the study of energy changes in reactions

the study of energy changes in reactions
Thermodynamics the study of energy changes in reactions

Energy Energy is the ability to do work or produce heat.
Energy is divided into two basic types: potential energy kinetic energy.

Potential Energy Potential energy is stored energy based on:
position - gravity-related, like a skateboarder at the top of a half-pipe or stretched rubber band chemical composition - like the chemicals in food or a battery

Kinetic energy Kinetic energy is energy of motion
This includes the constant, random motion of atoms and molecules Is proportional to temperature As temp increases, motion of particles increases →Thermal Expansion

Chemical Energy Form of potential energy based upon chemical composition Stored in bonds between atoms, molecules, and ionic crystals Potential to form new substance Do work produce or require heat Chemical energy of food powers your body Chemical energy of gas powers your car

Thermal Energy: Total internal energy of system: depends on both Heat and Temperature
Heat: Form of energy that flows from warmer to cooler object Temperature: measure of the average KE of the particles

Law of Conservation of Energy
Energy can be converted between the different forms of energy, but no matter how many times energy is converted the Law of Conservation of Energy States no energy can be created or destroyed

Heat (q): energy that moves from hot to cold
Will continue to flow until temps are equal Heat is quantified by measuring change in temp, Temp does not directly measure heat 20°C 10°C stored What is final temp?

calorie: To measure heat directly, the calorie (cal) was defined as the amount of heat needed to increase the temp of 1 gram of water by 1 °C.

1 Calorie (Cal) = 1000 calories (cal) = 1 kilocalorie (kcal)
Note: The Calories reported on food wrappers are measured as heat needed to increase the temperature of 1000 grams of water by 1 °C, 1 Calorie (Cal) = 1000 calories (cal) = 1 kilocalorie (kcal) Drinking a whole can of Pepsi actually means drinking: 250 Cal = 250,000 calories! Normal diet: 2,000 Cal = 2,000,000 cal

Calorimetry: measuring heat energy
Calorimeter: device used to measure heat lost or absorbed during chemical reaction, including energy from food

Joule: measure energy more accurately
Calories are still used in the nutrition field, but in chemistry we use the Joule (J). 1 cal = J Ex. 1) If my breakfast contains 230 Calories, how much energy in Joules will it supply?

Calories are still used in the nutrition field, but in chemistry we use the Joule (J). 1 cal = J Ex. 1) If my breakfast contains 230 Calories, how much energy in Joules will it supply? 230 Cal cal J 1 Cal cal

Calories are still used in the nutrition field, but in chemistry we use the Joule (J). 1 cal = 4.18 J Ex. 1) If my breakfast contains 230 Calories, how much energy in Joules will it supply? 230 Cal cal J = J 1 Cal cal = 9.6 x 105 J

Both the calorie and Joule are defined using water
But, other substances gain and lose heat at different rates Specific heat (c) – the amount of energy required to raise 1 gram of a any substance by 1°C Liquid water has a relatively high specific heat: loses & gains heat slowly – moderates earth’s climate

The better the conductor, the lower the specific heat

When a cake has been baked & removed from the oven it would cause severe burns to touch the metal pan, but the top of the cake can be touched to test if the cake is ready…Why?

The metal of the pan has a low specific heat b/c it can lose heat quickly
This heat would be absorbed by your finger!!! The cake contains a lot of trapped air and other materials with much higher specific heats. These objects lose heat slowly, allowing more time before your finger has absorbed enough heat to cause a burn.

Density of a cake is much less than the density of the metal pan, so for the same surface area touched by your finger there are a lot more metal atoms that transfer heat energy than cake molecules. So… the mass also has an affect on the amount of heat transferred

Plus… (old stuff) Metals are good conductors of heat and electricity While air is a poor conductor of heat and electricity (nonmetals)

To calculate the amount of heat gained or lost by an object
q = m × c × ∆T q = heat (J) m = mass (g) c = specific heat ∆T = Temp. final – Temp. initial, in °C J g °C

If the object is heating up:
the final temp is greater than the initial temp, so ∆T is positive and amount of heat gained is positive. + q

