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# Chapter 5 Written by JoAnne Swanson University of Central Florida - PowerPoint PPT Presentation

Thermodynamics. Chapter 5 Written by JoAnne Swanson University of Central Florida. Topics. Types of energy and units of energy Exothermic vs. endothermic reactions Heat capacity and specific heat Energy transfers in changes of state Calorimetry Enthalpy and entropy Hess’s Law.

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Chapter 5

Written by

JoAnne Swanson

University of Central Florida

• Types of energy and units of energy

• Exothermic vs. endothermic reactions

• Heat capacity and specific heat

• Energy transfers in changes of state

• Calorimetry

• Enthalpy and entropy

• Hess’s Law

• There are two categories of energy:

• Kinetic energy.

• Potential energy

• Kinetic Energy is the energy of motion

• Potential energy is energy of position,

Chemical energy is a form of potential energy stored in the structure of a chemical substance due to its composition.

Ek = ½ (mv2) or mv2/2

mass is in Kg,

and Kg.m2/s2 = 1 Joule, the unit used to designate energy.

From this equation you should see that Ek increases with mass and with velocity

Ep = mgh = (mass, gravity, height)

(gravitational constant = 9.8 m/s2)

But Ep of submicroscopic substances (molecules, ions, etc.)

electrical charges between particles.

Electrical charge is designated by the symbol ‘Q’ and has the value of an electron charge = 1.60 x 10-19 C

The unit for electrical charge is the coulomb ( C).

Eel = kQ1Q2 / d where k=8.99x109Jm

C2

This equation shows the electrostatic forces of attraction between two particles, 1 and 2, are inversely proportional to the distance between the particles.

This differs from the nutritional calorie which is 1C = 1000c = 1Kcal

Thermochemistry is the study of heat transfers occurring in chemical and physical changes of substances.

• Thermal energy

• random motion of atoms and molecules.

The Universe is made up of the chemical and physical changes of substances.system and the surroundings

The system is the part of the universe that is of interest to us.

The surroundings is everything outsideof the system.

Energy is transferred as heat and work. chemical and physical changes of substances.

Work = Force x Distance

= pressure x volume

It usually pertains to either the electrical work (for instance in a battery) or mechanical work (like the pressure exerted on or by a gas in a reaction).

• The change in Internal Energy of a system = chemical and physical changes of substances.DE = q + w

• DE is the change in internal energy

• q = heat

• w = work

• Heat can flow into the system or out of the system.

• sign on q

• Work can be done on the system or by the system.

• sign on w

Exothermic vs. Endothermic Processes chemical and physical changes of substances.

• Any process that gives off heat to the surroundings is an Exothermic process.

• When a process chemical and physical changes of substances.absorbs heat from the surroundings it is an Endothermic process. When ice melts it absorbs heat from the surroundings, thus it is an endothermic process.

To measure the change in internal energy of a chemical reaction, the reaction is often carried out in a “bomb calorimeter”. There is no change in volume in this sealed container, therefore there is no work involved and

DE = qreaction (at constant volume)

The symbol reaction, the reaction is often carried out in a “bomb calorimeter”. There is no change in volume in this sealed container, therefore there is no work involved and DH is used to represent the change in heat into or out of the system. It is defined as the change in enthalpy.

Enthalpy and Internal Energy are nearly equal. Enthalpy takes into account reactions at constant pressure where an expanding gas does work on the surroundings.

Remember w = PV, but if w is done by the system, w = -PV

H = E +PV and DH = DE + DPV

w = - DPV so, DH = (q + w) – w

DH = q (at constant pressure)

• 1 mole gas at 25 takes into account reactions at oCand 1atm has a volume of 24.5L =( 24.5 L.atm / mol )

• 1 L.atm = 0.1013 kJ

• DPV makes little difference in the DH value, and even less difference for liquids or solids.

• why?

If 1 L takes into account reactions at .atm = 0.1013 kJ how many kJ are in the molar volume of a gas at 25oC?

