CT 1 CHEMICAL THERMODYNAMICS Syllabus : Fundamentals of thermodynamics : System and surroundings, extensive and intensive properties, state functions, types of processes. First law of thermodynamics - Concept of work, heat internal energy and enthalpy, heat capacity, molar heat capacity; Hess s law of constant heat summation; Enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization and solution. Second law of thermodynamics - spontaneity of processes; S of the universe and G of the system as criteria for spontaneity, G o (standard Gibbs energy change) and equilibrium constant.
CT CONCEPTS C1 Definitions of Some Basic Terms used in Thermodynamics : Internal Energy It is the energy associated with a system by virtue of its molecular constitution and the motion of its molecules. The contribution of energy due to molecular constitution is known internal potential energy and the contribution of energy due to the motion of molecules is called internal energy of a system. It is given by the sum of two types of energies. Determination of E : When a reaction is carried out in such a manner that the temperature and volume of the reacting system remain constant, then the internal energy change ( E) of the reaction is equal to the heat exchanged with the surrounding. Enthalpy When we deal certain process in open vessels (at constant pressure), it becomes essential to introduce in place of internal energy, a new thermodynamics function called heat enthalpy. This new function is denoted by H. H = E + PV The change in enthalpy of a given system is given as follows H = E + (PV), H = E + P V, H = E + n g RT, n g = n p (g) n R (g) Practice Problems : 1. For the reaction, Fe O 3 (s) + 3CO(g) 3CO (g) + Fe(s). Which of the following is correct? H 0 = E 0 + 3RT H 0 = E 0 + RT H 0 = E 0 H 0 = E 0 RT. The difference between the heats of reaction at constant pressure and constant volume for the reaction C 6 H 6 (l) + 15O (g) 1CO (g) + 6H O(l) at 5 0 C in kj is 7.43 3.7 3.7 7.43 [Answers : (1) c () a] Heat Capacity The heat capacity of a system is defined as the quantity of heat required for increasing the temperature of one mole of a system through 1 0 C. Heat capacity may be given as follows : Temperature Dependence of H and E (Kirchoff s Equation) dq C dt H H1 T T 1 C E E 1 P, CV T T1 Where C P and C V are change of molar heat capacities at constant pressure and at constant volume respectively. C ] CALORIMETRY An object undergoing a temperature change without a chemical reaction or change of state absorbs or discharge an amount of heat equal to its heat capacity times the temperature change. Heat exchange = (heat capacity) (temperature change)
C3 Calculation of Enthalpy of Reaction : 1. On the basis of enthalpy of formation value CT 3 H = H (product) H (Reactant) f f. On the basis of bond energies of reactants H = {sum of bond energies of reactants sum of bond energies of products} C4 HESS S LAW The total enthalpy change of a reaction is the same regardless of whether the reaction is completed in one step or in several steps. H = H 1 + H + H 3 C5 Physical Meaning of Enthalpy From the above discussions, it is clear that Enthalpy is the total heat stored with the system and it is an extensive property i.e, it depends upon the nature as well as on the amount of the substance. Like internal energy, its absolute value can not be calculated. We can only determine the change in enthalpy through the heat of reaction. If H R and H P are enthalpies of reactants and products then Hence, enthalpy is a state function. Exothermic and Endothermic reactions H P H R = H = heat of reaction Depending upon the nature of heat exchanged, the chemical reactions are of two types. C6 Bond Energy : Origin of Enthalpy change in a Reaction It is known that a chemical reaction involves the breaking of one or more bonds in the reactants and the formation of one or more bonds in the products. The bond breaking process is purely endothermic and the bond formation is an exothermic process. Therefore, the net heat change during any chemical reaction (enthalpy change) is given as : H (B.E.) actants Re (B.E.) Products C7 Calculation of Resonance Energy : If a compound undergo resonance its experimental heat of formation and theoretical heat of formation will be different. Subtraction of H f(exp) and H f(th) is equal to resonance energy. Different types of Heat of Reactions (i) Heat of Formation or Enthalpy of Formation It is the amount of heat evolved or absorbed when one g mole of a substance is formed from its constituent element e.g. Formation of CH 4 C(s) + O (g) CO (g) + Q 1
