OPEN CLOSED ISOLATED. Only energy can be exchanged between system & surrounding. eg. closed vessel

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1 THERMODYNAMICS Limitations :- (1) Not applicable on microscopic system like, change inside an atoms. Or system having few molecules only. (2) deals only with initial & final state (does not deal with path ) (3) Deals only with spontaneity of a process does not deal with how process occurred or by what rate it occurred (4) Laws of thermodynamics are applied only when a system is in equilibrium or moves form one equilibrium to another equilibrium System & surrounding : The part of universe which is under observations is called system and remaining universe constitutes the surrounding of that system. The universe = The system + the surrounding Type of system OPEN CLOSED ISOLATED Both energy & matter can be exchanged between system and surrounding eg open beaker Only energy can be exchanged between system & surrounding. eg. closed vessel No exchange of energy or matter. eg. Thermos flask or Insulated flask There is no perfectly isolated system in the universe, Boundary :- Anything which separate system & surrounding. Boundary can be real / imaginary, rigid / non rigid, conductor (Diathermic) / non conductor (adiavatic). Extensive property :- depend upon mass of system Eg. Internal energy, enthalpy, entropy Gibbs free energy, heat capacity, volume etc. If system is divided in to two part the properties which become half are extensive in nature. Intensive property :- Do not change on changing mass. e.g. E, density of Homogenous system, surface tension, reflective index, viscosity, melting point, Boiling point, Molarity, Normality, etc. Extensive properties which are measured for 1 mol also become intensive. eg Volume is intensive but molar volume is extensive. Heat capacity is extensive but molar heat capacity is intensive. Similarly enthalpy is extensive but molar enthalpy is intensive. The ratio of two extensive variables gives intensive variable. Eg. Mass and volume are both extensive but mass/volume that is density is intensive. Other examples are molarity, normality State functions : The state of a thermodynamic system is described by its measurable or macroscopic properties. The state functions are those thermodynamic functions which depend only on the initial and final states and not on how the state is reached. Examples of thermodynamics state functions are : BMC CLASSES Page No - 32

2 Internal energy (U), enthalpy (H), entropy (S), free energy (G), volume (V) etc. For any state function, we can calculate the change in function ( ). For example, change in volume. V = V final V initial For a cyclic process, the change in state function is equal to zero because initial and final state are same. Path Dependent Functions. The thermodynamic functions which depend on the path are known as path dependent functions. The important path dependent functions are heat(q) and work(w). For example, the amount of work done in climbing a mountain peak depends on the path chosen. Reversible process. In a reversible process, the change is carried out so slowly in infinite number of steps so that the system and the surroundings are always in equilibrium. The reversible processes are ideal processes and are not realized in actual practice. Irreversible Process. A process which is not exactly reversed i.e., the system does not pass through the same intermediate states as in the direct process is known as irreversible process. An irreversible process cannot be reversed without the help of an external agency. All natural processes are irreversible. For example, falling of water from a hill top, germination of seeds, diffusion of gases etc, are all irreversible processes. Problems 1. IITJEE 2010 (Multiple choice) Among the following, the intensive property is (properties are) (a) molar conductivity (b) electromotive force (c) resistance (d) heat capacity Ans : (a), (b) 2. IITJEE 1993 (Multiple choice) Identify the intensive property from the following (a) Enthalpy (b) Temperature (c) Volume (d) Refractive Index. Ans : (d) 3. IITJEE 2001 In thermodynamics a process is called reversible when (a) surrounding and system change into each other (b) there is no boundary between system and surroundings (c) surroundings are in equilibrium with system (d) system change into surroundings spontaneously Ans : (c) 4. IITJEE 2001 Which of the following statements is false? (a) Work is a state function (b) Temperature is a state function (c) Change of state is completely defined when initial and final states are specified (d) Work appears at the boundary of the solution. Ans : (a) work is a path function BMC CLASSES Page No - 33

3 Isothermal T = 0, U = 0, q = -w Isobaric :- Isochoric :- Adiavatic :- P = 0, H = q V = 0, W = 0, U = q q = 0, U = W Adiavatic reversible T V γ-1 = constant P V γ = constant T Y P 1-γ = constant W = C v / R (P 1 V 1 P 2 V 2 ) or P 1 V 1 P 2 V 2 γ - 1 W = C v (T 1 -T 2 ) or C v T Isothermal reversible ;- W = nRT log V 2 /V 1 W = nRT log P 1 /P 2 Cyclic E = 0 H = 0 S = 0 G = 0 Exercise on type of Process (1) :- An ideal diatomic gas is caused to pass through a cycle shown on the P V diagram in figure where V 2 = 3V 1. If P 1, V 1 and T 1 specify state 1, then the temperature of state 3 is :- Isothermal Pressure Adiavatic Isometric Volume (a) (T 1 /3) 1.4 (b) T 1 /(3) 1.4 (c) T 1 /(3) 0.4 (d) can not be determined Solution:- State 1 P 1 V 1 T 1 State 3 V 2 = 3V 1 Adiavatic expansion TV Y 1 = constant T 1 /T 2 = (V 2 /V 1 ) Y 1 (diatomic gas y = 1.40) T 1 /T 2 = (V 2 /V 1 ) T 1 /T 2 = (3V 1 /V 1 ) 0.4 T 2 = T 1 /(3) 0.4 hence, (c) (2) :- A given mass of gas expands from the state A to the state B by three path 1, 2, 3 as shown in figure, W 1, W 2 & W 3 respectively are work done by the gas along three paths. What is relation between W 1, W 2 & W 3. A P 1 2 B V 3 BMC CLASSES Page No - 34

