Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

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1 CT 1 THERMODYNAMICS 6.1 Thermodynamcs Terms : Q. Defne system and surroundngs. Soluton : A system n thermodynamcs refers to that part of unverse n whch observatons are made and remanng unverse consttutes the surroundngs. The surroundngs nclude everythng other than the system. System and the surroundngs together consttute the unverse. Q. To what type of system the followng belong? () Tree () Pond () Anmals (v) Tea placed n a kettle (v) Tea placed n thermos flask (v) Tea placed n a cup. Soluton : () Open system () open system () open system (v) closed system (v) solated system (v) open system. Q. What are dfferent types of system? Soluton : Open System : In an open system, there s exchange of energy and matter between system and surroundngs. The presence of reactants n an open beaker s an example of an open system. Closed System : In a closed system, there s no exchange of matter, but exchange of energy s possble between system and the surroundngs. The presence of reactants n a closed vessel made of conductng materal e.g., copper or steel s an example of a closed system Isolated System : In an solated system, there s no exchange of energy or matter between the system and the surroundngs. The presence of reactants n a thermo flask or any other closed nsulated vessel s an example of an solated system. Q. What are state varables or state functons? Soluton : The state of a thermodynamcs system s descrbed by ts measurable or macroscopc (bulk) propertes. We can descrbe the state of a gas by ts pressure (p), volume (V), temperature (T), amount (n) etc. Varable lke p, V, T are called state varables or state functons because ther values depend only on the state of the system and not on how t s reached. Q. Defne nternal energy. Soluton : It s the total energy of the system whch may be chemcal, electrcal, mechancal or any other type of energy you may thnk of the sum of all these s the energy of the system. In thermodynamcs, we call t the nternal energy, U of the system, whch may change, when () heat passes nto or out of the system () work s done on or by the system () matter enters or leaves the system. Q. Can the absolute value of nternal energy be determned? Why or why not? Soluton : No, because t s the sum of dfferent types of energes some of whch cannot be determned. Q. Defne adabatc process. Soluton : An adabatc change by defnton, s one whch does not allow any transfer of heat,.e., q = 0, t follows from the 1st law, U = W, du = dw. Q. Defne heat. Soluton : Change of nternal energy of a system by transfer of heat from the surroundngs to the system or vce-versa wthout expendture of work. Ths exchange of energy, whch s a result of temperature dfference s called heat, q. Q. Defne frst law of thermodynamcs and also wrte ts mathematcal expresson. Soluton : The energy of an solated system s constant. It s commonly stated as the law of conservaton of energy.e., energy can nether be created nor be destroyed. U = q + w. Change n nternal energy s equal to heat added to the system + work done on the system. We take water at temperature, T A n a contaner havng thermally conductng walls, say made up of copper and enclose t n a huge heat reservor at temperature, T B. The heat absorbed by the system (water), q can be measured n terms of temperature dfference, T B T A. In ths case change n nternal energy, U = q, when no work s done at constant volume. The q s postve, when heat s transferred from the surroundngs to the system and q s negatve when heat s transferred from system to the surroundngs.

2 The general case CT Let us consder the general case n whch a change of state s brought about both by dong work and by transfer of heat. We wrte change n nternal energy for ths case as : U = q + w The energy of an solated system s constant. It s commonly stated as the law of conservaton of energy.e., energy can nether be created nor be destroyed. Q. Express the change n nternal energy of a system when () No heat s absorbed by the system from the surroundngs, but work (w) s done on the system. What type of wall does the system have? () No work s done on the system, but q amount of heat s taken out from the system and gven to the surroundngs. What type of wall does the system have? () w amount of work s done by the system and q amount of heat s suppled to the system. What type of system would t be? [NCERT Solved Example 6.1]. Soluton : () U = w ad, wall s adabatc () U = q, thermally conductng walls () U = q w, closed system. Work We wll consder only mechancal work.e., pressure-volume work. A process or change s sad to be reversble, f a change s brought out n such a way that the process could, at any moment, be reversed by an nfntesmal change. A reversble process proceeds nfntely slowly by a seres of equlbrum states such that system and the surroundngs are always n near equlbrum wth each other. Processes other than reversble processess are known as rreversble processes. In chemstry, we face problems that can be solved f we relate the work term to the nternal pressure of the system. We can relate work to nternal pressure of the system under reversble condtons by wrtng equaton as follows : w rev vf pexdv (pn v vf v dp) dv Snce dp dv s very small we can wrte w rev vf v p n dv Now, the pressure of the gas (p n whch we can wrte as p now) can be expressed n terms of ts volume through gas equaton. For n mol of an deal gas.e., pv = nrt Therefore, at constant temperature (sothermal process), w rev vf vt nrt Vf =.303 nrt log V dv Vf nrtln V V nrt p. V Free expanson : Expanson of a gas n vacuum (p ex = 0) s called free expanson. No work s done durng free expanson of an deal gas whether the process s reversble or rreversble (equaton 6. and 6.3). Now, we can wrte equaton 6.1 n number of ways dependng on the type of processes. Let us substtute w = p ex V (eq. 6.) n equaton 6.1, and we get U = q p ex V If a process s carred out at constant volume (V = 0), then U = q V the subscrpt v n q V denotes that heat s suppled at constant volume. Isothermal and free expanson of an deal gas For sothermal (T = constant) expanson of an deal gas nto vacuum ; w = 0 snce p ex = 0. Also, Joule determned expermentally that q = 0; therefore, U = 0