(ii) Standard Heat of Formation (Q f ) A substance is said to be in its standard state when it is present at 98 K (i.e. 5 0 C) under one atmospheric pressure. Standard heat of formation is the amount of heat evolved or absorbed when one mole of substance is formed from its elements in their standard states i.e., at 98 K and 1 atm. Conventionally, the heat of formation of element in its elemental state is zero. Heat of Combustion CT 4 It is the amount of heat evolved when one mole of a compound (or any substance) is completely oxidised (or burnt) in the presence of oxygen. The heat of combustion is represented by an equation which refers the one mole of substance used in combustion e.g. (i) CH 4 (g) + O (g) = CO (g) + H O(l) + Q 1 (ii) C H 5 OH(l) + 3O (g) = CO (g) + 3H O(l) + Q 3 Calorific Value of a fuel It is the amount of heat or energy released when one gram of a fuel is completely burnt in oxygen. (e) Heat of Neutralisation It is the amount of heat change (evolved) when 1 g equivalent of an acid in its aqueous and dilute solution is completely neutralised by a base through its dilute solution* and vice versa. Various examples are : (i) HCl (aq) + NaOH(aq) NaCl(aq) + H O(l), H = 13.6 K cal/eq or 57.1 kj/eq. (ii) H SO 4 (aq) + NaOH(aq) Na SO 4 + H O, H = 7. K cal/ But if in this process, either a weak acid or a weak base is involved, the heat of neutralisation is always less than 13.6 kcal/eq. Heat of Hydrogenation It is the amount of heat change (evolved) when one mole of an unsaturated organic compound is completely hydrogenated. Heat of Hydration CH = CH (g) + H (g) CH 3 CH 3 (g) It is the amount of heat change when one mole of an anhydrous salt combines with the required number of moles of water to form its hydrate. e.g. CuSO (s) 5H O CuSO.5H O(s) 4 anhydrous Blue 4 (Colourless ) Q(78.55kJ) H hydration (Anhydrous salt) = H solu.(anhydrous salt) H solu. (Hydrated salt) (f) Heat of Solution It is the amount of heat change when one mole of a substance is dissolved in such a large quantity of solvent so that further dilution does not give any further heat change e.g., NH 4 Cl(s) + aq NH 4 Cl (aq) Q NaOH(s) + aq NaOH (aq) + Q
(g) Born Haber Cycle CT 5 Lattice energies are calculated by employing energies involved in the various steps leading to the formation of an ionic compound. These steps can be shown graphically as a cycle hence the name Born - Haber cycle. Let us take an example of formation of an ionic compound NaCl, by the reaction of solid Na and gaseous chlorine at 5 0 C and 1 atm. This process evolves 410.9 kj/mole Na (s) + ½Cl (g) NaCl(s), H = 410.9 kj/mole This reaction consists of no. of steps. According to the principle of conservation of energy, the algebraic sum of the individual energy changes during various steps must be same as that of the overall change in energy. The various steps can be represented by Born Haber Cycle as, Hence the overall change in the reaction is B.E. H S I E.A. U [EA and U is ve] Practice Problems : 1. H 0 f for CO (g), CO(g) and H O(g) are respectively 339.5, 110.5 and 40.8 kj, Mol 1. The standard enthalpy change in (kj) for the reaction CO (g) + H (g) CO(g) + H O(g) is 54.1 11.8 6.5 41.. Given that C + O CO ; H 0 = x kj CO + O CO ; H 0 = y kj then enthalpy of formation of CO(g) is y x (x y)/ (y x)/ (x y)/ 3. Let us study the formation of NaCl then H f of NaCl is S + ½D + IE + EA + U S + ½D + IE EA + U S + ½D IE + EA + U S + ½ D + IE EA U
CT 6 4. Heat of hydrogenation of cyclohexene is x and that of benzene is y. Hence resonance energy of benzene is x 3y x + y x y 3x y 5. At 5 0 C the standard enthalpies in kj, mol 1 for following two reactions 3 0 3 FeO3(s) C(s) CO(g) Fe(s) H 34. 1 C ( s) O (g) CO (g) H 0 = 393.5 4Fe(s) + 3O (g) Fe O 3 is calculated as 3( 393.5) (34.1) 393.5 34.1 3 3 ( 393.5) + 34.1 ( 393.5) 34.1 6. Heat of neutralisation of NaOH and HCl is 57.46 kj/equivalent. The heat of ionisation of water will be 57.46 kj/mol 57.46 kj/mol 114.9 kj/mol 114.9 kj/mol 7. Heat released in neutralization of strong acid and strong base is 13.4 kcal/mol. The heat released on neutralization of NaOH with HCN is.9 kcal/mol, then H 0 of ionization of HCN in water is 10.5 kcal 16.3 kcal 9.5 kcal 11.5 kcal 8. Based on the following thermochemical equations H O(g) + C(s) CO(g) + H (g); CO(g) + ½O (g) CO (g); H (g) + ½O (g) H O(g); C(s) + O (g) CO (g); The value of X will be H = 131 kj H = 8 kj H = 4 kj H = X kj 393 kj 655 kj + 393 kj + 655 kj 9. The H 0 for chloride ion from the following data is f ½ H (g) + ½ Cl (g) HCl (g) HCl(g) + aq H + (aq) + Cl (aq); H 0 f H+ (aq) = 0.0 kj H 0 f = 9.4 kj H 0 = 74.8 kj 17. kj 18.4 kj 19. kj 167. kj 10. The standard enthalpy of combustion at 5 0 C of H, C 6 H 10 and cyclohexane (C 6 H 1 ) are 41, 3800, 390 kj mol 1 respectively. The heat of hydrogenation of cyclohexene (C 6 H 10 ) is 11 kj 150 kj 11 kj none 11. The dissociation energy of methane is 360 kcal mol 1 and that of ethane is 60 kcal mol 1. The C C bond energy is 10 130 180 80 1. The enthalpy change for the following reactions at 5 0 C are given below : ½H (g) + ½O (g) OH(g); H = 10.06 kcal...(1) H (g) H(g); H = 104.18 kcal...() O (g) O(g); H = 118.3 kcal...(3)
