ffffoooo////uuuu ffffooooppppkkkkjjjjrrrr HHHHkkkkhhhh#### ttttuuuu]]]] uuuugggghhhh aa aa vvvvkkkkjjjjEEEEHHHHkkkk ss ss ddddkkkkeeee]]]] ffffooooiiiiffffrrrr nnnn[[ss[[ss kkkk NNNNkkkkMMssMMss +s+s +s+s rrrrjjqqjjqq rraarraa eeee////;;;;eeee eeeeuuuu ddddjjjj '''';;;;kkkkeeeeAAAA iiii##qq##qq """"kkkk ffffllllggaaggaa llllddaaddaa YYYYiiii ddddjjjj]]]] llllggggrrrr ss ss ffffooooiiiiffffrrrr vvvvuuuuddssddss ]]]] ^^^^ccccuuuukkkk^^^^ uuuu NNNNkkkkMMssMMss +s+s +s+s ////;;;;;;ss;;ss ddddkkkk]]ss]]ss jjjj????kkkkccqqccqq jjjj jjjjkkkk[[[[kkkk ss ss VVVVddssddss AAAAAAAA jjjjffffpppprrrr%%%% eeeekkkkuuuuoooo ////kkkkeeeeZZZZ iiiizzzz....kkkksrsrsrsrkkkk llllnnnn~~~~xxxxqqqq#### JJJJhhhh jjjj....kkkkNNNNkkkkssssMMMM++++nnnnkkkklllltttthhhh eeeeggggkkkkjjjjkkkktttt STUDY PACKAGEThis is TYPE 1 Package please wait for Type 2 Subject : CHEMISTRY Topic : THERMODYNAMICS 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 (cid:1) : (0755) 32 00 000, 98930 58881, www.tekoclasses.com THE KEY Y R T The subject of Thermodynamics deals basically with the interaction of one body with another in terms E M of quantities of heat & work. It may be defined as the branch of science which deals with energy O I changes associated with various physical & chemical processes. The entire formulation of thermodynamics H C is based on a few (Three) fundamental laws which have been established on the basis of human experience I O T of the experimental behaviour of macroscopic aggregates of matter collected over a long period of time. S 9 Thermochemistry is the branch of physical chemistry which deals with the thermal or heat changes 2 of caused by chemical reactions. It is based on first law of thermodynamics. There are two laws of2 e g thermochemistry: a P (a) Lavoisier & Laplace law (b) Hess’s law. TERMS & CONVENTIONS A number of terms & conventions are used in thermodynamics. L A P A System is defined as that part of the universe which is at the moment under investigation. O H m Universe less the system is defined as Surroundings. The actual or imaginary surface that separates the B s.co system from the surroundings is called the Boundary. 81 , e 8 ass TYPES OF SYSTEMS: 58 cl A system is said to be Isolated if it cannot exchange matter and energy with the surroundings (coffee in 0 o 3 w.tek aA t shyesrtmemos i sfl saaskid) .to be Closed if it can exchange energy but not matter. Coffee in a closed stainless steel 0 989 ww flask is an example. A system is said to be Open if it can exchange matter and energy both. A thermo 0, flask or a steel flask if not closed is an example. A system is said to be homogeneous when it is completely 00 e: uniform throughout, made up of one phase only, pure liquid. solid, gas. 00 sit A system is said to be heterogeneous if it consists of two or more phases, liquid in contact with vapour. 2 b 3 we 5)- m STATE OF A SYSTEM : 75 0 o The state of a system is defined by a particular set of its measurable properties. For example, we can ( e fr describe the state of a gas by quoting its pressure (P) , volume (V) and temperature (T) etc. PH: kag Variables like P, V , T are State Functions OR State Variables because their values depend only on the Sir) ac state of a system and not on how the state was reached. K. P State variables can be intensive or extensive. An intensive variable (eg. temperature, pressure, R. dy concentration) is one whose value is independent of the size of the system. An extensive variable (eg. S. tu volume , mass, surface area is one whose value is proportional to the size of the system. A ( S Y d RI oa Extensive Properties Intensive Properties KA l wn (Depend upon quantity of (Do not depend upon quantity of R. o Matter present and are additive) Matter present and are non additive) G D A Volume Molar volume H E U E Number of moles Density S FR Mass Refractive index or : t Free Energy G Surface tension c e Entropy S Viscosity Dir Enthalpy H Free energy per mole S, E Internal energy E Specific heat S S Heat capacity Pressure A L Temperature C O Boiling point, freezing point etc K E T THERMODYNAMIC PROCESS : Y R A thermodynamic process involves change of a system from one state to another state. T E M O TYPES: HI C A process is called Isothermal, if the temperature of the system remains constant during the change. It I O is carried out in a thermostat and in such a process the exchange of energy between the system and T S surroundings takes place. In such a process dT = 0 & dE = 0. 