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Internal Energy Depends on the Conditions

Internal Energy Depends on the Conditions. Although we have no way of knowing the precise value of the internal energy E of a system, E does have a fixed value for a given set of conditions.

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Internal Energy Depends on the Conditions

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  1. Internal Energy Depends on the Conditions Although we have no way of knowing the precise value of the internal energy E of a system, E does have a fixed value for a given set of conditions. Functions with a fixed value for a given P, T, V, and component composition are called state functions. E is a STATE FUNCTION.

  2. STATE FUNCTIONS The value of a state function depends only on itspresent condition, NOT on the history of the sample. The internal energy of the 25°C water does NOT depend on the way in which the water arrived in this state.

  3. STATE FUNCTIONS ΔE depends only on the internal energy of the initial and final states, not on how the change was accomplished. ΔE = Efinal- Einitial …AND ΔE = q + w 1 Path 1: ΔE = q1 Shorting a battery releases heat but does not allow it to do any useful work. Path 2: ΔE = q2 + w q and w are NOT state functions!

  4. Pressure-Volume Work Reactions can be set up so that the pressure remains constant and only the volume changes. In this case, the work done by the system is given by w = -PΔV If ΔV is positive, the system is doing work on the surroundings and w is negative. What is the system in this example? If ΔV is negative, the surroundings are doing work on the system and w is positive.

  5. P-V Work: w = -PΔV • A definition of work is the energy used to move an object against a force. The object has to MOVE. There must be a force opposing this movement. • For P-V work, the MOVEMENT is the change in volume ΔV. If ΔV = 0, there is no P-V work. • For P-V work, the opposing force is the pressure the system is working against. If there is no opposing pressure, there is no P-V work.

  6. P-V Work: Example 1 • Nine moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The volume changed by 215L and the temperature remained constant. Calculate the work done on the system in joules. • ∆V = 215L • The question is, what pressure do we use? • We use the pressure the system is working against, which is 0.98 bar. • w = -PΔV = -0.98 bar (215 L) = -211 L bar

  7. P-V Work: Example 1 Nine moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The volume changed by 215L and the temperature remained constant. Calculate the work done on the system in joules. w = -0.98 bar (215 L) = -211 L bar w = -21000 J Know this conversion factor: 1 L bar = 100 J

  8. P-V Work: Example 1 Nine moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The volume changed by 215L and the temperature remained constant. Calculate the work done on the system in joules. w = -21000 J Note: Since the work done on the system is negative, it is the system that is doing work on the surroundings.

  9. P-V Work: Example 2 • Nine moles of an ideal gas initially at 17.5 atm expand into a vacuum. The volume changed by 215L and the temperature remained constant. Calculate the work done on the system in joules. • ∆V = 215L • For a vacuum, P = 0. • w = -PΔV = 0

  10. P-V Work: Example 3 9.0 moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The temperature of the gas remained constant at 25ºC throughout the expansion. Calculate the work done on the system in joules. w = –P∆V

  11. P-V Work: Example 3 9.0 moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The temperature of the gas remained constant at 25ºC throughout the expansion. Calculate the work done on the system in joules. The gas is ideal, and we have P1, T, and n. The ideal gas law is PV = nRT.

  12. P-V Work: Example 3 9.0 moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The temperature of the gas remained constant at 25ºC throughout the expansion. Calculate the work done on the system in joules. For V2, we can use Boyle’s Law (P1V1 = P2V2), since T and n are constant:

  13. P-V Work: Example 3 Nine moles of an ideal gas initially at 17.5 atm expand against the atmosphere, which that day had a pressure of 0.98 bar. The temperature of the gas remained constant at 25ºC throughout the expansion. Calculate the work done on the system in joules. w = –P∆V = -0.98 bar (227.547 – 12.576) L = -210 L bar w = -21000 J

  14. P-V Work: Example 4 Nine moles of an ideal gas initially at 2.0 bar and 25ºC are in a rigid container. Outside the container, atmospheric pressure is 0.98 bar. The gas is heated to 500K. Calculate the work done on the system in joules. What happens to this gas? Its pressure increases…but does its volume? “Rigid” equals “constant volume.” ∆V = 0, so w = –P∆V = 0

  15. Enthalpy Enthalpy is a state function that accounts for heat flow in chemical changes occurring at constant pressure. The enthalpy (H) is defined in terms of the internal energy, pressure, and volume of a system: H ≡ E + PV ΔH = Δ(E + PV) = ΔE + Δ(PV) ΔH = ΔE + PΔV + VΔP When pressure is constant, ΔH = ΔE + PΔV

  16. Enthalpy Enthalpy is a state function that accounts for heat flow in chemical changes occurring at constant pressure. When pressure is constant, ΔH = ΔE + PΔV ΔH = ΔE + PΔV = qp + w + PΔV When pressure-volume (P-V) work is the only work done by system, = qp – PΔV + PΔV = qp qp designates the heat going into the system under the condition of constant pressure.

