sign in sign up
<<
Home , Heat of combustion

fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';keA iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM +s +s /;s;/;s; dks]dks] j?kqcjj?kqcj jk[ksjk[ks VsdAAVsdAA jfpr% ekuo /keZ iz.ksrk ln~xq# Jh j.kNksM+nklth egkjkt This is TYPE 1 Package STUDY PACKAGE please wait for Type 2 Subject : CHEMISTRY Topic : THERMO CHEMISTRY

R

Index 1. Key Concepts 2. Exercise I 3. Exercise II 4. Exercise III 5. Exercise IV 6. Answer Key 7. 34 Yrs. Que. from IIT-JEE 8. 10 Yrs. Que. from AIEEE Student’s Name :______Class :______Roll No. :______

ADDRESS: R-1, Opp. Raiway Track, New Corner Glass Building, Zone-2, M.P. NAGAR, Bhopal : (0755) 32 00 000, 98930 58881, www.tekoclasses.com FREE Download Study Package from website: www.tekoclasses.com . h etap cag fr h rato C reaction the for change The Q.9 Q.11 At 300 K, the standard enthalpiesstandardformat the of K, 300 At Q.11 . When Q.1 typeit's reaction& of Q.2 The standard of formation of CH of formation of heats standard The Q.2 . The of neutralization of NaOH & NH Q.4 .0Tesadr nhlyo etaiaino KOH ofneutralization of enthalpy standard The Q.10 . The heat of solution of anhydrous CuSOQ.5 . A cylinder of gas supplied by a company containsQ.8 . The heat of reaction Q.6 . Calculate standard heats of formation of carbonQ.7 . Calculate the enthalpy change when infinitely di Q.3 .2The heat liberated on complete combustionQ.12 of 7. at the same temperature. 26 days, what percentage of gas is wasted due to 2658 inkJ/mol . A normal family requires 20 MJ of ene − − H reaction at 77° − formation, Ca − equivalent of NHrespectively . What would be the enthalpy change if f abn s , upu () & (s) sulphur , (s) carbon of is normal (STP) conditions. the heat of hydration of CuSO − t() osat rsue&(ii) constant volume. & at:(i) constant pressure HCl are 6.82, 7.70 & 6.80 cal mol at constant volume & at 27º 241.6 kJ mol 2480 cal . g 57 Jml Cluae h ha o cmuto o C of of heat the Calculate kJ/mol. 55.7 408, 1108.76 kJ mol 2 , CH , 2+ (aq), (aq), CO

− 2 moles2 Cof 4

& C 393 − ∆ 2220 kJ mol H

& eq 3 f

2– − C . The molar heat capacities at constant pressure º for C 1 1 − − 4 (aq) and CaCO 1

. . Calculate amount of heat evolved by burning − 286 KJ mol Cl in solution ? Assume that NH and 1 respectively. are are 2

½H H 2 H 6 −

− 6 −

13.68 kcal . g 2 are completely are 3120burnt . Given 1 (g)+ ½ Cl 285.8, 285.8, .

C . Calculate the of at

− 4 1 . respectively . Calculate the heat of combust 3 (s) are – 129.80, – 161.65, – 288.5 kcal mol ∆

− −1 H 890.0 & 890.0 2 respectively. (g) XRIE I EXERCISE f º º

eq f or CO 4 → is – 15.9 kcal and that of CuSO − 1 respectively . Find the enthalpy of dissociation − HCl (g) at 27°C is 4 − − (g) , CO , (g) di 2 3 393.5 393.5 kJ/ respectively. of Heat combustion of di 8 g of benzene is 327 kJ . This heat has been measu (g) (g) & H 14 kg of . The heat of combustion of butane H 4 ion of C of ion lute solution of CaCl − − OH and NaCl are quantitatively obtained. one gram equivalent of NaOH is added to one gram 8 efficient combustion. 4 upie ( sulphide sulphide ( H + OH by HCl are with (a) HCN (b) HCl in dilute solution is solution dilutein HCl (b) HCN (a) with rgy per day of cooking. If the cylinder lasts for

kJ of heat isofheat liberated kJ Calculate of. heat the 2 2 2 O (g) (g) (g) and H and (g)

6 ( H l l 2 ) ) are ) . Given the standard heat of combustion → H 5 COOH(s), CO COOH(s), l 6 ) are : : are ) g. h has f obsin of combustion of heats The (g). − − at at 27°C for , chlorine & 1 m C

− 22.1 k cal . Calculate the heat of 395 &

13680 calories and 2 2 2 H 3 3 and Na O ( O 4 of measured under 6 − .5H g + CH + (g) 393.3, 393.3, − l ) are are ) 286 kJ respectively. 2 O is 2.8 kcal. Calculate 2 2 constant pressure. 2 (g) & H& (g) ion of benzoic acid –1 CO repectively. − 76.2 , , 76.2 3 4 − mixed g t 25º at (g) 9.2 and 293.72 2 − O

12270 cal −

∆ ( of H CN 398.8 , 398.8 l H ) are are )

f ° for isC red is ; TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 2 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .4The molar enthalpy of vaporizationQ.14 of benzene Q.15 When 12.0 g of carbon reacted with to fo to oxygenwith reacted carbon of g 12.0 When Q.15 .81.00 Q.18 .3A cylinder of gas is assumedQ.13 to contains 11.2 k .6If the enthalpy of formation of HCl (g) and ClQ.16 .9Two solutions initially at 25°C were mixedQ.19 in a .0When 1.0 g of CQ.20 .70.16 g of methane was subjected to combustion a Q.17 .1A quantity of 1.92 g of methanolQ.21 was burnt in a .2The enthalpy of dissociation of PHQ.22 energyBond for butane. [Specific Calcul heat kJ/°C. of capacity 2.02 of was H vessel inner the of capacity increased by 4.2 °C. If the quantity of surro order to vaporise 7.8 g of the sample at its boilinmolar internal change? For how long would a per day for cooking, how long will the cylinder las 0.667 K. What was the mole percent of CH ∆ ∆ 75.0 kcal of heat was liberated and no carbon remai bond enthalpy of the P–P bond? calorimeter system (including water) was found to r of the calorimeter. acid? Assume density of solutionmixing 1.0 temperatureg/cm increased to 26.2 °C. How much 400 h ml of 0.2 M weak monoprotic acid solution. The complete combustion of the methane COto constant pressure in a calorimeter which together w enthalpy of solution of hydrogen chloride gas. 10.0 kJ/°C. Calculate the enthalpy of combustion of calorimeterincreases water byIf°C.heat the 1.56 The thermal capacity of calorimeter system is 17.7 methane at (i) constant volume (ii) constant Hº H f º comb