29 A process is called Adiabatic, if the system does not exchange energy with surroundings. Such a process 3 of e is carried out in perfectly insulated containers. During it the temperature of the system may change. In ag P such a process dQ = 0. A process carried out at a constant pressure is called an isobaric process. In such a process dP = 0. A process in which the volume of the system remains constant is called an isochoric process, whereby dV = 0. L A P O m CYCLIC PROCESS : BH s.co ∆WEh e=n 0a s&y s∆teHm =u n0d.ergoes a number of different processes and finally returns to its initial state. 81 , e 8 ss 8 a 5 cl REVERSIBLE PROCESS : (QUASI-STATIC) 0 o 3 k A process which is carried out so slowly that the system and the surroundings are always in equilibrium 9 w.te is known as a Reversible Process. If this condition does not hold good, the process is said to be, 0 98 ww Irreversible. 0, In a reversible process the driving force is infinitesimally larger than the opposing force. If the driving 0 0 : force is made infinitely smaller than opposing force, the system can be brought back without producing 0 e 0 sit any permanent change. 2 b 3 we A process which proceeds of its own i.e. without any external help, is called as Spontaneous Process 5)- m (or a natural process). 75 o Internal Energy (Intrinsic Energy) E – Every system having some quantity of matter is associated with a (0 fr definite amount of energy, called internal energy. H: e P g E = E + E + E + E + E + ..... ) ka Translational Rotational vibrational Bonding Electronic Sir ac K. P It is a state function & is an extensive property. R. dy ∆E = Efinal – Einitial ; ∆E = qv S. u ( t A d S ZEROTH LAW OF THERMODYNAMICS RIY a It states that, two systems in thermal equilibrium with a third system, are also in thermal equilibrium with A o K wnl each other. R. o G D A FIRST LAW OF THERMODYNAMICS H E U E It is law of conservation of energy. Mathematically, this law is written as : S FR ∆E = q + w, where ∆E is change in internal energy of the system and is a state function, q is the transfer or : of heat from / to the system and w is the work involved (either done on the system or by the system) . t c e According to IUPAC , heat, added to the system and work done on the system are assigned positive r Di values as both these Modes increase the internal energy of the system. S, E S TYPES OF WORK : S A L Two TYPES of work normally come across in chemistry. These are Electrical Work in system involving C ions, while the Mechanical Work is involved when a system changes its volume in the, presence of an O K externally applied pressure (i.e. pressure volume work). It is especially important in system containing E T gases. If a system expands from a volume V to V at constant pressure P, then the first law equation becomes, Y 1 2 R ∆E = q – P ∆V (∆V= V – V ) ....(1) T 2 1 E M O For a process carried at constant volume A B = q (heat absorbed at constant volume) HI C I O Work = Intensity factor X capacity factor T S w – by the system (Expansion) negative 29 of w – on the system (compression) positive 4 e q → absorbed by the system positive ag P q → given out by the system negative Work done in irreversible process w = – P ∆V Ext (Expansion) L V A 2 P Work done in isothermal reversible process w = – 2.303 nRT log O V H m 1 B s.co (maximum work) (Expansion) = – 2.303 nRT logPP1 81 , sse 2 88 a nR 5 cl work done is adiabatic reversible process w = [T – T ] 0 ko γ−1 