If the object is cooling down:
the final temp is less than the initial temp, so ∆T is negative and the amount of heat lost is negative. - q

If ∆T is positive the answer for heat is positive: heat is gained.
If ∆T is negative the answer for heat is negative: heat is lost. - q + q

Add the following to your notes
Specific heats for water: for steam = 1.97 J/goC for liquid = 4.18 J/goC for ice = 2.09 J/goC Specific heats (c) for various metals: aluminum = J/goC copper = J/goC iron = J/goC lead = J/goC mercury = J/goC

Ex. 2) What is the heat absorbed by 100
Ex. 2) What is the heat absorbed by 100. g of water to raise it from 25 °C to 75 °C? q = m × c × ∆T q = ________ x ________ x (________ - _______) q = _______________

Ex. 2) What is the heat absorbed by 100. g of water to raise it from 25 °C to 75 °C? q = m × c × ∆T q = _100.g__ x _4.18 J/goC_ x (75oC_- 25oC) q = __ ____ __

Ex. 2) What is the heat absorbed by 100. g of water to raise it from 25 °C to 75 °C? q = m × c × ∆T q = _100. g__ x _4.18 J/goC _ x (75oC - 25oC) q = __ J___ = 2.1 x 104 J or J

Ex. 3) What is the heat given off by 100. g of iron to cool from 50
Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = ________ x ________ x (________ - _______) q = _______________

Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = x J/goC x ( 20. oC – 50. oC) q =

Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = g x J/goC x ( 20.oC – 50. oC) q = 100.g x J/goC x (-30. oC) q =

Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = x x ( 20. – 50.) q = x x (-30.) q = J = J or x 103 J

Heat Movement The Law of Conservation of Energy states that heat energy cannot just appear out of nowhere and disappear to nowhere. If an object warms up, that heat had to be absorbed from somewhere and if an object cools down, that heat had to be given off to something.

In the Earth’s ecosystem heat is gained by the sun’s radiation and given off to space at night.
This is why a cloudy night is typically warmer than a clear night – the heat gets trapped in by the clouds.

Heat can move from one object to another by three ways:
Radiation Convection Conduction

The energy from the sun reaches earth by radiation - electromagnetic waves:
The infrared part of the electromagnetic spectrum is the “heat” form of radiation. Ex: heat from fire, light bulb, coil on electric stove

Convection is the movement of heat by circulating materials, so anything that is a fluid (gases and liquids) can move heat by convection, often called convection currents. Ex. Most air-conditioning systems, the ocean currents, and the warm and cold “fronts” the weather-person talks about on the news.

Conduction is the movement of heat between objects that are touching.
Most common form of heat movement when cooking. The food touches a pan that’s touching a burner. As the pan absorbs heat from burner the food absorbs heat from the pan.

When you touch a hot pan heat moves from the pan into your hand which then burns you.
Often the lasting effects of a burn can be minimized if you can quickly get the heat to move out of your hand by running cold water over your hand for a long time.

Heat energy will always naturally move from an object with higher heat to an object with lower heat, regardless of the method of heat transfer

The Law of Conservation of Energy also tells us that the amount of heat lost by the hot object must be equal to the amount of heat gained by the cold object. 20°C 10°C 15°C

- (mlost × clost × ∆Tlost ) = mgained × cgained × ∆Tgained
Hot becomes cooler, and cold becomes warmer. qlost = qgained - (mlost × clost × ∆Tlost ) = mgained × cgained × ∆Tgained

Phases or States of Matter: Solid, Liquid, Gas
Sometimes when an object gains or loses heat, it is not just the temperature that changes but the phase of the object can change. The phase depends on how much heat energy is in the object, often measured by the temperature. The solid will have lower temperature than the liquid form, which will have lower temperature than the gas (vapor) form.

Energy in states of matter
Hi energy & velocity - Low attractive forces & density Hi temp Gas Energy Liquid Energy Released In Solid Low energy & velocity - Hi attractive forces & density Low temp Exothermic Endothermic

Phase changes occur when sufficient heat energy is added or removed
+q: heat is added or absorbed – Endothermic –q: heat is lost or released – Exothermic

Phase changes occur when sufficient heat energy is added or removed to change the object’s phase.
Melting is the solid phase changing to the liquid phase Freezing is the liquid phase turning to a solid phase.