l (CO sample of a mixture of CH (CH 2 ) ) = 4 − ) ) = 95 kcal mol − 215 k cal mol 6 H 12 − 1 O , , 2 O = 4.18 J/g°C] 6 ∆ (s) is burned in 4 H − ( g) & O 1 f º . 3

(CO) (CO) = is 954 kJ/mol and that of P 3 , and specific heat of solution 4.2 J/g-K. Neglect 4 2 in the original mixture? (g) measured at 25º 2 − g of butane. If a normal family needs 20,000 kJ of & H constant pressure calorimeter. The temperature of – 24kcal mol adiabatic constant pressure calorimeter. One conta at its boiling point (353 K) is 30.84 kJmol pressure . (aq) are –92.3 kJ/mole and – 167.44 kJ/mol, find t

oxygen in a bomb calorimeter, the temperature of th g point ? t t if the enthalpy of combustion, ith its contents had a heat capacity of 1260 cal/K. t t 27º 2 unding the inner vessel was 2000 ml and the heat ned . Calculate the mass of oxygen which reacted. ise by 0.5º O caused a temperature rise in the calorimeter of kJ K fructose at 298 K. capacity calorimeterthe of isand contents its rm CO & CO& CO rm 12 volt source need to supply a 0.5 A current in eat is evolved in the neutralization of 1 mole of

other other contain 100 ml of 0.80 M NaOH. After C in a bomb calorimeter . The temperature of − t te et f obsin f . of combustion of heat the ate 1 . (R = 8.313 mol − 1 .

C . Calculate the heat of combustion of

C & 740 torr was allowed reactto at 2 H 2 4 at 25º at is 1.485 M J mol −

1 C & constant pressure, constant & C K − ∆ 1 ) H = –2658 kJ/mole –1 –1 . . What is the heat capacity What is the energy water Th e i ns he e

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 3 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .4The enthalpy change for the following process a Q.24 .7The heat of combustion of is 312 kcal Q.27 Using the bond enthalpy data given Q.23 below, calcu .8Using the given data calculate enthalpyQ.28 of form .5Te oyeiain f tyee o ier polyet linear to of polymerisation The Q.25 .1Cesium chloride is formed accordingQ.31 to the foll .6From the following data :Q.26 .0Calculate the electron affinityQ.30 of fluorine a BornHabercycle S (g) (iii) C (ii) i C (i) Find the enthalpy of S–S bond from Q.29 the followin h vle ae n J mol kJ in are values the od – C=C CH H–H C–H C C= C–C CH 410. follows: 336.81 606.68 KJ/mol KJ/mol Bond Enthalpy Bond Data: affinity of chlorine are 81.2, 243.0, 375.7 and The enthalpy of sublimation of Cs, enthalpy of dis − idn nryo :C binding energy of : C 150.0 150.0 & 51.5 k cal respectively and C Given: Enthalpy of dissociation of nitrogen = 945.58Enthalpy KJ/mo of sublimation of graphite = 719.65 KJ/mol [Na(g)] [Na(g)] = 494, the the enthalpy of polymerisation per mole of e n CH O = O = 495.0 ; H H ; =495.0 = O O Enthalpy of formation of CH Cs(s) + 0.5Cl formation of CsCl is Calculate C–C bond energy in CC of bond dissociation for C=C & C Calculate the (i) C–H bond enthalpy = 414.22 KJ/mol 68.38 68.38 kcal respectively, calculate C 2 2 2 2 H H H H 4 5 5 6 (g) 4 2 (g) (g) + H – – S S – C – – S C = CH ∆ ..(–)= 104.1 kcal/mole B.E. (H–H) → → sub sub 2 ()= 171.8 kcal/mole C(s) 2 C(g) + 4H(g) → 2 2 (g) 2C(g) + 6H(g) (g) (g) H (–CH ∆ ∈ 5 → (g) H 2 C → H − vap C C 5 − (g) 388.6 kJ mol 2 2 [Na(s)] = 101 . Lattice energy of NaF(s) =

H − CsCl(s) . ; (ii) CH 6   (g) ∈ 2 H=434 ;C H = 413.4 ; H = 435.8 ; − 3 ) CN = 87.86 KJ/mol ,Enthalpy of formation of C 1 C n t 25º at where n has a large integral value . Given that ≡ ∆ ∆ 2 N r r H H = 676 kcal/mole H = 396 kcal/mole 6 6 − − & heat of formation of C C at 298 1 . Calculate the lattice energy of CsCl. ≡ 

C bond energy . Given that heat of atomisation of ∆ ∆ ∆ C . . C tom using the following data . Make Born H H H H bond energy is 93.64 k cal. ° ° ° f f f = 222.8 kJ/mol = – 201.9 kJ/mol = – 147.2 kJ/mol ∆ owing equation

ation of K are H − sociation of chlorine, ionization energy of Cs & el late the enthalpy change for the reaction. ∆ g data. 348.3 kJ mol thylene at 298 diss . If heat of formation of CO H t t 25°C and under constant pressure at 1 atm are as  sub l ; Enthalpy of dissociation of H

C = 347.0 ; C = O = 728.0 ; =728.0 C= O ; C= 347.0 + (F of C = 718.4.