2 1 93 w.te C 0 98 ww γ = CP = Poisson's ratio 0, V 0 0 : ENTHALPY : 0 e 0 sit Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been 2 m web found usHe f=u lE to + d PeVfin (eB ay ndeewfin sittaioten )functioorn Enth∆aHlp y= (∆HE) a+s P: ∆V + V∆P 755)- 3 or ∆H = ∆E + P ∆V (at constant pressure) combining with first law. Equation (1) becomes 0 o ( fr ∆H = q H: e p P g Hence transfer of heat at constant volume brings about a change in the internal energy of the system ) ka whereas that at constant pressure brings about a change in the enthalpy of the system. Sir ac The difference between ∆H & ∆E becomes significant only when gases are involved (insignificant in K. dy P solids and liquids) and is given by: ∆H = ∆E + (∆n) RT , where ∆n is the INCREASE in the number of S. R. u moles of the gases involved (i.e. Total number of moles of product gases less the total number of moles ( t A S of reactant gases). Y d RI a A o K wnl FACTORS AFFECTING ∆∆∆∆H OF THE REACTIONS ARE : R. o (i) Temperature (ii) Physical states of reactants & products G D (iii) Allotropic forms of elements & (iv) Pressure & volume (in case of gases) A H E U E S FR Two Types of Reactions may be distinguished : or : (i) Exothermic Reactions : For these ∆H is negative, which implies that t c e ∑H (products) < ∑H (reactants) Dir S, E (ii) Endothermic Reactions : For these ∆H is positive, which implies that S S A ∑H (products) > ∑H (reactants) L C O K E T HESS’S LAW OF CONSTANT HEAT SUMMATION : Y R According to Hess’s law (a consequence of first law), if a set of reactants is converted into a set ofT E product by more than one sequence of reactions, the total enthalpy change will be the same for everyM O sequence. HI C As such, the chemical equations can be treated ordinary algebraic expressions and can be added or I O subtracted to yield the required equation. The corresponding enthalpy changes are also manipulated inT S the same way to obtain the enthalpy change for the desired equation. 29 of 5 e APPLICATIONS OF HESS’S LAW : g a P It helps us in Calculation of : (i) Heat of formation (∆H ) of many substances which cannot be synthesised directly from their elements. f (ii) Bond energies. (iii) Enthalpy changes of slow reactions and L A (iv) Enthalpy of transformation, say from one allotropic form to the other. P O H m B s.co CIt Ois NnoVt EpoNsTsibIOle NtoA dLet VerAmLinUe EthSe OabFso MluOte LvaAluRe EofN eTntHhaAlpLy PoIf Ea Ssu :bstance and further it also depends on 81 , e 8 ss the conditions under which its determination is carried out. It is therefore necessary to choose some 8 a 5 cl standard conditions for reporting the enthalpy data. 0 o 3 k Conventionally, the enthalpy of every element in its most stable state of aggregation at 1 atm. 9 w.te (101.325 k Pa) and 298 K is assigned a zero value. 0 98 ww Based on the above convention, the relative values of “Standard molar enthalpies” (∆H°) of other 0, 0 substances are obtained and it is obvious that in terms of ∆Ho values, the enthalpy change of any 0 : f 0 e 0 sit reaction is given as : 2 b 3 we ∆H° = ∑∆Hof (products) – ∑∆Hof (reactants) 5)- m 75 sum of standard enthalpies sum of standard enthalpies 0 fro i.e. ∆H° = of formation of product  – of formation of reactants  H: ( e P g ) ka Reactions are frequently classified according to type of thermochemical purpose and the enthalpies of Sir ac K. P reactions are given different names. R. dy A balanced chemical equation which expresses the heat changes taking place in a reaction as well as the S. u physical states of various reactants and products is known as a thermochemical equation. ( t A S Y d dq ∂E ∂H RI loa Heat capacity C = dT ; CV = ∂T ; Cp = ∂T KA wn V P R. Do q = C (T2 – T1) for 1 mole; q = nC (T2 – T1) for n moles AG E ∆H2 −∆H1 ∆E2−∆E1 UH E Kirchoffs Equation: = ∆C ; = ∆C S FR T2 −T1 P T2−T1 V or : t c e Trouton’s Rule : Entropy of vaporization of non-associated or non-dissociated liquid is constant & r Di may be taken as about 87.3 J k–1mol–1. S, E S S A L C O K E T Thermochemical Equations : An equation which indicates the amount of heat change in the reaction. Y R These can be added, subtracted or multiplied whenever required. T E M The various named Enthalpies are defined as the Enthalpy change when ....... O I H Enthalpy of reaction : "quantities of substances indicated in the balanced equation react completely to C I O form the product." T S Enthalpy of formation : “one mole of the substance is formed directly from its constituent elements." 