  17. Enthalpy When a chemical reaction is performed under conditions of constant pressure, the enthalpy can be used to determine whether the reaction is exothermic or endothermic: ΔH = qp ΔH < 0 exothermic ΔH > 0 endothermic System System heat qp<0 heat qp>0 Surroundings Surroundings work is P-V only work is P-V only

  18. Enthalpy of Reaction 2H2(g) + O2(g)  2H2O(l) ΔH(25°C) = ? ΔH = Hfinal – Hinitial This is the same as ΔH = Hproducts – Hreactants The enthalpy change which accompanies a reaction is called the enthalpy of reaction or heat of reaction. 2H2(g) + O2(g)  2H2O(l) ΔH(25°C) = -571.7 kJ The balanced reaction and its associated ΔH constitute a thermochemical equation.

  19. Enthalpy Diagram 2H2(g) + O2(g)  2H2O(l) ΔH(25°C) = -571.7 kJ The enthalpy change which accompanies a reaction can also be represented in an enthalpy diagram. The product is lower in enthalpy than the reactants, so the reaction is exothermic. Enthalpy

  20. Enthalpy is an Extensive Property 2H2(g) + O2(g)  2H2O(l) ΔH(25°C) = -571.7 kJ A thermochemical equation shows ΔH for the reaction as written. In this case, burning two moles of H2(g) in O2 (under conditions of constant P) would release 571.7 kJ of heat. How much heat is released when 0.275 g of H2 is ignited with O2 in a constant-pressure system? 0.275 g H2(g) x 1 mol H2 x -571.7 kJ = -39.0 kJ 2.016 g H2 2 mol H2 ΔH = -39.0kJ 39.0 kJ of heat is released.

  21. Enthalpy for the Reverse Reaction 2H2(g) + O2(g)  2H2O(l) ΔH(25°C) = -571.7 kJ A thermochemical equation shows ΔH for the reaction as written. Reversing the reaction also reverses ΔH. “reverses” means “changes the sign of” 2H2O(l) 2H2(g) + O2(g)ΔH(25°C) = 571.7 kJ

  22. Enthalpy for a Reaction Depends on the States of the Reactants and Products 2H2(g) + O2(g)  2H2O(l)ΔH(25°C) = -571.7 kJ 2H2(g) + O2(g)  2H2O(g)ΔH(25°C) = -483.6 kJ but H2O(l) H2O(g)ΔH(25°C) = 44.01 kJ ?? -483.6 – (-571.7) = 88.1 kJ (not 44.01) ?? Remember, ΔH is an extensive property!

  23. Relating Heat Flow to Temperature Change and vice versa Heat capacity (C) relates the temperature change (ΔT) of the system to the heat that goes into the system (q): q = C ΔT The heat capacity of a substance is the amount of heat it takes to raise the temperature of the substance 1 K. Units: J/K

  24. Molar Heat Capacity and Specific Heat • Molar heat capacity is the amount of heat it takes to raise the temperature of one mole of a substance 1 K. Units: J/mol·K • Specific heat is the amount of heat it takes to raise the temperature of 1 gram of a substance 1K. Units: J/g·K • These are readily tabulated, whereas heat capacity is not.

  25. Using Specific Heat to Calculate the Heat Capacity C To use q = C ΔT starting with specific heat, you have to calculate the heat capacity C. C = (specific heat)(mass) The heat capacity of 16.00 g of water, which has a specific heat of 4.184 , is C = (4.184 )(16.00 g) = 66.94 J/K

  26. Using Molar Heat Capacity to Calculate the Heat Capacity C To use q = C ΔT starting with the molar heat capacity, you have to calculate the heat capacity C. C = (molar heat capacity)(moles) The molar heat capacity of water is 75.37 . The heat capacity of 45 g of water is C = (75.37 )(45 g)(1 mol/18.015 g) = 190 J/K

  27. q = CΔT Example 1 How much heat would it take to raise a liter of room temperature water to 100°C, given that the specific heat of water is 4.184 ? dwater(25°C) = 0.9969g/mL Room temperature is 25°C. ΔT = (100 – 25) = 75°C = 75K C = (specific heat) (mass of water) = (4.184 J ) (0.9969 g x 1000 mL x 1 L) gK mL 1L = 4171.03 J/K q = C ΔT = (4171.03 J/K)(75K ) = 312,827 J = 310 kJ (2 significant figures)

  28. q = CΔT Example 2 How much heat would it take to raise a liter of room temperature water to 100°C, given that the molar heat capacity of water is 75.37 ? dwater(25°C) = 0.9969g/mL C = (molar heat capacity)(moles of water) = (75.37 J ) (0.9969 g x 1000 mL x 1 L x 1 mol H2O) molK mL 1L 18.015 g = 4170.76 J/K q = C ΔT = (4170.76 J/K)(75K ) = 312,807 J = 310 kJ (2 significant figures)

  29. Calorimetry The measurement of heat flow is calorimetry. The apparatus used is a calorimeter. Calorimetry experiments let us calculate ΔH. • What we actually measure are two temperatures, Ti and Tf, then we calculate ΔT. • To convert ΔT to the enthalpy change of the system, we need to know the heat capacity or specific heat of the system.