590 & + 2 ) = 160, 160, = ) yee s ersne b te reaction the by represented is hylene 2 − H

(g) . [All values in kJ mol

1 331 KJ mol

. The energy change involved in the K . 6 ( g) ∆ − 7K/o 87 431.79 KJ/molKJ/mol

H 894. f º (NaF(s)) = = (NaF(s)) º − 1 respectively. Calculate 2 2 the average enthalpies & H H 6 = – 83.68 KJ/mol

− 2 = 435.14 KJ/mol 2 Haber's cycle. All O O are − − 1

] 7, I.E. 571, C & H are − 94.38 94.38 & ectron

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 4 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .4Calculate the heat of combustion of methylQ.34 alco .2The Born Q.32 .5Calculate the enthalpy of combustion Q.35 of benzene Q.33 The enthalpy The offormation ofethane, ethylene a Q.33 EnergyResonance of C–H bond is given as equal to + 410.87 KJ/mol. odC Bond of vaporisation of methyl alcohol = 35. Find out the electron affinity of chlorine in kJ mo ( i i i ) Resonance energy of CO i Resonance energy of benzene( (i) ( i v ) Latent heat of vaporisation of water = 40.6 kJ mol i)Enthalpy of hydrogenation of cyclohexene( (ii) ( v ) Energy kJ mol –2275.2 and –5506 KJmol ∆ ∆ ∆ Hº Hº Hº − Haber cycle for rubidium chloride (RbCl) is given b f f f

of CO of H of C − 1 6 2 H O( 2 (g) (g) = 12 l ) ) = ( 1 5. 6. 9 711 494 464.5 351.5 414 l ) ) =

2 − –1 − = −

C H respectively. Calculate the resonance energy of be 285.8 285.8 kJ/mol −

393.5 393.5 kJ/mol −

156 kJ/mol

143 kJ mol l ) =

− − − 1

O O

152 kJ/mol hol at 298 K from the following data l ( − l 1 nd benzenend from–2839.2, atoms are gaseous the − ) on the basis of the following data : 5 kJ mol . l 1 ) = . −

119 kJ/mol

− −

1 O H . elow (the are in k cal mol

=

C O nzene. The bond enthalpy

=

O − 1 )

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 5 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .As per reaction, N 5. 4. For For any reaction, 4. 8. For For reaction 2A(g) + B(g) 8. .If heat content of X is smaller than that of Y th The combustion of a2. substance is always______items:appropriate with blanks the 1. in Fill Q.1 10. For For a particular reaction 10. .Heats of combustion of methane, carbon and hydrog 6. 3. C C (Diamond) 3. 9. .Resonance energy 7.is always negative. Combustion reactions6. are always exothermic. Heat of neutralisation5. of weak acid-strong base i Integral heat of 4. solution is same as heat of hydr The enthalpies of 3. elements are always taken to be The 2. 1. statementsasTrue orFalse: following the Mark Q.2 0 For the reaction, 10. 8. The heat of combustion of graphite and carbon moncombustionandgraphitecarbonof of heat The 8. .At 298 K, the bond energies of C–H, C–C, C = C an 9. .The heat of neutralisation of 1 mole of HClO 7. rigid, thermally conducting walls. then for reactiion _____kJ mol ∆ C H – – 283 kJ mol 435 kJ mol formation of CH at at constant temperature, ∆ E E = 0 Combustion of benzene in a sealed container t 3 2 H H C = CH ° f ∆ 8 (C, (C, diamond) (g) + 5O H of a reaction is independent of temperature. 2 –1 (g) (g) + H . The value of enthalpy change for the reactions –1 –1 → . . Thus, heat of formation of in kJ 2 . (g) (g) 4 is ______. ∆ 2 3 2 C (Graphite) + 2 ≠ H° = ______. → (g) (g) + 2O (g) (g) C (g) 0. → 3CO ∆ ∆ → H H – E E = → 2 H (g) (g) 2 (g) (g) + 4H ∆ ∆ RFCEC TEST PROFICIENCY A(g) + 3 C – CH U U is______. H H + P. 3 C(g) → x 2NO kJ. The heat change in this process is called_____ ∆ 2 B 2 3 ∆ O O ( V (g) (g) at 298 K will be______. 2 H H = – (g) (g) (g) – 66 kJ the value of l ) 4 with 1 gm-equivalent of NaOH in aqueous solutions en the process X ∆ ation of substance. H H = s not constant. x zero. kJ d H–H bonds are respectively 414, 347, 615 and 2 en are –212, –94, –68 kcal mol x _. hat is immersed in a water bath at 25°C and has kJ. oxide respectively are – 393.5 kJ molkJ oxiderespectively 393.5 – are → mol ∆ H Y is ______. –1 f f of NO is______. 2 is ______. –1 . . The heat of ______. –1 and is

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 6 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com . Calculate the mass of mercury which can be liber Q.1 . Although C Q.4 . From the following data of Q.3 . Calculate the heat produced when 3.785 of Q.8 Q.2 For For reduction of ferric oxide by hydrogen, Fe Q.2 i keeping the first term [i.e. C (i) . h etape o nurlzto o a tog aci strong a of neutralization of enthalpies The Q.7 a Find the expression for the amount of heat requi (a) i i keeping all the terms . of the above expres (iii) . A person takes 15 breaths per minute . The volum Q.6 . An intimate mix of ferric oxide & Al is used Q.5 (b) For CO For (b) ∆ water vapour at 25º at vapour water 1273 K, keeping the reaction temperature constant.Calculate the mole composition of the mixture of st ∆ ∆ & value per CC of the mix . Heats of ∆ − C variation with temperature as well . This relation ∆ − solutionacontaining to oneequivalent ofHAone & and C(s) C(s) + H and to to T − a osatpesr (b) constant volume conditions Given : (a) constant pressure HgO with 41.84 kJ of heat at density of Fe C is used in the combustion of , how 21% much of of oxygent by volume . The exhaled air co percent error in case (i) . 500 K in case (i) and case (ii) . (number of moles accuracy. heat heat is evolved .