3 CT 3 Equaton 6.1, U = q + w can be expressed for sothermal rreversble and reversble changes as follows : 1. For sothermal rreversble change q = w = p ex (V f V ). For sothermal reversble change Vf q w nrtln =.303 nrt log V V f. V 3. For adabatc change, q = 0, U = w ad Q. Two ltres of an deal gas at a pressure of 10 atm expands sothermally nto a vacuum untl ts total volume s 10 ltres. How much heat s absorbed and how much work n the expanson? [NCERT Solved Example 6.] Soluton : We have q = w = p ex (10 ) = 0(8) = 0. No work s done; no heat s absorbed. Q. Consder the same expanson, but ths tme aganst a constant external pressure of 1 atm. [NCERT Solved Example 6.3] Soluton : We have q = w = p ex (8) = 8 ltre-atm Q. Consder the same expanson, to a fnal volume of 10 ltres conducted reversbly. [NCERT Solved Example 6.4] Soluton : We have q = w = Enthalpy, H (a) A useful new state functon 10 log = 16.1 ltre-atm. We know that the heat absorbed at constant volume s equal to change n the nternal energy.e., U = q V. But most of chemcal reactons are carred out not at constant volume, but n flasks or test tubes under constant atmospherc pressure. We need to defne another state functon whch may be sutable under these condtons. We may wrte equaton (6.1) as U = q p pv at constant pressure, where q p s heat absorbed by the system and pv represent expanson work done by the system. Let us represent the ntal state by subscrpt 1 and fnal state by We can rewrte the above equaton as U U 1 = q p p (V V 1 ) On rearrangng, we get q p = (U + pv ) (U 1 + pv 1 ) (6.6) Now we can defne another thermodynamc functon, the enthalpy H [Greek word enthalpen, to warm or heat content] as : H = U + pv (6.7) so, equaton (6.6) becomes q p = H H 1 = H Although q s a path dependent functon, H s a state functon because t depends on U, p and V, all of whch are state functons. Therefore, H s ndependent of path. Hence, q p s also ndependent of path. For fnte changes at constant pressure, we can wrte equaton 6.7 as H = U + pv Snce p s constant, we can wrte H = U + pv It s mportant to note that when heat s absorbed by the system at constant pressure, we are actually measurng changes n the enthalpy. Remember H = q p, heat absorbed by the system at constant pressure. H s negatve for exothermc reactons whch evolve heat durng the reacton and H s postve for endothermc reactons whch absorb heat from the surroundngs. At constant volume (V = 0), U = qv, therefore equaton 6.8 becomes H = U = q V

