1 Exercise 2.19a pg 86

In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps. Numbers without underlines (including final answers) are shown with the proper number of sig figs. 1 Exercise 2.19a pg 86 Given The standard enthalpy of combustion of cyclopropane is 2091 kj mol 1 at 25 C. formation of its isomer propene is +20.42 kj mol 1. Additionally, the enthalpy of In terms of given variables, this is written: r H = 2091 kj mol 1 cyclopropane combustion f H = +20.42 kj mol 1 propene enthalpy of formation Find Calculate... (a) The enthalpy of formation of cyclopropane (b) The enthalpy of isomerization from cyclopropane to propene Strategy We begin by constructing the balanced chemical equation for combustion of cyclopropane 2C 3 H 6 + 9O 2 (g) 6H 2 O(l) + 6CO 2 (g) r H = 2 2091 kj mol 1 For any such reaction the reaction enthalpy can related to the enthalpy of products and reactants by (Equation 2.32 on pg 68 in your text book) which for this reaction is r H = Products ν p f H p Reactants ν r f H r r H = 6 f H (H 2 O(l)) + 6 f H (CO 2 (g)) 2 f H (C 3 H 6 ) + 9 f H (O 2 (g)) } {{ } } {{ } products reactants Using the fact that the standard state of oxygen is O 2 (g) ( f H (O 2 (g)) = 0) the following expression for f H (C 3 H 6 ) is obtained through rearrangement 1

f H (C 3 H 6 ) = 3 f H (H 2 O(l)) + 3 f H (CO 2 (g)) 1 2 rh = 3 285.83 kj mol 1 + 3 393.51 kj mol 1 1 2 2091 kj mol 1 2 = 52.98 kj mol 1 With this information we can now calculate the enthalpy of isomerization of cyclopropane to propene. The balanced equation for this process is The enthalpy of this reaction is simply given by C 3 H 6 (cyclopropane) C 3 H 6 (propene) r H =? r H = f H (propene) f H (cyclopropane) = 20.42 kj mol 1 52.98 = 32.56 Solution The enthalpy of formation of cyclopropane is f H (C 3 H 6 ) = 53 kj mol 1. The isomerization enthalpy of cyclopropane to propene is r H = 33 kj mol 1. 2

2 Exercise 2.20a pg 86 Given When a 120 mg sample of naphthalene, C 10 H 8 (s), was burned in a bomb calorimeter the temperature rose by 3.05 K. Additionally, we are to predict the temperature rise for the combustion of 10 mg sample of phenol in the same calorimeter (the enthalpy of combustion of phenol is c H (C 6 H 5 OH(s)) = 7061 kj mol 1 ). In terms of given variables, this is written: m C10 H 8 (s) = 120 mg T = 3.05 K m C6H 5OH(s) = 10 mg c H (C 6 H 5 OH(s)) = 7061 kj mol 1 Find Calculate the calorimeter constant. By how much will the temperature rise after the phenol sample is burned? Strategy To determine the calorimeter constant C we ll need to determine how much heat was released in this combustion reaction and for this we ll need to know the combustion enthalpy of naphthalene. We ll start by constructing the balanced chemical equation for the combustion of naphthalene. We can calculate the enthalpy of this reaction from C 10 H 8 (s) + 12O 2 (g) 4H 2 O(l) + 10CO 2 (g) r H =? r H = Products ν p f H p Reactants ν r f H r = 4 f H (H 2 O(l)) + 10 f H (CO 2 (g)) f H (C 10 H 8 (s)) + 12 f H (O 2 (g)) = 4 285.83 kj mol 1 + 10 393.51 kj mol 1 +78.53 kj mol 1 12 0 = 5156.950 kj mol 1 To calculate the heat released in this specific combustion we ll next need to know how many moles of naphthalene were burned. n = 120 mg C 10 H 8 (s) 1 g 1 mol 1000 mg 128.17 g = 9.363 10 4 mol C 10 H 8 (s) 3

We can now easily calculate the heat produced q p in the combustion of this quantity of moles. q p = n r H = 9.363 10 4 mol 5156.950 kj mol 1 = 4.883 kj The calorimeter absorbed this heat q a = q p = 4.883 kj. As we know the temperature change in the calorimeter resulting from the absorption of this heat, we can calculate the calorimeter constant C from q = C T. C = q a T 4.883 kj = 3.05 K = 1.601 kj K 1 As we now know the calorimeter constant we can go ahead and determine the calorimeter temperature change for burning the phenol sample. The balanced chemical equation for this reaction is and the enthalpy of this reaction is C 6 H 5 OH(s) + 7O 2 (g) 3H 2 O(l) + 6CO 2 (g) r H =? r H = Products ν p f H p Reactants ν r f H r = 3 f H (H 2 O(l)) + 6 f H (CO 2 (g)) f H (C 6 H 5 OH(s)) + 7 f H (O 2 (g)) = 3 285.83 kj mol 1 + 6 393.51 kj mol 1 165.0 kj mol 1 7 0 = 3053.55 kj mol 1 To determine the heat, we ll need the moles in this combustion. n = 10 mg C 6 H 5 OH(s) 1 g 1 mol 1000 mg 94.11 g = 1.06 10 4 mol C 10 H 8 (s) Now we ll calculate the heat produced in combustion. q p = n r H = 1.06 10 4 mol 3053.55 kj mol 1 = 0.324 kj And lastly we ll determine the temperature change induced in the calorimeter by absorbing this heat. T = q a C 0.324 kj = 1.601 kj K 1 = 0.202 K 4