1302.7 k cal/mol . H H Hº H Hº Hº should be at the most P P 6900 calories . In what ratio is the base distribut 13680 cal/equivalent & = a + bT + cT [Fe f f f º º (Al º º H COº 2 298 K, while 2 O 2 = C(s) C(s) + 1/2O 2 ∆ O( 2 , a = 5.0 , b = 15 x 10x15 b= , 5.0 = a , 3 2 − H O ] ] = 104.5, C (g) (g) = 35.1 kJ . The reaction was found to be too exother l f 3 ) ) = P º (HgO, s) = ) ) = is usually assumed to be constant, for more accura 2 O − − − 68.32 k cal mol ∆ 2 3 94.05 k cal mol

2 399 399 k cal/mole ; O(g) — O(g) = 5.2 g/cc ; density of Al = 2.7 g/cc. H + dT com 2

(g) (g) — C. The density of is 0.7025 gm/ml. Heat of c ofgm/ml. Heat is0.7025 octane densityTheof C. P of sucrose is = [Fe(s)] [Fe(s)] = 25.5, C 3 − cal mol − → 90.8 kJ mol − ∆

2900 cal/equivalent respectively . When one equival → P

H of the following reactions 26 kJ . At what temperature is it possible ? CO(g) + H =a i)keeping the first two terms . (ii) = a] CO(g) CO(g) ; − − − 5 − 1 1 . Find the heat required to raiseits temperature Find to requiredheat. the 1 K XRIE II EXERCISE − ∆ 1 − − H ; ; ∆ . 1 6000 kJ mol 2 &M & f H H = –110 KJ (g) (g) ; º º (Fe P sion . Note that each successive term introduc [H as solid rocket fuel . Calculate the fuel va 2 2 2 formation & densities are : ∆ O( red to raise the temperature of 1 mole of gas from ∆ ∆ O O is given by ; = 1) . Case (ii) gives a more accurate value . Fin H H = 132 KJ H H 3 e of air inhaled in each breath is 448 ml and conta 3 ated ated from HgO at 25º l octane octane (C ntains 16% of oxygen by volume . If all the oxygen

ed between HA & HB ? (Hg) (Hg) = 200.6 g mol (s) + 3 )] )] = 75.3, C ) ) = he latter is burnt in the body per day & how muc f f º º H º CO(g) = eam and oxygen on being passed over coke at H & wae ai H b NO are NaOH by HB acid weaker a & HA d − − 1 2 equivalent enthalpy the of HB, change was

. Take temperature to be 300 K throughout. 199 kcal / mole, O(g) O(g) =

H 2 8 (g) H mic to be convenient . It is desirable that − 18 P − 26.41 k cal mol → [H ) ) reacts with oxygen formto CO & te te calculations we must consider its 57.79 k cal mol 2 (g)] (g)] = 28.9 (all in J/mol) 2 −

1 Fe(s) + 3

. C by the treatment of excess ombustion of Combustionof . ent of NaOH is added −

H − 1 1 2 O( from 300 K tofromK300 l ; ) ; lue per gm es higher 8 H d het 18 T ins 1 is K h

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 7 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com . Using the data (all values are in kJ/mol at 25ºQ.9 .2Calculate the enthalpy change for the reaction Q.12 .0Using bond energy data, calculate heat Q.10 of forma Q.14 The standard enthalpy of combustion of sucrose of combustion of enthalpy standard The Q.14 .3By using the followingQ.13 data draw an appropriate .1Use the following data to calculate the Q.11 enthalp 85 kcal/mol & bond dissociation energy of F enthalpy change of solution of NaI(s) = + * lattice energy of NaI = * i)As(s) + (ii) (in kJ mol (i) C Calculate the following bond energies Bond energy :of O = O = 498.94 The average Xe lattice energy of NaCl = * v)H (vi) ( v ) HCl(g) + aq AsCl(iv) (iii) ∆ 5 ∆ ie C Given : enthalpy change of hydration of Na * of sucrose to lactic acid ? i As (i) ∆ ∆ enthalpy change of solution of NaCl(s) = * Comment on the difference in their of values. (i) the chloride ion ; (ii) the iodide io

Hº Hº for C Hº Hº Hº C(s) + 4 H f combustion combustion CO

−

(ii) C C 2 1 − 2 2 (s) H 1 (g) (g) + 2 (g) = (g) of energy released as heat) of complete aerobic ox O 2 (graphite) 2 (acetylene) = () = 3 (g) + 3 (g) (g) ( (s) (s) + [(3 H l ) +) H(3 C(s) C − 2 3 − 2 1 F bond energy is 34 kcal/mol, first I.E. of Xe is 2 →  

Cl 9. ; 393.5 O 2 1 2 → H ; H = 98.8 k cal C=8 a 4 a & C C= 147 =cal k ; = k C 83 cal 2 → Cl (g) (g) (g) (g) H HCl 2 C(g) = 171 k cal . → 2 (g) − O O + aq) 2 → → ∆ 2 C =

C 598; 1559.8 O + aq)] H − H → AsCl

a); (aq) (g) f 297; 1299.7

º º for lactic acid, CO − 2 HCl = + C CH − | O

(iii) C H

− 699 kJ/mol.

→ — ( 3 3

772 kJ/mol. l 3

( ; ) 716.68 ; Bond Bond energy of H ; 716.68

g ; (g) l H → ; ) C H | 3 2 H = CH AsO n. + 3 (g) = AsO 3 3 2

C) given below : 2 is 38 kcal/mol. energy cycle & calculate the enthalpy change of hy (aq) (aq) + 3 HCl(aq) ; y of formation of As (g) XeF tion of isoprene . 3 3

− 2 2 kJ/mol. −  and H a); (aq)

2 kJ/mol.