Melting and freezing both occur at the melting point ~ the temp at which a solid becomes a liquid.
If heat is being absorbed (+q), then melting is occurring. If heat is given off (-q), then freezing is happening.

Boiling (vaporizing) is the liquid phase changing to the gas phase, and
Condensation is the gas phase changing to the liquid phase. Boiling and condensing occur at the boiling point ~ the temp at which a liquid becomes a gas.

If heat is being absorbed (+q), then boiling is occurring.
If heat is given off (-q), then condensing is occurring. Sometimes liquids evaporate instead of boiling.

Evaporation is boiling, but occurs below the boiling point
Evaporation is boiling, but occurs below the boiling point. Evaporation can only happen at the surface of the liquid (boiling happens everywhere inside) and is a cooling process.

There are two other phase change possibilities, but they are rare in everyday life.
Sublimation is the solid phase changing directly to the gas phase (no liquid in-between) Deposition is the gas phase changing directly to the solid phase. Ex. Dry ice, CO2 A +q would indicate sublimation A –q would indicate deposition.

From this heat curve (for water) we can see how the temperatures change and the phases change. Notice that the temperature does not change during a phase change (the two flat lines).

q = heat (add “-“ sign for freezing), (J)
The heat equation q = m × c × ∆T will not work during a phase change, so new equations are needed during a phase change bc heat is still needed For melting or freezing q = m × Hf q = heat (add “-“ sign for freezing), (J) m = mass, (g) Hf = heat of fusion, J g

q = heat (add “-“ sign for condensing), (J) Hv = heat of vaporization,
For boiling/condensing q = m × Hv q = heat (add “-“ sign for condensing), (J) m = mass, (g) Hv = heat of vaporization, J g

Ex. 4) Calculate the heat needed for the heat curve shown below for 150. g of water.
Hfwater = 334 J/g Hvwater = 2260 J/g c ice = 2.09 J/g °C c water = 4.18 J/g °C c steam = 1.97 J/g °C

Step 1: Ice from -40 °C to 0 °C. q1 = m × c × ∆T q1 = ________ x ________ x (________ - _______) q1 = _______________

Step 1: Ice from °C to 0 °C. q1 = m × c × ∆T q1 = 150. g x 2.09 J/goC x ( 0oC – (-40.0oC)) q1 = J

Step 2: Turning ice into liquid water
q2 = m × Hf q2 = ___________ x ___________ q2 = _______________

Step 2: Turning ice into liquid water
q2 = m × Hf q2 = 150. g x 334 J/g q2 = J

Step 3: Water from 0 °C to 100 °C.
q3 = m × c × ∆T q3 = ________ x ________ x (________ - _______) q3 = _______________

Step 3: Water from 0 °C to 100 °C.
q3 = m × c × ∆T q3 = 150. g x 4.18 J/goC x ( 100oC – 0oC) q3 = J

Step 4: Turning liquid water into water vapor (steam).
q4 = m × Hv q4 = ___________ x ___________ q4 = _______________

Step 4: Turning liquid water into water vapor (steam).
q4 = m × Hv q4 = 150. g x 2260 J/g q4 = J

Step 5: Water vapor from 100 °C to 140 °C
q5 = m × c × ∆T q5 = ________ x ________ x (________ - _______) q5 = _______________

Step 5: Water vapor from 100 °C to 140.0 °C
q5 = m × c × ∆T q5 = 150. g x 1.97 J/goC x ( 140.0oC – 100oC) q5 = J

Total heat involved = q1 + q2 + q3 + q4 + q5

Total heat involved = 12400 J + 50100 J + 62700 J + 339000 J + 12200 J

Total heat involved = 12400 J + 50100 J + 62700 J + 339000 J + 12200 J
Total heat = J (with proper addition sig figs = J)