Solution The calorimeter constant is C = 1.6 kj K 1. Burning phenol will lead to a temperature increase of T = 0.2 K. 5

3 Exercise 2.24a pg 87 Given For the reaction C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(g), r U = 1373 kj mol 1 at T = 298 K. In terms of given variables, this is written: r U = 1373 kj mol 1 T = 298 K Find Calculate r H of this reaction. Strategy The following equation relates the enthalpy change of a reaction to the internal energy change (Equation 2.21 on pg 58 in your text book). H = U + n g RT where n g RT is the change in quantity of gas molecules of the reaction. In analyzing this reaction C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(g) we see that 5 moles of gas are produced (2 moles of CO 2 (g) and 3 of H 2 O(g)) while 3 are consumed (O 2 (g)). Therefore n g = 2 and r H is easily calculated as r H = r U + n g RT = 1373 10 3 J mol 1 + 2 mol gas 1 mol reaction 8.314 J K 1 mol 1 298 K = 1.36804 10 6 J mol 1 = 1368.04 kj mol 1 Solution r H = 1368 kj mol 1 6

4 Exercise 2.26a pg 87 Given Consider the reaction 2NO 2 (g) N 2 O 4 (g) at T = 100 C (T = 373 K) and the information in table 2.8 Find Calculate the standard reaction enthalpy of this reaction at this temperature. Strategy For this problem, we ll use the fact that enthalpy is a state function to create a series of processes identical to this reaction at the non-standard temperature. By summing summing the change in enthalpy over these steps we ll arrive at the enthalpy of this reaction at the non-standard temperature. These processes can be summarized as cooling 2 moles of NO 2 (g) from T = 373 K to the standard temperature of T = 298 K. At this standard temperature the 2 moles of NO 2 (g) can be reacted to form 1 mole of N 2 O 4 (g) and we can calculate the reaction enthalpy at this standard temperature. Lastly, we ll heat the 1 mole of N 2 O 4 (g) from T = 298 K to T = 373 K. Summing the enthalpy changes over these three processes will give us the enthalpy difference between 2 moles of NO 2 (g) and 1 mole of N 2 O 4 (g) at T = 373 K and this is the enthalpy difference of the reaction of interest. This is summarized as 2 mol NO 2 (g) at T = 373 K cool H cool 2 mol NO 2 (g)att = 298 K rh (at T = 298 K) 1 mol N 2 O 4 (g)att = 298 K H heat 1 mol N 2 O 4 (g)att = 373 K r H (at T = 373 K) = H cool + r H (at T = 298 K) + H heat The enthalpy of cooling or heating a substance can be determined from the substance s constant pressure heat capacity C p. H = Tf T i For this problem we ll assume heat capacities are temperature independent over the temperature range of interest and this simplifies the calculations. C p dt 7

Tf H = C p dt T i Tf = C p dt T i = C p (T f T i ) = C p T Using heat capacities from our textbook we can calculate the enthalpy of heating the 2 moles of NO 2 (g) and cooling 1 mole of N 2 O 4 (g). H cool = ν C p T (Cool NO 2 (g)) = 2 37.20 J K 1 mol 1 (298 K 373 K) = = 5.5800 kj mol 1 H heat = ν C p T (Heat N 2 O 4 (g)) = 1 77.28 J K 1 mol 1 (373 K 298 K) = = 5.7960 kj mol 1 (Note that in these last two expressions the reaction coefficients are exact quantities and don t affect number of significant figures.) In the original solution sets the temperatures were swapped which changed the signs of these enthalpies. The enthalpy of the reaction at the standard temperature of T = 298 can be calculated from enthalpies of formation. r H = Products ν p f H p Reactants ν r f H r = 1 9.16 kj mol 1 2 33.18 kj mol 1 = 57.2000 kj mol 1 Summing these enthalpies gives the reaction enthalpy at the non-standard temperature r H (at T = 373 K) = H cool + r H (at T = 298 K) + H heat = 5.5800 kj mol 1 + 57.2000 kj mol 1 + 5.7960 kj mol 1 = 56.9840 kj mol 1 Solution r H (at T = 373 K) = 56.98 kj mol 1 8