24.5 L.atm x 0.1013 kJ = 2.5 kJ

1 L.atm

see next example

Examples: given the values for PV, calculate the change in enthalpy for the combustion of methane where DE = -885 kJ

PV at 25oC

CH4(g) 2.5 KJ/ mol

O2(g) 2.5 KJ/ mol

CO2(g) 2.5 KJ/ mol

H20(l) 0.0018 KJ/ mol

DH = DE + DPV

(DPV = PV products – PV reactants)

continued………

CH enthalpy for the combustion of methane where 4(g) + 2 O2 (g)  CO2(g) + 2 H2O (l)

DPV = PV CO2 + 2PV H2O –[PV CH4 + 2PV O2

DPV = [2.5 KJ + 2(0.0018 KJ)] – [2.5 KJ + 2(2.5 KJ) ]

DPV = -5.0 KJ

DH = -885 KJ - 5.0 KJ = -890 KJ

Carried out at constant pressure, DH = -885 KJ

Not a big difference when DPV is added

• EXAMPLE 2: enthalpy for the combustion of methane where

• Calculate the kinetic energy in joules, of a 45 g golf ball moving at 61 m/s.

• Convert this energy to calories.

• What happens to the energy when the ball strikes a tree?

• Since 1J = 1kg m2/s2

• change g to kg.

• 45 g = 0.045 kg

• E enthalpy for the combustion of methane where k = mv2 / 2 = 0.045 kg x (61m/s)2

• 2

• = 84 kg m2= 84 J

• s2

• 84 J x 1cal = 20 cal

• 4.184 J

• c. When the ball hits the tree, velocity = 0 and so does Kinetic energy. Kinetic Energy is transferred as potential energy to the tree and the deformed golf ball.

Example 4: enthalpy for the combustion of methane where

Calculate DE when a balloon is inflated completely by heating it. The volume changes from 4.00 x 106 L to 4.50 x 106 L, with the addition of 1.3 x 108 J of heat. The balloon expands against an internal pressure of 1.0 atm. DE = q+w

w = - DPV because ________________

q is positive because _________________

D enthalpy for the combustion of methane where V = 4.50 x 106L - 4.00 x 106L = 5.00 x 105 L

-PDV = 5.00 x 105 L x (-1.0 atm) =5.00 x 105 Latm = -5.07 x 107 J

DE = 1.3 x 108J – 5.07 x 107J = 7.9 x 107 J

Example 3: enthalpy for the combustion of methane where

Calculate DE when a gas is compressed from 6.0 x 104L to 5.3 x 104L under an external pressure of 1.3 atm. The amount of heat transferred to the surroundings was 4.3 x 102 J.

DE = w + q, where w = +DPV why?

DV = 5.3 x 104L - 6.0 x 104L = - 7.0 x 103L

DPV = (1.3 atm)(- 7.0 x 103L) = -9.1 x 103Latm

-9.1 x 10 enthalpy for the combustion of methane where 3 Latm x 0.1013 kJ = -9.2 x 102 KJ

Latm

In this case since the gas was compressed, w = + and since heat was transferred to the surroundings, q = - .

DE = w – q, = -9.2 x 105 J – 4.3 x 102 J =

= - 9.2043 x 105 J

= - 9.2 x 105 J

The enthalpy of reaction is the enthalpy for the combustion of methane where difference between the enthalpies of the products and reactants.

DH(rxn) = S DHf(products) – S DHf(reactants)

energy expelled enthalpy for the combustion of methane where

Ea

Ea

products

Energy

reactants

Energy

products

reactants

reaction progression

reaction progression

Endothermic

Exothermic

The diagram illustrates energy enthalpy for the combustion of methane where givenoff when bonds are madebetween H2 and O2 , (a). Energy is absorbed when bonds are brokenin HgO , (b).

The units of energy are Kilojoules and Calories. Enthalpy of reaction will be expressed as Kilojoules (Kj).

It is important to note the following when calculating the enthalpy of reaction:

• The enthalpy of a substance in its standard state is equal to zero.

