CE 1 ENERGETICS 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 + PV, H = E + n g RT, n g = n p n R 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. Practice Problems : 1. For the reaction, Fe O 3 (s) + 3CO 3CO + 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 (l) + 15O 1CO + 6H O(l) at 5 0 C in kj is 7.43 3.7 3.7 7.43 3. The combustion of one mole of benzene takes place at 98 K and 1 atm. After combustion, CO and H O(l) are produced and 367.0 kj of heat is liberated. Calculate the standard enthalpy of formation, f H 0 of benzene. Standard enthalpies of formation of CO and H O (l) are 393.5 kj mol 1 and 85.83 kj mol 1 respectively. [Answers : (1) c () a (3) f H 0 = 48.51 kj mol 1 ]
CE 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. Practice Problems : Heat exchange = (heat capacity) (temperature change) 1. Calculate the number of kj of heat necessary to raise the temperature of 60.0 g of aluminium from 35 0 C to 55 0 C. Molar heat capacity of aluminium is 4 J mol 1 K 1.. 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 98 K and 1 atmospheric pressure according to the equation C (graphite) + O CO During the reaction, temperature rises from 98 K to 99 K. If the heat capacity of the bomb calorimeter is 0.7 kj K 1, what is the enthalpy change for the above reaction at 98 K and one atm? 3. Calculate the energy needed to raise the temperature of 10.0 g of iron from 5 0 C to 500 0 C if specific heat capacity of iron is 0.45 J ( 0 C) 1 g 1. What mass of gold of specific heat capacity 0.13 J ( 0 C) 1 g 1 can be heated through the same temperature difference when supplied with the same amount of energy as in? [Answers : (1) q = n C T, 1.07 kj () heat lost by the system = the heat gained by calorimeter, q = C V T, 0.7 kj. Negative sign indicates the exothermic nature of the reaction. Thus. U for the combustion of 1 g of graphite = 0.7 kj K 1. For combustion of 1 mole of graphite = 1 1.0g mol 0.7 kj 1g Energy needed = m c t, 34 g] =.48 10 kj mol 1 (3) Energy needed = m c t,.1 10 3 J C3 Calculation of Enthalpy of Reaction : 1. On the basis of enthalpy of formation value 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 H P H R = H = heat of reaction
Hence, enthalpy is a state function. Exothermic and Endothermic reactions Depending upon the nature of heat exchanged, the chemical reactions are of two types. CE 3 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 CO + 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 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 + O = CO + H O(l) + Q 1 (ii) C H 5 OH(l) + 3O = CO + 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. 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 :
(e) CE 4 (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 + H CH 3 CH 3 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., Born Haber Cycle NH 4 Cl(s) + aq NH 4 Cl (aq) Q NaOH(s) + aq NaOH (aq) + Q 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 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 CE 5 B.E. H S I E.A. U [EA and U is ve] Practice Problems : 1. H 0 f for CO, CO and H O are respectively 339.5, 110.5 and 40.8 kj, Mol 1. The standard enthalpy change in (kj) for the reaction CO + H CO + H O 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 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 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 Fe(s) H 34. 