29 of Enthalpy of combustion : “one mole of the substance undergoes complete combustion” (it is always e 6 g negative) Pa Calorific Value : “it is the amount of heat given out by complete combustion of unit weight of a solid or liquid or unit volume of a gas”. Enthalpy of solution : “one mole of the substance is completely dissolved in a large excess of the given L A solvent under given conditions of temperature and pressure”. P O H m Enthalpy of neutralisation : “one gram equivalent of an acid is neutralised by one gram equivalent of B s.co a base in fairly dilute solution” . 81 , sse Enthalpy of hydration : “one mole of an anhydrous (or a partly hydrated salt) combines with the 88 cla required number of mole of water to form a specific hydrate”. 0 5 o 3 k Enthalpy of sublimation : “one mole of a solid is directly converted into its vapour at a given temperature 9 w.te below its melting point” . 0 98 ww Enthalpy of fusion : “one mole of the solid substance is completely converted into the liquid state at its 0, 0 melting point” . 0 : 0 site Enthalpy of vaporisation : “one mole of a substance is converted from the liquid state to its vapour 2 0 b state at its boiling point”. 3 we 5)- m Resonance Energy = Observed heat of formation - Calculated heat of formation 75 0 o ( fr BOND ENTHALPIES’ (BOND ENERGIES) : H: e P g The bond enthalpy of a diatomic molecule (H , C1 , O ) is equal to its dissociation energy and is defined ) ka as “the enthalpy change involved in breaking2 the b2on2d between atoms of a gaseous molecule” (Bond Sir ac K. P breaking is an endothermic process). Average bond enthalpy (energy) is the average value of bond R. dy energy obtained from molecules that contain more than one bond of that type. S. u ( t ∆H of molecules A S f Y d Av. BE = no. of bonds RI a A o K wnl BE is an additive property. R. o G D A H E U E SECOND LAW OF THERMODYNAMICS S FR The essence of first law is that all physical and chemical processes take place in such a manner that the or : total energy of the universe remain constant. t c e However, it is observed that all processes have a natural direction ,i.e. a direction in which they take r Di place spontaneously. First law fails to answer this. Another feature of the spontaneous processes is that S, they proceed only until an equilibrium is achieved. E S The direction of a spontaneous process and that it eventually reaches equilibrium, can be understood on S A the basis of entropy concept introduced through the second law of thermodynamics. L C O K E T ENTROPY AND SPONTANEITY: Y R Entropy (denoted by S) in s state function. When the state of a system changes, entropy also changes. T E q M The change of entrops ∆S is defined by, ∆S = rev , where q means that the heat is being supplied O T rev HI C “Isothermally ” and “Reversibly” (JK–1). I O T One can think entropy as a measure of the degree of randomness or disorder in a system. The greater 9 S 2 the disorder, in a system, the higher is the entropy. of 7 A useful form of 2nd law of thermodynamics is : e g a “The entropy of the universe increases in the course of every spontaneous (natural) change”. P OR “For a spontaneous process in an isolated system, the change in entropy is positive”. When a system is in equilibrium the entropy is maximum. So mathematically L A ∆S = 0 (at equilibrium) P O H m B s.co SStEaCteOmNenDts L :AW : 81 , e 8 ss(i) No cyclic engine is possible which take heat from one single source and in a 8 a 5 cl cycle completely convert it into work without producing any change in 0 o 3 k surrounding. 