System 1Slide 2Slide 3ThermochemistryEndothermic ReactionExothermic ReactionCalorimetryheat capacitiesSlide 9Slide 10Slide 11Slide 12Slide 13System 1P1 = 6.0 atm V1 = 0.4 LPext = 1.5 atm, How much work will be done as the gas expands against the piston?P2 =V2 = P1 V1 = P2 V2(6.0 atm)1.6 Lw = -Pext V -(1.5 atm) -1.8 L atm(-1.8 L atm)PV = nRT1.5 atm (101.3 J/L atm) = -182 Jw =(0.4 L) = (1.5 atm) (V2)(1.6 - 0.4)L=T = 298 KSystem 2How much work is done when the stopcock is opened?P1 = 6 atm V1 = 0.4 LT1 = 298 KP2 = V2 = 1.2= T2P1 V1 = P2 V21.5 atmw = -Pext V= -(0 atm)w = 01.2 Lvacuum0.4 Lideal gas6.0 atm(1.6 L - 0.4 L)+ 0.4 L= 1.6 LSystem 1 System 2P1 = P2 =V1 = V2 =T1 = T2 =Same initial and final conditions E will be the sameIdeal gases :E = q + w = 0w = -182 J q = +182 J w = 0 q = 0all State functions are the same6.0 atm 6.0 atm0.4 L 0.4 L298 K 298 K 01.5 atm 1.5 atm1.6 L 1.6 Lif T = 0, E =P1 = P2 =V1 = V2 =T1 = T2 =ThermochemistryState 1 = State 2 =reactants productsE = Eproducts - EreactantsEndothermic reactionq a) <b) >c) =reaction lowered T of systemw00Ba(OH)2•8H2O (s)2NH3(g)+ 2NH4SCN (s) + Ba(SCN)2(l)+ 10H2O(l)a) <b) >c) =Endothermic ReactionTprod < TreactK.E.prod < K.E.reactP.E.prod > P.E.reactproducts less stable than reactantsExothermic ReactionTprod > TreactK.E.prod > K.E.reactP.E.prod < P.E.reactproducts more stable than reactants q 0w 0<<C12H22O11 (s) + KClO3(s) mix of gasesCalorimetryHow is heat measured?It isn’tTemperature measured T (K) C = heat capacity (J/K)= q (J)C= heat to raise T 1ox specific heat capacity gJ / Kmolar heat capacity molJ / Kheat capacities substance specific m.w. molarheat capacity (g/mol) heat capacity (J / K g) (J / K mol)Al (s)Fe (s) 0.440.8926.98 24.055.8524.8H2O(s) 36.5H2O(l) CCl4(l) 75.2133.0Calorimetry 1. Measure T (Tfinal - Tinitial)2. Convert to qT (K) x mass (g) =x C (J/K g) q (J)q is a path function E = q + w q =EE - w + Pext Vq = E + PextVHq = E + PVAt constant P, H =  E + P VqqEnthalpy HH = E + PV +VPP = 0H = E + PVE + PVpp= = HHqqvv= = EEgasesE = K.E.trans+ K.E.rot+ K.E.vib+ P.E.IMF+ P.E.bondq = C T= HT TTCP =qP= ET TCV =qVH= E + PVT TCP = CV +RTT= CV + RCV =3/2 RCP = 5/2 RIdeal1000 J of heat2.00 mol ArT1 = 298 Kfind T2at constant volumeat constant pressureCV = 3/2 RCP = 5/2 R1000 J 3/2 x8.314 x 10-3 x (T2 –T1)1000 J x 5/2 x8.314 x 10-3 x (T2 –T1)= 2 x = 2 T2 = T2 = 338 K322 KE = n CV T= 2 x 3/2 R x 338-298 = 1000 JE = n CV T= 2 x 3/2 R x 322-298 = 600 Jfind Eheat and ideal gasesPV1.02.010.020.0 30.0A CBDIIIIIIIVIqP =5/2R(TC-TA)= 10.1 kJw =-4.05 kJII q3/2R(TB-TC)= -4.55 kJw = -Pext V = 0 kJV =III qV =3/2R(TD-TA)= -1.52 kJw = -Pext V = 0 kJIVq5/2R (TB-TD)= 5.07 kJw = -Pext V = -2.03 kJP =E = 1.52 kJE = 1.52 kJ-Pext V =E = 3/2R(TC-TA)= 6.07 kJH = 5/2R(TB-TC)= -7.59 kJH = 5/2R(TD-TA)= -2.54 kJ= H= E= EE = 3/2R(TB-TD)= 3.04 kJ= HH = 2.51 kJH = 2.53