CT 7 The O H bond energy in hydroxyl radical is 11.31 kcal 116.15 kcal 110.11 kcal 111.3 kcal [Answers : (1) b () c (3) d (4) d (5) a (6) b (7) a (8) a (9) d (10) c (11) d (1) a] C8 C9 First law of Thermodynamics Energy may be converted from one form to another, but it is impossible to create or destroy it. There are various ways of enhancing the first law of thermodynamics. Some of the selected statements are given below : Mathematical Formulation of the First Law Suppose a system absorbs a quantity of heat q and its state change from A to B. This heat is used up. i) In increasing the internal energy of the system i.e., E = E B E A ii) iii) iv) To do some external work w by the system on its surroundings. From the first law, we get. Heat obsorbed by the system = its internal energy + work done by the system. q = E + w...1 The sign convention : According to latest S.I. convention, w is taken as negative if work is done by the system whereas it is taken as positive if work is done on the system. When heat is given by the system to surrounding it is given as negative sign. When heat is absorbed by the system from the surrounding then positive sign is given q and w are not state function because changes in their magnitude is dependent on the path by which the change is accomplished. Mathematically q & w are not exact differential and we always write the inexact - differential by q, w etc. For a cyclic process, the change in the internal energy of the system is zero because the system is brought back to the original condition. dq dw i.e. the total work obtained is equal to the net heat supplied. v) In an isolated system, there is no heat exchange with the surrounding i.e. dq = 0 de + dw = 0 or dw = de C10 Work in Reversible Process : a) Expansion of a gas Suppose n moles of a perfect gas is enclosed in a cylinder by a friction less piston. The whole cylinder is kept in large constant temperature bath at TK. Any change that would occur to the system would be isothermal. w = P. dv
CT 8 Let the gas expand from initial volume V 1 to the final volume V, then the total work done (w) v v1 PdV Work done in isothermal reversible expansion of an ideal gas : The small amount of work done, dw when the gas expands through, a small volume dv, against the external pressure, P is given by dw = PdV Total work done when the gas expands from initial volume V 1 to final volume V will be W v v1 v PdV nrt, W dv [ T = constant], V v1 W = nrt ln V V =.303 nrt V1 V1 log =.303 nrt log P P 1 The ve sign indicates work of expansion. Note : Work in the reversible process is the maximum and is greater than that in the irreversible process. Practice Problems : 1..5 mol of ideal gas at atm and 300k expands isothermally to.5 times of its original volume against the external pressure of 1 atm. The calculated value of q, w and E are 4.7 kj, 0 and + 4.7 kj + 4.7 kj, 0 and 4.7 kj 4.7 kj, 0 and 0 0, 0 and + 4.7 kj [Answers : (1) a] Adiabatic Process (Reversible) : An adiabatic change by definition, is one which does not allow any transfer of heat, i.e., q = 0, it follows from the 1st law, U = W du = dw Let only mechanical work of expansion or contraction is involved, dw = PdV. Moreover, du = C v dt PV = constant C v dt = PdV Similarly TV 1 1 RT T = constant P T P 1 = constant Adiabatic work : dw = C V dt = C V (T T 1 ) = C V (T 1 T ) Where T 1, T are initial and final temperatures. For 1 mole of gas T = PV/R Hence adiabatic work
CT 9 P1 V1 PV CV W CV (P1V1 P1V ) R R R or P1 V1 PV W 1 (e) Isochoric Process : dw = 0 Limitations of first Law of Thermodynamics 1. This law fails to tell us under what conditions and to what extent it is possible to bring about conversion of one form of energy into the other.. The first law fails to contradict the non-existence of a 100% efficient heat engine or a regrigerator. C11 Second law of thermodynamics It has been stated in several forms as follows. i) All the spontaneous process are irreversible in nature. ii) iii) The entropy of universe is always increasing in the course of every spontaneous process. Spontaneous or natural process are always accompanied with an increasing in entropy.