4 CT 4 The dfference between H and U s not usually sgnfcant for systems consstng of only solds and/or lquds. Solds and lquds do not suffer any sgnfcant volume changes upon heatng. The dfference, however, becomes sgnfcant when gases are nvolved. Let us consder a reacton nvolvng gases. If V A s the total volume of the gaseous reactants, V B s the total volume of the gaseous products, n A s the number of moles of gaseous reactants and n B s the number of moles of gaseous products, all at constant pressure and temperature, then usng the deal gas law, we wrte, and Thus, or or pv A = n A RT pv B = n B RT pv B pv A = n B RT n A RT = (n B n A )RT p(v B V A ) = (n B n A )RT pv = n g RT Here, n g refers to the number of moles of gaseous products mnus the number of moles of gaseous reactants. Substtutng the vlaue of pv from equaton 6.9 n equaton 6.8, we get H = U + n g RT The equaton 6.10 s useful for calculatng H and U and vce versa. Q. If water vapour s assumed to be a perfect gas, molar enthalpy change for vapoursaton of 1 mol of water at 1 bar and C s 41 kj mol 1. Calculate the nternal energy change, when () 1 mol of water s vaporsed at 1 bar pressure and C. () 1 mol of water s converted nto ce. [NCERT Solved Example 6.5]. Soluton : () kj mol 1 () U = kj mol 1 ] Extensve and Intensve Propertes : An extensve property s a property whose value depends on the quantty or sze of matter present n the system. For example, mass, volume, nternal energy, enthalpy, heat capacty, etc. are extensve propertes. Those propertes whch do not depend on the quantty or sze of matter present are known as ntensve propertes. For example temperature, densty, pressure etc. are ntensve propertes. A molar property, m, s the value of an extensve property of the system for 1 mol of the substance. If n s the amount of matter, m s ndependent of the amount of matter. Other examples are molar volume, V n m and molar heat capacty, C m. Heat Capacty Ths heat appears as a rse n temperature of the system n case of heat absorbed by the system. The ncrease of temperature s proportonal to the heat transferred q = coeff T The magntude of the coeffcent depends on the sze, composton and nature of the system. We can also wrte t as q = C T The coeffcent, C s called the heat capacty. The relatonshp between C p and C v for an deal gas At constant volume, the heat capacty, C s denoted by C v and at constant pressure, ths s denoted by C p. Let us fnd the relatonshp between the two. We can wrte equaton for heat, q at constant volume as q v = C v T = U at constant pressure as q p = C p T = H The dfference between C p and C v can be derved for an deal gas as : For a mole of an deal gas, H = U + RT H = U + (pv) = U + (RT) = U + RT

5 On puttng the value of H and U, we have C p T = C v T + RT C p = C v + R C p C v = R CT 5 Q. 1g of graphte s burnt n a bomb calormeter n excess of oxygen at 98 K and 1 atmospherc pressure accordng to the equaton C (graphte) + O (g) CO (g) Durng the reacton, temperature rses from 98 K to 99 K. If the heat capacty of the bomb calormeter s 0.7 kj/k, what s the enthalpy change for the above reacton at 98 K and 1 atm? [NCERT Solved Example 6.6] Soluton : H = U = kj mol 1 Enthalpy Change, r H of a reacton Reacton Enthalpy The enthalpy change accompanyng a reacton called the reacton enthalpy. The enthalpy change of a chemcal reacton, s gven by the symbol r H r H = (sum of enthalpes of products) (sum of enthalpes of reactants) ahproducts b H reactants (Here symbol ( sgma) s used for summaton and a and b are the stochometrc coeffcents of the products and reactants respectvely n the balanced chemcal equaton. For example, for the reacton CH 4 (g) + O (g) CO(g) + H O(l) H r ahproducts b H reactants = [H m (CO, g) + H m (H O, l)] [H m (CH 4, g) + H m (O, g) where H m s the molar enthalpy. Enthalpy change s a very useful quantty. Standard enthalpy of reactons The standard enthalpy of reacton s the enthalpy change for a reacton when all the partcpatng substances are n ther standard states. The standard state of a substance at a specfed temperature s ts pure form at 1 bar. Enthalpy changes durng phase transformatons H O(s) H O(l); fus H = 6.00 kj mol 1 The enthalpy change that accompanes meltng of one mole of a sold substance n standard state s called standard enthalpy of fuson or molar enthalpy of fuson, fus H. Meltng of a sold s endothermc, so all enthalpes of fuson are postve. H O(l) H O(g); vap H = kj mol 1 vap H s the standard enthalpy of vaporzaton. Amount of heat requred to vaporze one mole of a lqud at constant temperature and under standard pressure (1 bar) s called ts standard enthalpy of vaporzaton or molar enthalpy of vaporzaton, D vap H. Standard enthalpy of sublmaton, sub H s the change n enthalpy when one mole of a sold substance sublmes at a constant temperature and under standard pressure (1 bar). The strong hydrogen bonds between water molecules hold them tghtly n lqud phase. For an organc lqud, such as acetone, the ntermolecular dpole-dpole nteractons are sgnfcantly weaker. Thus, t requres less heat to vaporse 1 mol of acetone that t does to vaporze 1 mol of water.