4 - Work is a path function hence W 1 W 2 W 3 Work is slop of PV graph hence it is clear W 1 < W 2 < W 3 (3) :- A sample of 2 kg of helium (ideal) is taken through the process ABC and another sample of 2Kg of the same gas is taken through the process ADC, then the temperature of state A & B are B C 10 P (10 4 N/m 2 ) 5 A D V (m 3 ) - At state A PV = nrt T = PV/nR = 5 x 10 4 x x = K At stage B P is double Hence T B = 2 x T A = 241 K (4) : Two mols of a perfect gas undergoes the following processes (a) a isobaric expansion from 1.0 atm, 20 lit to 1.0 atm, 40lit (b) a isochoric change of state from 1.0 atm, 40 lit to 0.5 atm, 40 lit (c) a reversible isothermal compression from 0.5 atm, 40 lit to 1.0 atm 20 lit (i) (ii) (i) sketch with labels each of the process on the same P-V diagram calculate the total work & total heat change q invalved in the above Process. PV = nrt 1.0 atm A B T = PV/ nr = (1 x 20) / (2 x 0.082) 0.5 = k atm 20 lit 40 lit (ii) work from A B W = - P V = - 1 (20) = -20 lit atm = -20 x = J Work from B C = 0 (isobaric ) Work from C A = nrt log V 2 / V 1 = x 2 x 8.31 x log 20/40 PV = nrt BMC CLASSES Page No - 35

5 = J Total work = W 1 + W 2 + W 3 = = J Cyclic hence E = 0 E = q + w q = - w = 622 J (5) IITJEE 2010 (integer type) One mole of an ideal gas is taken from a to b along two paths denoted by the solid and he dashed lines as shown in the graph below. If the work done along the solid line path is w s and that along the dotted line path is w d, then the integer closest to the ratio w d /w s is W d irreversible P external V = W 1 + W 2 + W = ( 4 1.5) + ( 1 1) + ( ) = 8.67 lit. atm W S reversible nrt log V 2 /V 1 = P initial V initial (as PV =nrt ) = log 5.5/0.5 = 4.8 lit. atm W d /W S = 8.67 / 4.8 = 2 Internal Energy : U or E The sum of all kinds of energies of a system ie. Chemical, electrical, mechanical (kinetic & potential) etc, is called as internal energy. Internal energy of a system may change when (a) Heat passing into or out of the system (b) Work is done on or by the system (c) Matter leaves or enters the system. First law of thermodynamics Energy can be neither created nor destroyed. If both the system and the surroundings are taken into account it also be stated as. The internal energy U of an isolated system is constant. U U Total = U System + U Surrounding = 0 Means according to first law U System = U Surrounding U for a gas can be increased by heating it in a flame or by doing compression work on it. U = q + w BMC CLASSES Page No - 36

6 E or U : - Change in internal energy : * HEAT OF REACTION MEASURED AT CONSTANT VOLUME. * For (i) Isothermal process (But the phase of substance should not change.) (ii) Cyclic process E or U = 0 * According to first law of thermodynamics E = q + w [Using this formula the most important considerations are (i) sign convension (ii) unit of q & W ] eg. Q :- 500 J of heat was given to a system by which its volume increased from 10 lit ot 20 lit at 1 atmospheric pressure calculate change in interal energy. - q = +500 (Heat gives to system hence +ve) E = q + W ( W = - P V) = ( -1 {20-10}) = = 490 This method can not be taken as W = - P V = - 1 (20-10 ) = -10 lit atm not -10 J while q is in Joule. Hence when E = q + W is applied W should be calculated separately and converted into Joule. W = - P V = = - 10 lit atm = J = J E = q + W = = J * In Bomb calorimeter the heat released is also called E as bomb calorimeter works at constant volume. Q : - When 3.2 gm of C 2 H 6 (g) was combusted in a bomb calorimeter the temprature of calorimeter increased by 0.5 C if thermal capacity of calorimeter is 200 kj / k find heat of combustion of C 2 H 6 at constant volume. - Heat released = Thermal capacity T (Rise in temp) = = 100 kj. E & H should be answerd in kj / mol hence 3.2 gm = 3.2 / 32 = 0.1 mol 0.1 mol gives = 100 kj 1 mol = 100 / 0.1 = 1000 kj E = kj / mol Remember heat released is always taken ve. ISOTHERMAL PROCESS AND E : - Isothermal means constant temperature. At constant temperature E is considered to be zero but only when th state of substance does not chane eg. Q : -Pressure of 2 gms of H 2 (g) is reduced from 10 atm to 1 atm at a constant temprature of 300 k, the gas behave idealy find (i) W (ii) q (iii) E BMC CLASSES Page No - 37