CT 10 INITIAL STEP EXERCISE 1. In thermodynamics, a process is called reversible when surroundings and system change into each other there is no boundry between system and surroundings the surroundings are always in equilibrium with the system the system changes into the surroundings spontaneously. For a liquid enthalpy of fusion is 1.435 Kcal mol 1 and molar entropy change is 5.6 Cal., mol 1, k 1, The m.pt of the liquid is 0 0 C 73 0 C 173K 100 0 C 3. The enthalpy of certain reaction at 73 K is 0.75 kj. The enthalpy of same reaction at 373 k will be (heat capacities of reactants and products are same) 0.75 kj 075 kj 373 zero 0.75 kj 73 4. Match the column 1 with column and pick up the correct alternate Column 1 Column I. For a spontaneous process a. B.E. II. For endothermic process b. III. Bond dissociation energy c. IV. For solids and liquids in a d. (reactants) B.E. (products) H = E G must be ve H (products) > H thermochemical reaction. (reactants) I - c, II - a, III - d, IV - b I - b, II - d, III - a, IV - c I - c, II - d, III - b, IV - a I - c, II - d, III - a, IV - b 5. Which of the following statement is incorrect An exothermic reaction is nonspontaneous at high temperature An endothermic reactions is at very low temperature such that T S < H in magnitude the process is spntaneous. exothermic process are spontaneous at low temperature endothermic process are spontaneous at high temperature. 6. Energy required to dissociate 4 g of gaseous hydrogen into free gaseous atoms is 08 kcal at 5 0 C. The bond energy of H H bond will be 104 kcal 10.4 kcal 1040 kcal 104 kcal 7. The final temperature in an adiabatic expansion 8. For a reaction : greater than initial temperature same as the initial temperature half of the initial temperature less than the initial temperature A + B C + D at 300 K temperature K = 10, the value of G and G 0 are 0 and 11.48 kj 0 and 11.48 kj 11.48 kj and 0 11.48 kj and 0 9. Which of the following conditions are not favourable for the feasibility of a reaction? H = +ve, T S = +ve and T S > H H = ve, and T S = +ve H = ve, T S = ve and T S < H H = +ve, T S = +ve and T S < H 10. Standard molar enthalpy of formation of CO is equal to zero the standard molar enthalpy of combustion of gaseous carbon the sum of standard enthalpies of formation of CO and O the standard molar enthalpy of combustion of carbon (graphite)
11. The enthalpy change for which of the following processes represents the enthalpy of formation of AgCl. Ag + (aq) + Cl (aq) AgCl(s) Ag(s) + ½Cl (g) AgCl(s) AgCl(s) + Ag(s) ½ Cl (g) Ag(s) + AuCl(s) Au(s) + AgCl(s) 1. In which case of mixing of a strong acid and strong base, each of 1N concentration, temperature increase is the highest? 0 ml acid and 30 ml alkali 10 ml acid and 40 ml alkali 5 ml acid and 5 ml alkali 35 ml acid and 15 ml alkali 13. The standard heat of formation of sodium ions in aqueous solution from the following data Heat of formation of NaOH(aq) at 5 0 C = 470.7 kj Heat of formation of OH (aq) at 5 0 C = 8.8 kj 51.9 41.9 kj 41.9 kj 51.9 kj 14. Molar heat capacity of water in equilibrium with ice at constant pressure is Zero infinity 40.45 J k 1 mol 1 75.48 J k 1 mol 1 15. The heats of neutralization of four acids A, B, C and D when neutralized against a common base are 13.7, 9.4, 11. and 1.4 kcal respectively. The weakest among these acids is A B C D 16. The word standard in standard molar enthalpy change implies temperature 98 K pressure 1 atm temperature 98 K and pressure 1 atm all temperatures