9 w.te(ii) Efficiency of Carnot engine working reversibly is maximum. 0 98 ww Carnot cycle 0, 0 : AB – Iso. Rev.Exp. w = – nRT ln V2 0 0 site AB 2 V1 2 0 web BC – Ada. Rev. Exp. wBC = CV (T1– T2) 5)- 3 m V  75  4  0 o CD – Iso. Rev. Comp. w = – nRT ln   ( fr CD 1 V3  H: e P g DA – Iso. Rev. Comp. w = C (T – T ) ) ka DA V 2 1 Sir Pac Carnot efficiency η =−wqTotal = T1T−T2 = q1q+q2 R. K. dy 2 2 2 S. u ( t CARNOT CYCLE : A S Y oad q1 +q2 = 0 for rev. cycle KARI wnl T1 T2 R. o G D q q A E Irreversible engine T1 + T2 < 0 UH E 1 2 S FR ∫ qrev = 0 ⇒ qrev is a state function. tor : c T T e r Di ∆S = ∫dqrev S, E T S S Also ∆S + ∆S = 0 for rev. process A syt surr L ∆S + ∆S > 0 for irrev. process C syt surr O ⇒ ∆S + ∆S ≥ 0 ( In general ) K syt surr E T Y ENTROPY CHANGE (General Expression ): R T E T V M 2 2 ∆S = nC ln + nR ln O V T V I 1 1 H C Change in state function for various processes. I O Reversible irreversible isothermal expansion and contraction : (ideal gas ) ST 9 2 ∆E = 0; ∆H = 0; ∆S = nR ln VV2 e 8 of 1 ag P Isobaric heating or cooling : ∆E = C ∆T V ∆H = C ∆T = q P P L T  A  2  P ∆S = nC ln  O P T  H m 1 B classes.co Isochori∆∆c EHhe =a=t iCCngVP o∆∆rTT c o=o qliVng : 0 58881 , w.teko ∆S = nCV lnTT12  0 9893 ww Adiabatic process : 0, 0 ∆E = C ∆T 0 : V 0 e ∆H = C ∆T 0 sit P 2 b T V 3 we ∆S = nC ln 2 + nR ln 2 for irreversible process 5)- m V T1 V1 75 0 o ∆S = 0 for reversible adiabatic compression and expansion. ( e fr Gibb's function : PH: kag G = H – TS Sir) ac at constant T and pressure K. P ∆G = ∆H – T∆S R. dy S. tu ∆G = (∆H – T∆S) ≤ 0for rev. process. A ( S Y d (–∆G)T, P = work done by system max. non P – V RI a A o ∆G K l wn T = – (∆Ssyst + ∆Ssurr..) R. o G D ∆G < 0 for spontaneous process A H E ∆G = 0 for equilibrium. U E S FR or : t c e r Di S, E S S A L C O K E T GIBBS FREE ENERGY (G) AND SPONTANEITY: Y R A new thermodynamic (state) function G , the Gibbs free energy is defined as : T E G = H – TS or ∆G = ∆H – T ∆S (at constant temperature and pressure) M O For a spontaneous reaction ∆G must be negative. The use of Gibbs free energy has the advantage that HI C it refers to the system only (and not surroundings also as in entropy). I O To summaries, the spontaneity of a chemical reaction is decided by two factors taken together: T S (i) the enthalpy factor and (ii) the entropy factor. 29 of The equation ∆G = ∆H – T ∆S takes both the factors into consideration. 9 e The most favorable situation for a negative value of ∆G is a negative value of ∆H and a positive value ag P of ∆S. However a large negative value of ∆H may outweigh an unfavorable ∆S value and a large value of ∆S may outweigh an unfavorable value of ∆H. STANDARD FREE ENERGY CHANGE (∆∆∆∆G°) : L A The standard free energy change ∆G° is defined as the free energy change for a process at a specified P O m temperature in which the reactants in their standard state are converted to the products in their standard BH s.co sLtiaktee .t hIte i ss tdaenndoatredd e bnyth ∆aGlp°y. of formation of an element “the standard free energy of formation of an 81 , e 8 ss element in its standard state is zero”. And so ; 8 a 5 ocl ∆Go =∑∆Go (products) – ∑∆Go (reactants) 30 k r f f 9 w.te The standard free energy change. ∆G° is related to the equilibrium constant keq by the relation; 0 98 ww It can be∆ sGho°w =n – t h2a.t3 f0re3e R eTne lroggy kcehqa.nge for a process is equal to the maximum possible work that can be 0, 0 : derived from the process i.e. 0 0 e 0 sit ∆G° = Wmax (for a reversible change at constant pressure and temperature) 2 web ∆InG c a=s e– onfF aE galv ,a wnihce creel l E, free = e en.emrg.fy. cohfa tnhgee c, e∆llG ; Fis =re Flaatreadd taoy tchoen eslteacntrtiacnadl work done in the cell. 5)- 3 m cell cell 75 n = number of electrons being transferred in the chemical process 0 o ( ge fr So ∆G = – nFEocell, where