Lecture 2: Enthalpy Reading: Zumdahl 9.2, 9. Outline Definition of Enthalpy (ΔH) Definition of Molar Heat Capacity (C v and C p ) Calculating using C v and C p Changes in ΔE and ΔH as well as q and w for different paths Problems: Z9.11,Z9.2-25(no calculations required), Z9.27, Z9.29 1
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 Each quantity can be determined by measuring the system, so they are quantities that depend on the state of the system. What does depend on the state of the system mean? Can determine change (eg DE) by looking at system before and after, not during the change This is not true of q and w, you must know how the system changes (during the change) to determine q and w. State function depend on the state of the system defined by P,V and T for a gas. 2
Definition of Enthalpy (cont.) Consider a process carried out at constant pressure. (Such as all the reactions in our body and in the laboratory, usually.) If work is of the form -PΔV, then: Δ E = q+ w= q PΔV q p is heat transferred at constant pressure. If: Δ P = 0 ( isobaric) then qp q Δ = Δ +Δ ( ) = Δ + Δ = H E PV q P V P V q Δ H = q P Therefore, for reactions run at constant pressure the heat energy that flows (in or out), which can be measured, is equal to the reaction enthalpy, which is a state function. P P
Changes in Enthalpy Consider the following expression for a chemical reaction: ΔH = H products -H reactants If ΔH >0, then q p >0. The reaction is endothermic If ΔH <0, then q p <0. The reaction is exothermic Endo- means into; exo- means out from. 4
Enthalpy Changes: Pictorially Similar to previous discussion for Energy. Entha lpy Entha lpy ΔH < 0 ΔH > 0 H initial H final H final H initial q out q in Heat comes out of system, enthalpy decreases (ex. Cooling water). Heat goes in, enthalpy increases (ex. Heating water) Is it possible to have a reaction go that takes heat (I.e. gets cold)? 5
Heat Capacity: The relation of internal energy to temperature Recall from Chapter 5 (section 5.6): (KE) average = /2RT (ideal monatomic gas) Temperature is a measure of molecular motion. 2 2 E = PV = nrt Then for a change in temperature, the energy change is: ( ) Δ E = n 2 R ΔT In thermodynamic terms, an increase in system temperature corresponds to an increase in system kinetic energy ( i.e., T is proportional to E), and DE depends only the the temperature change (and nothing else). This relation is distinct from the first law and complementary to it. Both forms of DE hold. 6
Qualitative Problems we can answer (Z9.22-26) DE depends only on DT, what about DH? Δ H = Δ E+Δ PV =Δ E+Δ nrt ( ) ( ) ( ) ( ) Δ H = n 2R Δ T + nrδ T = n 2R+ R ΔT So yes, DH depends only on DT as well. The proportionality is larger (add nr) Therefore both E and H are state functions (unlike q and w). Because: If you know the temperature before and after the change, the amount of heat and work flow do not matter and changes in P and V alone do not matter. If a reaction is exothermic, what precautions do you need to make when it happens? [Hint: Cool it.] For the formation of (2 moles of) NO (from air) DH=180kJ. What does this imply about it happening? Why high temperatures? 7
How Energy and Enthalpy connect to reactions Z9.25. Have the reaction to form water: ( ) ( ) ( ) 2H g + O g 2HO Δ H= 572kJ 2 2 2 How much heat do you get if you form 1 mole of water? How about how much heat if you make 1.8 grams of water? [Hint convert grams to moles and scale the heat proportionally.] What is the heat for the reaction to make the two gasses from water? [Whatever you do to the reaction, do it to the energy/enthalpy for the reaction. In this case change direction, which is change DH sign.] How much heat did the Hindenberg give off when it burned? [Hint: Convert volume to number of moles. The above reaction tells you the heat per mole of H 2 burned is 286kJ. 8
Heat Capacity at Constant V (isochoric) Δ V = 0 How much energy (at constant volume) in the form of heat is required to change the gas temperature by an amount ΔT? V V ( R) No PdV work was allowed. See Table 9.1 ( ) Δ E = q PΔ V = q = n R ΔT C q nδt The heat capacity, C v = /2 R,is the amount of heat required to raise the temperature by one degree (of one mole of a monoatomic ideal gas at constant volume). C v is referred to as the constant volume (molar) heat capacity. V V = 2 2 9