and all pressures 17. The factor which does not influence the heat of reaction is the physical state of reactants and products. the temperature of the reaction the method by which the final products are obtained whether the reaction is carried out at constant pressure or constant temperature CT 11 18. The bond energy of H is 104.3 kcal/mol. It means that 104.3 kcal heat is needed to break one bond to form two atoms 104.3 kcal is required to break 6.0 10 3 molecules into atoms of hydrogen 104.3 kcal is required to break 3.015 10 3 hydrogen molecules to 6.0 10 3 hydrogen atoms none of these 19. H (g) + ½O (g) H O(l) H = 68.39 kcal K(s) + H O(l) + aq KOH(aq) + ½H (g) KOH(s) + aq KOH (aq) H = 48 kcal H = 14 kcal The heat of formation of KOH(s) is (in kcal) 68.39 + 48 14 68.39 48 + 14 68.39 48 + 14 68.39 + 48 + 14 0. Heat of neutralization of a strong dibasic acid in dilute solution by NaOH is nearly 7.4 kcal/eq 13.7 kcal/mol 13.7 kcal/eq 13.7 kcal/mol 1. At 5 0 C and 1 atm, which one(s) of the following of any has a non-zero H 0 f? Fe O C(s) Ne
CT 1 FINAL STEP EXERCISE 1. If 1.00 k cal of heat is supplied to 1. litre of oxygen in a cylinder at 1.00 atm, the volume increases to 1.5 lit. So E for this process is 0.993 K cal 0.0993 K cal 1.0073 K cal 1.00073 K cal. The enthalpy change involved in the oxidation of glucose is 880 kj mol 1. Twenty five percent of this energy is available for muscular work. If 100 kj of muscular work is needed to walk one kilometer, then the maximum distance that a person will be able to walk after eating 10 g of glucose is 4.8 km 3.8 km 1. km 8 km 3. The H 0 f (KF, s) is 563 kj mol 1. The ionization energy of K(g) is 419 kj mol 1 and the enthalpy of sublimation of potassium is 88 kj mol 1. The electron affinity of F(g) is 3 kj mol 1 and F F bond enthalpy is 158 kj mol 1. The lattice energy of KF(s) is 87 kj mol 1 87 kj mol 1 45 kj mol 1 550 kj mol 1 4. A couple sitting in a warm room on a winter day takes ½ kg of cheese sandwitches (an energy intake of 8130 kj for both). Suppossing that none of the energy is stored in body, the mass of water would they need to perspire in order to maintain their original temperature is (the enthalpy of vaporization of water is 40.65 kj mol 1 ) 3.6 gm.5 gm 1. gm 5. gm 5. The enthalpy of formation of H O(l) is 85.7 kj mol 1 and enthalpy of neutralization of a strong acid and a strong base is 56.07 kj mol 1. The enthalpy of formation of OH ions is 9 kj/mol 9 kj/mol 111 kj/mol 155 kj/mol 6. The heat of formation of anhydrous Al Cl 6 from : (i) (ii) (iii) (iv) Al(s) + 6HCl(aq) Al Cl 6 (aq) + 3H (g); H = 39.76 kcal H (g) + Cl (g) HCl(g); H = 44.0 kcal HCl(g) + aq HCl (aq); H = 17.3 kcal Al Cl 6 (s) + aq. Al Cl 6 (aq) H = 153.69 kcal 31.99 k cal 31.99 k cal 14.88 k cal.41 k cal 7. Study following figure and match the column 1 with column and pick up the correct answer Column 1 Column I. Isothermal expansion (i) Curve bc II. Adiabatic expansion (ii) Curve da III. Isothermal compression (iii) Curve cd IV. Adiabatic compression (iv) Curve ab I (iv), II (i), III (iii), IV (ii) I (i), II (ii), III (iii), IV (iv) I (iv), II (i), III (ii), IV (iii) I (i), II (iv), III (iii), IV (ii) 8. At 0 0 C. ice and water are in equilibrium and H = 6.0 kj mol 1 for the process, H O(s) H O(l) The value of S and G for the conversion of ice into liquid water are 1.9J, K 1 mol 1 and 0 0.19 J K 1 mol 1 and 0 1.9J, K 1 mol 1 and 0 0.019 J K 1 mol 1 and 0 9. Given H 0 ioniz (HCN) = 45. kj mol 1 and H 0 ioniz (CH 3 COOH) =.1 kj mol 1. Which one of the following facts is true? pk a (HCN) = pk a (CH 3 COOH) pk a (HCN) > pk a (CH 3 COOH) pk a (HCN) < pk a (CH 3 COOH) pk a (HCN) = (45.17/0.7)pK a (CH 3 COOH)
