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# Physical Chemistry I CHM 3410

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This 55 page Class Notes was uploaded by Angelo Cassin on Monday October 12, 2015. The Class Notes belongs to CHM 3410 at Florida Atlantic University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/221666/chm-3410-florida-atlantic-university in Chemistry at Florida Atlantic University.

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Date Created: 10/12/15

Standard Enthalpies of Formation Enthalpy changes for reactions can be used to assign enthalpies values to substances and to the ions of aqueous solutions Standard Enthalpies of Compounds from the Formation Reactions The standard enthalpy of formation of any substance is de ned as the enthalpy of that substance compared to the enthalpies of the elements from which it is formed Thus the enthalpy change for the reaction in which a compound formed from its elements is therefore the standard enthalpy of formation of that compound For eg C graphite 02 g quot9 CO2 g The measured value for AH ie the enthalpy change at 25 C is 393 51 kJ mol39l This value gives us the enthalpy of 1 mol of CO2 g compared with the enthalpy ofl mol ofC graphite and 1 mol ofO2 g Thus the standard enthalpy of formation of CO2 g at 250C is 393 51 kJ mol39l The definition of the standard enthalpy of formation gives a value of zero for the standard enthalpy of formation of each of the elements in its stable or designated form ie For eg C graphite 02 g quot9 CO2 g AHOf C 0 AHOfOZ 0 AHOfCOZ 39351 kJ mol391 The standard enthalpy of formation is represented as AHOf Where the degree sign denotes the standard conditions ie Pressure 1 bar and Temperature 25 C The subscript f stands for formation Standard Enthalpies of Formation from Other Chemical Reactions Reactions other than those in which the elements give the compounds can be used to deduce the standard enthalpy of formation of a compound Problem 1 Calculate the standard enthalpy of formation of CH3OH l at 25 C from the AH value ofl28l2 kJmol391 that would be obtained for the reaction in which 1 mol of CH3OH is formed from hydrogen and carbon monoxide The standard enthalpy of formation of carbon monoxide is 11052 kJmol39l 2H2 g co g quot9 CH3OH 1 AHOf0 AH0f 11052 AH0f7 The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHO 12812 kJmol391 therefore AHO AHOf products AHOf reactants 12812 kJmol39l AH0fCH3OH 11052 kJmol39l AHOfCH3OH 23864kJmol391 Problem 2 A rocket engine which produces benign products depends on the oxidation of hydrazine by nitrogen dioxide The equation for the reaction is 2 N2H4 g t 2 N02 g quot9 3 N2 g t 4 H20 g If the standard enthalpies of formation AHOf for N2H4 N02 and H20 are given as 9535 kJmol39l 3310 kJmol39l 24l83 kJmol391 respectively Calculate the change in enthalpy for gaseous reactants AHO 2 N2H4 g 2 N02 g 9 3 N2 g 4 H20 g AHOf 29535 AH0f 23310 AHOf 0 AHOf 4 24183 The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHOf products 0 4 24183 kJmol39l 96732 kJmol39l AHOfreactants 2 9535 kJmol39l 2 3310 kJmol39l 19070 kJmol391 6620 kJmol391 25690 kJmol39l therefore AHO AHOf products AHOf reactants 96732 kJmol391 25690 kJmol39l 122422 kJmol391 Standard Enthalpies of Formation of Ions in Aqueous Solutions Many chemical reactions involve ions in aqueous solutions Thus the data for standard enthalpies for the formation of pure substances can be used to nd the enthalpy change for any reaction involving pure substances or aqueous ionic species For eg A chemical process involving hydrogen gas dissolved in a large amount of water Since HCl is strong electrolyte the solution contains hydrated H ie H3O and Cl39 ions No HCl molecules are present The reaction can be then represented as follows H20 HCl g quot9 H aq Cl aq The symbol aq implies that the ion is present in large amounts of water The process of dissolving