Relationship between△Hand△U AH=A+△(PV) If the reaction only has solid and liquid, A(Pv is small, So △H全△U If the reaction involves gas reactants and/or products Initial state P1V1=no1RT Final state: P2v2=n2RT So: A(PV=P2v2-P1v1=norT- noRT=Ana RT Where,△no=zn g, products ∑ g, reactants
Relationship between ΔH and ΔU 16 H = U + (PV) If the reaction only has solid and liquid, (PV) is small, So: H U If the reaction involves gas reactants and/or products: Initial state： P1V1 = ng1RT Final state： P2V2 = ng2RT So: (PV) = P2V2 – P1V1 = ng2RT – ng1RT = ng RT where, ng = ∑ ng, products – ∑ ng, reactants
Practice Q4: At 1.0 atm. 298 K. the combustion of benzene is C6H6()+7.502(9)→3H2O()+6CO2(g) f△ U of this reaction is32639 KJ/mol, What is△H? Solution AH=AU+△(P)=AU+An只RT An,=6-7.5=-1.5 mol, (only count gas species △nRT=-15X8.314×298=-3.72×103Jmo AH=A∪U+△n。RT=-32639+(-372)=-32676 KJ/mol
Practice Q4: At 1.0 atm, 298 K, the combustion of benzene is: C6H6 (l) + 7.5 O2 (g) 3 H2O (l) + 6 CO2 (g) If ΔU of this reaction is -3263.9 kJ/mol, what is ΔH? Solution: H = U + (PV) = U + ngRT ng = 6 – 7.5 = – 1.5 mol, (only count gas species!) ngRT = – 1.5 x 8.314 x 298 = – 3.72×103 J/mol H = U + ngRT = – 3263.9 + (– 3.72) = – 3267.6 kJ/mol 17
Thermal chemical equation EXample 1: H2(9)+ 1202(9)H2o( AHm (298K=-2858 kJ/mol EXample2:Hgo(s)→Hg(+佐O2(9) △H0 m(298K +90.7 kJ/mol Pay attention to the sign of heat Vs. AHmo m(298K EXothermic (9>0): AH <0; Endothermic( q<0): AH >0 Enthalpy Enthalpy ∑ Reactants Products ΣH reactants 18
Thermal Chemical Equation 18 Example 1: H2 (g) + ½ O2 (g) → H2O (l) Hm 0 (298K) = –285.8 kJ/mol Example 2: HgO(s) → Hg(l) + ½ O2 (g) Hm o (298K) = + 90.7 kJ/mol Pay attention to the sign of heat vs. Hm o (298K) Exothermic (q > 0): H < 0; Endothermic (q < 0): H > 0
Practice Q: The thermal chemical equation of C10Hg combustion is C10Hg(S)+12O2(9)→10c02(9)+4H2O( △H0=-5.13X103kJ/mo How much heat is released if 10 gram of C1oH(s)is fully burnt?(MWo C10H8 128g/mo) Solution: nC10H8=10/128=0.078 mol Ip =AHren C10H8 =5.13X103×0.078=4.02X102kJ
Practice 19 Q：The thermal chemical equation of C10H8 combustion is: C10H8 (s) + 12 O2 (g) → 10 CO2 (g) + 4 H2O (l) Hrxn o = –5.13 x 103 kJ/mol How much heat is released if 10 gram of C10H8 (s) is fully burnt? (MWC10H8 = 128 g/mol) Solution： nC10H8 = 10 / 128 = 0.078 mol qp = Hrxn o x nC10H8 = 5.13 x 103 x 0.078 = 4.02 x 102 kJ