ANSWERS (INITIAL STEP EXERCISE) 1. c. a 3. a 4. d 5. b 6. a 7. d 8. b 9. d 10. d 11. b 1. c 13. c 14. b 15. b 16. b 17. c 18. b 19. b 0. c CT 13 ANSWERS (FINAL STEP EXERCISE) 1. a. a 3. a 4. a 5. b 6. a 7. a 8. c 9. b 1. b AIEEE ANALYSIS [00] 1. For the reactions, C + O CO ; Zn + O ZnO; Zn can oxide carbon H = 393 J H = 41 J oxidation of Zn is not feasible carbon can oxidise Zn oxidation of carbon is not feasible. If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then? H is negative, S is positive H and S both are negative H and S both are positive H is positive, S is negative 3. A heat enginer absorbs heat Q 1 at temperature T 1 and heat Q at temperature T. Work done by the engine is J (Q 1 + Q ). This data does not violate 1 st law of thermodynamics violates 1 st law of thermodynamics violates 1 st law of thermodynamics if Q 1 is negative violates 1 st law of thermodynamic if Q is negative 4. The heat required to raise the temperature of body by 1 0 C is called thermal capacity water equivalent specific heat none of these
CT 14 AIEEE ANALYSIS [003] 5. The enthalpy change for a reaction does not depend upon the difference in initial or final temperature of involved substances the physical states of reactants and products use of different reactants for the same product the nature of intermediate reaction steps 6. In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dg) and change in entropy (ds), satisfy the criteria (ds) V, E = 0, (dg) T, P > 0 (ds) V, E < 0, (dg) T, P < 0 (ds) V, E > 0, (dg) T, P < 0 (ds) V, E = 0, (dg) T, P = 0 7. The internal energy change when a system goes from sate A to B is 40 kj/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy? Zero 40kJ > 40 kj < 40 kj 8. If at 98 K the bond energies of C H, C C, C = C and H H bonds are respectively 414, 347, 615 and 435 kj mol 1, the value the enthalpy change for the reaction H C = CH (g) + H (g) H 3 C CH 3 (g) at 98 K will be 15 kj + 50 kj 50 kj + 15 kj AIEEE ANALYSIS [004/005/006] 9. An ideal gas expands in volume from 1 10 3 m 3 to 1 10 m 3 at 300 K against a constant pressure of 1 10 5 Nm. The work done is 900 J 900 kj 70 kj 900 kj [004] 10. The enthalpies of combustion of carbon and carbon monoxide are 393.5 and 83 kj mol 1 respectively. The enthalpy of formation of carbon monoxide per mole is 110.5 kj 675.5 kj 676.5 kj 110.5 kj [004] 11. Consider the reaction : N + 3H NH 3 carried out at constant temperature and pressure. If H and U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true? H = 0 H = U H < U H > U [005] 1. The exothermic formation of ClF 3 is represented by the equation Cl (g) + 3F (g) ClF 3 (g); H = 39 kj Which of the following will increase the quantity of ClF 3 in an equilibrium mixture of Cl, F an ClF 3? Increase the temperature Removing Cl Increasing the volume of the container Adding F [005] 13. If the bond dissociation energies of XY, X and Y (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and f H for the formation of XY is -00 kj mole 1. The bond dissociation energy of X will be 100 kj mol 1 00 kj mol 1 300 kj mol 1 400 kj mol 1 [005]