7 - Pressure is decreasing continuously hence W rev = nrt log P1 P 2 10 = log 1 (n = 2/2 = 1 mol. Because work is measured in J or in cal hence. R = (if in s) or 1.98 (if in cal)). = J W = kj As Isothermal process and state is not changing hence E = 0 E = q + w q = - W = - ( kj) = kj q = 5.74 kj If In Isothermal conditions state of substance get change than E 0 eg. Q : - Calculate (i) W (ii) E for the conversion of 36 gm of water into steam at a tempature of 100 C and at a pressure of 1 atmostpher. Latent heat of vaporisation of water is 40.7 kj / mol. - Asd H 2 O (l) H 2 O (g) Inspite of isothermal conditions E 0 W = - P V = - n RT n = number of gaseous mols of product number of gasesous mols of reactant = 1-0 = 1 As 2 mols of gas (36 gm ) hence n = 1 2 = 2 W = - n RT = = J = - 6.2kJ q = kj = 81.4 kj E = Q + W = (- 6.2) = 75.2 kj H CHANGE IN ENTHALPY (Heat of reaction at constant pressure) * Heat given to a substance at costant pressure & temperature to change the state (phase) is considered H. Q. : kj of Heat is given to vaporise 1 mol of H 2 O at 100 C and 1 atmospheric pressure, calculate H & E of the process. - H = kj H = E + nrt E = H - nrt (going for this formula remember there are several chances of mistakes eg. H = in kj R = J mol -1 k -1 ) BMC CLASSES Page No - 38

8 H 2 O (l) H 2 O (g) n = 1 E = = E = kj H in Bomb calorimeter Heat change in bomb calorimeter is E (as calorimeter works on constant volume) H = E + n RT Before using this formula remember (i) sign of E (-ve if released) (ii) n = number of gaseous mols of products number of gaseous mols of reactants. [in calculating n, H 2 O should be considered liquid if temp is < 373 k.] (iii) most important is unit of E & nrt should be same. Q : gm of isobutane (C 4 H 10 )gas was combusted in a bomb calorimeter at 300 k.the temperature of calorimeter increases by 0.4 C, if thermal capacity of calorimeter is 80 k cal k -1 find. (i) Heat of combustion of isobutane at 300 k and constant volume. (ii) Heat of combustion of isobutane at 300 k and constant pressure. - n = 11.6 / 58 = 0.2 mols Heat released = = 32 k cal E = - {32 1}/ 0.2 = kcal (Heat relesased ) C 4 H / 2 O 2 4 CO H 2 O (g) (g) (g) (l) (Beacuse at 300 K) n = 4 ( ) = H = E + n RT = ( -3.5 x 1.98 x 300 ) 1000 = = H = k cal / mol ENTHALPY AND TEMPRATURE :_ Kirchoff equation C p = H 2 H 1 T 2 T 1 Where H 1 & H 2 are enthalpies at T 1 & T 2 temprature respetively. Q : - Enthalpy of vaporization of SO 2 at -10 C is 5950 cal/ mol calculate its value at - 25 C. C P SO 2 (l) = 206 cal / mol : C p SO 2 (g) = 93 cal / mol - SO 2 (l) SO 2 (g) C P = C P SO 2 (g) C P SO 2 (l) = = C p = H 2 H 1 T 2 T 1 T 1 = - 25 C = = 248 K H 1 =? T 2 = -10 C = = 263 K BMC CLASSES Page No - 39

9 H 2 = = 5950 H = 5950 H = 5950 H 1 ; H 1 = 7645 cal / mol. Standard enthalpy of formation : H f standard enthalpy of formation ( H f ) is considered to be zero for an element in its most stable naturally existing state at room temperature. eg H f O 2 (g) = 0 (naturally existing form at 298 of oxygen) H f O 3 (g) 0 (naturally existing stable state of oxygen is O 2 ) H f C graphite = 0 H f C Diamond 0 (naturally existing stable of carbon is diamond) H f Br 2 (l) = 0 H f Br 2 (g) 0 (Br 2 exist as liquid at room temperature) H f Hg(s) 0 H f Hg(l) = 0 (Hg exist as liquid at room temperature) Similarly H f = 0 for Al(s), As grey (s), Ba(s), B(s), Ca(s), Ce(s), Cl 2 (g), Cu(s), D 2 (g), F 2 (g), H 2 (g), H + (aq), I 2 (s), Pb(s), Mg(s), N 2 (g), P(s) white, K(s), Si(s), Na(s), S(s)rhombic, Sn(s)white, Zn(s) * For H + (aq) all H f, G & S are considered zero. * For all the element in which H f is zero G f is also zero but S is not equal to zero. * H f H 2 O 0 Because it is a compound and H f is zero only for element not for compound. * To form equation of standard enthalpy of formation of a substance 1. First write one molecule of product. 2. Write reaction in the form of element only eg H f CH 4 C graphite + 2H 2 CH 4 (g) H = H f CH 4 (g) H f N 2 O 3 (g) N 2 + 3/2 O 2 N 2 O 3 (g) H = H f N 2 O 3 (g) (g) Not N 2 + O 3 (g) N 2 O 3 (g) H H f N 2 O 3 as naturally existing form of oxygen is O 2 not O 3 O 3 O 2 Not 2N 2 + 3O 2 2 N 2 O 3 H H f N 2 O 3 Because here two moles of N 2 O 3 are forming Problem IITJEE 2010 (Multiple choice) The species which by definition has zero standard molar enthalpy of formation is 298 K is (a) Br 2 (g) (b) Cl 2 (g) (c) H 2 O (g) (d) CH 4 (g) Ans : (b) BMC CLASSES Page No - 40