hydrogen chloride in water is very exothermic If the reactions were carried out at a pressure of 1 bar and at 25 C the thermal surroundings will gain 7585 Id of energy for each mole of HC1 that is dissolved Thus the enthalpy change for the above reaction is AHO 7585 kJmol391 The standard enthalpy of formation of HC1 g is 9231 kJmol39l The standard enthalpy of formation for the reaction can be written as H20 HC1 g quota H aq C1quot aq AHOf 9231 AH0fH aq AHOf Cl39 aq The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHOf products AHOf H aq AHOf Cl39 aq and AHOfreactant AHOfHCl g 9231kJmol391 therefore we can write AHO AHOf products AHOf reactants 7585 kJmol391 AHOfH aq AHOfCl39 aq 923l kJmol39l AHOfH aq AHOfCl39 aq l67l6 kJmol391 This gives a value of the standard enthalpy of formation of the pair of ions H and Cl39 in aqueous solution The value for the standard enthalpy of formation of Hl ions in dilute solution is set to be equal to zero Hence AHOftH aq 0 Thus we can now calculate the value for the standard enthalpy of formation of Cl39 ions AHOfH aq AHOfCl39 aq l67l6 kJmol391 AHOfCl39 aq l67l6 kJmol391 AHOfH aq AHOfCl39aq l67l6 kJmol391 0 AHOfCl39aq l67l6 kJmol391 Problem 3 Dissolving 1 mol ofKCl in a large amount of water at 25 C produces an enthalpy change of 1723 kJmol39l The standard enthalpy of formation of KCl s is 43675 kJmol39l The standard enthalpy of Cl39 aq is l67l6 kJmol39l Calculate the standard enthalpy of formation of the aqueous potassium Kl ion in the aqueous solution Thus the enthalpy change for the reaction is AHO 1723 kJmol391 The standard enthalpy1fformation for the reaction can be written as KCl 5 mine K aq Cl39 aq AHOf 43675 AH0fK aq AHOf 167 16 The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHOf products AHOf K aq AHOf Cl39 aq AHOf K aq l67 l6 kJmol39l and AHOfreactant AHOfKCl s 43675 kJmol391 therefore we can write AHO AHOf products AHOf reactants 1723 kJmol39l AH0fK aq 16716 kJmol39l 43675 kJmol39l AHOfK aq 25256 kJmol391 Problem 4 Calculate the enthalpy change for the reaction in which the addition of carbon dioxide to an aqueous solution containing Ca2 ions brings about the precipitation of calcium carbonate Use the standard enthalpy of formation for Ca2 aq CO2 g CaCO3 s and H20 1 as 64283 kJmol39l 39351 kJmol39l 120692 kJmol39l 28583 kJmol391 respectively The reaction can be written as Ca2 211 C02 g H20 1 quot9 CaC03 S H 211 AHOf 64283 39351 28583 kJmol 120692 0 AHO AHOf products AHOf reactants 120692 kJmoll 422217 kJmoll 1525 kJmol391 Distribution Over Quantum States Using the kineticmolecular derivation of PV nRT we saw AVg translational energies per molecule kT 2 and AVg translational energies per mole RT 2 In order to understand how the molecules are distributed in each allowed energy states we must first look at their probability of distribution Let us consider an example of throwing marble randomly in to a box with equal sized compartment Let us consider the fourmarble two compartment distribution as shown below Refer Figure 213 The number of ways W of distribution of these marbles is given as W 4 l234 The same analogy can be extended to the distribution of molecules through out the set of quantum states W This gives an expression with which we can calculate the number of ways a particular distribution can be achieved This number is proportional to the probability of distribution Qualitative Consideration Let us find out the values of W for various distributions subject to following constraints 1 The total number of particles is fixed 2 The system of the particles has some fixed total energy eg The most probable distribution of 14 particles with an average energy as 2 units Since N and Ni values are small we can calculate the value of W The largest value for W is obtained when the particles are spread out as much as possible Refer figure 214 e g The most probable distribution of 14 particles with varying