1 C ( s) O CO H 0 = 393.5 4Fe(s) + 3O 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 + C(s) CO + H ; CO + ½O CO ; H + ½O H O; C(s) + O CO ; 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 + ½ Cl HCl HCl + aq H + (aq) + Cl (aq); H 0 f H+ (aq) = 0.0 kj H 0 f = 9.4 kj H 0 = 74.8 kj CE 6 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 + ½O OH; H = 10.06 kcal...(1) H H; H = 104.18 kcal...() O O; H = 118.3 kcal...(3) The O H bond energy in hydroxyl radical is 11.31 kcal 116.15 kcal 110.11 kcal 111.3 kcal 13. U 0 of the combustion of methane, graphite and dihydrogen at 98 K are, 890.3 kj mol 1, 393.5 kj mol 1 and 85.8 kj mol 1 respectively. Enthapy of formation of CH 4 will be (i) 74.8 kj mol 1 (ii) 5.7 kj mol 1 (iii) +74.8 kj mol 1 (iv) +5.6 kj mol 1 14. Standard vaporisation enthalpy of benzene at its boiling point is 30.8 kj mol 1, for how long would a 100 W electric heater have to operate in order to vaporise a 100 g sample of benzene at its boiling temperature? [Power = energy/time and 1 W = 1 J s 1 ] [Answers : (1) b () c (3) d (4) d (5) a (6) b (7) a (8) a (9) d (10) c (11) d (1) a (13) (i) (14) Heat required to vaporize 78 g benzene = 30.8 kj mol 1. Heat required to vaporize 100 g benzene = 30.81000J 100 39487 J, 100 J heat is given in one second, 6.6 min.] 78 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 CE 7 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 Practice Problems : 1. Choose the correct answer. A thermodynamics state function is a quantity used to determine the heat changes whose value is independent of path used to determine pressure volume work whose value depends on temperature only..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) b () a] 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
Let the gas expand from initial volume V 1 to the final volume V, then the total work done (w) CE 8 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. 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 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 CE 9 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. Practice Problems : 1. For the process to occur under adiabatic conditions, the correct condition is (i) T = 0 (ii) p = 0 (iii) q = 0 (iv) w = 0. If water vapour is assumed to be perfect gas, molar enthalpy change for vaporisation of 1 mol of water at 1 bar and 100 0 C is 41 kj mol 1. Calculate the internal energy change when 1 mol of water is vaporized at 1 bar pressure and 100 0 C 1 mol of water is converted into ice. [Answers : (1) (iii) () H O(l) H O, U = H n g RT, ng = 1, U = 37.904 kj mol 1 H O(l) H O(s), pv = n g RT = 0, U = 41.00 kj mol 1 ] C11 Second law of thermodynamics It has been stated in several forms as follows. All the spontaneous process are irreversible in nature or the entropy of universe is always increasing in the course of every spontaneous process or spontaneous or natural process are always accompanied with an increasing in entropy. Entropy of a system is a measure of randomness or disorder of the system and is denoted by the symbol S. It is a state function. The change in entropy of a system is given by S = S final S initial. qrev For a reversible reaction S is related to q and T as : S and the total entropy change (S T total ) for the system and surroundings of a spontaneous process is given by S total = S system + surr > 0. G is the net energy available to do useful work and is thus a measure of free energy. G is related with S and H as G = H TS. The above equation is called Gibbs equation. G gives a criteria for spontaneity at constant pressure and temperature (i) If G is negative (< 0), the process is spontaneous. (ii) If G is positive (> 0), the process is non-spontaneous. r is also related with equilibrium constant of the reaction k as follows : r = RT ln K or r =.303 RT log K. Practice Problems : 1. Calculate r G 0 for conversion of oxygen to ozone. 