Heat Capacity at Constant P What about at constant pressure? In this case, PdV type work also occurs: The change in enthalpy with temperature is known, and the heat released at constant pressure in the enthalpy. Therefore we can define a heat capacity at constant pressure as: P ( ) q =Δ H = n R+ R ΔT Physically Cp is greater than Cv: When warming an ideal gas the heat goes into increasing the energy in the translational modes. Now at constant pressure the heat is also used to expand the cylinder and do productive work on the surroundings (work is negative). So the gas can absorbs more heat (under constant P conditions) for the same temperature rise (than under constant V conditions). 2 q = ( ) n T 2 + = + Δ P CP R R CV R 10
C v for Monatomic Gases 1 2 2 E = mv = nrt 2 What are the energetic degrees of freedom for a monatomic gas? y x Ans: Just translations, which contribute /2R to C v. 11
C v for Polyatomics I z I y What are the energetic degrees of freedom for a polyatomic gas? N 2 C v = 5/2 R (approx.) I z I y I x Ans: translations, rotations, and vibrations. All of which may contribute to C v (depends on T). NO 2 C v = 7/2 R (approx.) 12
Variation in C p and C v Ar, He, Ne 12.47 20.8 H 2 2 R = 12.47 J mole K C v C p 20.54 28.86 CO 2 28.95 7.27 Units: J/mol. K Monatomics: C v = /2 R C p = 5/2 R Polyatomics: C v > /2 R C p > 5/2 R But.C p = C v + R The idea is called the equipartition of energy which means that each possible way of moving (mode) gets about ½ R for the molar heat capacity. (See Table 9.1). 1
Keeping Track: Table 9.2 of Text is great. See also Table 9.1 for real numbers R~8.J/Kmol Ideal Monatomic Gas (IMG) C v = /2R C p = C v + R = 5/2 R All Ideal Gases (IG) (regardless of process) ΔE = nc v ΔT Polyatomic Gas C v > /2R C p > 5/2 R The heat capacity will be a given quantity or you are told it is an IMG. ΔH = nc p ΔT C p = C v + R Notice: For the same process the enthalpy change is larger than the energy change. 14
Example (A part of Z9.29) What is q, w, ΔE and ΔH for a process in which one mole of an ideal monatomic gas with an initial volume of 15 liters and pressure of.0 atm is heated until a volume of 55 liters is reached with pressure unchanged? A gas is expanding by heating. What are the signs of the different terms? P init = atm P final = atm V init = 15 l V final = 55 l T init =? K T final =? K Before you start; realize you may not need the initial and final temperature separately; only the difference. The process is a DP=0 process. DE and DH are determined only by DT. It is a IMG, so you know the two heat capacities. On a test the first thing you write down is DE and DH in terms of DT, which you then need to find. Then ask, can I determine the work, and finally you find the heat by rearranging the first law. 15
Problem Z9.29 R=8.1451 J/(mol-K) = 0.08206 (liter-atm)/(mol-k) Adapt PV=nRT to the change for the process: PDV =nrdt Now we can determine DT. Δ V = 55 15 = 40 ( ) ( atm)( 40l) Δ T = = 1460K (1 mol) (.0821latm. ) mol. K 16
Example (cont.) From the temperature change, and IMG, 1 mole, get change in energy and enthalpy; and the definition of PdV work and the first law give the work and heat for this process. ( J )( ) Δ E = ncvδ T = (1 mol) 12.5 1460K = 18.2kJ mol. K 5 Δ H = ncpδ T = Δ E = 0. kj w= P ( )( 40 )( 101.J ) extδ V = atm l = 12kJ latm. ( ) q=δe w= 18 12 = 0J It is no coincidence that q and DH are the same: This is a constant pressure process for which q=q p =DH. 17
Problem Z9.29 Two Paths We have done step one (from A to C) of the PV diagram. The goal is to go from A to B by two different paths. The first (blue) path takes you through C and the second (red) path through D. Need to compute DE, DH, w and q for each step. The goal is to realize that DE and DH when we sum steps 1 and 2 will give the same result as when we sum steps and 4. They are path independent. w and q will be different for the two different paths. Notice that the work on the path through D must be larger than the work on the path through C because the DV is the same but the pressure is higher. There is no work in either steps 2 or as DV=0. 18
The second step in A-C-B path We have done an isobaric process; so what is different for an isochoric process? I.G. difference equation is a bit different VΔP=nRΔT Solve for ΔT. There is no work: w=0; ΔE and ΔH are computed the same way, based on the temperature change. It is a constant volume change so q=q V =ΔE 19