Thermodynamics Cartoon courtesy of NearingZero.net

I. Introduction Thermochemistry: Study of heat flow in chemical reactions. Heat (Q) – The energy that transfers from one object to another, because of a temperature difference between them. flows from warmer  cooler object Ethalpy ( Δ H) – The heat released or absorbed. Used interchangeably with “Q”! Energy units: Joules (J),Kilojoules (kJ),calories (cal), kilocalories (kcal) Food: “C” 90 0 C

Exothermic and Endothermic Processes In studying heat changes, think of defining these two parts: – the system - the part of the universe on which you focus your attention. – the surroundings - includes everything else in the universe.

Heat (Enthalpy) Change Two types of heat change exist. Endothermic: Exothermic: Processes in which energy is absorbed into the system as it proceeds, and surroundings become colder. Processes in which energy is released as it proceeds, and surroundings become warmer. + q system - q surroundings - q system + q surroundings

Activation Energy (E A ) The minimum energy required to initiate a chemical reaction. Examples: Flame, spark, high temperature, radiation are all sources of activation energy

Endothermic Reactions *** Where does the released energy come from and the absorbed energy go? Breaking and forming bonds

Exothermic Reactions

8 Exothermic and Endothermic Every reaction has an energy change associated with it. Exothermic reactions release energy, usually in the form of heat. Endothermic reactions absorb energy. Energy is stored in bonds between atoms.

HEATHEATHEATHEATHEATHEATHEATHEAT

The Effect of a Catalyst Catalyst: A substance that speeds up a reaction without being consumed. It lowers the activation energy! Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

Endothermic Reaction with a Catalyst

Exothermic Reaction with a Catalyst

Hess’s law Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. For example: C + O 2  CO 2 The book tells us that this can occur as 2 steps C + ½O 2  CO  H  = – kJ CO + ½O 2  CO 2  H  = – kJ C + CO + O 2  CO + CO 2  H  = – kJ I.e. C + O 2  CO 2  H  = – kJ Hess’s law allows us to add equations. We add all reactants, products, &  H  values. We can also show how these steps add together via an “enthalpy diagram” …

For the equation C + O 2  CO 2 What is the ∆H? C + ½ O 2  CO  H  = – kJ CO + ½ O 2  CO 2  H  = – kJ C + O 2  CO 2  H  = – kJ

For the reaction GeO(s) + ½ O 2 (g)  GeO 2 (s) What is the ∆H? GeO(s)  Ge(s) + ½ O 2 (g)  H  = kJ Ge(s) + O 2 (g)  GeO 2 (s)  H  = – kJ GeO(s) + ½ O 2 (g)  GeO 2 (s)  H  = – kJ

For the reaction NO(g) + ½ O 2 (g)  NO 2 (g) What is the ∆H ? NO(g)  ½ N 2 (g) + ½ O 2 (g)  H  = – kJ ½ N 2 (g) + O 2 (g)  NO 2 (g)  H  = kJ NO(g) + ½ O 2 (g)  NO 2 (g)  H  = – kJ

Hess’s law: Example We may need to manipulate equations further: 2Fe + 1.5O 2  Fe 2 O 3  H  =?, given Fe 2 O 3 + 3CO  2Fe + 3CO 2  H  = – kJ CO + ½ O 2  CO 2  H  = – kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 1.5O 2  Fe 2 O 3 3CO O 2  3CO 2  H  = – kJ 2Fe + 3CO 2  Fe 2 O 3 + 3CO  H  = kJ CO + ½ O 2  CO 2  H  = – kJ  H  = – kJ For more lessons, visit

PRACTICE HESS’S LAW PROBLEMS ON WORKSHEET

Change in Enthalpy (____) = _______ content = Heat of _______ 1. ____ = Heat of __________ - Heat of __________ 2. Exothermic reactions: _____ = ____ 3. Endothermic reactions:_____ = ____ 4.Units of _____: 5. Example: Combustion of methane (_____) ____= -802 kJ/mol 6.Example: Decomposition of limestone (_______) _____= kJ/mol  H Heat reaction  Hproductreactants  H– + or CH 4  H kJ CaCO 3  H CaCO kJ CaO + CO 2 CH 4 + 2O 2 CO H 2 O