10 RESONANCE ENERGY In case of certain compound, the observed heat of formation differs significantly from the calculated bond energy data. The difference between the observed value of heat of formation and the value obtained from bond energy data is called the resonance energy of the compound. Resonance energy is measure of extra stability of the compound. Resonance energy is measure of extra stability of the compound. For example, the observed heat of formation of CO 2 is 94 kcal and the value calculated from bond energy data is 64 kcal, therefore, resonance energy of CO 2 = = 30 kcal Q :- Calculate the resonance energy of N 2 O from the following data H o f of N 2 O = 82 kjmol -1 Bond energies of N N, N = N, O = O and N = O bonds are 946, 418, 498 and 607 kj -1 respectively. - N 2 + ½ O 2 N 2 O N N + ½ (O = O) N = N = O H =[{(N N) + ½ ( O = O) {(N = N) + ( N = 0)}] = ( ) ( ) = 170 R.E = H f Cal ~ H f ob = 88 kj Q : - The standard molar enthalpies of formation of cyclohexane (l) and benzene (l) at 25 C are and +49 kj mol -1 respectively. The standard enthalpy of hydrogenation of cyclohexene (l) at 25 C is 119 kj mol -1. Use these data to estimate the magnitude of the resonance energy of benzene. -Standard enthalpy of hydrogenation of cyclohexene = kj mol -1. This is the enthalpy of hydrogenation of one double bond (present in cyclohexene). If benzene is considered as cyclohextriene, the enthalpy of reaction + 3 H 2 (g) H of this hydrogenation = 3 (-119) = kj mol -1 Thus, the observed heat of hydrogenation of benzene ( H ob ) = kj mol -1 On the basis of above thermochemical equation H cal = [ H f of cyclohexane (l) ( H f of benzene (l) + 3 H f of H 2 )] = [-156 ( )] = -205 kj mol -1 Resonance energy = H ob - H cal = [-357 (-205)] kj mol -1 = kj mol -1 Q : - Calculate the electronegativity of fluorine from the following data : E H H = kcal mol -1, E F F = 36.6 kcal mol -1 E H F = kcal mol -1, X H = Let the electronegativity of fluorine be X F. Applying Pauling s equations X F X H = [E H F (E F- F x E H - H ) 1/2 ] 1/2 In this equation, dissociation energies are taken in kcal mol -1. X F 2.1 = [ (104.2 x 36.6) 1/2 ] 1/2 X F = 3.87 Q : - Calculate the electronegativity of carbon from the following data : E H H = kcal mol -1, E C C = 83.1 kcal mol -1 E C H = 98.8 kcal mol -1, X H = 2.1 BMC CLASSES Page No - 41

11 - Let the electro negativity of carbon be X C. Applying Pauling s equations X C X H = [E C H (E C - C x E H - H ) 1/2 ] 1/2 X C 2.1 = [ 98.8 (83.1 x ) 1/2 ] 1/2 X C = 2.59 Q : - Ionisation potential and electron affinity of fluorine are and 3.45 ev respectively. Calculate the electronegativity of fluorine. - According to Mulliken equation X = IP + EA when both IP and EA are taken in ev. 5.6 X F = = Q : - The electron affinity of chlorine is 3. ev. How much energy in kcal is released when 2 g of chlorine is completely converted to Cl - ion in a gaseous state? - Cl + e Cl ev x kcal Energy released for conversion of 2 g gaseous chlorine into Cl - ions = 3.7 x x 2 = 4.8 kcal 35.5 Q : - The first ionistation potential of Li is 5.4 ev and the electron affinity of Cl is 3.6 ev. Calculate H in kcal mol -1 for the reaction Li(g) + Cl (g) Li + + Cl - Carried out at such low pressures that resulting ions do not combine with each other. - The overall reaction is written into two partial equations. Li(g) Li + + e E 1 = 5.4 ev Cl(g) + e Cl - E 2 = ev H = E 1 - E 2 = = 1.8 ev = 1.8 x kcal mol -1 = kcal mol -1 BMC CLASSES Page No - 42

12 THERMODYNAMICS - II SPANTANIOUS PRECESS An irreversible process which can be reversed only by some external agency. OR A process which either start by its own or by any external factor but remain continue till the end is called a spontaneous process. Eg. 1. Melting of ice above 0 C 2. Combustion of L.P.G. 3. Combustion of coal 4. Dissolution of salt in water 5. Neutralization reaction when aced and base are mixed. 6. Mixing of two gases (diffusion) 7. Flow of heat from a region of higher temperature to a region of lower temp. 8. Evaporation of water etc. NON SPONTANIUS PROCESS A process which occur only when external work is carried out. A non spontaneous process only remain continue till external factor is working. eg 1. In a refrigerator or air conditioner heat flow from a region of lower temp oto higher temp and it take place only till electric current is passed. 2. electrolysis of water take place only in presence of electric work WHAT WAS THE NEED OF SECOND LAW OF THERMODYNAMICS OR WHAT WAS THE NEED OF TERM ENTROPY : H change in enthalpy can not be the criterion for spontaneity of a process because (I) THERE ARE SEVERAL SPONTANEOUS PROCESS IN WHICH H IS NEGATIVE. eg 1. Combustion of L.P.G., C.N.G. etc CH O 2 CO H 2 O H = ve 2. Neutralization of Acid/Base HCl + NaOH NaCl +H 2 O H = ve 3. Formation of NH 3 N H 2 2NH 3 H = ve 4. Combustion of Hydrogen H O2 H 2 O H = ve BUT (II) THERE ARE SEVERAL SPONTANEOUSLY PROCESS IN WHICH H IS +VE. Eg 1. Melting of ice above 0 C H 2 O(s) H 2 O(l) H = +ve 2. Evaporation of H 2 O H 2 O(l) H 2 O(g) H = +ve 3. Dissolution of salt in water H 2 O NaCl(s) NaCl(aq) H = +ve MEANS It becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity but it is not true for all cases. To explain spontaneity an another term was require ENTROPY, certainly second law of thermodynamics. BMC CLASSES Page No - 43

