Energy Thermodynamics

Law of Conservation of Energy

Energy can be converted from one form to another but can neither be created or destroyed.

(E_universe is constant)

Energy is….

·        The ability to do work.

·        Conserved.

·        Made of heat and work.

·        A state function.

·        Independent of the path, or how you get from point A to point B.

·        Work is a force acting over a distance.

·        Heat is energy transferred between objects because of temperature difference.

The Universe

·        Is divided into two halves.

·        The system and the surroundings.

·        The system is the part you are concerned with.

·        The surroundings are the rest.

·        Exothermic reactions release energy to the surroundings.

·        Endothermic reactions absorb energy from the surroundings.

The Two Types of Energy

Potential: due to position or composition-can be converted to work.

Kinetic: due to motion of the object.

KE = ½ mv²

(m = mass, v = velocity)

Direction

·        Every energy measurement has three parts.

1.     A unit (Joules of calories).

2.     A number (how many).

3.     And a sign to tell direction.

·        Negative-exothermic

·        Positive-endothermic

Same rules for heat and work

·        Heat given off is negative.

·        Heat absorbed is positive.

·        Work done by system on surroundings is negative.

·        Work done on system by surroundings is positive.

·        Thermodynamics-The study of energy and the changes it undergoes.

First Law of Thermodynamics

·        The energy of the universe is constant.

·        Law of conservation of energy.

·        Q = heat (J)

·        W = work

·        DE = q + w

·        Take the systems point of view to decide signs.

What is work?

·        Work is a force acting over a distance.

·        W = F x Dd

·        P = F/area

·        D = V/area

·        W = (P x area) x D(V/area) = PDV

·        Work can be calculated by multiplying pressure by the change in volume at constant pressure.

·        Units of liter – atm   L-atm

Work needs a sign

·        If the volume of a gas increases, the system has done work on the surroundings.

·        Work is negative

·        W = -PDV

·        Expanding work is negative.

·        Contracting, surroundings do work on the system, w is positive.

·        1 L-atm = 101.3 J

Examples

1.  What amount of work is done when 15 L of gas is expanded to 25 L at  

2.4 atm pressure?

2. If 2.36 J of heat are absorbed by the gas above, what is the change in energy?

Answers

1. 15Là 25L

P = 2.4 atm.

W= -PDV    W= (-2.4)(25-15) = -24L-atm. OR –2400J

To change –24L-atm to –2400J you must multiply by 101.3 since 1 L-atm = 101.3J.

2.     DE = q + w

      = 2.36J – (-2400J )

      = -2397.64J = -2398J

Enthalpy

·        Abbreviated H

·        H = E + PV (that’s the definition)

·        At constant pressure.

·        DH = DE + PDV

·        The heat at constant pressure q_p c an be calculated from

·        DE  = q_p + w = q_p - PDV

·        q_p = DE + PDV = DH

Example

1.     For the reaction:

S (s) + O₂ (g) ® SO₂ (g)           DH = -296 kJ/mol

a.      How much heat is evolved when 275 g sulfur is burned in excess O₂?

b.     How much heat is evolved when 25 mol sulfur is burned in excess O₂?

c.     How much heat is evolved when 150. g sulfur dioxide is produced?

Answers

1.     a 

b.

c. 

Calorimetry

·        Measuring heat.

·        Use a calorimeter.

·        Two kinds

·        Constant pressure calorimeter (called a coffee cup calorimeter).

·        Heat capacity for a material, C is calculated.

·        C = heat absorbed / DT = Dh / DT

·        Specific heat capacity = C/mass

·        Molar heat capacity = C/moles

·        Heat = specific heat x m x DT

·        Make the units work and you’ve done the problem right.

·        A coffee cup calorimeter measures DH.

·        An insulated cup, full of water.

·        The specific heat of water is 1 cal./g °C.

·        Heat of reaction = DH = s.h. x mass x DT

Examples

1.     The specific heat of graphite is 0.71 J/g°C. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

2.     A 46.2 g. sample of copper is heated to 95.4 °C and then placed in a calorimeter containing 75.0 g. of water at 19.6 °C.  The final temperature of both the water and the copper is 21.8 °C. What is the specific heat of copper?

Calorimetry continued

·        Constant volume calorimeter is called a bomb calorimeter.

·        Material is put in a container with pure oxygen.  Wires are used to start the combustion. The container is put into a container of water.

·        The heat capacity of the calorimeter is known and tested.

·        Since DV = 0, PDV = 0, DE = q.

Properties

·        Intensive properties not related to the amount of substance:

Density, specific heat, temperature.

·        Extensive property-does depend on the amount of stuff.

·        Heat capacity, mass, heat from a reaction.

Hess’s Law

·        Enthalpy is a state function.

·        It is independent of the path.

·        We can add equations to come up with the desired final product, and add the DH.

·        Two rules:

1.     If the reaction is reversed the sign of DH is changed.

2.     If the reaction is multiplied, so is DH.

Standard Enthalpy

·        The enthalpy change for a reaction at standard conditions (25°C, 1atm., 1 M solutions)

·        Symbol DH°

·        When using Hess’s Law work by adding the equations up to make it look like the answer.

·        The other parts will cancel out.

Examples

1.     Given:          DH° = -1300.kJ, DH° = -394kJ, DH° = -286kJ

Calculate DH° for this reaction.

2.     Given:          DH° = +77.9kJ, DH° = +495kJ, DH°= +435.9kJ

Calculate DH° for this reaction.

Standard Enthalpies of Formation

·        Hess’s Law is much more useful if you know lots of reactions.

·        Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.

·        Standard states are 1 atm., 1 M and 25°C

·        For an element it is 0.

·        There is a table in Appendix 4 (pg A22)

·        Need to be able to write the equations.

·        What is the equation for the formation of NO₂?

·        ½ N₂ (g) + O₂ (g) ® NO₂ (g)

·        Have to make one mole to meet the definition.

·        Write the equation for the formation of methanol CH₃OH.

Since we can manipulate the equations…

·        We can use heats of formation to figure out the heat of reaction.

·        Let’s do it with this equation.

·        C₂H₅OH + 3O₂ (g) ® 2CO₂ +3H₂O

·        Which leads us to this rule.

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