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Lecture 2: Enthalpy • Reading: Zumdahl 9.2, 9.3 • Outline – Definition of Enthalpy (H) – Definition of Heat Capacity (Cv and Cp) – Calculating E and H using Cv and Cp What is Enthalpy? Thermodynamic Definition of Enthalpy (H) H = E + PV E = energy of the system P = pressure of the system V = volume of the system Consider a process carried out at constant pressure: If work is of the form w= -PV, then: E = qp + w = qp - PV Rearranging, E + PV = qp where qp is heat transferred at constant pressure. Recall that H = E + PV Then H = E + (PV) Since (PV) = PV + VP = PV (P constant) Substituting, But we just showed So finally, H = E + PV E + PV = qp H = qp In words, the change in enthalpy (H) is equal to the heat transferred at constant pressure (qp) What about changes in enthalpy? Consider the following expression for enthalpy change in a chemical process: H = Hprods - Hreacts If H >0, then qp >0, and the reaction is endothermic. If H <0, then qp <0, and the reaction is exothermic. Picturing Enthalpy Changes Enthalpy (Similar to previous discussion for energy changes) Hinitial q out Hfinal Enthalpy H < 0 Hfinal Hinitial H > 0 q in If heat comes out of system, the enthalpy decreases (ex. cooling water). If heat goes into the system, the enthalpy increases (ex. heating water) Heat capacity at constant volume • Recall from Chapter 5 (section 5.6): (KE)ave = (3/2) RT (for an ideal monatomic gas) • Temperature is a measure of molecular speed. • In thermodynamic terms, an increase in system temperature corresponds to an increase in average kinetic energy of a gas ( i.e., T is proportional to KE) Heat capacity at constant volume • (KE)ave = 3/2 RT (ideal monatomic gas) • How much energy in the form of heat is required to change the gas temperature by an amount T? Heat required = 3/2RT = 3/2R (for T = 1K) • Therefore, Cv = 3/2 R is the heat required to raise one mole of an ideal gas by 1K at constant volume. Cv is referred to as the constant volume heat capacity. Q: Heat capacity at constant P? What about at constant pressure? In this case, (PV)-type work can also occur: PV = nRT PV = nRT = RT (for 1 mole) = R (for T = 1 K) Cp = “heat into translation” + “work” = Cv + R = 5/2R (for an ideal monatomic gas) Cv for Monatomic Gases z 1/2 mv 2 = 3/2 RT x y Q: What are the energetic degrees of freedom for a monatomic gas? Ans: Just translations(x,y,z), which contribute (3/2)R to Cv. Cv for Polyatomics Iz Iy N2 C v = 5/2 R (approx.) Iy Iz Ix 3 NO2 C v = 7/2 R (approx.) Q: What are the energetic degrees of freedom for a polyatomic gas? Ans: translations, and rotations, and vibrations… all of which may contribute to Cv (depends on T). Variation in Cp and Cv Ar, He, Ne H2 CO2 Units: J/mol. Cv Cp 12.47 20.8 20.54 28.86 28.95 K 37.27 • Monatomics: Cv = 3/2 R Cp = 5/2 R • Polyatomics: Cv > 3/2 R Cp > 5/2 R But, Cp = Cv + R (always!) Heating an ideal monatomic gas at constant volume Recall So Recall So, Eave = 3/2 nRT (av trans. energy) E = 3/2 nRT = (3/2R)nT Cv = 3/2 R E = nCvT Q: Why Cv? We envision heating our system at constant volume. As such, all heat goes towards increasing E (no work is done, since no volume change is possible) What if we heat the ideal gas at constant pressure? In this case, we have a volume change and (PV) work can occur: recall Cp = Cv + R q p = n Cp T q p = n (Cv + R) T recall E = nCvT so q p = E + nRT = E + PV q p = H = n Cp T Keeping Track Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R All Ideal Gases E = nCvT H = nCpT Example What is q, w, E and H for a process in which one mole of an ideal monatomic gas with an initial volume of 5 liters and pressure of 2.0 atm is heated until a volume of 10 liters is reached with pressure unchanged? Pinit = 2 atm Pfinal = 2 atm Vinit = 5 l Vfinal = 10 l Tinit = ? K Tfinal = ? K Using the ideal gas law PV = nRT PV = nRT, so we can determine T. V = (10 L - 5 L) = 5 L Substituting values, (2 atm (5 L ( (1 mol) .0821 L atm mol K 121.8 K T So, now we can calculate ∆E, ∆H, w and q: ( E nCv T (1 mol) 12.5 J mol.K (121.8 K 1522.5 J ( H nC p T (1 mol) 20.8 J mol.K (121.8 K =2533.4 J ( w Pext V (2 atm (5 L 101.3 J L.atm = -1013.0 J q E w 1522.5 J- (-1013.0 J 2535.5 J