13 ENTROPY : S It is the measurement of the degree of randomness or disorder of a system. Greater the entropy more the system is disordered (Chaotic) * Entropy is a state function S = S f Si * When a system is in equilibrium its entropy is maximum and at equilibrium S = O S univ = S Total = S system + S Surrounding for Irreversible process S uni > O for Reversible process at equilibrium S uni = O 2 nd Law of Thermodynamics : S uni = S syst + S surrounding for a spontaneous process Entropy of universe always increases. 3 rd Law of Thermodynamics : Entropy of a perfectly crystalline solid is considered to be zero at zero Kelvin temp (absolute zero) From the 3 rd law it is obvious that S can not be zero for any substance because S means slandered entropy i.e measured at 298 K and entropy can be considered zero at zero Kelvin only. S +ve : The process in which randomness increases 1. Solid liquid 2. Liquid gas 3. Solid gas 4. A reaction in which n > O 5. Mixing of two or more gases/liquid/solid in which reaction is not taking place. eg. Diffusion of gases 6. Increase in temperature/decreases in pressure 7. If temp and pressure both are changed find value of P T if it increases than S = +ve 8. Breaking of a bond. 9. Dissolution of a solid e NaCl (s) NaCl (aq) 10. Decomposition of a solid into two or more solid/ l /where n can not calculated Pb 3 O 4 (s) 2 PbO + pbo 2 (s) eg (2) 2HCl H 2 + Cl 2 (g) (g) (g) S ve : Apposite of process in which S is +ve particularly gas liquid or solid liquid solid, mixing of two or more reacting gases/ substance. eg. Mixing of H 2 + He S = +ve Mixing of H 2 & Cl 2 S = ve reacting (s) Crystallization NaCl (aq) NaCl(s) etc. * for an isothermal reversible process S = q rev T BMC CLASSES Page No - 44

14 GIBBS FREE ENERGY : If a reaction is carried out in a isolated system or thermally insulated container than spontaneity means positive value of S uni. BUT Most of reaction are carried out in open or closed system then both enthalpy and entropy charge take place hence both should be considered to explain spontaneity. GIBBS FREE ENERGY (G) IS THE ENERGY OF A SYSTEM WHICH CAN BE CONVERTED INTO USEFUL WORK AT A GIVEN TEMPERATURE G = H TS G = G f G in (state function, extensive property) G = H (TS) at constant temp G = H T S * G = T S TOTAL Let at constant temp (T) and constant pressure (P) q foule of heat is given to system Constant pressure hence S H = q YST q H system = + q EM H system = q q H S system = = T T H S surrounding = T 1 S Total = S System + S Surrounding 2 By equation 1 & 2 H S Total = S System + T T S Total = S System H T S Total = G G = T S Total ( G = H T S) * G = W non expansion or G = W max According to first law of thermodynamics U = q +W W = W expansion + W non expansion U = q + W expansion + W non expansion q = U W expansion W non expansion = U ( P V) W non expansion = U + P V W non expansion H = U + P V q = H W non expansion 1 q S =, q = T S T T S = H W non expansion W non expansion = H T S W non expansion = G BMC CLASSES Page No - 45

15 OVERALL CRITERION FOR SPONTANEITY For spontaneous process G = ve If G = O system is in equilibrium If G < O Process is irreversible (Spontaneous) If G > O Process is non spontaneous EXOTHERMIC REACTION ( H = ve) (i) if S = +ve reaction is spontaneous at all temp. (ii) if S = ve, G = H T S G = H T ( S) G = H + T S Can be negative when T S < H means if H = ve, S = ve reaction is spontaneous only when T S < H certainly low temp favour spontaneity for such reaction. ENDOTHERMIC REACTION ( H = +ve) (i) if S = +ve G = + H T (+ S) G = H T S Can be negative only when T S > H mean if for a reaction H = +ve & S = +ve reaction can be spontaneous if T S > H mean high temp will favour spontaneity (ii) if S = ve H = +ve, S = ve Reaction can not be spontaneous at any condition G = +OH T ( S) G = + H + T S Can not be negative Conclusion (i) H = ve, S = +ve spontaneous always (ii) H = ve, S = ve spontaneous only when H > T S (low temp) (iii) H = +ve, S = +ve spontaneous only when H < T S (High temp) (iv) S = +ve, S = ve can never be spontaneous G & EQUILIBRIUM At equilibrium G = O but G O (usually) G = RT log K C G = RT log K G = G + RTl n K P G & emf G = n F E Cell COUPLING OF REACTION : A B G 1 = +ve (small value) C D G 2 = ve (large value) Coupling A B G = G 1 + G 2 = ve C D Both reaction take place spontaneously eg. Metallurgical operations Reduction of metal oxide into metal ( G = +ve) + Oxidation of carbon/carbon mono oxide. BMC CLASSES Page No - 46