average energy of the system All distributions have same general shape Lower the average energy the more particles are crowded in lower energy state Higher the average energy more particles are spread out into higher energy level Refer Figure 215 Boltzmann Distribution This a mathematical expression for a large number of molecules which gives the most probable distribution The Boltzmann distribution gives the number if particles per state at one energy 8i compared to that of another energy 81 as 818 kT Ni e J More conveniently we compare the population of the states at one energy level with the population of the sates of lowest energy level Then we get 1i N0 6SiSokT 1 Thus according to Boltzmann distribution The population depends upon the ratio of 8i 80 to kT The population of ith state is always less than that of the lowestenergy state At low temperature The eXponential factor makes population decrease rapidly with increasing energy and particles are crowded into the states of lower energy At high temperature The eXponential factor falloff is less rapid and the particles can spread out into the higherenergy states Problem 1 Use the Boltzmann distribution expression to calculate the ratio of the number of molecules per quantum state at an energy equal to l2kT the average energy for onedimensional motion to the number per quantum state at the lowest allowed energy N No 26SiSo kT 1 8 12 kT 80 o N N0 e4121ltT01ltT 1 Ni N0 e412 Ni N0 O606 In most cases we are given the total number of particle ie N The Boltzmann distribution relates the number of particles Ni in the quantum states at any energy to the number N0 in the states at the lowest energy Ni N0 e 48180 Ni N0 e 48180 The total number of particles is given as N 2 Ni N Z N0 6 481 80kT N No 2 e ver 1 N No q 2 Where q is the partition function and its given as q Z 6 481 80kT 3 2 e 81 80kT 2 e 82 80kT 2 e 783 80kT N0Nq 4 This relation gives us how many molecules are in the lowestenergy state if we know the total number of molecules in a given sample Onedimensional Translational Motion The Boltzmann distrubution can be used to find the distribution of the molecules of a gas over the allowed energies and speeds Let us consider the molecule of a gas with mass m moves along in X direction and is confined to a linesegment of length a Then its translational energy is given as Snx nX2 h2 8 m a2 where nX 123 The energy difference is given as Snx Sno nX2 h2 8 m a2 n02 h2 8 m a2 nX2 hz8 m a2 12 h2 8 m a2 nX2 1h28ma2 Since most of the nX values are very much larger than 1 and thus le2 1 E nxz 8nX Sno E nX2 h2 8 m a2 5 The partition function q from equation 3 can be written as 8 8 15ka no1ltT Summation can be replaced by integration since summation is over so many closely spaced energy levels Integration can be evaluated to give l n 12 qx 2 112 8ma2 6 Thus the partition function q for a particle of a gas with mass m moving in Xdirection confined to a linesegment of length a at any temperature can be calculated From equation 4 we can write NoNqx lt7gt Thus now we can calculate the number of molecules present in the lowest energy state Problem 2 Calculate the value for qx and the value for N0 for 1mol sample of nitrogen gas in 1L container at 25 C Mass of one N2 molecule m 0028 kg 602 x 1023 465 x 103926 kg Assuming the container to be cubic the length of container is a 10 cm 01 m k 138066 x 103923 Jk391 T 25 27315 29815 K h 6626 x 103934 J s l n 12 qX 523 X 109 2 h2 8ma2 and N0 N qX With a total 602 x 1023 molecuels in this 1mol sample N0 602 X1023523 X 109 115 gtlt1014 Distribution of Quantum States The number of molecules per unit I lX interval can be expressed as dN dI lX Then The fraction of the molecules per unit I lX interval as dN an N Then The Boltzmann distribution gives the number of molecules in a unit quantumnumber interval as Nn dN 111 N0 e SnxSOVkT The fraction of the molecules per unit quantum state