  30. Constant-Pressure Calorimetry At constant pressure, the heat flow for a reaction measured in a calorimeter equals the change in enthalpy for the reacting system: qrxn = ΔH The design of the calorimeter is such that no heat escapes the system. The heat generated by the reaction qrxn being studied will be absorbed solely by the solution, the temperature of which we can measure. constant-pressure “coffee cup” calorimeter

  31. Constant-Pressure Calorimetry to find the specific heat of a metal The nested styrofoam cups make the calorimeter an isolated system, so qcalorimeter = 0. When a hot Cu block is put in water, it loses heat. Its T goes down. The water absorbs the heat. Its T goes up. qcalorimeter = qCu + qwater = 0 So, - qCu = qwater The process is over when Tf(Cu) = Tf(H2O). We can compute qwater using q = CΔT for the water and then find the specific heat of the Cu using q = CΔT for the Cu. constant-pressure “coffee cup” calorimeter -(specific heat of metal)(mass of metal)(ΔTmetal) = (specific heat of water)(g of water) (ΔTwater)

  32. Constant-Pressure Calorimetry to find the specific heat of a metal A 98.5 g block of metal initially at 95.0°C is put in 100. 0 g of water at 22.0°C in a coffee-cup calorimeter. The final T of the system is 31.2°C. Calculate the specific heat of the metal. Because this is in a calorimeter: - qmetal = qwater - Cmetal ΔTmetal = Cwater ΔTwater - Cmetal (31.2 - 95.0) = - Cwater(31.2 - 22.0) - Cmetal (-63.8 °C) = Cwater(9.2 °C) Cmetal = Cwater(9.2 °C) / (63.8 °C) = Cwater(0.1442) = (specific heat of water)(mass of water)(0.1442) = (4.184 J/g*K)(100.0 g)(0.1442) = (418.4 J/K)(0.1442) = 60.33 J/K

  33. Constant-Pressure Calorimetry to find the specific heat of a metal A 98.5 g block of metal initially at 95.0°C is put in 100. 0 g of water at 22.0°C in a coffee-cup calorimeter. The final T of the system is 31.2°C. Calculate the specific heat of the metal. Cmetal= 60.33 J/K (specific heat of metal)(mass of metal) = 60.33 J/K (specific heat of metal)(98.5 g) = 60.33 J/K specific heat of metal = 60.33 J/K / 98.5 g specific heat of metal = 0.61

  34. Constant-Pressure Calorimetry to find the heat of solution Example: 8.52 g LiCl is mixed with 80.00 g of water in a constant-pressure calorimeter. The initial temperature of the water is 22.30°C. The final temperature of the salt water solution is 31.64°C. Assume the specific heat of the LiCl solution is the same as that of water. Calculate the molar heat of solution of LiCl. LiCl(s)  LiCl(aq) ΔH = ? - or - LiCl(s)  Li+(aq) + Cl-(aq) ΔH = ?

  35. Constant-Pressure Calorimetry to find the heat of solution Example: 8.52 g LiCl is mixed with 80.00 g of water in a constant-pressure calorimeter. The initial temperature of the water is 22.30°C. The final temperature of the salt water solution is 31.64°C. Assume the specific heat of the LiCl solution is the same as that of water. Calculate the molar heat of solution of LiCl. LiCl(s)  LiCl(aq) ΔH = ? kJ Since the T rose, we know this process is exothermic. So the liquid inside the calorimeter absorbed the heat from the process of LiCl dissolving. qliquid = (4.184 J ) (88.52 g )(9.34 °C) g K = 3459.23 J The heat needed to dissolve the LiCl is -3459.23 J.

  36. Constant-Pressure Calorimetry to find the heat of solution Example: 8.52 g LiCl is mixed with 80.00 g of water in a constant-pressure calorimeter. The initial temperature of the water is 22.30°C. The final temperature of the salt water solution is 31.64°C. Assume the specific heat of the LiCl solution is the same as that of water. Calculate the molar heat of solution of LiCl. The heat needed to dissolve the 8.52 g of LiCl is -3459.23 J. 8.52 g LiClis 8.52 / 42.394 = 0.2010 mol. ΔH = -3459.23 J / 0.2010 mol = -17210 J/mol = -17.2 kJ/mol For 1 mole of LiCl, ΔH is -17.2 kJ/mol x 1 mol = -17.2 kJ The thermochemical equation that shows the molar heat of solution of LiCl is LiCl(s)  LiCl(aq) ΔH = -17.2 kJ

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