16 I.I.T (1) :- The process A B is a difficult process hence carried out in several step as follows C D A B Given S (A C) = 50e.u. ; S (C D) = 30e.u. S (B D) = 20e.u. where e.u. is entropy unit then S (A B) is BMC CLASSES Page No - 47

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21 NCERT PROBLEMS 1. Express the change in internal energy of a system when (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have? (ii) No work is done on the system, but q amount of heat is taken out form the system and given to the surroundings. What type of wall does the system have? (iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be? E system =? (i) q = 0, W = +Ve E = q + W E = W, Adiavatic walls. ( q = 0) (ii) W= 0, q = ve E = q + W (iii) W = ve, q = +ve E = q + W = q W E = q (Negative) Walls through which heat can flow conducting walls. Conducting walls, closed system. 2. Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion? In vacuum Pext = 0 W = Pext V = 0, W = 0 Isothermal expansion E = 0 E = q + W 0 = q + 0, q = 0 3. Consider the same expansion, but this time against a constant external pressure of 1 atm. W = Pex V = 1 (10 2) = 8 lit atm 4. Consider the same expansion, to a final volume of 10 litres conducted reversibly V2 W rev = nrt log V1 as temperature and number of mols are not given 10 = log 2 but PV = nrt = (0.6990) nrt = PV = lit atm = 10 2 = 20 E = 0 E = q + W, q = W = ( ) = lit atm = = J 5. If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol water at 1 bar and 100 C is 41 kj mol 1. Calculate the internal energy change, when (i) 1 mol of water is vaporised at 1 bar pressure and 100 C. (ii) 1 mol of water is converted into ice. H = 41 kj, E =? H 2 O (l) H 2 O(g) n = 1 0 = 1 BMC CLASSES Page No - 52

22 H = E + nrt E = H nrt As H is in kj and = R = J/mol/K 1000 hence R = kj mol 1 K = kj/mol (ii) H 2 O (l) H 2 O(s) (It is an exothermic process) H = ve (Value of enthalpy of fusion is not given) As volume H 2 O(l) is approximately same to H 2 O(s) means nrt = P V would be considered zero E = H But H should be ve 6. 1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C (graphite) + O 2 (g) CO 2 (g) During the reaction, temperature rises from 298 to 299 K. if the heat capacity of the bomb calorimeter is 20.7 kj/k. What is the enthalpy change for the above reaction at 298 K and 1 atm? A bomb calorimeter works at constant volume hence heat released per mol = E Heat released = Thermal capacity T = = 20.7 kj 1 gm graphite = 12 1 mol E = = 248 kj (As heat is released E = ve) C gr + O 2 CO 2 (g) n = 1 1 = 0 (g) H = E + nrt H = E = 248 kj/mol 7. A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at 100 C. vap H for water at 373K = kj mol 1 H vap of H 2 O = kj mol 1 Means H 2 O (l) H 2 O (g) H = kj 1 mol = 18 gm water will require kj heat H = E + nrt H 2 O (l) H 2 O(g) n = 1 0 = 1 E = H nrt = As H is in 1000 kj, R = = kj 8. The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO 2 (g) and H 2 O (1) are produced and kj of heat is liberated. Calculate the standard enthalpy of formation, f H of benzene. Standard enthalpies of formation of CO 2 (g) and H 2 O(1) are kj mol 1 and kj mol 1 respectively kj of heat is released H = 3267 kj BMC CLASSES Page No - 53

23 C 6 H O 2 6 CO 2 + 3H 2 O 2 H = (6 H f CO H f H 2 O) (H f C 6 H 6 ) 3267 = 6 ( 393.5) + 3 ( ) (x) x = kj Alternatively Enthalpy of combustion of benzene = 3267 kj 1 C 6 H O 2 6 CO 2 + 3H 2 O H = Enthalpy of formation of CO 2 = kj 2 Cgr + O 2(g) CO 2(g) H = kj Enthalpy of formation of H 2 O(l) = kj 3 H 2(g) + ½ O 2(g) H 2 O(l) H = kj H f C 6 H 6 =? 4 6 C gr + 3H 2(g) C 6 H 6 H =? To form equation No 4 from (1), (2) & (3) multiply equation No. 2 by six and equation number 3 by three and add them. Substract equation number one from resulting equation. H f benzene = 6 H H 3 H 1 = kj 9. Predict in which of the following, entropy increases/decreases : (i) A liquid crystallizes into a solid. (ii) Temperature of a crystalline solid is raised from 0 K to 115 K. (iii) 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O (g) (iv) H 2 (g) 2H (g) (i) liquid solid (Entropy decreases S = ve) (ii) Temperature is increased means randomness will increase S = +ve (iii) 2 NaHCO 2 (s) Na 2 CO 3 (s) + CO 2(g) + H 2 O (g) n = 2 0 = 2 As n is +ve entropy will increases (a substance is decomposing means S = +ve) (iv) H 2(g) 2H (g) n = 2 1 = 1 S = +ve 10. For oxidation of iron. 4Fe (s) (g) 2Fe 2 O 3 (s) Entropy change is JK 1 mol 1 at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous? ( r H for this reaction is J mol 1 ) H = J/mol, S = 549 J K 1 mol 1 G = H T S = ( 549) = = J As G is ve reaction is spontaneous. BMC CLASSES Page No - 54