6 CT 6 Q. A swmmer comng out from a pool s covered wth a flm of water weghng about 18g. How much heat must be suppled to evaporate ths water at 98 K? Calculate the nternal energy of vaporsaton at C. vap H for water at 373K = kj mol 1 Soluton : kj mol 1 Standard enthalpy of formaton The standard enthalpy change for the formaton of one mole of a compound from ts elements n ther most stable stated of aggregaton (also known as reference states) s called Standard Molar Enthalpy of Formaton. It symbol s f H, where the subscrpt f ndcates that one mole of the compound n queston has been formed n ts standard state from ts elements n ther most stable states of aggregaton. The reference state of an element s ts most stable state of aggregaton at 5 0 C and 1 bar pressure. For example, the reference state of dhydrogen s H gas and those of doxygen, carbon and sulphur are O gas, C graphte and S rhombc respectvely. Some reactons wth standard molar enthalpes of formaton are gven below, H (g) + ½O (g) H O(l); f H = 85.8 kj mol 1 C(graphte, s) + H (g) CH 4 (g); f H = kj mol 1 C(graphte, s) + 3H (g) + ½O (g) C H 5 OH(l); f H = 77.7 kj/mol 1. By conventon, standard enthalpy for formaton, f H, of an element n reference state,.e., ts most stable state of aggregaton s taken as zero. r H a f H (products) b f H (reactan ts) Hess s Law of Constant Heat Summaton If a reacton takes place n several steps then ts standard reacton enthalpy s the sum of the standard enthalpes of the ntermedate reactons nto whch the overall reacton may be dvded at the same temperature. Consder the enthalpy change for the reacton C(graphte, s) 1 O(g) CO(g); rh? It s possble to calculate the enthalpy change for the above reacton. Let us consder the followng reactons : C(graphte, s) + O (g) CO (g); r H = kj mol 1 CO(g) + 1 O (g) CO (g); r H = 83.0 kj mol 1 () () CO(g) + CO(g) + 1 O (g); r H = kj mol 1 () Addng equaton () and (), we get the desred equaton, C(graphte, s) + 1 O (g) CO (g); for whch D r H = ( ) = kj mol 1. Standard enthalpy of combuston (symbol : c H ) Combuston reactons are exothermc n nature. Standard enthalpy of combuston s defned as the enthalpy change per mole (or per unt amount) of a substance, when t undergoes combuston and all the reactants and products beng n ther standard states at the specfed temperature.

7 CT 7 C 4 H 10 (g) + 13 O (g) 4CO (g) + 5H O(l); c H = kj mol 1. Q. The combuston of one mole of benzene takes place at 98 K and 1 atm. After combuston, CO (g) and H O (l) are produced and kj of heat s lberated. Calculate the standard enthalpy of formaton, f H of benzene. Standard enthalpes of formaton of CO (g) and H O(l) are kj mol 1 and kj mol 1 respectvely. [NCERT Solved Example 6.8] Soluton : kj mol 1 Enthalpy of atomzaton (symbol; a H ) Consder the followng example of atomzaton of dhydrogen H (g) H(g); a H = kj mol 1 You can see that H atoms are formed by breakng H H bonds n dhydrogen. The enthalpy change n ths process s known as enthalpy of atomzaton, a H. It s the enthalpy change on breakng one mole of bonds completely to obtan atoms n the gas phase. The other examples of enthalpy of atomzaton can be CH 4 (g) C(g) + 4H(g); a H = 1665 kj mol 1 Note that the products are only atoms of C and H n gaseous phase. Now see the followng reacton : Na(s) Na(g); a H = kj mol 1 In ths case, the enthalpy of atomzaton s same as the enthalpy of sublmaton. Δ H r bond enthalpes bond enthalpes products reactants Enthalpy of Soluton (symbol : sol H ) Enthalpy of soluton of a substance s the enthalpy change when one mole of t dssolves n a specfed amount of solvent. The enthalpy of soluton at nfnte dluton s the enthalpy change observed on dssolvng the substance n an nfnte amount of solvent when the nteractons between the ons (or solute molecules) are neglgble. When an onc compound dssolves n a solvent, the ons leave ther ordered postons on the crystal lattce. There are now more free n soluton. But solvaton of these ons (hydraton n case solvent s water) also occurs at the same tme. The enthalpy of soluton of AB(s), sol H, n water s, therefore, determned by the selectve values of the lattce enthalpy, lattce H and enthalpy of hydraton of ons, hyd H as sol H = lattce H + hyd H. Lattce Enthalpy The lattce enthalpy of an onc compound s the enthalpy change whch occurs when one mole of an onc compound dssocates nto ts ons n gaseous state. Na + Cl (s) Na + (g) + Cl (g); lattce H = +788 kj mol 1 Snce t s mpossble to determne lattce enthalpes drectly by experment, we use an ndrect method where we construct an enthalpy dagram called a Born-Haber Cycle.