390 kJ/mol. 4 H=14kcl; H = 104 k cal is

→

= ∆ ∆ ∆ −

O (iv) C = C (v) C (v) (iv) C = C O 2

Hº Hº Hº O O is molkJ 5645 Xe f combustion combustion of H idation compared to anaerobic hydrolysis − + 694, + F 79 kcal/mol, electron affinity of F is 2 2 O O( − 3 () = (ethene) = + F − . 395.0 and l ) =) − ∆ ∆ ∆ ∆ ∆ ∆ 1 2 H H = 7550 + cal H H = H = H = H = H = . ht i h advantagethe is What . + F.

− −

H H = 435.94 285.8 − − − − − cal68360 cal22000 cal17315 cal17580 cal71390 − −

286.0 respectively. 1410.9 −

1192.3

≡ dration

C

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 8 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .5In the combustion of solid naphthaleneQ.15 (C (ii) ( i ) .8Te nhly f omto o C of formation of enthalpy The Q.18 .9During one of his adventures, ChachaQ.19 Chaudhary .6Calculate the proton affinity of Q.16 NH Q.17 The standard enthalpy of formation of FeO of formation of enthalpystandard The Q.17 exhaling take? wasted in thinking of a way to rescueSabu shouldrescuehim. Chacha Chaudhary within At6 minute maxim need to suck out air in order to save Chacha Chaudh If the safe level of CO in the atmosphere is less t respectively. A mixture of two oxides contains FeO forB.E. C– C bond = 345.5 kJ mol Enthalpy of atomization of C = 715.5 kJ molEnthalpy of formation of CO at constant pressure. An initial sample of aircave droppedtaken to halffrom its theinitial cave value measur caveof onetill atmos the pressure was again airone andatmosphere. exhale Ea it out in the hesurroundings. started sucking In the them poisonous air out of the cav initial mixture ?changed into a 1 : 2 mole ratio mixture, how much o All data are at 25º Lattice energy of NH k cal/mol & addition to O two hundred years back. The air inside was poisonou Electron affinity of Cl = 348Ionization energy of H = 1310 CH Enthalpy of formation of HLatent heat of sublimation of naphthalene are = evolved.72.0 kJ Calculate resonance energy of naphthal ∆ ∆ H H 3 ° ° dissociati formation − O − ∆ H–H bond = 436 kJ/mol C – H bond = 413.0 kJ mol C = C bond = kJ617.0 mol CH H on c omb − : NH 2 3

68 k cal/mol respectively, calculate : H an d N is (CO) (CO) = − 2

g 28; = 218 (g) 348 k cal/mol . Given that the enthalpies of format 3

C. g 6; = 46 – (g) 2 . Sabu, being huge, could not enter the cave. So, i 4 Cl (s) = – 683 − 280kJ.mol 2 O O = –286.0 kJ mol 2 = –393.5 kJ mol 3 2 (g) from the following data (in kJ/mole) H − –1 –1 1 –1 5 . . Neglect any use of Graham's Law. OH( 10 H l ) is ) –1 8 ) at 298 K and atmospheric pressure 5157 kJ/mol of eantime, fresh air from the surroundings effused in –1 –1 ∆ ed 11.2 mL at STP and gave 7J on complete combustio e by mouth. Each time he used to fill his lungs wit ∆ ∆ phere. − han 0.001% by volume, how many times does Sabu H for the isomerisation of to methoxymethan got trapped in an underground cave which was sealed H H ch time Sabu sucked out some air, the pressure in t

Fe& /mol 6 clml h etap o obsin ofcombustion of enthalpy The . cal/mol k 66 & F e um, how much time should each cycle of inhaling ene.Given ° f ° dissociati ar y ? f thermal energy will be released per mole of the : NH s, having some amount of carbon monoxide in 2 2 O O selse he willdie. Precious seconds 80 are 3 on 4 3 in the mole ratio 2 : 1 . If by oxidation, it is Cl(s) = –314 is : Cl

− 2

65 kcal molkcal 65 g = 124 (g) ion of CO n order to save Chacha Chaudhary, 2 (g) and H − 1 and − 197kcalmol 2 O( l ) are h cave to the − heat

94 he − e. − n 1

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 9 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .0Fe Q.20 I.E I.E I.E Enthalpy of sublimation, Prove that value of x is 3 with the help of given d Lattice Energy of Fe I.E I.E B.D.E B.D.E of O Enthalpy of formation, mixed oxide. From the data given below, x 2 1 3 2 1 O of O of O of Fe = 2960 kJ/mol. of Fe = 1560 kJ/mol. of Fe = 760 kJ/ mol. 4 a mixed oxide of iron consists of only Fe 2– 2– = 142 kJ/mol. = –844 kJ/mol. 2 = 490 kJ/mol. x O ∆ 4 = 18930 kJ/mol. ∆ H H f (Fe sub (Fe) = 390 kJ/mol. x O 4 ) ) = –1092 kJ/mol. +2 ata. & F e +3 ions, with the ratio of Fe +2 : F e +3 = 1:2 in the