5 Exercise 2.29a pg 87 Given We are given the following enthalpies... Enthalpy of sublimation of Mg(s) sub H = +167.2 kj mol 1 Mg(s) Mg(g) sub H = +167.2 kj mol 1 First ionization enthalpy of Mg(g) ion1 H = 7.646 ev Mg(g) Mg + (g) ion1 H = 7.646 ev Second ionization enthalpy of Mg(g) ion2 H = 15.035 ev Mg + (g) Mg +2 (g) ion2 H = 15.035 ev Dissociation enthalpy of Cl 2 (g) dis H = +241.6 kj mol 1 Cl 2 (g) 2Cl(g) dis H = +241.6 kj mol 1 Electron gain enthalpy of Cl(g) elec H = 3.78 ev Cl(g) Cl (g) elec H = 3.78 ev Enthalpy of solution of MgCl 2 (s) sol H = 150.5 kj mol 1 MgCl 2 (s) Mg +2 (aq) + 2Cl (aq) sol H = 150.5 kj mol 1 Enthalpy of hydration of Cl (g) hyd H = 383.7 kj mol 1 Cl (g) Cl (aq) hyd H = 383.7 kj mol 1 Find Calculate the enthalpy of hydration hyd H of for Mg +2 (g). Strategy For this problem we need to construct a thermodynamic cycle which includes the unknown process; hydration of Mg +2 (g), Mg +2 (g) hyd H. If we know the enthalpies all other processes in this cycle, then we can solve for the single unknown enthalpy. Such a cycle is shown is Figure 5 where one mole of Mg +2 (g) and two moles of Cl (g) are solvated through two different pathways. These sum of enthalpies over each of these two paths are thereby equal. For this cycle we found it necessary to include a process that was not given to us by the textbook: the enthalpy of formation of MgCl 2 (s) from Mg(s) and Cl 2 (g). We can calculate the enthalpy for this process from its balanced chemical equation The enthalpy of this reaction is calculated as Mg(s) + Cl 2 (g) MgCl 2 (s) 9

Figure 1: Thermodynamic cycle diagram r H = Products ν p f H p Reactants ν r f H r = f H (MgCl 2 (s)) f H (Mg(s)) f H (Cl 2 (g)) = 641.32 kj mol 1 0 0 = 641.32 kj mol 1 This gives us the enthalpy of the additional process we need: The enthalpy of formation of MgCl 2 (s) f H = 641.32 kj mol 1 Mg(s) + Cl 2 (g) MgCl 2 (s) f H = 641.32 kj mol 1 To reiterate, we had to obtain this quantity because it doesn t appear possible to complete this problem with only the given information. The themodynamic cycle diagram in the figure also illustrates the fact that we can t complete the cycle unless we link the MgCl 2 (s) to the corresponding stable elements. The formation reaction is precisely the missing link, and the problem statement didn t have to provide it because we are able to obtain it using information from the Table 2.8. If we equate the sum of the enthalpies of the two solvation paths in this thermodynamic cycle we get the following equation: 10

2 hyd H (Cl (g)) + hyd H (Mg +2 (g)) = [ ion1 H (Mg(g)) + ion2 H (Mg + (g)) + sub H (Mg(s)) ] [ 2 elec H (Cl (g)) + dis H (Cl 2 (g)) ] This equation can be solved for our unknown hyd H (Mg +2 (g)) + f H (MgCl 2 (s)) + sol H (MgCl 2 (s)) hyd H (Mg +2 (g)) = [ ion1 H (Mg(g)) + ion2 H (Mg + (g)) + sub H (Mg(s)) ] [ 2 elec H (Cl (g)) + dis H (Cl 2 (g)) ] + f H (MgCl 2 (s)) + sol H (MgCl 2 (s)) + 2 hyd H (Cl (g)) = [ 737.722 kj mol 1 + 1450.650 kj mol 1 + 167.2 kj mol 1] [ 2 364.71 kj mol 1 + 241.6 kj mol 1 + ] + 641.32 kj mol 1 + 150.5 kj mol 1 + 2 383.7 kj mol 1 = 3346.012 kj mol 1 where the energies in electron-volts have been converted by the following expression ion1 H = ion2 H = elec H = 7.646 ev 1.60218 10 19 J 1 kj 6.0221 10 23 electrons 1 ev 1000 J 1 mol 15.035 ev 1.60218 10 19 J 1 kj 6.0221 10 23 electrons 1 ev 1000 J 1 mol 3.78 ev 1.60218 10 19 J 1 kj 6.0221 10 23 electrons 1 ev 1000 J 1 mol = 737.722 kj mol 1 = 1450.650 kj mol 1 = 364.71 kj mol 1 Solution hyd H (Mg +2 (g)) = 3346.0 kj mol 1 11

6 Exercise 2.30a pg 87 Given When a certain freon used in refrigeration was expanded adiabatically from an initial pressure of p i = 32 atm and a T i = 0 C to a final pressure of p f = 1.00 atm, the temperature fell by T = 22 K. In terms of given variables, this is written: p i = 32 atm T i = 0 C p f = 1.00 atm T = 22 K Find Calculate the Joule-Thomson coefficient, µ, at T = 0 C assuming it remains constant over this temperature range. Strategy The Joule-Thomson coefficient is defined as µ = ( ) T p H If we assume µ is constant over the range of temperature and pressures changes then the derivative can be replaced with finite differences. µ = T p Substituting in our temperature and pressure changes gives µ = 22 K 1.00 atm 32 atm = 0.7097 K atm 1 Solution µ = 0.71 K atm 1 12