3 O O3 at 98 K If K p for this conversion is.47 10 9.. Find out the value of equilibrium constant for the following reaction at 98 K. NH 3 + CO NH CONH (aq) + H O (l) Standard Gibbs energy change, r G 0 at the given temperature is 13.6 kj mol 1 3. At 60 0 C, dinitrogen tetroxide is fifty per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. 4. How is the change in free energy related to spontaneity? For a reaction : M 1 0 1, rs 0.07kJK (at 1 atm) 0 1 O(s) M(s) O; rh 30kJ mol Calculate up to what temperature the relation would not be spontaneous. [Answers : (1) 163 kj mol 1 () K =.4 10 (3) 763.8 kj mol 1 (4) 48.57 K]
CE 10 SINGLE CORRECT CHOICE TYPE 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. 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 (reactants) B.E. (products) H = E G must be ve H (products) > H thermochemical reaction. (reactants) 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 TS < 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, TS = +ve and TS > H H = ve, and TS = +ve H = ve, TS = ve and TS < H H = +ve, TS = +ve and TS < 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 AgCl(s) AgCl(s) + Ag(s) ½ Cl 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 CE 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 + ½O H O(l) H = 68.39 kcal K(s) + H O(l) + aq KOH(aq) + ½H 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. 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 3. 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
4. The H 0 f (KF, s) is 563 kj mol 1. The ionization energy of K is 419 kj mol 1 and the enthalpy of sublimation of potassium is 88 kj mol 1. The electron affinity of F 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 5. 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 6. 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 7. The heat of formation of anhydrous Al Cl 6 from : (i) (ii) (iii) (iv) Al(s) + 6HCl(aq) Al Cl 6 (aq) + 3H ; H = 39.76 kcal H + Cl HCl; H = 44.0 kcal HCl + 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 CE 1 9. 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 30. 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 (SINGLE CORRECT CHOICE TYPE) 8. 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) 1. c. a 3. a 4. d 5. b 6. a 7. d 8. b 11. b 1. c 13. c 14. b 15. b 16. b 17. c 18. b 1. b. a 3. a 4. a 5. a 6. b 7. a 8. a 9. d 19. b 9. c 10. d 0. c 30. b
CE 13 EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE Comprehension-1 An ideal monatomic gas at 1 bar and 73.15 K is allowed to expand adiabatically against a constant pressure of 0.315 bar until it doubles its volume. 1. What is the change in molar volume?.71 L mol 1 3.71 L mol 1 4.71 L mol 1 5.71 L mol 1. How much work is done on the gas in this process? 615.4 J mol 1 715.4 J mol 1 815.4 J mol 1 915.4 J mol 1 3. What is the final temperature? 7.4 K 37.4 K 47.4 K 57.4 K 4. What is the change in the molar internal energy of the gas? The value of 3 C V is R. 615.4 J mol 1 715.4 J mol 1 815.4 J mol 1 915.4 J mol 1 Comprehension- Liquid water is vaporized at 100 0 C and 1.013 bar. The heat of vaporization is 40.69 kj mol 1. 