can be expressed as dN an N N0 N e SnxkaT 8 For onedimensional translational motion from equation 7 N N0 qx And from equation 5 8nX Sno 11X2 h2 8 m a2 Then Fraction per unit nX interval dN an N N auxSQkT 1 qx e SmSo kT 2 h2 8ma2 2 ln kT lt9gt Problem 3 Use the equation for fraction per unit nx interval to calculate he number of molecules per unit quantumnumber interval at nx 0 for he system of 1mol sample of nitrogen in a 1L container at 25 C Mass of one N2 molecule m 0028 kg 602 x 1023 465 x 103926 kg Assuming the container to be cubic the length of container is a 10 cm 01 m k 138066 x 103923 Jk391 T 25 27315 29815 K h 6626 x 103934 J s Thus the fraction per unit interval at nx0 is l dN 2 hZ8ma2 2 N an HFO l 7 kT 191 gtlt103910 This is the fraction of the molecules that are in the first quantumnumber unit interval The number of molecules in a lmol sample that are in this interval is obtained by multiplying this fraction by Avogadro s number Thenweget l dN NAgtlt NAxl9lgtltlO3910 N dnx nx0 602 X 1023 191 X 1010 115 X 1014 Distribution Over Energies To obtain the distribution molecules over the allowed energies we convert the derivative dNan which gives the number of molecules per unit interval nx to dNdSX which gives the number of molecules per unit energy interval 8x We can write dNdsx dNdnx andex We have ex 11X2 h2 8 m a2 nX2 8X 8 m a2 h2 nX 5 2J2Jm ah nX 2ah 2m8X andsX ah 2 mSX 10 Now the fraction per unit energy interval 8x on be written as Fraction per unit energy interval lN dNdsx lN dNdnx andsx From equation 9 and 10 1 1 esX kn m Distribution over Speeds We know 8X 12 m UX2 Differentiate with respect ox We get dSX d1X 12 In duxz doX m1X Wax 1 1 Using equation 11 and 12 we can write Fraction per unit speed interval lN dNdex e12mux2 kT Fall 2001 Dr Naneep B Malkar CHM 3410 Exam 11 Name Soc Sec N0 Closed Notes Closed Book Calculator Time 50 min 101901 h 662 gtlt103934 J S c 299 gtlt108 ms391 R 8314 J K391 mol39l Solve any four questions 25 points each 1 Use the expression of LZ h 2 11 i d do for the 2 component of the angular momentum to nd the expression for the angular momentum for a particle on a ring described by the wave function G A elm with m 3 and m 3 Fall 2001 Dr Naneep B Malkar 2 Write position operator and momentum operator for one dimensional system for deducing physical properties and derive an operator for kinetic energy using a momentum operator Fall 2001 Dr Naneep B Malkar 3 If the total energy of N molecules is given as 8X 00 N Total Energy quotTX esxkT dSX nkT 8X 0 Show that using y 8X it is equal to 12 N kT Fall 2001 Dr Naneep B Malkar 4 Calculate the values of most probable speed at and root mean square speed 42 for carbon dioxide at 25 C Fall 2001 Dr Naneep B Malkar 5 Calculate qms for He molecules at 1 bar of pressure and at 25 C Fall 2001 Dr Naneep B Malkar 6 What is the thermal energy of lmol of rigid ie nonVibrating diatomic molecules of gas at 298 K Fall 2001 Dr Naneep B Malkar 7 Explain and Derive an expression 1 10 6km for BeerLambert Law Fall 2001 Dr Naneep B Malkar 8 The infrared absorption spectrum of HCl gas shows an absorption band centered at 2885 cm39l Calculate the force constant of the bond in the HCl molecule Standard Enthalpies of Formation Enthalpy changes for reactions can be used to assign enthalpies values to substances and to the ions of aqueous solutions Standard Enthalpies of Compounds from the Formation Reactions The standard enthalpy of formation of any substance is de ned as the enthalpy of that substance compared to the enthalpies of the elements from which it is formed Thus the enthalpy change for the reaction in which a compound formed from its elements is therefore the standard enthalpy of formation of that compound For eg C graphite 02 g quot9 CO2 g The measured value for AH ie the enthalpy change at 25 C is 393 51 kJ mol39l This value gives us the enthalpy of 1 mol of CO2 g compared with the