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The Nature of Energy The First Law of Thermodynamics Enthalpy Enthalpies of Reaction Calorimetry

CHM 101 – Chapter Five. The Nature of Energy The First Law of Thermodynamics Enthalpy Enthalpies of Reaction Calorimetry Hess’ Law Enthalpies of Formation. CHM 101 – Chapter Five. The Nature of Energy. Chemical reactions frequently involve large exchanges of energy.

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The Nature of Energy The First Law of Thermodynamics Enthalpy Enthalpies of Reaction Calorimetry

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  1. CHM 101 – Chapter Five • The Nature of Energy • The First Law of Thermodynamics • Enthalpy • Enthalpies of Reaction • Calorimetry • Hess’ Law • Enthalpies of Formation CHM 101 - Reeves

  2. CHM 101 – Chapter Five The Nature of Energy Chemical reactions frequently involve large exchanges of energy Sometimes, the reaction releases large amounts of energy to the surroundings in the form of heat and light. CHM 101 - Reeves

  3. CHM 101 – Chapter Five The Nature of Energy Chemical reactions frequently involve large exchanges of energy Less commonly, chemical reactions absorb energy from the surroundings. CHM 101 - Reeves

  4. CHM 101 – Chapter Five Forms of Energy Kinetic Energy: Energy associated with motion: Potential Energy: Energy associated with position: CHM 101 - Reeves

  5. CHM 101 – Chapter Five Kinetic Energy & Units of Energy What is the kinetic energy of a 1500 lb automobile traveling at 100 mph (45 m/s)? The energy unit Joule is defined as: Thus, the energy of the automobile is CHM 101 - Reeves

  6. CHM 101 – Chapter Five The First Law of Thermodynamics The total amount of energy in the universe is constant. Energy cannot be created or destroyed. If no energy crosses the boundary, the total energy of the system cannot change. CHM 101 - Reeves

  7. CHM 101 – Chapter Five The First Law of Thermodynamics system surroundings Energy that crosses the boundary from the surroundings into the system increases the energy of the system and is considered positive. Energy can only cross the boundary in the form of heat (q) or work (w). Heat flowing into the system or work done on the system is defined as positive. Heat flowing out of the system or work done by the system is defined as negative. CHM 101 - Reeves

  8. CHM 101 – Chapter Five The First Law of Thermodynamics Thus the First Law is: What is the change in the energy of a system if 1 kJ of heat is absorbed by the system and 500J of work are done by the system? CHM 101 - Reeves

  9. CHM 101 – Chapter Five Functions of State A function of state is one that depends only on the state of the system, as defined by its contents and conditions such at temperature and pressure. Internal energy (E) is a function of state. The change of a state variable such as DE depends only on the initial and final states of the system, NOT on the path. CHM 101 - Reeves

  10. A Dh CHM 101 – Chapter Five Work • Work is defined as force times distance in the direction of the force. Since work done by the system is negative: • If the system expands or contracts against a pressure P, then "PV" work will result. Consider the expansion of the cylinder below. CHM 101 - Reeves

  11. However, most processes in chemistry occur at constant pressure, where PV work can occur. Thus we define enthalpy (H) as Then at constant pressure: CHM 101 – Chapter Five Enthalpy Work may take many forms including mechanical, electrical, etc. In this course, we will not explicitly consider systems that involve work in these forms. Thus, if a process takes place at constant volume, DV = 0 Then, the change in DE can be determined directly by measuring the heat that flows in a constant volume process. CHM 101 - Reeves

  12. CO2(g) + 2H2O(l) CH4(g) + 2 O2(g)DH = 890 kJ CHM 101 – Chapter Five Enthalpy Like internal energy (E), enthalpy (H) is a state function. Thus, to carry out the reverse reaction, 890 kJ must be put into the system, an endothermic process. CHM 101 - Reeves

  13. CH4(g) + 2 O2(g) CO2(g) + 2H2O(l) CHM 101 – Chapter Five Enthalpy Determine the enthalpy change when 1.5 g of CH4 is burned in excess oxygen. DH = -890 kJ What mass of oxygen would be consumed if this reaction released 100 kJ of heat? CHM 101 - Reeves

  14. CHM 101 – Chapter Five Calorimetry If a process such as a chemical reaction is carried out at constant pressure, the heat (q) is a direct measure of the enthalpy change (DH) If the system is surrounded by a water bath (surroundings), the flow of heat will result in a temperature change of the water bath. CHM 101 - Reeves

