N E) 0.86 N. v = ω k = 80 = 32 m/s. Ans: (32) 2 = 0.77 N

A) 0.77 N B) 0.24 N C) 0.63 N D) 0.31 N E) 0.86 N. v = ω k = 80 = 32 m/s. Ans: (32) 2 = 0.77 N Q1. A transverse sinusidal wave travelling n a string is given by: y (x,t) = 0.20 sin (2.5 x 80 t) (SI units). The length f the string is 2.0 m and its mass is 1.5 g. What is the magnitude f the tensin

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t eep this site up and bring yu even mre cntent cnsider dnating via the lin n ur site. Still having truble understanding the material? Chec ut ur Tutring

More information

Solutions to the Extra Problems for Chapter 14

Solutions to the Extra Problems for Chapter 14 Slutins t the Extra Prblems r Chapter 1 1. The H -670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +

More information

Supporting information

Supporting information Electrnic Supplementary Material (ESI) fr Physical Chemistry Chemical Physics This jurnal is The wner Scieties 01 ydrgen perxide electrchemistry n platinum: twards understanding the xygen reductin reactin

More information

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions? 1 CHEM 1032 FALL 2017 Practice Exam 4 1. Which f the fllwing reactins is spntaneus under nrmal and standard cnditins? A. 2 NaCl(aq) 2 Na(s) + Cl2(g) B. CaBr2(aq) + 2 H2O(aq) Ca(OH)2(aq) + 2 HBr(aq) C.

More information

Revision: August 19, E Main Suite D Pullman, WA (509) Voice and Fax

Revision: August 19, E Main Suite D Pullman, WA (509) Voice and Fax .7.4: Direct frequency dmain circuit analysis Revisin: August 9, 00 5 E Main Suite D Pullman, WA 9963 (509) 334 6306 ice and Fax Overview n chapter.7., we determined the steadystate respnse f electrical

More information

Hess Law - Enthalpy of Formation of Solid NH 4 Cl

Hess Law - Enthalpy of Formation of Solid NH 4 Cl Hess Law - Enthalpy f Frmatin f Slid NH 4 l NAME: OURSE: PERIOD: Prelab 1. Write and balance net inic equatins fr Reactin 2 and Reactin 3. Reactin 2: Reactin 3: 2. Shw that the alebraic sum f the balanced

More information

Process Engineering Thermodynamics E (4 sp) Exam

Process Engineering Thermodynamics E (4 sp) Exam Prcess Engineering Thermdynamics 42434 E (4 sp) Exam 9-3-29 ll supprt material is allwed except fr telecmmunicatin devices. 4 questins give max. 3 pints = 7½ + 7½ + 7½ + 7½ pints Belw 6 questins are given,

More information

GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW

GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW GASES Pressure & Byle s Law Temperature & Charles s Law Avgadr s Law IDEAL GAS LAW PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2 Earth s atmsphere: 78% N 2 21% O 2 sme Ar, CO 2 Sme Cmmn Gasses Frmula Name

More information

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS CHPTER 6 / HRVEY. CHEMICL B. THERMODYNMICS ND C. MNUPULTING CONSTNTS D. CONSTNTS FOR CHEMICL RECTIONS 1. Precipitatin Reactins 2. cid-base Reactins 3. Cmplexatin Reactins 4. Oxidatin-Reductin Reactins

More information

CHE 105 EXAMINATION III November 11, 2010

CHE 105 EXAMINATION III November 11, 2010 CHE 105 EXAMINATION III Nvember 11, 2010 University f Kentucky Department f Chemistry READ THESE DIRECTIONS CAREFULLY BEFORE STARTING THE EXAMINATION! It is extremely imprtant that yu fill in the answer

More information

Physics 2B Chapter 23 Notes - Faraday s Law & Inductors Spring 2018

Physics 2B Chapter 23 Notes - Faraday s Law & Inductors Spring 2018 Michael Faraday lived in the Lndn area frm 1791 t 1867. He was 29 years ld when Hand Oersted, in 1820, accidentally discvered that electric current creates magnetic field. Thrugh empirical bservatin and

More information

Instructions: Show all work for complete credit. Work in symbols first, plugging in numbers and performing calculations last. / 26.

Instructions: Show all work for complete credit. Work in symbols first, plugging in numbers and performing calculations last. / 26. CM ROSE-HULMAN INSTITUTE OF TECHNOLOGY Name Circle sectin: 01 [4 th Lui] 02 [5 th Lui] 03 [4 th Thm] 04 [5 th Thm] 05 [4 th Mech] ME301 Applicatins f Thermdynamics Exam 1 Sep 29, 2017 Rules: Clsed bk/ntes

More information