5. The value of w rev per mole is.10 kj mol 1 3.10 kj mol 1 4.10 kj mol 1 5.10 kj mol 1 6. The value of q per mole is 10.69 kj mol 1 0.69 kj mol 1 30.69 kj mol 1 40.69 kj mol 1 7. The value of U is 17.59 kj mol 1 7.59 kj mol 1 37.59 kj mol 1 47.59 kj mol 1 8. The value of H is 10.69 kj mol 1 0.69 kj mol 1 30.69 kj mol 1 40.69 kj mol 1 Comprehension-3 A mole of argon is allowed to expand adiatabically and reversibly from a pressure of 10 bar and 98.15 K to 1 bar. 9. The final temperature is 116.70 K 117.70 K 118.70 K 119.70 K 10. The work is done on the argon is 1138 J mol 1 38 J mol 1 3338 J mol 1 4438 J mol 1 MATRIX-MATCH TYPE Matching-1 Column - A Column - B (A) For a spontaneous (P) B.E. process (reactants (products) (B) For endothermic process (Q) H = E (C) Bond dissociation energy (R) G must be ve (D) For solids and liquids (S) H in a thermochemical (products > reaction. H (reactants Matching- Study following figure : Column - A Column - B (A) Isothermal expansion (P) Curve bc (B) Adiabatic expansion (Q) Curve da (C) Isothermal compression (R) Curve cd (D) Adiabatic compression (S) Curve ab Matching-3 Column - A Column - B (A) Automobile engine (P) Electrical energy to light energy (B) Photo electric cell (Q) Mechanical energy to heat energy (C) Friction (R) Chemical energy to mechanical energy (D) Fluorescent tamp. (S) light to electric energy
MULTIPLE CORRECT CHOICE TYPE 1. Which of the following are endothermic reactions? Combustion of methane Decomposition of water Dehydrogenation of ethane to ethene Conversion of graphite to diamond. Evaporation of water is an endothermic process but take place spontaneously because TS > H entropy increases overall tendency of driving forces favours the spontaneity G = H TS = +ve value 3. Variation of H and E with temperature is given by kirchoff s equation which is H H1 T T 1 E 1 E T Cp Cv H = H 1 + CP(T T 1 ) H G S T 4. For which of the following substances the value of H 0 f = 0? Water Aluminium Br (l) Br 5. H 0 f of water is 86 kj, mol 1, what are true? heat of combustion of hydrogen is 86 kj mol 1 heat of hydration of oxygen is 86 kj mol 1 86 kj of heat required to dissociate 1 mol of liquid H O in to H and O gases. Combination of one mole of O with sufficient H to form liquid water involves release of 86 kj of heat 6. Which of the following are not true? Entropy is a state function and intensive property naturally occurring spontaneous processes always irreversible complete conversion of heat into work is impossible entropy of perfectly crystalline solids is zero at 0 0 C temperature. CE 14 7. Work done (w) in isothermal reversible expansion of ideal gas is given by w nrtin w E q w nrt In w = PV Assertion-Reason Type V V 1 P P Each question contains STATEMENT-1 (Assertion) and STATEMENT- (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (A) (B) (C) (D) 1 Statement-1 is True, Statement- is True; Statement- is a correct explanation for Statement-1 Statement-1 is True, Statement- is True; Statement- is NOT a correct explanation for Statement-1 Statement-1 is True, Statement- is False Statement-1 is False, Statement- is True 1. STATEMENT-1 : Heat absorbed during isothermal expansion of ideal gas in vacuum is zero. STATEMENT- : Volume occupied by molecular of ideal gas is zero.. STATEMENT-1 : Reversible process and cyclic process have same meaning. STATEMENT- : H cycle and E cycle is zero 3. STATEMENT-1 : The dissolution of NH 4 Cl in water is endothermic even through it is spontaneous process STATEMENT- : The bonds in NH 4 Cl are weak. 4. STATEMENT-1 : Enthalpy change for a process is a constant value no matter, the process is complete in one step or more STATEMENT- : Enthalpy is a state function 5. STATEMENT-1 : Neither q nor w is a state function but q +w is a state function. STATEMENT- : E is a state function