• Enthalpy is dependent on the quantity of the substance therefore the enthalpy of a substance must be multiplied by its coefficient in the chemical equation.

Enthalpy of changed.formation values of compounds are found in tables of thermodynamic data.

The enthalpy of formation is defined as the energy associated with the formation of 1 mole of a substance from its elements in their standard states.

Na(s) + 1/2 Cl2 (g) NaCl

The formation equation for sodium chloride

Ex. 1 changed.

Determine the enthalpy for the following chemical reaction:

CO2 (g) + 2 H2O (l) -----> 2 O2 (g) + CH4 (g)

Use appendix C pg. 1041 in your text to find the individual enthalpy values.

CO2 (g) = -393.509, H2O (l) = -285.83

CH4 (g) = -74.81 , O2(g)= 0

CO2 + 2 H2O  2 O2 + CH4

DH =SDHf(products) – S DHf(reactants)

DH = (0 + -74.81)-(-393.509 + 2(- 285.83))

= -74.81 + 965.17 = 890.36 Kj

HESS’S LAW enthalpy values.

ENTHALPY IS A STATE FUNCTION. ENTHALPY OF A REACTION WILL BE THE SAME NO MATTER WHAT PATH IS TAKEN TO ARRIVE AT THE PRODUCTS.

This means that if we need to calculate the enthalpy of a reaction for which we do not know the enthalpies of formation, we can algebraically manipulate other reactions to arrive at the enthalpy for the desired reaction.

Ex. Determine the reaction for which we do not know the enthalpies of formation, we can DHrxn for

Sn(s) + 2 Cl2 (g) SnCl4(l)

Given the following:

Sn(s) + Cl2 (g) SnCl2(s) DH1= -349.8 kJ

SnCl2(s) + Cl2 (g) SnCl4(l) DH2= -195.4 kJ

Sn (s) + 2 Cl2 (g) SnCl4(l) DH3= ?

DH1 + DH2 = DH3 = -545.2 kJ

Ex. reaction for which we do not know the enthalpies of formation, we can Calculate theDHrxn for

S (s) + O2(g) SO2(g)

Given

S (s) + 3/2 O2(g) SO3(g) DH= -395.2 kJ

SO2(g) + O2(g) 2 SO3(g) DH= -198.2 kJ

NOTICE THAT WE NEED SO2(g) TO BE ON THE PRODUCT SIDE. THEREFORE WE MUST REVERSE THE SECOND RXN

CONTINUED

S reaction for which we do not know the enthalpies of formation, we can (s) + O2(g) SO2(g)

S (s) + 3/2 O2(g) SO3(g) DH= -395.2 kJ

2SO3(g) O2(g)+ 2 SO2(g) DH= +198.2 kJ

IT IS ALSO NECESSARY TO DIVIDE EQUATION 2 BY 2, SO THAT WE HAVE 1SO2(g) IN THE PRODUCT. WE MUST ALSO DIVIDE THE DH +198.2 kJ / 2 = 99.10 kJ

Now add the reactions and the enthalpies. reaction for which we do not know the enthalpies of formation, we can

S (s) + 3/2 O2(g) SO3(g) DH= -395.2 kJ

SO3(g) 1/2 O2(g)+ SO2(g) DH= +99.1 kJ

S (s) + O2(g) SO2(g)

DH= -296.1 kJ

The enthalpy of formation of reaction for which we do not know the enthalpies of formation, we can a substance in a chemical reaction can also be calculated if the enthalpy of reaction is known and the enthalpies of the other substancesin the reaction areknown. Simply call the unknown enthalpy ‘X’ and plug in all known values to solve for ‘X’.

example on transparency

Specific heat and heat Capacity reaction for which we do not know the enthalpies of formation, we can

Heat capacity is the amount of heat required to raise the temperature of a given quantity of a substance by 1o C.

Specific Heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1o C. J/goC

Molar heat capacity is J / mol oC

• a substance’s reaction for which we do not know the enthalpies of formation, we can ability to absorb heat and to store heat.