  15. CHM 101 – Chapter Five Calorimetry Heat and temperature are NOT the same; the amount of heat (q) required to induce a temperature change (DT) in a substance depends on: • The mass (m) of the substance, • The size of the temperature change (DT), and • The specificheat capacity (CP) or specific heat of the substance. CPis the amount of heat required to raise the temperature of one gram of the substance by one degree centigrade. The specific heat capacity of water is 4.184 CHM 101 - Reeves

  16. CHM 101 – Chapter Five Calorimetry If 200 g of copper metal (CP = 0.39 ) at 100oC is immersed in 100 g of water (CP = 4.184 ) at 20oC in an insulated container, what it the final temperature? CHM 101 - Reeves

  17. CHM 101 – Chapter Five Calorimetry If 200 g of copper metal (CP = 0.39 ) at 100oC is immersed in 100 g of water (CP = 4.184 ) at 20oC in an insulated container, what it the final temperature? CHM 101 - Reeves

  18. CHM 101 – Chapter Five Calorimetry If 1.00 g of hydrogen gas were burned in an open container surrounded by a 10.0 kg water bath originally at 25.00 oC, the water temperature would rise to 28.42 oC. What is the enthalpy change for the combustion of hydrogen? 2 H2(g) + O2(g) 2 H2O(l) CHM 101 - Reeves

  19. CHM 101 – Chapter Five Hess’ Law When two moles of carbon is burned in limited oxygen, carbon monoxide is formed. How might the enthalpy change of this reaction be determined? The enthalpy change cannot be measured directly because combustion in limited oxygen produces both CO(g) and CO2(g). CHM 101 - Reeves

  20. CHM 101 – Chapter Five Hess’ Law 2 C(s) + O2(g) 2 CO(g) DH = ? While this reaction cannot be carried out cleanly, two related combustion reactions can be done: 1) The combustion of carbon in excess oxygen produces carbon dioxide exclusively. 2) The combustion of carbon monoxide produces carbon dioxide exclusively. CHM 101 - Reeves

  21. C(s) + O2(g) CO2(g) CO(g) + O2(g) CO2(g) CHM 101 – Chapter Five Hess’ Law 2 C(s) + O2(g) 2 CO(g) DH = ? Hess’ Law: If two or more equations can be added together to produce a desired reaction, the enthalpy of the desired reaction will be the sum of their enthalpies. DH = -394 kJ DH = -283 kJ CHM 101 - Reeves

  22. CHM 101 – Chapter Five Hess’ Law The combustion of nitrogen monoxide forms nitrogen dioxide. Calculate the heat of combustion of nitrogen monoxide given the following: CHM 101 - Reeves

  23. CHM 101 – Chapter Five Enthalpy of Formation Hess' Law can be formalized by defining Enthaply of Formation, which is the enthalpy change (DHf0) that occurs when one mole of a substance is formed from its elements in their standard state at 25oC. The Standard state of an element is its most stable form at 25oC. For example, carbon is a solid (graphite), oxygen and hydrogen are both diatomic gases. Thus for methane (CH4) the enthalpy of formation is represented by the equation CHM 101 - Reeves

  24. CHM 101 – Chapter Five Enthalpy of Formation Enthalpies of Formation have been determined for a large number of compounds, and are found in tabular form in chemistry reference texts CHM 101 - Reeves

  25. CHM 101 – Chapter Five Enthalpy of Formation Using Hess’ Law, it can be shown the Enthalpies of Formation ( ) can be combined to determine the enthalpy change ( ) for any chemical reaction. Thus: CHM 101 - Reeves

  26. CHM 101 – Chapter Five Enthalpy of Formation Calculate the enthalpy change (DH) the reaction of gaseous hydrogen peroxide (H2O2) with hydrogen to form liquid water. H2O2(g) + H2(g) 2 H2O(l) = -435.5kJ CHM 101 - Reeves

  27. CHM 101 – Chapter Five Enthalpy of Formation To see how enthalpies of formation can be applied, consider the combustion of methane: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) First, break the reactants down into their elements: CHM 101 - Reeves

  28. CHM 101 – Chapter Five Enthalpy of Formation CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) Then, make the products from those elements: CHM 101 - Reeves

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