Heat & Chemical Reactions
Thermochemical Equations: A balanced chemical equation that includes all physical states (s, l, g) and change in enthalpy (H) (heat content). 4 Fe(s) + 3O2(g) → 2Fe2O3(s) kJ Heat can be written as either a reactant or product Every chemical reaction involves a change in energy: heat and/or light . Heat released

If a chemical reaction feels cold:
absorbing heat. endothermic reaction – energy in heat energy is written on the reactant side q is positive (+q) 27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq) Heat Added

If a chemical reaction feels warm,
giving off heat. exothermic reaction - energy out heat energy is written on the product side q is negative (-q) Heat released 4 Fe(s) + 3O2(g) → 2Fe2O3(s) kJ

Ex. 5) If 515 kJ were needed in the dissolving process, a) How many moles of ammonium nitrate were originally used? b) How many grams would that be? 27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq)

Ex. 6) How many Joules of heat are released when 77 grams of iron reacts with excess oxygen?
4 Fe(s) + 3O2(g) → 2Fe2O3(s) kJ

solid + energy (heat) → liquid
Endothermic (a reactant) solid + energy (heat) → liquid Heat flows into the system from the surroundings. This is why your skin gets cold when you hold ice; the heat is flowing out of your skin to melt the ice. System: ice Surrounding: your skin

Exothermic (a product)
Iron + oxygen → iron(III) oxide + energy (heat) heat is released from the system and is absorbed by the surrounding environment which is why it’s used in instant hot pads. System: hot pad Surrounding: your skin

Enthalpy (H): total heat energy of system
∆H = change in enthalpy Exothermic Endothermic ∆H is negative ∆H is positive Energy of system Energy of system decreased increased decreased

To separate water into hydrogen and oxygen (electrolysis) electricity is often used.
2H2O  2H2 + O2 Is electrolysis of water endothermic or exothermic? Where does the energy “go” in the equation?

To separate water into hydrogen and oxygen (electrolysis) electricity is often used.
Is electrolysis of water endothermic or exothermic? Where does the energy “go” in the equation? ΔH + 2H2O  2H2 + O2

The enthalpy of reaction (∆H) can be calculated to determine if heat is released or gained.
If ΔH is +, then heat is absorbed, so it is an endothermic reaction If ΔH is -, then heat is released, so it is an exothermic reaction

ΔH = ∑∆nHfo(products) - ∑∆nHfo(reactants)
∆Hfo = standard enthalpy (heat) of formation from the formation of 1 mole of compound

Ex. 7) Find the ∆Hrxn for liquid ethanol, C2H5OH, undergoing combustion:
C2H5OH O2 → 2CO H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH kJ/mol O2 0 kJ/mol CO kJ/mol H2O kJ/mol

Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH, undergoing combustion:
C2H5OH O2 → 2CO H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH kJ/mol O2 0 kJ/mol CO kJ/mol H2O kJ/mol ∆Hfo(reactants): ∆Hfo(products): ΔH = ∆Hfo(products) - ∆Hfo(reactants)

C2H5OH O2 → 2CO H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH kJ/mol O2 0 kJ/mol CO kJ/mol H2O kJ/mol Per mole ∆Hfo(reactants): ( ) + 3(0) = kJ ∆Hfo(products): 2( ) +3( ) = ( ) = kJ ΔH = ∆Hfo(products) - ∆Hfo(reactants) = ( ) – ( ) = kJ

C2H5OH O2 → 2CO H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH kJ/mol O2 0 kJ/mol CO kJ/mol H2O kJ/mol Per mole ∆Hfo(reactants): ( ) + 3(0) = kJ ∆Hfo(products): 2( ) +3( ) = ( ) = kJ ΔH = ∆Hfo(products) - ∆Hfo(reactants) = ( ) – ( ) = kJ Neg = Exothermic

An additional way to show a change in enthalpy without a calculation is with an enthalpy diagram.
What is the key to knowing if the diagram shows exothermic or endothermic? Energy in or out

An additional way to show a change in enthalpy without a calculation is with an enthalpy diagram.
What is the key to knowing if the diagram shows exothermic or endothermic? Energy in or out Energy in Energy out

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