• For example, a metal _______________

• ________________________________

• A liquid like water (which contains strong intermolecular forces, hydrogen bonds), __________________________________________________________________

The reaction for which we do not know the enthalpies of formation, we can metal has a ______________ and the water has a ________________.

The specific heat capacity ( C ) of water is an important and easy number to remember ; C = 1 cal / g oC ,

(or 4.184 J / goC )

Since oC and Kelvin have the same size increments, they are interchangeable in these equations.

Calorimetry reaction for which we do not know the enthalpies of formation, we can is a technique used in the lab to measure the change in enthalpy of a reaction. One apparatus used is the Bomb Calorimeter. It is a Heavy walled, steel container which has a known specific heat.

• Types of Calorimetry problems we will cover: reaction for which we do not know the enthalpies of formation, we can

• simple change in water temperature

• change in state of water

• a rxn in a coffee cup calorimeter

• a rxn in a bomb calorimeter

• a mixture of hot substance with cold substance in a calorimeter.

• Equations needed: reaction for which we do not know the enthalpies of formation, we can

• q = C x m x DT

• q = qice + DHf + qwater + DHv + qsteam

• qrxn = DHrxn = -qwater

• DHrxn = -[qwater + qbomb], or = -(Ccal x DT)

• qcold = -qhot

Some needed values: reaction for which we do not know the enthalpies of formation, we can

specific heat of water(l) = 4.184 J/gK

specific heat of water(s) = 2.05 J/gK

specific heat of water(g) = 2.01 J/gK

DHf = enthalpy of fusion = energy involved in a change of state from liquid to solid or vice versa = 333 J/g

DHv = enthalpy of vaporization = energy involved in a change of state from liquid to vapor or vice versa = 2444 J/g

The heat of the reaction (q) is equal to the reaction for which we do not know the enthalpies of formation, we can negative of the heat absorbed by the(bomb plus the water surrounding the bomb).

qrxn = DHrxn = - (qbomb + q water )

q reaction for which we do not know the enthalpies of formation, we can is used to represent the quantity and direction of heat transferred, using a calorimeter.

Necessary Equations: reaction for which we do not know the enthalpies of formation, we can

q = C x m x DT

q = (specific heat)(mass)(change in Temp.)

q = (J/gK)(g)(DT)

q(rxn) = - (q of the water + q of the bomb)

Subliminal message..... reaction for which we do not know the enthalpies of formation, we can

Wake up !!!

1. reaction for which we do not know the enthalpies of formation, we can Calorimetry

A 466g sample of water is heated from 8.50oC to 74.60oC. Calculate the amount of heat absorbed by the water.

q = (sp. heat)(mass)(DT)

q = (4.184 J/goC)(466g)(74.60-8.50oC)

q = 1.29 x 105 J = 129 KJ

change of state of water example on transparency. reaction for which we do not know the enthalpies of formation, we can

coffee cup calorimeter on transparency

Ex. 3 reaction for which we do not know the enthalpies of formation, we can

1.435 g of Naphthalene (molar mass =128.2) was burned in a bomb calorimeter. The temp. rose from 20.17oC to 25.84 oC. The mass of the water surrounding the calorimeter was 2000.g and the heat capacity of the bomb was 1.80 KJ/oC. Calculate the heat of combustion of Naphthalene.

q(rxn) = -(q water + q bomb) reaction for which we do not know the enthalpies of formation, we can

q(water) = (2000g)(4.184 J/goC)(5.67oC)

= 4.74 x 104

q(bomb) = (1.80 x 103J / oC)(5.67oC)

= 1.02 x 104J

q(rxn) = -(4.74 x 104 J + 1.02 x 104 J)

= -5.76 x 104 J

MORE EXAMPLES WILL BE GIVEN IN CLASS. reaction for which we do not know the enthalpies of formation, we can

SINCE I HAD TO RUSH TO GET THESE NOTES ON LINE, THEY WERE NOT COMPLETELY EDITED. THERE IS SOME REPETITION, and they may need some corrections.