Enthalpy (5.3) • Objectives – Today I will be able to: • Define state function • Calculate the enthalpy of a system • Informal assessment – monitoring student interactions and questions as they complete the practice problems • Formal assessment – analyzing student responses to the practice problems Lesson Sequence • • • • Evaluate: Warm Up Explain: Enthalpy Elaborate: Enthalpy Calculations Evaluate: Closure Warm Up Calculate the change in internal energy for a process in which a system absorbs 30 J of heat from the surroundings and does 44 J of work on the surroundings. Answer • E=q+w • q = 30 J because heat was absorbed • w = -44 J because work was done by the system • E = 30 J + (-44 J) • E = - 14 J Objectives • Today I will be able to: – Define state function – Calculate the enthalpy of a system Homework • Organic Functional Groups Quiz – Thursday, October 2 • Bring textbook to exchange • Finish practice problems Agenda • • • • Warm Up Enthalpy Notes Practice Problems Exit Ticket Enthalpy (5.3) What two components make up the total energy of a system? Work • Mechanical work is the focus for chemical and physical changes • Associated with a change in volume • Constant pressure is maintained Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Work cont. • Work involved in the expansion or compression of gases is called pressure – volume work w = - PΔV • Units: L-atm • Conversion factor 1 L-atm = 101.3 J Practice Problem • 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? Answer • w = -PΔV • W = - (1.35 atm)(0.730 L) = -0.9855 L-atm • -0.9855 L-atm (101.3J / 1 L-atm) = -99.8 J • W = -99.8 J Practice Problem 2 Answer Enthalpy (H) • Internal energy plus the product of the pressure and volume of a system H = E + PV • The equation is used to account for the absorption/release of heat and work during a chemical or physical change • Relates mainly to heat flow Enthalpy is a state function • State function – A property of a system that is determined by specifying the systems condition or state – Value of a state function depends only on the present state of the system, not on the path the system took to reach the state Example Potential energy of hiker 1 and hiker 2 is the same eventhough they took different paths. 6.7 Which of the following variables are examples of state functions? • • • • • ΔE q w H PV Enthalpy Change (ΔH) • Change in heat exchange between a system and its surroundings at constant external pressure ΔH = ΔE + PΔV Keep in mind… • ΔH = ΔE + PΔV • ΔH = (qp + w) – w • ΔH = qp • For most reactions the difference between ΔH and Δ E is small because there is not a lot of work • If PΔV is small it can be ignored from calculations Closure • Complete practice problems: – 5.31, 5.32, 5.37,