enthalpy ofl mol ofC graphite and 1 mol ofO2 g Thus the standard enthalpy of formation of CO2 g at 250C is 393 51 kJ mol39l The definition of the standard enthalpy of formation gives a value of zero for the standard enthalpy of formation of each of the elements in its stable or designated form ie For eg C graphite 02 g quot9 CO2 g AHOf C 0 AHOfOZ 0 AHOfCOZ 39351 kJ mol391 The standard enthalpy of formation is represented as AHOf Where the degree sign denotes the standard conditions ie Pressure 1 bar and Temperature 25 C The subscript f stands for formation Standard Enthalpies of Formation from Other Chemical Reactions Reactions other than those in which the elements give the compounds can be used to deduce the standard enthalpy of formation of a compound Problem 1 Calculate the standard enthalpy of formation of CH3OH l at 25 C from the AH value ofl28l2 kJmol391 that would be obtained for the reaction in which 1 mol of CH3OH is formed from hydrogen and carbon monoxide The standard enthalpy of formation of carbon monoxide is 11052 kJmol39l 2H2 g co g quot9 CH3OH 1 AHOf0 AH0f 11052 AH0f7 The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHO 12812 kJmol391 therefore AHO AHOf products AHOf reactants 12812 kJmol39l AH0fCH3OH 11052 kJmol39l AHOfCH3OH 23864kJmol391 Problem 2 A rocket engine which produces benign products depends on the oxidation of hydrazine by nitrogen dioxide The equation for the reaction is 2 N2H4 g t 2 N02 g quot9 3 N2 g t 4 H20 g If the standard enthalpies of formation AHOf for N2H4 N02 and H20 are given as 9535 kJmol39l 3310 kJmol39l 24l83 kJmol391 respectively Calculate the change in enthalpy for gaseous reactants AHO 2 N2H4 g 2 N02 g 9 3 N2 g 4 H20 g AHOf 29535 AH0f 23310 AHOf 0 AHOf 4 24183 The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHOf products 0 4 24183 kJmol39l 96732 kJmol39l AHOfreactants 2 9535 kJmol39l 2 3310 kJmol39l 19070 kJmol391 6620 kJmol391 25690 kJmol39l therefore AHO AHOf products AHOf reactants 96732 kJmol391 25690 kJmol39l 122422 kJmol391 Standard Enthalpies of Formation of Ions in Aqueous Solutions Many chemical reactions involve ions in aqueous solutions Thus the data for standard enthalpies for the formation of pure substances can be used to nd the enthalpy change for any reaction involving pure substances or aqueous ionic species For eg A chemical process involving hydrogen gas dissolved in a large amount of water Since HCl is strong electrolyte the solution contains hydrated H ie H3O and Cl39 ions No HCl molecules are present The reaction can be then represented as follows H20 HCl g quot9 H aq Cl aq The symbol aq implies that the ion is present in large amounts of water The process of dissolving hydrogen chloride in water is very exothermic If the reactions were carried out at a pressure of 1 bar and at 25 C the thermal surroundings will gain 7585 Id of energy for each mole of HC1 that is dissolved Thus the enthalpy change for the above reaction is AHO 7585 kJmol391 The standard enthalpy of formation of HC1 g is 9231 kJmol39l The standard enthalpy of formation for the reaction can be written as H20 HC1 g quota H aq C1quot aq AHOf 9231 AH0fH aq AHOf Cl39 aq The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHOf products AHOf H aq AHOf Cl39 aq and AHOfreactant AHOfHCl g 9231kJmol391 therefore we can write AHO AHOf products AHOf reactants 7585 kJmol391 AHOfH aq AHOfCl39 aq 923l kJmol39l AHOfH aq AHOfCl39 aq l67l6 kJmol391 This gives a value of the standard enthalpy of formation of the pair of ions H and Cl39 in aqueous solution The value for the standard enthalpy of formation of Hl ions in dilute solution is set to be equal to zero Hence AHOftH aq 0 Thus we can now calculate the value for the standard enthalpy of formation of Cl39 ions AHOfH aq AHOfCl39 aq l67l6 kJmol391 AHOfCl39 aq l67l6 kJmol391 AHOfH aq AHOfCl39aq l67l6 kJmol391 0 AHOfCl39aq l67l6 kJmol391 Problem 3 Dissolving 1 mol ofKCl