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 10 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com Q.4 50.0 50.0 mL of 0.10 M HCl is mixed with 50.0 mL of 0 Q.4 Q.3 For which of the following change Q.1 . NH Q.8 . If x Q.7 Q.2 . The enthalpy of neutralisation of a weak acid Q.5 in . For the allotropic change represented by the eq Q.6 mol (A) H (C) (C) 2H(g) ∆ (C) (C) The heat of formation of NCl H N (A) –2.5×10 3.0°C. Calculate the enthalpy of neutralization per (C) more than in the second case by 0.95 kJ (D) (D) less (B) more (C) more than in the second case by 0.95 kJ (A) less than in the second case by 1.9 kJ (C) (C) C(s) + O 6 g of diamond and 6 g of graphite are separately b (A) – 228.5 kJ mol (A ) x vaporisation of water, estimate the standard enthal strong base is – 57.3 kJ equiv (A) A 0()1 C 0(D) 25 (C) 20 (B) 15 (A) 10 acid to be monobasic)? (A) H enthalpy of ionization of the acid is 1.5 kJ mol ∆ r 2 2 H H of which of the following reactions is zero? (g) Cl+ (g) 3H+ (g) 1 3 –1 f 0 ∆ ∆ , , x (g) +(g) 3Cl 1 of water is – 285.8 kJ mol , , H H 2 2 + (g) + I (g) (g) 2 ∆ f f and x = x = – = H 2 2 → → f 0 2 ∆ –2x 2 (g) (g) of OH H 2 (g) (g) 2 ∆ kJ (B) (B) –1.3×10 kJ 2 (g) (g) 3 2 1 (g) H are enthalpies of H–H, O=O and O–H bonds respectiv (g) (g) 3 – l 2H 2H +x l 1 2HCl ;(g) + → 2NH → l 4 ∆ –1 – + + ion will be NCl (g) + 2e H (g) + 2e 2 ∆ 2 H CO 2 2HI (g) (B) (B) HCl (aq)+ NaOH(aq) 2HI (g) (B) x (B) 228.5 kJ mol 3 − 2 (g); (g); 3 (g) (g) + 3HCl (g); – 2 3 − 2 g (D) N (g) –1 – 1 ∆ 2 3 – ∆ 3 ∆ + , what is the % ionization of the weak acid in mola (g) in the terms of H H ∆ H x –1 3 2 H 3 2 2 . . If enthalpy of neutralisation of monoacid strong 3 ∆ –2x XRIE III EXERCISE H 2 2 kJ (C) (C) –8.4×10 kJ ≠ ∆ 3 –x –1 E? 4 ∆ –1 H 1 M solution with a strong base is – 56.1 kJ mol and enthalpy of neutralization of the strong acid uation C (graphite) (C) (C) 114.25 kJ mol (D) H (B) 2H(g) + aq (D) None (B) C)( x 1 py of combustion of hydrogen mole of HCl. ∆ urnt to yield CO H ∆ .10 .10 M NaOH. The solution temperature rises by 1 1 H 2 2 + , (g) + 3H (g) (g) + aq ∆ f than in the second case by 11.4 kJ = = x 2 H than in the second case by 11.4 kJ 2 2 –x ∆ and 1 H kJ (D) (D) –6.3×10 kJ 3 +x 1 2 → → + (g) ∆ 4 H –1 → 2 ∆ → , , the heat liberated in first case is 3 2H 2H is H 2 2 (D) (D) –114.25 kJ mol (D) 2x C (diamond), ely, and x + + − 2NH (aq) (aq) + 2e (aq) (aq) + 2e → 2 3 ∆ 3 3 H –x r solution (assume the ( g) NaCl(aq) + H 3 1 4 – is the enthalpy of – – 1 1 x base is –57.3 kJ 2 kJ 2 ∆ –x H = 1.9 kJ. If 4 –1 –1 2 . If the with a O( l )

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 11 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com . Ethanol can undergoes decomposition to form two Q.9 .2The reaction CH Q.12 .0Reactions involving gold haveQ.10 been of particula .1(i) Cis Q.11 (iv) Enthalpy of combustion of 1 (v) 9(v) (A) 70 kCal (B) 80 kCal (C) 67.75 kCal (D) (D) 57.75 kCal (C) 67.75 kCal (B) 80 kCal (A) 70 kCal From the given data, what is the bond energy of Cl— (iii) Trans (ii) Cis if the molar ratio of C 4 4 moles of HCl . What is the percentage conversion (A) 65.98 kJ (B) 48.137 kJ (C) 48.46 kJ (D) (D) 57.22 kJ (C) 48.46 kJ (B) 48.137 kJ (A) 65.98 kJ Au(OH) (A) 0.5 % (B) 0.6 % (C) 5 % (D) (D) 50 % (C) 5 % (B) 0.6 % (A) 0.5 % (vi) Enthalpy of combustion of trans 2 In an experiment there was an absorption of 0.44 kc Au(OH) (A) The value of decomposition of 1 mole of ethano

− ∆ 1.0 , 1.8 (B) 1.8, -1.0 (C) –5, 9 (D) (D) –2, 3.6 (C) –5, 9 (B) 1.8, -1.0 1.0 , 1.8 − H − 2 C 1 3 3 2 + 5 2 − + 4 HBr + 4 HCl − H − butene 2 butene 5 ∆ OH (g) − H butene is more stable than cis ∆ 1 4 H & (g) (g) + Cl → → → 2 → 0= ∆ 2 → trans H H 1 4 HAuCl 2 HAuBr

to CH

− 2 in (g) (g) butene,

Kcal/mole are − 2

→ 3 4 4 − − CHO is 8 : 1 in a set of product gases, then the en + 3 H + 3 H butene, butene, l is CH ∆ ε ε ε ε Bond H Cl—Cl C—H H—Cl C—Cl − 2 2 2 butene, 3 O , O , x : y = 9 : 5 Cl(g) + HCl(g) has ∆ ∆ H = H Ene rgy r interest to a chemist . Consider the following re − 1 Bond kCal 2 − 103 ∆ 84 649.8 kcal/mol y x − H = sets of products butene. of HAuBr Cl bond al when one mole of HAuBr − ∆ ∆ 647.0 kcal/mol. H =H H =H ∆ − − 4 H = –25 kCal.

into HAuCl 36.8 36.8 kcal 28 kcal 4 ? ergy involved in the 4 4 was mixed with actions,