CE 15 (Answers) EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE 1. a. b 3. d 4. b 5. b 6. d 7. c 8. d 9. c 10. b MATRIX-MATCH TYPE 1. [A-R, B-S, C-P, D-Q]. [A-S, B-P, C-R, D-Q] 3. [A-R, B-S, C-Q, D-P] MULTIPLE CORRECT CHOICE TYPE 1. b, c, d. a, b, c 3. a, b, c 4. b, c 5. b, c 6. a, d 7. a, c ASSERTION-REASON TYPE 1. B. 3. C 4. A 5. A INITIAL STEP EXERCISE (SUBJECTIVE) 1. Use the Born-Haber cycle and the following data to calculate the electron affinity of chlorine. H f (RbCl) = 10.9 kcal/mol Ionization energy of Rb = 95 kcal/mol H sub (Rb) = + 0.5 kcal/mol B.E. (Cl ) = 54 kcal/mol Lattic energy of RbCl = 166 kcal/mol. Calculate H f (C 4 H 10 ) from the following data : H f (C ) = 0 kcal/mol H f (C 3 H 8 ) = 4 kcal/mol B.E. (C C) = 8 kcal/mol B.E. (C H) = 99 kcal/mol 3. Calculate the resonance energy of C 6 using kekule formula for C 6 from the following data : (i) H f 0 for C 6 = 358.5 kj mol 1 (ii) (iii) Heat of atomisation of C = 716.8 kj mol 1 Bond energy of C H, C C, C = C and H H are 490, 340, 60, 436.9 kj mol 1 respectively. 4. Using the data given below, calculate the bond enthalpy of C C and C H bonds. H 0 comb (ethane) = 1556.5 kj mol 1 H 0 (propane) = 117.5 kj mol 1 comb H C(s) C = 719.7 kj mol 1 Bond enthalpy of H H = 435.1 kj mol 1 H 0 (H O, l) = 84.5 kj mol 1 f H 0 (CO, g) = 393.3 kj mol 1 f 5. The molar heats of combustion of C H, C(graphite) and H are 310.6, 94.05 and 68.3 kcal respectively. Calculate heat of formation of C H. 6. At 300 K, the standard enthalpies of formation of C 6 H 5 COOH(s), CO and H O(l) are 408, 393 and 86 kj mol 1 respectively. Calculate the heat of combustion of benzoic acid at (i) constant pressure and (ii) constant volume. 7. Compute the heat of reaction (at constant volume) at 5 0 C of the following reaction : 4 NH 3 + 5O 6H O(l) + 4NO Given : At 5 0 C, H f (NH 3, g) = 46. kj mol 1 H f (NO, g) = 90.4 kj mol 1 H f (H O, l) = 85.9 kj mol 1 8. 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, what is the maximum distance that a person will be able to walk after eating 10 g of glucose? 9. Determine enthalpy change for C 3 H 8 + H C + CH 4 at 5 0 C using heat of combustion values under standard conditions. Compounds H CH 4 H 0 in kj/mole 85.8 890.0 Compounds C C(graphite) H 0 in kj/mole 1560.0 393.5 The standard heat of formation of C 3 H 8 is 103.8 kj mol 1
10. Calculate H 0 for chloride ion from the following f data : ½ H + ½ Cl HCl H 0 f = 9.4 kj HCl + aq H (aq) + Cl (aq); H 0 f H+ (aq) = 0.0 kj H 0 = 74.8 kj 11. The H 0 f (KF, s) is 563 kj mol 1. The ionization energy of K is 419 kj mol 1 and the enthalpy of sublimation of potassium is 88 kj mol 1. The electron affinity of F is 3 kj mol 1 and F F bond enthalpy is 158 kj mol 1. Calculate the lattice energy of KF(s). 1. Calculate the resonance energy of N O from the following data H 0 f of N O is 8 kj mole 1. N N 946 kj mole 1 N N 418 kj mole 1 O = O 498 kj mole 1 N = O 607 kj mole 1 13. The standard heat of formation of CH 4, CO and H O are 76., 394.8 and 41.6 kj mol 1 respectively. Calculate the amount of heat evolved by burning 1 m 3 of methane measured under normal conditions. 