Eocell is the standard cell potential. ) PH: dy Packa Clausius Claperyon’s Equation : log pp12 = 2.3∆0H3R T11 −T12 (For liquid ⇔ gas equilibrium) S. R. K. Sir u p & p are vapour pressure at T & T ( t 1 2 I 2 A S Y d RI a THIRD LAW OF THERMODYNAMICS A o K ownl “aAbsto albusteo lzuetreo z eervoe,r yth cer eynsttarollpinye o sfo ali pde irsf ienc tal ys tcartyes otafl lpineerf seucbt sotradnecre a ins dta iktse enn atsr ozperyo s”h, owuhldic bhe m zeearnos. that at G R. D By virtue of the third law, the absolute value of entropy (unlike absolute value of enthalpy) for any pure A H E substance can be calculated at room temperature. U E S FR or : The standard absolute entropy of, a substance” So, is the entropy of the substance in its standard at t c 298K and 1 atm. e r Di Absolute entropies of various substances have been tabulated and these value are used to calculate S, entropy changes for the reactions by the formula; E S ∆S° = ∑S° (products) – ∑S° (reactants) AS L C O K E T EXERCISE-I Y R T E First law : Heat (q), work (w) and ∆∆∆∆U, ∆∆∆∆H M O I H Q.1 In which of the following changes at constant pressure is work done by system on surrounding? By the C I O surrounding on system? T S Initial state Final state 9 2 (i) H O (g) → H O (l) of 2 2 0 (ii) H2O (s) → H2O (g) ge 1 (iii) H O (l) → H O (s) a P 2 2 (iv) 2H (g) + N (g)→ 2NH (g) 2 2 3 (v) CaCO (s) → CaO (s) + CO (g) 3 2 Q.2 The gas is cooled and loses 65 J of heat. The gas contracts as it cools and work done on the system L A P equal to 20 J is exchanged with the surroundings. What are q, w and ∆E ? O H m B s.coQ.3 ∆THhe =e n–t h0a.3lp1y K chJ.a Wnghea fto irs tthhee r∆eaEc.tion of 50 ml of ethylene with 50.0 ml of H2 at 1.5 atm pressure is 81 , e 8 ssQ.4 The enthalpy of combustion of glucose is – 2808 KJmol–1 at 25°C. How many grams of glucose do you 8 a 5 cl need to consume [Assume wt = 62.5 Kg]. 0 o 3 w.tek((ab)) ttoo cclliimmbb aa fmligohutn otaf isnt aoirf sa rltisitiungde t h3r0o0u0g hM 3?M. 0 989 ww Assume that 25% of enthalpy can be converted to useful work. 0, 0 : Q.5 What is ∆E when 2.0 mole of liquid water vaporises at 100°C ? The heat of vaporisation , ∆H vap. of 0 0 site water at 100°C is 40.66 KJmol–1. 2 0 b 3 weQ.6 If 1.0 k cal of heat is added to 1.2 L of O2 in a cylinder of constant pressure of 1 atm, the volume 5)- m increases to 1.5 L. Calculate ∆E and ∆H of the process. 75 0 o ( frQ.7 When the following reaction was carried out in a bomb calorimeter, ∆E is found to be – 742.7 kJ/mol of H: e NH CN (s) at 298 K. P kag 2 3 Sir) ac NH CN (s) + O (g) → N (g) + CO (g) + H O (l) K. P 2 2 2 2 2 2 R. dy Calculate ∆H298 for the reaction. S. u ( t A SQ.8 When 1 mole of ice melt at 0°C and at constant pressure of 1 atm. 1440 calories of heat are absorbed Y ad by the system. The molar volumes of ice and water are 0.0196 and 0.0180 litre respectively. Calculate ARI lo ∆H and ∆E for the reaction. K wn R. oQ.9 Water expands when it freezes. Determine amount of work in joules, done when a system consisting of G D A 1.0 L of liquid water freezes under a constant pressure of 1.0 atm and forms 1.1 L of ice. H E U E S FRQ.10 eLnimereg yis w mhaedne 1 c.o0m0 mmeorlec ioalfl yso blyid d CecaoCmOp o(sVit i=o n3 4o.f2 l immle) satbosnoer bCsa 1C7O73.9. WkJh oaft hise atht ea ncdh adnegceo imn pinotseersn aatl or : t 3 c 25°C against a pressure of 1.0 atm to give solid CaO. (Volume = 16.9 ml) and CO (g) (V = 24.4 L). e 2 Dir Q.11 One mole of solid Zn is placed in excess of dilute H2SO4 at 27 °C in a cylinder fitted with a piston. Find S, E the value of ∆E, q and w for the process if the area of piston is 500 cm2 and it moves out by 50 cm S S against a pressure of 1 atm during the reaction. The heat given to surrounding is 36.5 KJ. A L Zn(s) + 2H+ (aq) l Zn2+ (aq) + H (g) C 2 O K Q.12 Two mole of ideal diatomic gas (C = 5/2 R) at 300 K and 5 atm expanded irreversly & adiabatically E V,m T to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w, ∆H & ∆V.