Chapter 5 (5.3 & 5.4) Thermochemistry: Consolidated Presentation Chapter 5 (5.3 & 5.4) Thermochemistry: Enthalpy & James F. Kirby Quinnipiac University Hamden, CT (Edited by E.Schneider, 10/15)

Work: In Chemistry Work = Pressure x Volume

w = −PV

Enthalpy Process at constant pressure majority of processes we study Only work generally done in these processes is pressure–volume work So, we account for heat flow during the process by measuring the enthalpy of the system. Enthalpy (H) is the internal energy plus the product of pressure and volume: At one point in time! H = E + PV

Derivation

Endothermic and Exothermic A process is endothermic when H is positive. A process is exothermic when H is negative.

Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants

Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Vocab Review Extensive property - property that changes when the size of the sample changes. Examples are mass, volume, length, and total charge. Intensive property – property that doesn't change when you take away some of the sample. Examples are temperature, color, hardness, melting point, boiling point, pressure, molecular weight, and density.

Enthalpy: Summary Enthalpy is an extensive property. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. H for a reaction depends on the state of the products and the state of the reactants.

Sample Exercise 5.1 Describing and Calculating Energy Changes Practice Exercise 2 What is the kinetic energy, in J, of : (a) an Ar atom moving at a speed of 650 m ⁄ sec (b) a mole of Ar atoms moving at 650 m ⁄ sec? (Hint: 1 amu = 1.66 ×10−27 kg.)

Sample Exercise 5.2 Relating Heat and Work to Changes of Internal Energy Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4 and react to form a solid product C(s): A(g) + B(g) → C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? Solution Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so q is negative and w is positive: q = −1150 J and w = 480 kJ. Thus, ΔE = q + w = (−1150 J) + (480 J) = −670 J The negative value of ΔE tells us that a net quantity of 670 J of energy has been transferred from the system to the surroundings.

Sample Exercise 5.3 A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 0.250 L, and the final volume is 0.980 L. If the piston expands against a constant pressure of 1.35 atm, how much work (in J) is done? (1 L-atm = 101.3 J) Solution The volume change is ΔV = Vfinal − Vinitial = 0.980 L − 0.250 L = 0.730 L Thus, the quantity of work is w = −PΔV = −(1.35 atm)(0.730 L) = −0.9855 L-atm Converting L-atm to J, we have

Sample Exercise 5.4 Determining the Sign of ΔH Indicate the sign of the enthalpy change, ΔH, in the following processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic: (a) An ice cube melts (b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O. Solution In (a) the water that makes up the ice cube is the system. The ice cube absorbs heat from the surroundings as it melts, so ΔH is positive and the process is endothermic. In (b) the system is the 1 g of butane and the oxygen required to combust it. The combustion of butane in oxygen gives off heat, so ΔH is negative and the process is exothermic.

Sample Exercise 5.5 Relating ΔH to Quantities of Reactants and Products How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? (Use the information given in Equation 5.18.) Solution Analyze Our goal is to use a thermochemical equation to calculate the heat produced when a specific amount of methane gas is combusted. According to Equation 5.18, 890 kJ is released by the system when 1 mol CH4 is burned at constant pressure. Plan Equation 5.18 provides us with a stoichiometric conversion factor: (1 mol CH4 −890 kJ). Thus, we can convert moles of CH4 to kJ of energy. First, however, we must convert grams of CH4 to moles of CH4. Thus, the conversion sequence is Solve By adding the atomic weights of C and 4 H, we have 1 mol CH4 = 16.0 CH4. We can use the appropriate conversion factors to convert grams of CH4 to moles of CH4 to kilojoules: The negative sign indicates that the system released 250 kJ into the surroundings.

Sample Exercise 5.5 Relating ΔH to Quantities of Reactants and Products Continued Practice Exercise 1 The complete combustion of ethanol, C2H5OH (FW = 46.0 g ⁄ mol), proceeds as follows: C2H5OH(l) + 3→O2(g) 2CO2(g) + 3H2O(l) ΔH = −555 kJ What is the enthalpy change for combustion of 15.0 g of ethanol? (a) −12.1 kJ (b) −181 kJ (c) −422 kJ (d) −555 kJ (e) −1700 kJ Practice Exercise 2 Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l) → 2 H2O(l) + O2(g) ΔH = −196 kJ Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure.

In what ways is the balance in your checkbook a state function? It is determined by many factors and involves data such as the number of checks, date, and payee. It is a numerical value calculated from known amounts of checks. It is calculated from multiple transactions. It only depends on the net total of all transactions, and not on the ways money is transferred into or out of the account. Answer: d

In what ways is the balance in your checkbook a state function? It is determined by many factors and involves data such as the number of checks, date, and payee. It is a numerical value calculated from known amounts of checks. It is calculated from multiple transactions. It only depends on the net total of all transactions, and not on the ways money is transferred into or out of the account. Answer: d

2 H2(g) + O2(g) ⟶ 2 H2O(g) ΔH = –483.6 kJ [5.17] If the reaction to form water were written H2(g) + O2(g) ⟶ H2O(g), would you expect the same value of ΔH as in Equation 5.17 (above)? Why or why not? Yes, because the reactants and products are the same. No, because only half as much matter is involved. Yes, because mass does not affect enthalpy change. No, because enthalpy is a state function. Answer: b

If the reaction to form water were written H2(g) + O2(g) ⟶ H2O(g), would you expect the same value of ΔH as in Equation 5.17? Why or why not? Yes, because the reactants and products are the same. No, because only half as much matter is involved. Yes, because mass does not affect enthalpy change. No, because enthalpy is a state function. Answer: b

If the battery is defined as the system, what is the sign on w in part (b)? a. w > 0 b. w = 0 c. w < 0 d. Cannot be determined Answer: c

If the battery is defined as the system, what is the sign on w in part (b)? a. w > 0 b. w = 0 c. w < 0 d. Cannot be determined Answer: c

If the amount of zinc used in the reaction is increased, will more work be done by the system? Is there additional information you need in order to answer this question? a. If Zn(s) is the limiting reactant, then more gas is formed and more work will be done. b. If Zn(s) is the limiting reactant, then less gas is produced and less work will be done. c. If HCl(aq) is the limiting reactant, then addition of Zn(s) increases the amount of gas formed and more work is done. d. If HCl(aq) is the limiting reactant, then addition of Zn(s) decreases the amount of gas formed and less work is done. Answer: a

If the amount of zinc used in the reaction is increased, will more work be done by the system? Is there additional information you need in order to answer this question? a. If Zn(s) is the limiting reactant, then more gas is formed and more work will be done. b. If Zn(s) is the limiting reactant, then less gas is produced and less work will be done. c. If HCl(aq) is the limiting reactant, then addition of Zn(s) increases the amount of gas formed and more work is done. d. If HCl(aq) is the limiting reactant, then addition of Zn(s) decreases the amount of gas formed and less work is done. Answer: a

Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is, E = q + w.

Work Defined as: Work = Force * distance (∆h) Pressure = Force / Area (example units “psi”) so, Force = Pressure * area Substitute: Work = P * area * ∆h Work = P x Volume

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −PV

Change in Enthalpy When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) This can be written H = E + PV

Enthalpy E = q + w AND w = −PV Substitute these into the change in enthalpy expression: H = E + PV H = (q + w) − w H = q At constant pressure, the change in enthalpy is the heat gained or lost.