2. Lecture 2: Enthalpy Reading: Zumdahl 9.2, 9.3 Outline –Definition of Enthalpy (  H) –Definition of Heat Capacity (C v and C p ) –Calculating  E and  H using C v and C p

3. Definition of Enthalpy Thermodynamic Definition of Enthalpy (H): H = E + PV E = energy of the system P = pressure of the system V = volume of the system

4. Definition of Enthalpy (cont.) Consider a process carried out at constant pressure. If work is of the form -P  V, then:  E = q p + w = q p - P  V  E + P  V = q p q p is heat transferred at constant pressure.

5. Definition of Enthalpy (cont.) Recall: H = E + PV  H =  E +  PV) =  E + P  V (P is constant) = q p Or  H = q p The change in enthalpy is equal to the heat transferred at constant pressure.

6. Changes in Enthalpy Consider the following expression for a chemical process:  H = H products - H reactants If  H >0, then q p >0. The reaction is endothermic If  H <0, then q p <0. The reaction is exothermic

7. Enthalpy Changes Pictorally Similar to previous discussion for Energy. Heat comes out of system, enthalpy decreases (ex. Cooling water). Heat goes in, enthalpy increases (ex. Heating water)

8. Heat Capacity at Constant V 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 system kinetic energy ( i.e., T is proportional to E)

9. Heat Capacity at Constant V (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 = 3/2R  T = 3/2R (for  T = 1K) Therefore, C v = 3/2 R is the heat required to raise one mole of an ideal gas by 1K at constant volume. C v is referred to as the constant volume heat capacity.

10. Heat Capacity at Constant P What about at constant pressure? In this case, PV type work can also occur: P  V = nR  T = R  T (for 1 mole) = R (for  T = 1 K) C p = “heat into translation” + “work” = C v + R = 5/2R (for an ideal monatomic gas)

11. C v for Monatomic Gases What are the energetic degrees of freedom for a monatomic gas? Ans: Just translations, which contribute 3/2R to C v.

12. C v for Polyatomics What are the energetic degrees of freedom for a polyatomic gas? Ans: translations, rotations, and vibrations. All of which may contribute to C v (depends on T). 3

13. Variation in C p and C v Monatomics: –C v = 3/2 R –C p = 5/2 R Polyatomics: –C v > 3/2 R –C p > 5/2 R –But….C p = C v + R

14. Energy and C v Recall from Chapter 5: E ave = 3/2 nRT (average trans. energy)  E = 3/2 nR  T  E = n C v  T (since 3/2 R = C v ) Why is it C v ? We envision heating our system at constant volume. As such, all heat goes towards increasing E (no work).

15. Enthalpy and C p What if we heated our gas at constant pressure? Then, we have a volume change such that work occurs. q p = n C p  T = n (C v + R)  T =  E + nR  T =  E + P  V =  H or  H = nC p  T

16. Keeping Track Ideal Monatomic Gas C v = 3/2R C p = C v + R = 5/2 R Polyatomic Gas C v > 3/2R C p > 5/2 R All Ideal Gases  E = nC v  T  H = nC p  T

17. 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 l and pressure of 2.0 atm is heated until a volume of 10 l is reached with pressure unchanged? P init = 2 atm V init = 5 l T init = ? K P final = 2 atm V final = 10 l T final = ? K

18. Example (cont.) Since P  V = nR  T, we can determine  T.  V = (10 L - 5 L) = 5 L And:

19. Example (cont.) Given this: