Chemistry 3202 Squires Thermo Notes 2014 (Text Ch 16 and 17, page 624)

Thermodynamics is the study of energy changes that happen when matter changes, when it melts or boils (physical change)or burns (chemical change). Thermo is greek for hot, dynamics means power derived from movement.

Energy - the ability to do work, “en” is greek word for in, & ergon is for work. Energon = energy, or “work in”.This energy can be changed from one form to another, for example electrical energy can be changed to light.

2 Types of energy:1) Kinetic - Greek work for movement is kinema (like cinema for moving ) where

temperature is used to describe the average kinetic energy of particles2) Potential - stored energy in the bonds of compounds, or the intermolecular forces

between molecules in phase changes .

Energy UnitsThe unit is the joule (J) & kilojoule = 1000 J & calories (still is for food only food is called calories but is actually a Kilocalorie.

Physical ChangeExample using a Physical Change of state: H2O(l) + energy (40.7 KJ/mol)

H2O(g)

The energy, in this case, is absorbed and the reaction is thus endothermic. For the process:

H2O(l) H2O(s) + energy (6.02 KJ/mol) (I call this freezing) The reverse is weirdly called fusion, or melting

In this process, freezing, heat is lost and the phase change is exothermic. It involves IMF’s and is a potential energy change.

Chemical Change- where the chemicals changeUnlike in a physical chage, where the chemicals stay the same, as in melting and boiling.In a chemical change, one substance(s)(reactants) with a specific amount of energy, is changed into another substance(s)(products) with a different amount of energy.

-Energy changes involve changes in bonding; thus the energy involved is potential energy.-Changes in potential energy are due to changes in the covalent and ionic bonds. Actually, reactant bonds break and rearrange to make new product bonds.-For all reactions, energy is absorbed for bonds to break reactant bonds and is released when new product bonds form.

Conversion from one form of energy to another obeys the first Law of Thermodynamics, no energy is lost or made whenwe change forms of energy, say the chemical change of the burning of coal in a power station into:

Thermal (heat) energy from the burning

Electrical energy from the spinning turbines coils of wire (in a magnetic field) sent to your home.

Mechanical (kinetic + potential) energy of the trbines Light energy for you to read this by Chemical energy again , if you are reading this by a battery charged laptop!

Example of a chemical change:1. Propane burns: C3H8(g) + 5O2(g) 3CO2 + 4 H2O(g) + Energy (Exothermic Reaction)

2. Photosynthesis (MEMORIZE THIS ENDOTHERMIC REACTION) . It is often asked.6CO2(g) + 6 H2O(g) + 3177.6 kJ C6H12O6(l) +

6O2(g)

Photosynthesis is an endothermic reaction due to the reaction requiring 3177 kJ of energy for the formation of one mole of glucose. The energy is stored in the glucose as chemical potential energy.The heat of the reaction can also be written OUTSIDE of te equation and then a sign (here positive for Endo thermic) is used:

6CO2(g) + 6 H2O(g) C6H12O6(l) + 6O2(g) ΔH = +3177 kJ

The opposite of photosynthesis, cellular respiration (a form of burning), is accompanied by an energy release.

C6H12O6(l) + 6O2(g) 6CO2(g) + 6 H2O(g) + 3177 kJORC6H12O6(l) + 6O2(g) 6CO2(g) + 6 H2O(g) ΔH = -3177 kJ

Thus, in an endothermic reaction, the gaining of heat energy can be written in two different ways:(1) In the equation, on the reactant side or (2) Outside the equation as ΔH = +

In an exothermic reaction, the releasing of heat energy can be written:(1) In the equation, on the product side or (2) Outside the equation as ΔH = -

Historical Backgrounf to the development of the Joule

Energy is the ability to do work or heat. Physics students calculate work as: W=F x d (m x a x d). This is easy for Physics students to measure, heck , all they need is a meter stick and balance. In chemistry we have a hard time catching heat, and measuring it is even harder. Heat is not tangible, you can’t grab it.

James Joules experiment solves all that. He catches heat and converts the POTENTIAL energy of a raised brick into the KINETIC energy of a moving brick. This mechanically generated value, gives 4.184 J /g C (the mechanical equivalent of heat). (Note: In 1850, Joule published a refined measurement of 772 ft·lb or Btu (4.1 J/cal) or 772 pounds raised 1 foot generates 1 Fahrenheit temperature change in 1 pound of water. We use Metric today or 4.184 J/g C

The First Calorimeter (as seen in the film absolutezero.)

Temperature Meaasures Kinetic EnergyTemperature (T): a measure of the average kinetic energy of the particles. Or a thermometer is a molecular speedometer.

Molar Heat of Combustion

Burning one mole of a substance is called the molar heat of combustion. An Example: The molar heat of combustion of methane gas is 890.4 kJ.

CH4(g) + 2O2(g) CO2(g) + H2O(g) ΔH = -890.4 kJ

Systems, Surrounding, and Universe

The First Law of Thermodynamics Energy is not created or destroyed, it is transformed, and conserved, so this is also known as the Law of Conservation of Energy. Or, more simply, heat is transferred from a hotter object to () a colder object, again, a one way street. So if we have 2 blocks of Al (same mass) one at 80 oC touching the other one at 20 oC , they will equal out to 50 oC as the hot block losses it’s energy to the cold one, NOT the other way around. HOT TO COLD, a one way street.

The Second Law of Thermodynamics A rock will fall if you lift it up and then let go.  Hot frying pans cool down when taken off the stove.  Air in a high-pressure tire shoots out from even a small hole in its side to the lower pressure atmosphere.  Ice cubes melt in a warm room.  No matter how often you organize your “stuff”, your room, it eventually just ends up cluttered again!What’s happening in each of those processes? Energy of some kind is changing from being localized ("concentrated" in the rock or the pan) to becoming more spread out. A simple way of stating fundamental science behind the second law: Energy spontaneously disperses from being concentrated (or localized) to becoming spread out if it is not hindered from doing so. Entropy means being being scattered or disordered. A rock has potential energy localized in it when you lift it up above the ground. Drop it and that PE changes to kinetic energy (energy of movement), pushing air aside as it falls (therefore spreading out the rock’s KE a bit) before it hits the ground, dispersing a tiny bit of sound energy (compressed air) and causing a little heating of the ground it hits and the rock itself.  The potential energy that your muscles used in lifting it up is now spread out over in a little air movement and a little heating of the air and ground. Systems and SurroundingsSystem: The part being studied, the chemical.Surroundings: The parts of the universe, the surroundings, with which the system is in contact.

Types of Systems: NEED to look at 2 things, Energy and Matter (mass)1. Open Systems:Both energy and matter can be transferred between a system and its surroundings. Example: An open beaker of hot water loses heat to the surroundings

Upon cooling loses matter in the form of water vapor. System = water molecules and Surroundings = beaker + air near it

2. Closed Systems:Energy can be exchanged between the system and its surroundings, but matter cannot.

Example: A stoppered erlenmeyer flask of hot water loses heat to the surroundings, however, water vapor cannot escape.System = water molecules and Surroundings = beaker + stopper + air near it

3. Isolated Systems:Neither energy nor matter is exchanged between the system and its surroundings. Thermally isolated, well now come to think of it, TOTALLY isolated.

Example: A vacuum flask or vacuum thermos of hot water approximates an isolated system.

SURROUNDINGSSystem = liquid contents and Surroundings = thermos + air near it

Note: A vacuum contains no matter, it cannot transfer heat by conduction or convection.

PHASE CHANGESA change in the state of matter without any change in the chemical compostion. Usually that is melting or boiling, however PHYSICAL changes may be sublimation, which occurs when you go from solid to liquid, quickly skipping through the liquid phase as with :

I2 (s) I2 (g) or dry ice, in smoke machines….. OR CO2(s)

dry ice CO2 (g)

PHYSICAL changes may also be simply dissolving NaCl (s) NaCl (aq) which further dissociates to Na+

(aq) + Cl - (aq)

In the lab we disslove NaOH all the time and you remember it gave of a LOT of heat.Ex: NaOH(s) Na+

(aq) + OH - (aq) + ENERGY

However, the major calculations, for physical changes involve Phase changes (… plus the Kinetic changes ) . When there is a phase change, there is no change in the kinetic energy of the

particles, or no temperature changes. A change in potential energy is taking place due to changes in the attractions between particles (IMF, intermoloceular forces, for molecular substances).

Heating Curve for Water

GENERAL STEPS:Leg A. Heating of Solid: As the temperature increases, kinetic energy increases and potential energy remains constant. The energy goes to Increased vibration of particles about their fixed positions in solid.

Leg B. Fusion (Melting) of Solid: Here the temperature remains constant and there is no change in the kinetic energy, however the potential energy increases. The added energy weakens the attractions between the particles, so they move more freely; now the substance changes into a liquid

Leg C. Heating of Liquid: The temperature again rises causing an increase in kinetic energy but the potential energy remains constant. The particles moving more freely, but the attractions are still strong enough to hold particles in the liquid phase.

Leg D. Vaporization (Boiling) of Liquid: Again, like #B, this is a phase change so the temperatue remains constant. This means the kinetic energy is constant and the potential energy is increasing. Again, the energy is used to overcome the attractions between particles as the particles break free and the substance changes into the gas phase.

Leg E. Heating of Gas: The temperature begins to rise above its boiling point causing the kinetic energy to increase (potential energy remains constant.) Adding heat speeds up the gas particles.

Heating Curve Notes

Below is a diagram showing a typical heating/cooling curve for water. It reveals a wealth of information about the structure and changes occurring in water as it is heated or cooled through all three phases of matter at different temperatures. At the top of the diagram are pictures representing the typical particle arrangement as substances change through their states.

1. Identify by letter (A-E) in which section the following are found:

a. ____B___ Freezing (if cooling) b. ___E____ Particles farthest apart

c. ___E____ Boiling d. ____A___ Particle motion is most restricted e. ____B___ Heat of fusionf. _B & D_ All areas where energy change is potential only

g. ____D___ Heat of vaporization h.___A____ Least kinetic energyj. _A,C,E_All areas where kinetic energy is changing

k ____E___ most potential energy

l. ____B,D___ All areas where phase changes occur

m. _A,C,E______ All areas in which the heat is making the particles move faster

n. ____B,D___ All areas in which the heat is breaking the attractions or bonds between particles

Calculating Kinetic Energy:

Kinetic energy involves movement, thus there will be a temp change as a thermometer is a molecular speedometer. To get energy we need 3 things, speed , type and number of molecules.q = mcT (or with mass and c put together and to be C in the equation q = CT )q = mct is for when WATER is given in the calorimeter, HOWEVER, when we look at the specific heat of the whole calorimeter, the water AND the calorimeter can itself, we look at heat capacity (C ).

Heat Capacity (C):The quantity of heat required to change the temperature of an object by 1 degree Celsius, or q = C∆t (really the same as q = mc∆t , just that mc are combined or already multipied for you)

specific heat (c) = q m-1 T-1 the amount of heat needed to increase 1 g of material by 1 oC, as in water,

= 4.184J g-1 oC-1 whereas:heat capacity (C) = q T-1 the amount of heat needed to increase an object by 1 oC (notice mass is

absent or incorporated into C)= J oC-1 increase an object by 1 oC (notice mass is absent or incorporated into C)

Example: A 1.32 g sample of sucrose, C12H22O11, is burned in a bomb calorimeter with a heat capacity of 9.43 x 103 J oC-1. If the temperature of the calorimeter changed from 25.00 oC to 27.31 oC, calculate the heat of the reaction in units of kJ mole-1 of sucrose.

C = 9.43 x 103 J oC-1 , (THIS IS NOT specific heat, it is HEAT CAPACITY) , T = 2.31 oCq = C x Tq= 21.8 kJ

moles sucrose = mass / molar mass= 1.32 g C12H22O11 / 342 g . mole-1

= 3.8596 x 10-3 mole C12H22O11

H = - 21.7833 kJ / 3.8596 x 10-3 mole C12H22O11

H = - 5643.8 kJ mole-1 or H = - 5.64 x 103 kJ mole-1

Molar Heat: The total energy of a system is called the enthalpy (H) of the system, also referred to as Molar Enthalpy ( ∆H):

q = n H ( the enthalpy change involving one mole of a substance, the units = kJ/mol. ) H rx = q / # of moles or H rx = mcT / # of moles

Example: Calculate the heat required to melt 50.0 g of ice at 0.0°C.

Answer

Notice the sign of the heat value. Melting (fusion) ice

melting is an endothermic process - energy is required to break the intermolecular forces that hold water molecules together a ice.

Types of Problems1) Physical Changes Ex: Hot metal is dropped into 1000.0 g ( always be sure to check the units of mass as they must match the units of specific heat, KJ with KJ, and J with J ) of water. If the temperature of the water rose from 22.1 oC to 35.9 oC, how much heat energy did the water gain?q = mcT where m = 1000.0 g c = 4.18 J g-1 oC-1 T = (35.9 oC – 22.1 oC) = 13.8 oCq = (1000.0 g)( 4.18 J g-1 oC-1)( 13.8 oC)q = 57.7 kJ or 5.77 x 104 Jq = 57684 J NOTE - this is an exothermic change as heat lost to fridge. Practice Problems: Page 636: #’s 7 – 10 I (in the book)2) Chemical Changes Example: A student placed 50.0 mL of 1.00 M HCl at 25.5 oC in a coffee cup calorimeter. To this, 50.0 mL of 1.00 M NaOH also at 25.5 oC was added. The mixture temperature rose to a maximum of 32.4 oC. Calculate the heat of the reaction (heat of neutralization).mAcid = 50.0 mL x 1.0 g mL-1 = 50.0 g we also have 50 grams base OR 100 grams acid and base which when reacted is water.c = 4.18 J g-1 oC-1 T = (32.4 oC – 25.5 oC) = 6.9 oC q = mcT q = (100.0 g water (acid and base combo))( 4.18 J g-1 oC-1)( 6.9 oC) and so q = 2.88 kJ

H per mole of HCl …n = C x v or n = ( 1.00 M )( 0.0500 L)n = 0.0500 mole HClH = - 2.88 kJ/ 0.0500 mole HCl = - 57.7 kJ mole-1

There are 3 types of changes in Thermodynamics you need to know:We are concerned with 3 types of changes in:

a) Physical changes - Tens of KJ/mol (LITTLE HEAT)b) Chemical Changes - Hundreds of KJ/mol ( A LOT OF HEAT)c) Nuclear changes - Millions and even Billions KJ/mol (HOOCHIE MAMA,

AWESOME HEAT)

physical H2(l) H2(g) 10 to 100 kJ/mole (Tens)

chemical 2 H2(g) + O2(g) 2 H2O(l) 100 to 10,000 kJ/mole (Thousands)

nuclear Hnucleus + Hnucleus Henucleus 1010 to 1012 kJ/mole (Billions)

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CHEMICAL CHANGE CALCULATIONS

Knowing the heat change associated with a chemical change involving one mole of a chemical allows you to calculate the heat change associated with any measured amount of that chemical.

Example : Calculate the heat change when 10.4 g of CaCl2 is formed from its elements . The molar heat for this FORMATION reaction can be obtained from molar heats of formation tables, but is given here as -795.4 KJ/mol. THIS is a CHEMICAL change

Ca(s) + Cl2 (g) CaCl2 (s)

Example : A 5.00 g sample of benzene was burned at SATP conditions resulting in the release of 209.2 kJ of heat. Calculate the molar heat of combustion for benzene.

The molar heat of combustion of benzene is -3267 kJ/mol.

Heat of Reaction and Simple Calorimetry

Heat changes associated with aqueous chemical reactions can be determined by simple calorimetry. This information can then be used to calculate heat of reaction.

HCl + NaOH HOH + NaCl

Example: A 75.00 mL sample of 0.500 M hydrochloric acid at 25.00°C was mixed with 75.00 mL of 0.500 M NaOH in a simple calorimeter. The highest temperature recorded after mixing was 27.19°C. Calculate the molar heat of reaction for hydrochloric acid.

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CHEMICAL CHANGE CALCULATIONS WITHOUT CALORIMETERS ( 3 METHODS)

1) FLIP 2) Math (formula) (ΔH°rxn = Σ nΔH°f, products - Σn ΔH°f, reactants) 3) Bond Energies

1) Hess's Law: (FLIP THE EQUATION METHOD)

Mr. Hess states, if two or more equations can be added together to produce an overall equation, the sum of the enthalpy terms equals the enthalpy change of the overall equation.

Ex: Given these thermochemical equations:

use Hess’s law of heat summation calculate the molar enthalpy for the incomplete combustion of graphite:

2) Hess's Law: (THE MATH METHOD)

There is a second way to use Hess' Law : ΔH°rxn = Σ nΔH°f, products - Σn ΔH°f, reactants

Problem #1: Calculate the standard enthalpy of combustion for the following reaction: C2H5OH (l) + (7/2) O2 (g) ---> 2 CO2 (g) + 3 H2O (l)

The ΔH°f values I will now give as, FOR 1 Mole: -393.5 for CO2, -286 for water, -278 for ethanol AND zero for ALL elements

ΔH°combustion = [ 2 (-393.5) + 3 (-286) ] minus [ (-278) + (7/2) (0) ]

Doing the math gives us ΔH°comb = -1367 kJ/mol of ethyl alcohol (C2H5OH). ____________________________________________________________________________________________________________________________________________

3) Bond Energies. The process of measuring the changes in the enthalpy of a system is called calorimetry. The change in enthalpy, H, is related to what is happening in the reaction.

Ex: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

We can burn methane and the heat it gives off can be absorbed and measured by a cup of water (calorimeter)

Bonds being broken (THIS ABSORBS ENERGY): : CH4(g) + 2 O2(g) Products4 C-H bonds are being broken AND 2 O-O bonds are being broken these require

energy

Bonds being formed (RELEASES ENERGY): Reactants CO2(g) + 2 H2O(l)

2 C=O bonds being formed AND 4 H-O bonds being formed these release energyAs the type and number of bonds being broken and formed change, so will the change in

enthalpy.H = Hbond energy reactants - H bond energy products

Calculating Bond Energies:

Bond breaking is endothermic, because energy is absorbed to separate bonded particles. The potential energy of the separated particles is greater than that of the bonded particles.

The amount of energy needed to separate one mole of a bonded species is called the bond energy. Keeping the sign convention in mind, all bond energies are positive!

Example: Using average bond energies, calculate the enthalpy change for the combustion of ethanol.

Enthalpy Diagrams

Potential energy diagrams, also known as enthalpy diagrams, show the relative potential energies of the species involved in a chemical (or physical) change. Molar enthalpy is the potential energy change associated with a physical or chemical change involving one mole of a substance. You may heard this as the heat of reaction. Consider the following EXOTHERMIC diagram.

The down arrow shows the reaction is exothermic. The potential energy of products are less than the reactants. The P.E. of the system decreases and is released to the surroundings as heat.

Now consider this ENDOTHERMIC diagram.

The enthalpy diagram for an endothermic reaction always shows the arrow pointing upwards. The potential energy of the system increases as energy is absorbed from the surroundings.

Fueling our Bodies STSE - Caloric IntakeBefore going further we must clarify a potentially confusing unit related to the Calories quoted on food labels. A calorie (cal) is the amount of energy required to increase the temperature of one gram of water by degree Celsius:

1 cal = 4.184 Jyet this is not the same "calorie" as the food Calorie (Cal). One food Calorie is equal to one kilocalorie(note the uppercase "C" in the unit for food Calorie

1 food Calorie = 1 Cal = 1000 cal = 1 kcal = 4.184 kJSample ProblemA 2.50g sample of bar is placed in a bomb calorimeter with a heat capacity of 13.17 kJ/oC. The temperature of the calorimeter contents increases by 3.86oC upon complete combustion of the sample.Determine: (a) the food value, and(b) number of Calories in a 23.0g serving (one granola bar).Solution(a) First, determine the heat of the calorimeter surroundings:

qsurr = CT = (13.17 kJ/oC ) (3.86oC) = 50.8 kJSecond, we could state the heat evolved by the system (the granola bar) as:

qsys = - qsurr = - 50.8 kJ

Next.FV = Food kJ = 50.8 kJ/2.50 g = 20.3 kJ/g

To calculate the total Calories in a 23.0g serving:Food kJ = FV x m = ( 20.3 kJ/g ) x ( 23.0g ) = 467 kJ.

In case you are "counting calories", since 1 Cal = 4.184 kJ, the same food energy in the old unit of Calories is:Cal = (467 kJ ) x ( 1 Cal / 4.184 kJ) = 112 Cal

The 23g granola bar contains 467 kJ, or 112 Calories. __________________________________________________________