Understanding Thermochemistry: Enthalpies & Laws

1. Chapter 7: Thermochemistry Juana Mendenhall, Ph.D. Assistant Professor Lecture 2 March 31 General Chemistry: Chapter 20

2. Objectives • Define work and it’s units. • Define, explain, and apply the first law of thermodynamics. • Apply the direct and indirect method of the standard enthalpy of formation and reaction. General Chemistry: Chapter 20

3. Standard Enthalpies of Formation Hf° • The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. • The standard enthalpy of formation of a pure element in its reference state is 0. General Chemistry: Chapter 20

4. Standard Enthalpy of Formation and Reaction • H is an extensive property. • Enthalpy change is directly proportional to the amount of substance in a system. aA + bB → cC + dD Hrxn = [cHf(C )+ dHf(D)] - [aHf(A) + bHf(B)] Hrxn = [nHf(products)] - [mHf(reactants)] Direct Method Example with C(graphite) General Chemistry: Chapter 20

5. Standard States General Chemistry: Chapter 20

6. Standard Enthalpies of Formation General Chemistry: Chapter 20

7. Indirect Determination of H:Hess’s Law • Hess’s Law states: when reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ • H changes sign when a process is reversed NO(g)→ ½N2(g) + ½O2(g)H = -90.25 kJ General Chemistry: Chapter 20

8. ½N2(g) + ½O2(g) → NO(g)H = +90.25 kJ NO(g) + ½O2(g) → NO2(g)H = -57.07 kJ Hess’s Law • Hess’s law of constant heat summation • If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N2(g) + O2(g) → NO2(g)H = +33.18 kJ General Chemistry: Chapter 20

9. Hess’s Law Schematically General Chemistry: Chapter 20

10. Functions of State • Any property that has a unique value for a specified state of a system is said to be a State Function. • Water at 293.15 K and 1.00 atm is in a specified state. • d = 0.99820 g/mL • This density is a unique function of the state. • It does not matter how the state was established. General Chemistry: Chapter 20

11. Work kg m m = J Fd w = [w ] = s2 7-4 Work • In addition to heat effects chemical reactions may also do work. • Gas formed pushes against the atmosphere. • Volume changes. • Pressure-volume work. General Chemistry: Chapter 20

12. Pressure Volume Work w = F  d = (m  g) h (m  g)  A h = A = PV w = -PextV General Chemistry: Chapter 20

13. Definition of Terms • State of a system: the values of all relevant macroscopic properties-example: energy, temp., pressure, volume. • State function: ppts that are determined by the state of the systems. • U is a function of state. • Not easily measured. • U has a unique value between two states. • Is easily measured. General Chemistry: Chapter 20

14. 7-5 The First Law of Thermodynamics • Internal Energy, U. • Total energy (potential and kinetic) in a system. • Translational kinetic energy. • Molecular rotation. • Bond vibration. • Intermolecular attractions. • Chemical bonds. • Electrons. General Chemistry: Chapter 20

15. First Law of Thermodynamics • A system contains only internal energy. • A system does not contain heat or work. • These only occur during a change in the system. • Law of Conservation of Energy • The energy of an isolated system is constant U = q + w General Chemistry: Chapter 20

16. First Law of Thermodynamics General Chemistry: Chapter 20

17. Sign conventions for work & heat General Chemistry: Chapter 20

18. Example The work done when a gas is compressed in a cylinder like that show in the figure is 462 J. during this process, there is a heat transfer of 128 J from the gas surroundings. Calculate the energy change for this process. General Chemistry: Chapter 20