14. The standard enthalpies of formation at 98 K for CCl 4, H O, CO and HCl are 106.7, 41.8, 393.7 and 9.5 kj mol 1, respectively. Calculate H 0 for the reaction 98 K CCl 4 + H O CO + 4HCl 15. Calculate the enthalpy change for the following reaction XeF 4 Xe + + F + F + F The average X F bond energy is 34 kcal/mol, first I.E. of Xe is 79 kcal/mol, electron affinity of F is 85 kcal/mol and bond dissociation energy of F is 38 kcal/mol. 16. Magnetite Fe 3 O 4 is reduced to iron by hydrogen or carbon monoxide as (i) (ii) Fe 3 O 4 (s) + 4H 3Fe(s) + 4H O Fe 3 O 4 (s) + 4CO 3Fe(s) + 4CO Calculate the enthalpy change involved during the reduction to get 1 g of iron in both the reaction mentioned above. Data given is : Substance : Fe 3 O 4 (s) CO H f (98 K) 1117.1 110.5 Substance : CO H O H f (98 K) 393.5 41.8 [kj/mol] CE 16 17. Using the bond enthalpy data given below, compute the enthalpy of formation of gaseous isoprene CH = C(CH 3 ) CH = CH B.E. (C H) = 413.4 kj mol 1 B.E. (C C) = 347.7 kj mol 1 B.E. (C = C) = 615.1 kj mol 1 Enthalpy of sublimation of C(graphite) = 718.4 kj mol 1 Enthalpy of formation of H = 18.0 kj mol 1 18. When moles of C are completely burnt, 319 kj of heat liberated. Calculate the heat of formation of C. H f for CO and H O(l) are 395 and 86 kj mol 1 respectively. 19. Calculate enthalpy change of the following reaction : H C = CH + H H 3 C CH 3 The bond energy of C H, C C, C = C, H H are 414, 347, 615 and 435 kj mol 1 respectively. 0. Estimate the average S-F bond energy in SF 6. The standard heat of formation values of SF 6, S and F are 1100, 75 and 80 kj mol 1 respectively. 1. Find the standard enthalpy change when acetylene is hydrogenated to ethane at 5 0 C using heat of combustion values under standard conditions. Compounds C(graphite) H C H 0 (kj/mol) 393.5 85.8 1560 The standard heat of formation of C H is 7 kj/mol.. When 1.0 g of carbon (graphite) reacted with oxygen to form CO and CO at 5 0 C and constant pressure, 313.8 kj of heat was released and no carbon remained. Calculate the mass of oxygen comsumed. Given: H 0 (CO, g) = 110.5 kj mol 1 f H 0 (CO, g) = 393.5 kj mol 1 f 3. Calculate the bond enthalpies of C H, C C and C = C bonds from the following data. H 0 (CH, g) = 74.9 kj mol 1 f 4 H 0 (C H, g) = 84.7 kj mol 1 f 6 H 0 (C H, g) = 5.3 kj mol 1 f 4 H 0 sub (C, graphite) = 718.4 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, what mass of water would they need to perpire in order to maintain their original temperature. The enthalpy of vaporization of water is 40.65 kj mol 1.
5. Calculate the heat of formation of liquid hydrazine N H 4 from the following data : 1. NH 3 + 3N O 4N + 3H O(l), H = 1010 kj. N O + 3H N H 4 (l) + H O(l), H = 317 kj 3. NH 3 + ½O N H 4 (l) + H O(l), H = 143 kj 4. H + ½O H O(l), H = 86 kj Also find the heat of combustion of hydrazine. 6. Calculate standard heat of formation of CS. Given that standard heat of combustion of C, S and CS are 393.3, 93.7 and 1108.76 kj mol 1. 7. 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. Calculate heat of hydrogenation of cyclohexene (C 6 H 10 ). 8. The bond dissociation energy of gaseous H, Cl and HCl are 104, 58 and 103 kcal mol 1 respectively. Calculate the enthalpy of formation of HCl gas. 9. The enthalpy change for the following reactions at 5 0 C are given below : ½H + ½O OH; H = 10.06 kcal...(1) H H; H = 104.18 kcal...() 1. A sample of solid napthalene C 10 H 8, weighing 0.600 g is burnt to CO and H O(l) in a constant volume calorimeter at T = 98 K. In this experiment the observed temperature rise of the calorimeter and its contents is.7 0 C. In a separate experiment, the total heat capacity of the calorimeter was found to be 556 cal/deg. What is E for the combustion of one mole of napthalene? What is H for this reaction. Also calculate the enthalpy of formation of napthalene. H f0 (H O, l) = 68.3 kcal/mol H f0 (CO, g) = 94.05 kcal/mol. The thermochemical equation for the dissociation of hydrogen gas into atoms may be written : H H; H = 436 kj/mol What is the ratio of the energy yield on combustion of hydrogen atoms to steam to the yield on combustion of an equal mass of hydrogen molecules to steam? H f [H O] = 41.81 kj/mol 3. An athelent takes 0 breaths per minute or room temperature. The air inhaled in each breath is O O; CE 17 H = 118.3 kcal...