Here are some hints and observations I made while doing these exercises: I'd be curious to know if your answer for a problem is much different than mine. Of course, rounding off discrepancies (as much as several percent) are to be expected.

Exercise

2.1a)       My answers w_earth = 98.1 J; w_moon = 16.0 J (Book values the same) Remember w = -FDl, where Dl is the distance over which the force applied.

2.3a)       My answer w = -101 J (Book value the same)

2.5a)       My answers DU = 1.247 kJ; q = 1.247 kJ; w = 0; p₂ = 1.33 atm (Book values the same)

2.6a)       My answer -88 J (Book value the same)

2.6b)       My answer -168 J (Book value the same)

2.8a)       My answers DH = -40.056 kJ mol^-1, DU = -36.8 kJ mol^-1, q= -40.056 kJ mol^-1, w= 3.30 kJ mol^-1 (Book value the same DH = -40.056 kJ mol^-1, DU = -37.55 kJ mol^-1, q= -40.056 kJ mol^-1, w= 3.10 kJ mol^-1) This is a transition from gas to liquid. The amount of heat needed to liquefy the gas is equal to the opposite of the amount it takes to vaporize it. The amount of pV work done is -P_extDV; and since the volume taken up by the gas is than the volume of the liquid, the pV work @ P_extV_gas.

2.10a)       My answer 84.8 MJ (Book value 85.0 MJ) Will need the value for D_trsH for Na (See table 2.3 in back of text book).

2.13a)       My answers -145 J [Using Ideal Gas Approximation] and -194 J [Using C_v=C_p-R; C_p =37.11 J K^-1 mol^-1 from table 2.5 in back of text book] (Book value -194 J) Remember for adiabatic processes w=C_vDT; also remember for adiabatic processes ( [Ti/Tf]) = ( [Vf/Vi]) ^1/_c where c=[(C_v)/R]. To obtain my second value, and the book value use the calculated value for C_v when finding final T and work. This illustrates just how different values calculated with IDG approximations can be from those calculated with real data.

2.16a)       My answer -125 kJ mol ^-1 (Book value the same). For combustion the products are CO₂ gas and H₂O liquid.

2.21a)       My answers DH= 5.4 kJ, DU= w = 4.1 kJ, q=0, p_final=5.2 atm and V_final=11.8 L (Book values the same). I did the problem a little differently than the solutions manual, in that I solved for DH by using the definition DH=C_pDT where C_p=C_v-R so C_p= (27.5 - 8.3145) J K^-1 mol^-1.

2.26a)       My answers DH= 5.4 kJ, DU= w = 4.1 kJ, q=0, p_final=5.2 atm and V_final=11.8 L (Book values the same) Pretty straight ahead, but remember to take into account n= 0.5 mol.

2.28a)       My answer D_rxnH= -126 kJ (Book values the same) Determine D_fH for hexene from the combustion data (i.e. C₆H₁₂+9O₂® 6CO₂+6H₂O). Next calculate the D_rxnH for hydrogenation.

2.29a)       My answers D_fH_cyclopropane = 53 kJ mol^-1; D_iH cyclopropane to propene = -33 kJ mol^-1 (Book values the same) Very similar to 2.28a).

2.32a)       My answers Calorimeter Constant = 1.58 kJ K^-1; DT = 2.1 K (Book values Calorimeter Constant = 1.58 kJ K^-1; DT = 2.05 K) Remember the calorimeter constant is equal to [(q_evolved)/(DT)]; so determine the amount of heat released from the combustion of napthalene and relate it to change in T to get the calorimeter constant. Then determine amount of heat evolved from combustion of phenol and determine the T change.

2.35a)       My answer D_fH = -383 kJ mol^-1 (Book value the same) Very similar to 2.28a).

2.38a)       Trivial Exercise, simply get the stoichiometric quantities in terms of mass balance. Remember, if a process has negative enthalpy its exothermic; positive enthalpy and its endothermic.

2.39a)a]       My answer D_rxnH = -57.2 kJ mol^-1 (Book value the same)

2.39a)b]       My answer D_rxnH = -176.0 kJ mol^-1 (Book value the same)

2.42a)a]       My answer D_fH [KClO₃(s)]= -392 kJ mol^-1 (Book value the same)

2.42a)b]       My answer D_fH [NaCO₃(s)]= -947 kJ mol^-1 (Book value the same) Both part a) and b) are very similar to 2.28a).

2.45a)       NOT AVAILABLE YET