14. The standard enthalpy of formation ( f H 0 ) 98 K for methane, CH 4 (g), is 78.4 kj mol 1. The additional information required to determine the average energy for C-H bond formation would be [006] the dissociation energy of hydrogen molecule, H the dissociation energy of H and enthalpy of sublimation of carbon latent heat of vapourization of methane the first four ionization energies of carbon and electron gain ehthalpy of hydrogen 15. ( H U) for the formation of carbon monoxide (CO) from its elements at 98 K is (R = 8.314 J K 1 mol 1 ) 477.57 J mol 1 138.78 J mol 1 138.78 J mol 1 477.57 J mol 1 [006] 17. Identify the correct statement regarding a spontaneous process : Exothermic processes are always spontaneous Lowering of energy in the reaction process is the only criterion for spontaneity For a spontaneous process in an isolated system, the change in entropy is positive Endothermic processes are never spontaneous 18. In conversion of lime-stone to lime, CaCO 3 (s) CaO(s) + CO (g) the values of H 0 and S 0 are +179.1 kj mol 1 and 160. J/K respectively at 98 K and 1 bar. Assuming that H 0 and S 0 do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is 845 K 1118 K 1008 K 100 K CT 15 16. The enthalpy changes for the following process are listed below : Cl (g) = Cl(g), 4.3 kj mol 1 I (g) = I(g), 151.0 kj mol 1 ICI(g) = I(g) + Cl(g), 11.3 kj mol 1 I (s) = I (g), 6.76 kj mol 1 Given that the standard states for iodine and chlorine are I (s) and Cl (g), the standard enthalpy of formation of ICI(g) is + 44.8 kj mol 1 14.6 kj mol 1 16.8 kj mol 1 + 16.8 kj mol 1 [006] AIEEE ANALYSIS [007] 19. Assuming that water vapour is an ideal gas, the internal energy change ( U) when 1 mol of water is vapourised at 1 bar pressure and 100 0 C, (Given : Molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kj mol 1 and R = 8.3 J mol 1 K 1 ) will be : 37.904 kj mol 1 41.00 kj mol 1 4.100 kj mol 1 3.7904 kj mol 1 ANSWERS AIEEE ANALYSIS 1. c. b 3. b 4. a 5. d 6. c 7. a 8. a 9. a 10. d 11. d 1. d 13. b 14. b 15. c 16. d 17. c 18. b 19. a
CT 16 TEST YOURSELF 1. Mathematical expression for Ist law of thermodynamic is q = E w q = E + w E = w q all. Which one of these we get during adiabatic expansion? T P 1 = constant V 1 T= constant T 1 P 1 = constant T P 1 = constant 3. C p /C v for the monoatomic gas 1.40 1.66 1.33 1 4. Which of the following is not the IInd law of thermodynamics? It is impossible to obtain work by cooling a body below it lowest temperature Heat cannot of itself pass from a colder body to hotter body without the intervention of external work. Spontaneous or natural processes are always accompanied with an increase in entropy. All the spontaneous processes are irreversible in nature 5. The S.I. unit of entropy change is J K mol 1 J K 1 mol 1 J 0 C 1 mol 1 J K 1 mol 6. If refrigerator door is kept open, then we get room cooled room heated more heat is passed out no effect on room 7. A carnot engine operates between temperatures T and 400 K (T > 400 K). If efficiency of engine is 5%, the temperature T is 400 K 500 K 533.3 K 600 K 8. The heat of reaction x for N + 3H NH 3 at 7 0 C is 91.94 kj. Heat reaction at 50 0 C is (C p for NH 3 = 37.07 Joules N = 8.45 Joules H = 8.3 Joules) 9.843 kj 75.340 kj 84.35 kj 80.305 kj 9. The heat of combustion of C H 4 (g), C H 6 (g) and H (g) are 1409.5, 1558.3 and 85.6 kj respectively. The heat of hydrogenation of ethylene is 36.8kJ 136.8 kj 36.8 kj 336.8 kj 10. Follow the reaction K(s) + H O(l) + aq. KOH (aq) + ½H (g) H (s) + ½O (g) + aq. H O(l) KOH(s) + aq. KOH (aq) 1. a. d 3. b 4. c 5. b H = 48.0 Kcal H = 68.4 Kcal H = 14.0 Kcal The heat of formation of KOH(s) is 53.47 kj 150.35 kj 15.3 kj 10.4 kj 6. b 7. c 8. a 9. b 10. d ANSWERS