in a large amount of water at 25 C produces an enthalpy change of 1723 kJmol39l The standard enthalpy of formation of KCl s is 43675 kJmol39l The standard enthalpy of Cl39 aq is l67l6 kJmol39l Calculate the standard enthalpy of formation of the aqueous potassium Kl ion in the aqueous solution Thus the enthalpy change for the reaction is AHO 1723 kJmol391 The standard enthalpy1fformation for the reaction can be written as KCl 5 mine K aq Cl39 aq AHOf 43675 AH0fK aq AHOf 167 16 The change in the enthalpy for the above reaction is AHO AHOf products AHOf reactants but AHOf products AHOf K aq AHOf Cl39 aq AHOf K aq l67 l6 kJmol39l and AHOfreactant AHOfKCl s 43675 kJmol391 therefore we can write AHO AHOf products AHOf reactants 1723 kJmol39l AH0fK aq 16716 kJmol39l 43675 kJmol39l AHOfK aq 25256 kJmol391 Problem 4 Calculate the enthalpy change for the reaction in which the addition of carbon dioxide to an aqueous solution containing Ca2 ions brings about the precipitation of calcium carbonate Use the standard enthalpy of formation for Ca2 aq CO2 g CaCO3 s and H20 1 as 64283 kJmol39l 39351 kJmol39l 120692 kJmol39l 28583 kJmol391 respectively The reaction can be written as Ca2 211 C02 g H20 1 quot9 CaC03 S H 211 AHOf 64283 39351 28583 kJmol 120692 0 AHO AHOf products AHOf reactants 120692 kJmoll 422217 kJmoll 1525 kJmol391 Fall 2001 Dr Naneep B Malkar CHM 3410 Exam 11 Name Soc Sec N0 Closed Notes Closed Book Calculator Time 50 min 101901 h 662 gtlt103934 J S c 299 gtlt108 ms391 R 8314 J K391 mol39l Solve any four questions 25 points each 1 Use the expression of LZ h 2 11 i d do for the 2 component of the angular momentum to nd the expression for the angular momentum for a particle on a ring described by the wave function G A elm with m 3 and m 3 Fall 2001 Dr Naneep B Malkar 2 Write position operator and momentum operator for one dimensional system for deducing physical properties and derive an operator for kinetic energy using a momentum operator Fall 2001 Dr Naneep B Malkar 3 If the total energy of N molecules is given as 8X 00 N Total Energy quotTX esxkT dSX nkT 8X 0 Show that using y 8X it is equal to 12 N kT Fall 2001 Dr Naneep B Malkar 4 Calculate the values of most probable speed at and root mean square speed 42 for carbon dioxide at 25 C Fall 2001 Dr Naneep B Malkar 5 Calculate qms for He molecules at 1 bar of pressure and at 25 C Fall 2001 Dr Naneep B Malkar 6 What is the thermal energy of lmol of rigid ie nonVibrating diatomic molecules of gas at 298 K Fall 2001 Dr Naneep B Malkar 7 Explain and Derive an expression 1 10 6km for BeerLambert Law Fall 2001 Dr Naneep B Malkar 8 The infrared absorption spectrum of HCl gas shows an absorption band centered at 2885 cm39l Calculate the force constant of the bond in the HCl molecule Problem 1 When Zinc is added to an acid solution such as hydrochloric acid the zinc reacts and hydrogen gas is evolved The reaction can be described by the equation Zn s 2H aq quot9 Zn2 aq H2 g The solution becomes warm showing that the reaction evolves heat ie it is an exothermic reaction Reconcile these experimental results given below and in so doing calculate values for the change in the intemal energy and the enthalpy of the system a If the reaction were carried out in a constantpressure calorimeter as in a simple Styrofoam cup experiment it would be found that the thermal surroundings gain 15389 kJ per mol of zinc that reacts blf the reaction were carried out in a constantvolume quotbomb calorimeterquot it would be found that the thermal surroundings gain 15637 kJ per mole of zinc reacted a For constantpressure processes AH AUtherm 15389 kJmol391 15389 kJmol39l Since AH AU Avg RT We can write AU AH Avg RT 15389 k mol39l 1 8314 JK39lmol39l 298 K 15389 k mol39l 1 8314 JKlmol1298 K 15389 k mol39l 2477572 Jmoll 15389 kJ mol39l 248 kJmol39l 15637 kJ mol391 b For constantvolume processes AU AUtherm 15637 kJmol391 15637 kJmol391 Since