24 Alternatively S Total = S system + S surrounding S system = 549 J K 1 mol 1 S surrounding = H surr = H sys T T 3 ( ) = = J 298 S Total = = J (+ve) means spontaneous 11. Calculate r G for conversion of oxygen to ozone. 3/2 O 2 (g) O 3 (g) at 298 K. if K p for this conversion is G = RT log Kp = log = J = 163 kj 12. Find out the value of equilibrium constant for the following reaction at 298 K. 2NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) Standard Gibbs energy change, r G at the given temperature is 13.6 kj mol 1. G = 13.6 kj Very important is G = RT log Kc Unit system = log Kc G is kj log Kc = 2.38 R is in J Kc = Just learn = J 13. At 60 C dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. N 2 O 4(g) 2 NO 2(g) α 2 α Total mols at equilibrium = 1 α + 2 α = 1 + α α = 50% = 0.5 P NO 2 = 2α P = P = P = 1+ α = α P N = P = P = P = 2O4 = 1+ α (P = 1atm) K p = ( ) 2 2 P NO = = = 1.33 P 1 N 3 2O2 4 3 G = RT log Kp = log 1.33 = J = kj BMC CLASSES Page No - 55

25 NCERT EXERCISE 1. Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only. (ii) State functions are independent of path. 2. For the process to occur under adiabatic, conditions the correct condition is: (i) T = 0 (ii) p = 0 (iii) q = 0 (iv) w = 0 (iii) Adiavatic q = 0 3. The enthalpies of all elements in their standard states are : (i) unity (ii) zero (iii) < 0 (iv) different for each element (ii) Enthalpy of formation of all elements in their naturally existing most stable form = 0 4. U of combustion of methane is X kj mol 1. The value of H is (i) = U (ii) > U (iii) < U (iv) = 0 CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O If Temp < 373 K H 2 O is taken as liquid then n = 1 3 = 2 H = E 2 RT = x 2RT H > E If temp 373 K H 2 O is taken as gas n = 3 3 = 0 H = E 5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, kj mol kj mol 1, and kj mol 1 respectively. Enthalpy of formation of CH 4 (g) will be (i) 74.8 kj mol 1 (ii) kj mol 1 (iii) kj mol 1 (iv) kj mol 1 CH 4 + 2O 2 CO 2 + 2H 2 O H = kj Enthalpy of combustion = (H f CO 2 + 2H f H 2 O) (H f CH 4 ) of C graphite = [ ( 285.8)] (x) = H f CH 4 x = ( ) similarly enthalpy of combustion of = 74.8 kj/mol (i) H 2 = H f H 2 O 6. A reaction, A + B C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (iv) possible at any temperature A + B C + D + q Heat means it is an exothermic reaction H = ve, S = +ve ] reaction possible at any temperature (iv) 7. In a process, 701 J of heat is absorbed by a system an 394 J of work is done by the system. What is the change in internal energy for the process? q = 701 J, W = 394 E = q + W = = 317 J BMC CLASSES Page No - 56

26 8. The reaction of cynamide, NH 2 CN (s), with dioxygen was carried out in a bomb calorimeter, and U was found to be kj mol 1 at 298 K. Calculate enthalpy change for the reaction at 298/ K. NH 2 CN(g) O2 (g) N 2 (g) + CO 2 (g) + H 2 O(l) U = kj, H =? NH 2 CN (g) O2(g) N 2(g) + CO 2(g) + H 2 O(l) n = = 0.5 H = U + nrt = ( 0.5) = = kj 9. Calculate the number of kj of heat necessary to raise the temperature of 60.0 g of aluminium from 35 C to 55 C. Molar heat capacity of Al is 24 J mol 1 K 1. Molar heat capacity means heat required to raise the temperature of 1mol by 1 C H = ncp T = (55 35) 27 = J = kj 10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 C to ice at 10.0 C. fus H = 6.03 kj mol 1 at 0 C. C p [H 2 O (l)] = 75.3 J mol 1 K 1 C p [H 2 O (s)] = 36.8 J mol 1 K 1 1 mol Water (10 C) H1 water (O C) H 1 = Cp H 2 O(l) T = = 753 J H 2 = 6.03 kj = 6030 J H 3 = Cp H 2 O (g) T = 36.8 ( 10) = 368 H = H 1 + H 2 + H 3 = ( 6030) + ( 368) = 7151 = kj H 2 Ice (O C) Enthalpy H 3 of fusion Ice ( 10 C) 11. Enthalpy of combustion of carbon to CO 2 is kj mol 1. Calculate the heat released upon formation of 35.2 g of CO 2 from carbon and dioxygen gas gm CO 2 = mol 44 H = = kj 44 BMC CLASSES Page No - 57

