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

Potential energy of hiker 1 and hiker 2 is the same eventhough they took different paths.

6.7

Example

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,