AH AU Avg RT 15637 k mol391 1 8314 JKlmoll 298 K 15637 k mol391 1 8314 JKlmol1298 K 15637 k mol39l 2477572 Jmoll 15637 kJ mol391248 kJmol39l 15389 kJ mol391 Problem 2 The cylinder containing 1 mol of liquid water at 100 C is heated until the liquid is converted to vapor Th cylinder is fitted with a piston which applies a pressure of 1 bar a How much energy is transferred to mechanical surroundings b If the enthalpy of vaporization of water is 40670 J what is the change in internal energy a AUmech Pext AV Pext V2 Paxt nRT P nRT 1 mol 8314 J mol391 11 373 K 310112 J b AH AU AU h AU AH AUmh 40670 J 3101 J 3756888 J 3757 k Enthalpy and Chemical reactions 1 The energy change that accompanies a chemical reaction depends upon whether each of the reagents is solid liquid or gas This is specified by adding 5 l or g after the formula 2 Energies are written based on the molar amounts of the reagents and not on 1 mol of any one of the reagents Reaction in which 1 mol of H20 is formed H2g 1202 g quot9 H20 1 AH 28583 kJ mol391 and Reaction in which 2 mole of H20 is formed 2H2 9 02 9 quot399 2H20 1 AH 57l66 kJ mol391 3 Combustion reactions are written as the combustion for 1 mol of the reactants C6H6 g 15202 g quot399 6C02 9 3H20 1 AH 330l48 kJ mol391 4 If reaction is carried out at standard pressure 1 bar and a temperature of 250C ie 29815 K These conditions are indicated by a subscript 298 suggesting a temperature value and a degree sign implying standard pressure H2g 1202 g quot9 H20 1 AHquot298 28583 kJ mol391 Indirect determination of Enthalpy change in chemical reactions Many reactions are not suitable for direct calorimetric study In such cases the internal energy change or enthalpy change can be obtained by an indirect method This procedure was developed by a scientist Hess in 1840 and hence called Hess Law of Heat Summation This is based on the First Law of Thermodynamics ie energy is a property and that changes in the energy depend only on the initial and final states and not on the path between these states For e g conversion of carbon in graphite fOI H l to its diamond form C graphite quotquot9 C diamond Since this reaction can be made to occur it is not suitable for direct calorimetric measurement But the combustion of both graphite and diamond can be carried out in a calorimeter These reactions and their enthalpy changes for the combustion reactions are written as C graphite 02 g quot9 CO2 g AH 298 39351 kJ mol391 C diamond 02 g quot9 CO2 g AH 298 39540 kJ mol391 The graphical display for the enthalpy change for both these reactions is shown as follows Refer to Figure 34 On the basis of this diagram we can write C graphite quot9 C diamond AH0298 189 kJ mol391 The another way of obtaining the result is by calcualtion 1 Addition The second equation is turn around and equations are added and common products are cancelled C graphite ltggt quot9 2 g AH 298 39351 kJ mol391 ko2ltggt quot9 C diamond 2 g AH 298 39540 k mol39l C graphite quot9 C diamond AH 298 39351 kJ mol391 39540 k mol39l 189 kJ mol391 2 Substraction The second equation is subtracted from first equation C graphite 02 g quot9 CO2 g AH 298 39351 kJ mol391 C diamond 02 g quot9 CO2 g AH 298 39540 kJ mol391 C graphite quot9 C diamond AHO298 39351 kJ mol391 39540 k mol39l 189 kJ mol391 Problem 3 Calculate the enthalpy change for the formation of methane from its elements according to the equation C graphite 2H2 g quot9 CH4 g The values for enthalpy changes for combustion of Cgraphite H2g and CH4 g in units of kilojoules per mole are 69351 28584 and 89035 The combustion reactions for Cgraphite H2g and CH4 g are written as follows A C graphite 02 g quot9 CO2 g AH 298 39351 kJ mol391 b H2g 1202 g quot9 H20 1 AHquot298 28583 kJ mol1 0 CH4 g 202 g quot9 CO2 g 2H20 1 AH 298 89035 kJ mol1 Then a 2b 01 C graphite 2H2 g quot9 CH4 g AH 298 7484 kJ mol1

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