27 12. Enthalpy of formation of CO(g),CO 2 (g) N 2 O(g) and N 2 O 4 (g) are 110, - 393, 81 and 9.7 kj mol 1 respectively. Find the value of r H for the reaction : N 2 O 4 (g) + 3CO(g) N 2 O(g) + 3CO 2 (g) N 2 O 4(g) + 3CO (g) N 2 O (g) + 3CO 2(g) H = (H f N 2 O + 3H f CO 2 ) (H f N 2 O 4 + 3H f CO) = ( ( 393)) ( ( 110)) = ( ) ( ) = kj 13. Given N 2 (g) + 3H 2 (g) 2NH 3 (g); r H = kj mol 1 What is the standard enthalpy of formation of NH 3 gas? For two mols of NH kj H f NH 3 = 92.4 = 46.2 kj Calculate the standard enthalpy of formation of CH 3 OH(l) from the following data: CH 3 OH (l) O 2 (g) CO 2 (g) + 2H 2 O(l) ; r H = kj mol 1 C(graphite) + O 2 (g) CO 2 (g) ; c H = kj mol 1 H 2 (g) O2 (g) H 2 O(l) ; f H = kj mol 1. H f CH 3 OH(l) Cgr + 2H 2 + ½ O 2 CH 3 OH H =? (1) CH 3 OH(l) 2 3 O2(g) CO 2(g) + 2H 2 O (l) H 1 = 726 kj (2) Cgr + O 2(g) CO 2(g) H 2 = 393 kj (3) H 2(g) + ½ O 2(g) H 2 O(l) H 3 = 286 kj To obtain above equation from equation (1), (2) & (3) Multiply equation No. (3) by two and add it with equation No. (2), substract equation No. (1) from this reaction H = (2( H 3 ) + H 2 ) H 2 ) H 1 = ( 393) ( 726) = 239 kj 15. Calculate the enthalpy change for the process CCl 4 (g) C(g) + 4 Cl(g) and calculate bond enthalpy of C Cl in CCl 4 (g). vap H (CCl 4 ) = 30.5 kj mol 1. f H (CCl 4 ) = kj mol 1. a H (C) = kj mol 1, where a H is enthalpy of atomization a H (Cl 2 ) = 242 kj mol 1 First find H of reaction CCl 4(g) C (g) + 4Cl (g) BMC CLASSES Page No - 58

28 (1) H vap of CCl 4 CCl 4 (l) CCl 4(g) H = 30.5 kj (2) H f CCl 4 (l) Cgr + 2 Cl 2(g) CCl 4 (l) H = kj (3) H f atom C(gr) Cgr C (g) H = 715 kj (4) H atom Cl 2 Cl 2(g) 2 Cl (g) H = 342 kj To obtain reaction CCl 4(g) C (g) + 4Cl (g) Multiply equation No. 4 by two and add it with equation No. 3. (5) C gr + 2Cl 2 C (g) + 4Cl (g) H = 2 (242) kj Add equation No. (1) & (2) (6) C(gr) + 2Cl 2(g) CCl 4(g) H = kj Subtract equation No. 6 from equation No. 5 H = 2(242) ( ) = = 1304 kj/mol. B. Energy C Cl bond 1304 = = 326 kj/mol 4 (As the equation obtained is enthalpy of atomisation of CCl 4 ) 16. For an isolated system, U = 0, what will be S? For a spontaneous process carrying out in a isolated system if U = 0 then certainly S would be driving force S = +ve means S > For the reaction at 298 K, 2A + B C H = 400 kj mol 1 and S = 0.2 kj K 1 mol 1 At what temperature will the reaction become spontaneous considering H and S to be constant over the temperature range. For spontaneity G = ve G = H T S O (minimum condition for spontaneity) T S = H H 400 T = = S 0.2 = 2000 K BMC CLASSES Page No - 59

29 18. For the reaction 2 Cl(g) Cl 2 (g), what are the signs of H and S? 2 Cl (g) Cl 2(g) Bond formation is exothermic H = ve n = 1 2 = 1 hence S = ve 19. For the reaction 2 A(g) + B(g) 2D(g) U = kj and S = JK 1. Calculate G for the reaction, and predict whether the reaction may occur spontaneously. 2 A (g) + B (g) 2 D (g) = n = 2 3 = 1 = kj H f = U + nrt G = H T S = ( 1) = = = kj Reaction is non spontaneous 20. The equilibrium constant for a reaction is is 10. What will be the value of G? R = JK 1 mol 1, T = 300 K. Kc = 10 G = RT log Kc = log 10 = = 5745 J = kj 21. Comment on the thermodynamic stability of NO(g), given 1 1 N2 (g) + O2 (g) NO(G) ; r H = 90 kj mol NO(g) O2 (g) NO 2 (g) ; r H = - 74 kj mol 1 As conversion of NO into NO 2 is exothermic NO would not stabilize in air. 22. Calculate the entropy charge in surroundings when 1.00 mol of H 2 O(l) is formed under standard condition. f H = kj mol 1. H f H 2 O = 286 kj Means H surr = +286 kj T = 298 S = H T 286 = = J mol 1 K 1 S surr = J mol 1 K 1 BMC CLASSES Page No - 60