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 12 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com i )If at initial temperature T (iv) .3From the following data at 25°CQ.13 ii What will be the final pressure in atm.(iii) i)What will be the maximum temperature attained i (ii) .4A mixture of hydrogen gas and the theoretical a Q.14 i)Match the gas and percentage of each gas in one (ii) i The amount of heat liberated when one litre of p (i) i The value of C (i) .5The commerical production of "WaterQ.15 gas" utilis exploded in a I N (I) (C) E O ttmn a : Statement (a) Which of the following statement(s) is/are correct: (A) A .,83()83 13()1.,1. (D) 11.3, 8. (C) 11.3, 11.3 (B) 8.3, 11.3 (A) (A) 8.3, 8.3 A ttmn B ttmn ,b (C) Statement b (B) Statement a, b, d (A) Statement c : Enthalpy of formationStatement of(d) OH(g) is : Enthalpy42 of formationStatement of (c) H(g) is –21 C Ic,(Id,(I-) I-)(D) (I-b), (II-d) (B) (I-b), (II-a) (C) (I-c), (II-d), (III-a), (IV-c) (A) (I-a), (II-b), (III-c), (IV-d) I)H H (IV) CO(III) (II) undergoes 75% conversion. Using the following data, (A) [Take air as 80% N H ttmn b : Statement (b) stoichiometric amount of air.(79% N H (A) E , then which option is true + CO(g). The heat required for this reaction is gen isreaction this for required heat The CO(g). + Given (i) C datas answer the questions that follow : ∆ 1 2 2 2 2 H H (g) (g) (g) (g) + f ≅ ≅ ≅ 2 [CO(g)] = –110.53 kJ/mol ; (g) + 85(B) 8.5 1 36k (B) 3.6 kJ 24 (B) 2940 K 1 → → = E > E a Percentage Gas Reaction 1 2 2 2 2 (d) O 2 2 O P 2()436 2H(g) 2()495 2O(g) 1 2 2 = 8.3 cal deg 2 closed rigid vessel O (g) P 2 of N (g) (g) ∆ ∆ → r r H° for the reaction OH(g) H° for the reaction H → 2 2 , , 20% O & H H ( c ) (b ) ( a ) 2 O g 42 OH (g) 1 –1 ()–242 O(g) E 2 mol ≅ ≅ ≅ O in the order N 1 is initial internal energy & at higher final tempe 76(C) 7.6 39k (C) 3.9 kJ 26 (C) 2665 K 2 by volume] ≈ ≈ ≈ ≈ –1 . If the process occurs under 9.7 7.7 36.4 23.1 ; (ii) C ∆ H 2 f by volume and 21% O [H P 2 = 11.3 cal deg O (g) 2 O(g)] = –241.81 kJ/mol ; 2 , , H → es the endothermic reaction C(s) + H ∆∆∆ (D) can't be compared from the given data(B) E ∆ → roduct gases are burnt at 373 K and one atm is mount of air at 25°C and a total pressure of 1 atm, litre product gases. 2 f the process occurs in adiabatic container. r O will be (in cal. deg. °kJ/mol H° ≅ ≅ ≅ H(g) + O(g) is 502 kJ/mol answer the question that follows : 2H(g) + O(g) is 925.5 kJ/mol 8 kJ/mol 5.46 (D) (D) 5.46 4.43 kJ (D) (D) 4.43 kJ 1900 K (D) (D) K 1900 2 kJ/mol > E erated by combustion of to CO to coal combustionbyof erated –1 mol 1 adiabatic condition –1 2 , , (III-d), (IV-c) , , (III-a), (IV-c) by volume). The superheated steam (D) Statement a, d , c ; ∆ H ∆ f [H –1 H mol f 2 [CO O(g)] = –57.8 Kcal ≅ ≅ ≅ 850. rature. rature. T 5.34 kJ 298 K –1 2 ) (g)]=–314.0 kJ/mol then using the given 3 2 O(g) 2 E 2 → is the final 2 using H 2 ( g ) is

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 13 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com . Diborane is a potential rocket fuel which underQ.5 Standard molar enthalpy of formation of COQ.4 Which of the following is not an endothermicQ.3 rea Q.2 Which of the following reactions defines Q.1 OBJECTIVE . The Enthalpy change involved in the oxidation of Q.9 Compute the heat of formation of liquid methylQ.8 a enthalpycalculatethe following data, the From Q.7 SFinenergybond S–F average the Estimate Q.6 SUBJECTIVE – 1100 , , 275 and – 1100 80 kJ/mol respectively. (D) Conversion of graphite to diamond(C) Dehydrogenation of ethene to ethylene(B) Decomposition of water (A) Combustion of methane 298 k. Theenthalpy k. 298 offormation of CO (C) (C) N (A) C standard enthalpy change (in kJ) for the reaction their standard states ; H vaporisation of liquid CH maximum distance a person will be ableavailbale to walk for muscular . work afteIf 100kJ of muscul (D) sum of molar enthalpies of formation of (C) COstandard and molar enthalpy of combustion of(B) standardgaseou molar enthalpy of combustion(A) ofzero carbon Average Bond C energies From the following data, calculate the enthalpy cha (A) 524.1 (B) 41.2 (C) –262.5 (D) (D) – 41.2 (C) –262.5 (B) 41.2 (A) 524.1 20.42 kJ/mol respectively. The enthalpy of isomeris ∆ H ° f for COfor 2 (diamond) (g) (g) + 3H B 2B(s) + 3H CO H 2B(s) + 2 2 H (g) (g) + 2 6 (g) + H 2 (g) (g) + 3O + C (g), CO(g) andHCO(g) (g), H 2 (g) (g) 1 2 3 2 2 2 O( O O (g) 2 2 (g) → 2 2 (g) (g) l (g) (g) (g) (g) ) ) 2 → (g) C → →

→ , , 218 KJ 3 → → 2NH − − OH = 38 kJ/

→ H, 415 KJ 356 O , KJ CO B H CO(g) + H B H 3 2 2 2 H B 2 O(g) 2 g (B) 1/2 H (g) 2

O(g) are – 393.5, –110.5 and –241.8 kJ molkJ –110.5and–241.8 393.5, – are O(g) / O O( 6

2 mol ; C, (g) O 3 XRIE IV EXERCISE (s) l 3 )

;

; (s) + 3H

mol . Heat of formation of gaseous atoms from the

; /

/ ; mol ; ∆

o ; O mol 2 H O (g) is 2 ° f 2 2

is equal to 715 KJ (g), H (g), ? goes combustion according to the reaction, 2 ∆ ∆ ∆ ∆ (D) (D) CO(g) + 1/2O ct ion? lcohol in kJ mol 6 O(g) . The . H = 44=HkJ H = 36=HkJ H = –286 kJ–286=H H = –1273kJ=H is nge for the combustion of diborane : ation of cyclopropane to is change for the combustion of cyclopropaneatcombustionof the for change r eating 120 g of glucose. O ar work ar iswork needed walk to one km.What is the (graphite)

s carbon 2 /

O 2 mol ; O, O, 249 KJ mol ; − ∆

H463 KJ, ( H 2 l ) and ) are propene(g) (g) + 1/2 F f ° −

values of SFvaluesof

2880 KJ mol − 1 ,

2 using the following data. Heat of (g) (g)