8 CT 8 Let us now calculate the lattce enthalpy of Na + Cl (s) by followng steps gven below : 1. Na(s) Na(g), sublmaton of sodum metal, sub H = kj mol 1. Na(g) Na + (g) + e 1 (g), the onzaton of sodum atoms, onzaton enthalpy H = 496 kj mol Cl (g) Cl(g), the dssocaton of chlorne, the reacton enthalpy s half the bond dssocaton enthalpy. 1 1 bond H 11kJ mol. A spontaneous process s an rreversble process and may only be reversed by some external agency. Entropy and spontanety Then, what drves the spontaneous process n a gven drecton? Let us examne such a case n whch H = 0.e., there s no change n enthalpy, but stll the process s spontaneous. Let us consder dffuson of two gases nto each other n a closed contaner whch s solated from the surroundngs as shown n fgure. The two gases, say, gas A and gas B are represented by black dots and whte dots respectvely and separated by a movable partton. When the partton s wthdrawn, the gases begn to dffuse nto each other and after a perod of tme, dffuson wll be complete. Let us examne the process. Before partton, f we were to pck up the gas molecules from left contaner, we would be sure that these wll be molecules of gas A and smlarly f we were to pck up the gas molecules from rght contaner, we would be sure that these wll be molecules of gas B. But, f we were to pck up molecules from contaner when partton s removed, we are not sure whether the molecules pcked are of gas A or gas B. We say that the system has become less predctable or more chaotc. We may now formulate another postulate : n an solated system, there s always a tendency for the systems energy to become more dsordered or chaotc and ths could be a crteron for spontaneouls change.

9 CT 9 q s s related wth q and T for a reversble reacton as : rev S. T The total entropy change (S total ) for the system and surroundngs of a spontaneous process s gven by Stotal = S system + S surr > 0 When a system s n equlbrum, the entropy s maxmum, and the change n entropy, S = 0. We can say that entropy for a spontaneous process ncreases tll t reaches maxmum and at equlbrum the change n entropy s zero. Snce entropy s a state property, we can calculate the change n entropy of a qsys.rev reversble process by Ssys. T We fnd that both for reversble and rreversble expanson for an deal gas, under sothermal condton, U = 0, but S total.e., S sys + S surr s not zero for rreversble process. Thus, U does not dscrmnate between reversble and rreversble process, whereas S does. Q. Predct n whch of the followng, entropy ncreases/decreases : () A lqud crystallzes nto a sold. () Temperature of a crystallne sold s rased from 0 K to 115 K. () NaHCO 3 (s) Na CO 3 (s) + CO (g) + H O(g) (v) H (g) H(g) [NCERT Solved Example 6.9] Soluton : () After freezng, the molecules attan an ordered state and therefore, entropy decreases. () At 0 K, the conttuent partcles are statc and entropy s mnmum. If temperature s rased to 115 K, these begn to move and oscllate about ther equlbrum postons n the lattce and system becomes more dsordered, therefore entropy ncreases. () Reactant, NaHCO 3, s a sold and t has low entropy. Among products there are one sold and two gases. Therefore, the products represent a condton of hgher entropy. (v) Here one molecule gves two atoms.e., number of partcles ncreases leadng to more dsordered state. Two moles of H atoms have hgher entropy than one mole of dhydrogen molecule. Q. For oxdaton of ron, 4Fe(s) + 3O (g) Fe O 3 (s) entropy change s JK 1 mol 1 at 98 K. Inspte of negatve entropy change of ths reacton, why s the reacton spontaneous? ( r H for ths reacton s J mol 1 ) [NCERT Solved Example 6.10] Soluton : 5530 JK 1 mol 1, JK 1 mol 1 Gbbs energy and spontanety : Nether decrease n enthalpy nor ncrease n entropy alone can determne the drecton of spontaneous change for these systems. For ths purpose, we defne a new thermodynamc functon the Gbbs energy of Gbbs functon, G, as G = H TS Gbbs functon, G s an extensve property and a state functon. The change n Gbbs energy for the system, G sys can be wrtten as G sys = H sys TS sys S sys T At constant temperature, T = 0 Gsys = H sys TS sys Usually the subscrpt system s dropped and we smply wrte ths equaton as G = H TS Now let us consder how G s related to reacton spontanety. We know, S total = S sys + S surr If the system s n thermal equlbrum wth the surroundng, then the temperature of the surroundng s same as that of the system. Also, ncrease n enthalpy of the surroundng s equal to decrease n the enthalpy of the system. Therefore, entropy change of surroundngs, S surr H T surr H T sys