(3) Calculate O H bond energy in hydroxyl radical. 30. Give below are some standard heats of reaction; Heat of formation of water = 68.3 kcal mol 1 Heat of combustion of acetylene = 310.6 kcal mol 1 Heat of combustion of ethylene = 337. kcal mol 1 Calculate the heat of reaction for the hydrogenation of acetylene at constant volume at 5 0 C. 31. 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. What is the enthalpy of formation of OH ions. 3. Calculate the heat of formation of anhydrous Al Cl 6 from : (i) (ii) (iii) (iv) Al(s) + 6HCl(aq) Al Cl 6 (aq) + 3H ; H = 39.76 kcal H + Cl HCl; H = 44.0 kcal HCl + aq HCl (aq); H = 17.3 kcal Al Cl 6 (s) + aq. Al Cl 6 (aq) H = 153.69 kcal FINAL STEP EXERCISE (SUBJECTIVE) 00 ml which contains 0% oxygen by volume, while exhaled air contains 10% oxygen by volume. Assuming that all the oxygen consumed if used for converting glucose into CO and H O(l), how much glucose will be burnt in the body in one hour and what is the heat produced. (Room temperature = 7 0 C and enthalpy of combustion of glucose is 8.5 kj mole 1 at 0 0 C). 4. In order to get maximum calorific output, a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of the fuel A burner which has been adjusted for methane as fuel (with x litre/hour of CH 4 and 6x litre/hour of O ) is to be readjusted for butane, C 4 H 10. In order to get the same calorific output what should be the rate of supply of butane and oxygen? Assume that losses due to incomplete combustion etc. are the same for both fuels and that the gases behave ideally. Heats of combustion : CH 4 = 809 kj/mol and C 4 H 10 = 878 kj/mol
5. The heat evolved on combustion of 1.00 g of starch, (C 6 H 10 O 5 ) x into CO and H O(l) is 4.18 kcal. Calculate the standard enthalpy of formation of 1.00 g of starch. Given : H 0 f H O(l) = 68.3 kcal mol 1, H 0 f CO = 94.05 kcal mol 1. 6. A natural gas may be assumed to be a mixture of methane and ethane only. On complete combustion of 10 L of the gases at STP, the heat evolved was 464.6 kj. Assuming H comb (CH 4, g) = 894 kj mol 1 and H comb (C, g) = 1560 kj mol 1, calculate the percentage of volume of each gases in the mixture. CE 18 7. 3.67 litre of ethylene and methane gaseous mixture on complete combustion at 5 0 C produces 6.11 litre of CO. Find out the amount of heat evolved on burning one litre of gaseous mixture. The heats of combustion of C H 4 and CH 4 are 143 and 891 kj mol 1 at 5 0 C. ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE) 1. H = 79 kcal/mol. H = 8 kcal/mol 3. 150.0 kj mol 1 4. B.E.(C C) = 414 kj mol 1 B.E.(C H) = 344.3 kj mol 1 4. 54. k cal/ mol 6. (i) 301 kj (ii) 3199.75 kj 7. 1156.6 kj mol 1 8. 4.8 km 9. 55.7 kj 10. 167. kj 11. 87 kj mol 1 1. 88 kj mol 1 13. 3579.64 kj 14. 173.4 kj mol 1 15. H = 9. kcal/mol 16. (i) 0.896 kj/gm (ii) 88.69 J/gm 17. H = 768.8 kjmol 1 18. 83.5 kj/mol 19. 15 kj 0. 309 kj 1. 311.4 kj. 7.5 gm 3. 416.3, 331.4, 591.1 kj mol 1 4. 3.6 gm 5. 50.5 kj, 6.5 kj 6. 18.0 kj 7. 11 kj 8. kcal 9. 11.31 kcal 30. 41.104 k cal 31. 9 kj/mol 3. 31.99 k cal. 1. E = 137.76 kcal/mol H = 138.95 kcal/mol H f (C10H8 ) = 5.17 kcal/mol ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)..8 3. 9.5 gm, 458.66 kj 4. 0.8 x lt hr 1 = rate of supply of butane 5.4 x lt hr 1 = rate of supply of O. 5. 1.41 kcal/gm 6. CH 4 = 78%, C = % 7. 50.91 kj