/ 2

mol (g) (g) → →

6 6 /

mol. − (g), S(g), and F (g) are (g) F and S(g), (g), CO 1 . 25% of this energy is HF(g) 2 –1 − (g) respectively. The

393.5, 393.5, − [JEE 1999] [JEE 2000] [JEE 2003] [JEE '97, 2 ] [ JEE '97, 5 ] [ JEE '98, 5 ] [JEE 99, 5] [JEE 2000] [JEE 1997]

33.0 kJ/mol. elements in −

8.8 & 8 285.

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 14 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com .3–2.8K/o Q.34 H Q.35 – 23.68 KJ/mol Q.33 Q.31 E Q.27 (i) 343.08 ; (ii) 891.2 Q.26 Q.1 Q.2 Q.1 .4Q.14 .223k o .3– 120.08 J/ml Q.23 B.E. (C–C) = 82 kcal/mole; Q.24 213 kJ / mol Q.22 Q.15 27.43 27.43 g O Q.15 . –1560.1 kJmol Q.9 Q.25 Q.18 10.0 10.0 mol % CH Q.18 Q.1 1 . –1. clQ.6 – 18.7 kcal Q.5 Q.28 .Tu .Fle8 re9 re10. False 5. True 9. True 4. True 8. True 3. False 7. False 2. False 6. True 1.False 4. 3. Heat of transition 5.+33 kJ mol 2. endothermic 1. exothermic (i) − − − 9. –125.0 kJ 10. 10. – 3RT –125.09. kJ − ∆

8JmlQ2 593M . 29 clQ.4 2.95 kcal Q.3 35.973 MJ Q.2 88kJ/mol 72 kJ mol 618.7 618.7 kJ mol 192.3 192.3 kJ mol f E= 27.91 KJ mol º(benzene) = 49 kJ mol

−

3201 kJ/mol 2 − 1 –1 − − –1 1 4 1 − ; (ii) 1 , , t= 514 sec. − 6. 6. –18 kcal mol E.A. = – 352 kJ mol Q.30 Q.32 277.5 kJ/mol Q.29 –2808 kJ/mole Q.20 –31.5 kJ/mole Q.19 Q.17 Q.16 – 75.14 kJ/mole

213kclQ7180 JQ818.95 Q.8 128.02 kJ Q.7 22.123 k cal − − 1

197 JmlQ.12 3199.75 kJ/mol , , H ∆ f

C ANSWER KEY ANSWER H[C RFCEC TEST PROFICIENCY − − º(benzene) = 669.7 kJ mol

C 90.75 kcal mol ≡ 2 C H XRIE I EXERCISE = 160.86 k cal 6 (g)] (g)] = –20.1 kcal/mole –1 −

− 3267.4 kJ mol .011.2Kcal. Q.10 .–72k 8.–110.5 7. –57.2 kJ 1 − 1 − (i)

237 JmlQ1 25.66 days Q.13 3273.77 kJ/mol

−

885 kJ/mol − 1 ∑ ν p

∆ H Q.21 (ii) °

)P( −

− − –726.6 kJ/mole

889.980 kJ/mol

∑ 1410 cal ν –1

r % ∆ H ° ( R )

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 15 of 16 THERMO CHEMISTRY FREE Download Study Package from website: www.tekoclasses.com ...... B Q.7 D C Q.13 Q.6 D C Q.12 Q.5 A Q.11 A Q.4 C Q.10 A B Q.3 Q.9 (i) B D Q.14 A Q.2 Q.8 D Q.1 (i) 13 times, (ii) 21.53 secQ.19 advantage = 5396 kJ.mol Q.14 Q.3 moleK 404 Q.2 (a) 92.435 g (b) 93.72 g Q.1 . . . . . –2035kJ mol Q.5 B Q.4 A Q.3 Q.7 4.82 km B Q.9 309.6 Q.2 kJ/mol Q.6 B Q.1 . C Q.9 .7Heat released= 13.4 kcal. per mol of Q.17 initial mi .222ka/o .3for Cl Q.13 292 kcal/mol Q.12 . 0.9346 k cal g Q.5 .0+ Q.10 . .7 .2 Q.8 0.371 : 0.629 Q.7 Q.4 (a) (a) (i)

06kclQ.11 20.6 k cal −

C = 97.81 97.81 =CC kJ,

(i i ) A a ∆ T T

(iii)A (iv) C Q.15 Q.15 (i) D A (ii) C (iii)A (iv) (ii) − 1 , , 3.94 kcal cm a ∆ T T +

−

H = 454.64C= =CH OkJ, kJ, 434.3=804.26=C= kJ − 2 b 1 (T −

2091.32 kJ mol 2 − 2 2 3 −

T XRIE III EXERCISE XRIE IV EXERCISE 1 XRIE II EXERCISE 2 ) ) .5–287.0 kJ mol Q.15 (b) 1000 cal, 1012 cal, 1.1858 tr Q.18 xture − − . 9.822 MJ/day evolved Q.6 1

154.68 k cal − − Q.8 −

384 kJ mol 666.81 k cal mol % % O –1 −

266 kJ mol 2 (g) (g) = 37.5, H − 1 , , for I .6– 718 kJ/mol Q.16

% % error − 1 , , 15.55 Mcal − − –1 1

− 22 , C ,

307 kJ mol 2 O(g) O(g) = 62.5 kcal mol

≡

C =C kJ733.48 − 1 − 1

TEKO CLASSES, Director : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL Page 16 of 16 THERMO CHEMISTRY