10 CT 10 S total S sys H T sys Rearrangng the above equaton : TS total = TS sys H sys For spontanous process, S total > 0, so TS sys H sys > 0 (H sys TS sys ) > 0 Usng equaton, the above equaton can be wrtten as G > 0 G = H TS < 0 H sys s the enthalpy change of a reacton, TS sys n the energy whch s not avalable to do useful work. So G s the net energy avalable to do useful work and s thus a measure of the free-energy. For ths reason, t s also known as the free energy of the reacton. G gves a crtera for spontanety at constant pressure and temperature. () If G s negatve (< 0), the process s spontaneous. () If G s postve (> 0), the process s non spontaneous. Gbbs Energy Change And Equlbrum : So, the crteron for equlbrum A + B C + D; s r G = 0 Gbbs energy for a reacton n whch all reactants and products are n standard state, r G s related to the equlbrum constant of the reacton as follows : or or 0 = r G + RT ln K r G = RT ln K We also known that r G =.303 RT log K r G = r H T r S = RT ln K Q. Calculate r G for converson of oxygen to ozone, 3/ O (g) O 3 (g) at 98 K, f K p for ths converson s [NCERT Solved Example 6.11] Soluton : 163 kj mol 1. Q. Fnd out the value of equlbrum constant for the followng reacton at 98 K. NH 3 (g) + CO (g) NH CONH (aq). + H O(l) Standard Gbbs energy change, r G at the gven temperature s 13.6 kj mol 1. [NCERT Solved Example 6.1] Soluton :.38 [Hence K = antlog.38 =.4 10 ] Q. At 60 0 C, dntrogen tetroxde s ffty percent dssocated. Calculate the standard free energy change at ths temperature and at one atmosphere. Soluton : 1.33 atm, kj mol 1.

11 CT 11

12 CT 1 NCERT EXERCISE 6.1 Choose the correct answer. A thermodynamcs state functon s a quantty () () () (v) A. () used to determne the heat changes whose value s ndependent of path used to determne pressure volume work whose value depends on temperature only 6. For the process to occur under adabatc condtons, the correct condton s A. () () T = 0 () p = 0 () q = 0 (v) w = The enthalpes of all elements n ther standard states are : A. () () unty () zero () < 0 (v) dfferent for each element 6.4 U 0 of combuston of methane s X kj mol 1. The value of H 0 s A. () () = U 0 () > U 0 () < U 0 (v) = The enthalpy of combuston of methane, graphte and dhydrogen at 98 K are, kj mol 1, kj mol 1, and 85.8 kj mol 1 respectvely, Enthalpy of formaton of CH 4 (g) wll be A. () (a) 74.8 kj mol 1 (b) 5.7 kj mol 1 (c) kj mol 1 (d) +5.6 kj mol A reacton, A + B C + D + q s found to have a postve entropy change. The reacton wll be () possble at hgh temperature () possble only at low temperature () not possble at any temperature (v) possble at any temperature A. (v), G < 0 or ve, whch s possble at any temperature 6.7 In a process, 701 J of heat s absorbed by a system and 394 J of work s done by the system. What s the change n nternal energy for the process? A. 307 J 6.8 The reacton of cyanamde, NH CN(s), wth doxygen was carred out n a bomb calormeter, and U was found to be 74.7 kj mol 1 at 98 K. Calculate enthalpy change of the reacton at 98 K. 3 NH CN(g) O(g) N(g) CO(g) HO(l) A kj mol Calculate the number of kj of heat necessary to rase the temperature of 60.0 g of alumnum from 35 0 C to 55 0 C. Molar heat capacty of alumnum s 4 J mol 1 K 1. A. q = n C T, 1.07 kj 6.10 Calculate the enthalpy change on freezng 1.0 mol of water at C to ce at C. fus H = 6.03 kj mol 1 at 0 0 C C p [H O (l)] = 75.3 J mol 1 K 1 C p [H O (s)] = 36.8 J mol 1 K 1 A. H = (1 mol of water of 10 0 C 1 mol of water at 0 0 C) + (1 mol of water at 0 0 C 1 mol ce at 0 0 C) + (1 mol ce at 0 0 C 1 mol ce at 10 0 C), H = C p [H O (l) T + H freezng + C p [H O (s)] T, kj mol 1

13 CT Enthalpy of combuston of carbon to CO s kj mol 1. Calculate the heat released upon formaton of 35. g of CO from carbon and doxygen gas. A kj 6.1 Enthalpes of formaton of CO(g), CO (g), N O(g) and N O 4 (g) are 110, 393, 81 and 9.7 kj mol 1 respectvely. Fnd the value of r H for the reacton : A kj mol 1 N O 4 (g) + 3CO(g) N O(g) + 3CO (g) 6.13 Gven : N (g) + 3H (g) NH 3 (g); r H 0 = 9.4 kj mol 1. What s the standard enthalpy of formaton of NH 3 gas? A. 46. kj mol Calculate the standard enthalpy of formaton of CH 3 OH (l) from the followng data : CH 3 l...(1) 0 1 3OH( ) O(g) CO(g) HO( l); rh 76kJ mol C(g) + O (g) CO (g); c H 0 = 393 kj mol 1...() A. 39 kj mol 1 H (g) + 1 O (g) H O(l); f H 0 = 86 kj mol 1...(3) 6.15 Calculate the enthalpy change for the process : CCl 4 (g) C(g) + 4 Cl(g) and calculate bond enthalpy of C Cl n CCl 4 (g) A. 36 kj mol 1 vap H 0 (CCl 4 ) = 30.5 kj mol 1 f H 0 (CCl 4 ) = kj mol 1 a H 0 (C) = kj mol 1, where a H 0 s enthalpy of atomsaton a H 0 (Cl ) = 4 kj mol For an solated system, U = 0, what wll be S? A. For an solated system, let us consder that two dfferent gases contaned separately n two bulbs and are connected by a stop cock and solated from the surroundngs. When the stop cock s opened gases mx up and the system becomes more dsordered.e., S > For the reacton at 98 K, A + B C H = 400 kj mol 1 and S = 0. kj K 1 mol 1 At what temperature wll the reacton become spontaneous consderng H and S to be constant over the temperature range. A. Frst calculate the temperature at whch the reacton would be n equlbrum,.e., G = 0. For G to be negatve, T should be greater than 00 K 6.18 For the reacton, Cl (g) Cl (g), what are the sgns of H and S? A. The gven reacton s an example of bond formaton, hence n ths energy s released.e., H s ve. Further, the number of moles decreases on the rght hand sde from to 1.e., S s also ve 6.19 For the reacton A(g) + B(g) D (g); U 0 = 10.5 kj and S 0 = J K 1 Calculate G 0 for the reacton and predct whether the reacton may occur spontaneously. A kj, As G 0 s +ve, the reacton would not occur spontaneously

14 CT The equlbrum constant for a reacton s 10. What wll be the value of G 0? R = J K 1 mol 1, T = 300 K. A J 6.1 Comment on the thermodynamc stablty of NO(g), gven : (g) O(g) NO(g); rh 90 kj mol N 1 0 NO(g) O(g) NO(g); rh 74kJ / mol A. In the frst reacton, energy s absorbed, therefore NO(g) s unstable. In the second reacton, energy s released, therefore NO s stable 6. Calculate the entropy change n the surroundngs when 1.00 mol of H O (l) s formed under standard condtons : f H 0 = 86 kj mol 1. A. 960 J K 1 mol 1

15 CT 15 ADDITIONAL QUESTIONS AND PROBLEMS Q.