Determination of ______? 1. Experimentally 2. Using Heats of __________ 3. Using ______ Law HH Formation Hess’s 3 Techniques

Calculating _____ Experimentally Heat exchange: Heat released in a chemical reaction = Heat absorbed by water in calorimeter Q released= Q absorbed Simple Calorimeter: Heat of solutionHeat of combustion HH ^ Thermometer Styrofoam cup Solution can

Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1  C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1  F

The Joule The unit of heat used in modern thermochemistry is the Joule. 1 joule = calories

A Bomb Calorimeter

A Cheaper Calorimeter

Heat Capacity/Specific Heat (c) Heat Capacity/ Specific Heat - the amount of heat needed to increase the temperature of an object exactly 1 o C. – Depends on both the object’s mass and its chemical composition. – Units are either J/(g o C) or cal/(g o C)

Specific Heat cont. * Specific heat of water: ________ = ___________ * Specific heat of iron: __________ Example 1: compare heating an empty stainless steel pan to a pan full of water… Example 2: Which has a higher heat capacity – water or sand? Example 3: Which as a higher heat capacity – cheese or pizza crust? 4.For a constant amount of heat (Q), as specific heat decreases, the change in temperature ___________. 5. Conversion between J and cal: ______________________ 1 cal/g o C J/g o C J/g o C It takes less heat to raise the temp. of iron than water. Water: It takes a lot of heat to warm water. The crust increases J = 1 cal

Calculations Involving Specific Heat c = Specific Heat Capacity q = Heat lost or gained  T = Temperature change OR

Table of Specific Heats Page 501

- Page 510

Sample Problem #1 What is the temperature change of 2.0 L of water heated with a 200 Calorie (kcal) candy bar?  T = ? = 200,000 cal 2.0 L L mL 1 mL 1 g = 2000 g Q = 200 kcal m = 2.0 L c = 1 cal/g o C Q = mc  T  T = 100 o C

2. How much heat energy (kJ) must be added to change the temperature of mL of water from 20.0  C to 85.0  C? Q = mc  T Q = ? m = g c = J/g o C  T = final (T final ) - initial (T initial )  T = 85.0 o C – 20.0 o C = 65.0 o C Q = (150.0 g)(4.184 J/g o C)(65.0 o C) Q = 40,794 J Q = 40.8 kJ

3. What is the final temperature of mL of 10.0  C water if 1500 calories of heat is absorbed? Q = mc  T Q = 1500 cal m = mL c = 1 cal/g o C  T = T final - T initial = g  T = 15 o C T f =  T + T i T f = 15 o C o C T f = 25 o C T f =  T o C

Experimental Setup: Place a 150 g block of aluminum into a pot of boiling water (100.0 o C). Once the block is hot, place it into mL of 10.0  C water. As the block cools, the water warms up. The final temperature is 21.7  C for the mixture. What is the specific heat of aluminum (c Al )? Heat lost by aluminum = Heat gained by water Q H (Hot object) = Q C (Cold object) T H = o CT C = 10.0 o CT F = 21.7 o C m Al = 150 gm w = gc w = J/g o C mc  T= mc  T m Al c Al (T hot – T final )= m w c w (T final – T cold ) (150 g Al)(c Al )(100.0 o C – 21.7 o C) = (200.0g H 2 O)(4.184 J/g o C)(21.7 o C – 10.0 o C) (150 g Al)(c Al )(78.3 o C) (11745 g o C)(c Al ) = (200.0 g H 2 O)(4.184 J/g o C)(11.7 o C) = J c Al = J/g o C % error = = 7.3 %

E. Sample problems: 1. A student places an 85.5 g piece of metal at  C into 122 mL of 16.0  C water. If the final temperature is 20.2  C, what is the specific heat of the metal? Heat lost by the metal = Heat gained by water Q H (Hot object) = Q C (Cold object) T H = o CT C = 16.0 o CT F = 20.2 o C m ? = 85.5 gm w = 122 gc w = J/g o C mc  T= mc  T m ? c ? (T hot – T final )= m w c w (T final – T cold ) (85.5 g)(c ? )(100.0 o C – 20.2 o C) = (122 g H 2 O)(4.184 J/g o C)(20.2 o C – 16.0 o C) (85.5 g)(c ? )(79.8 o C) ( g o C)(c ? ) = (122 g H 2 O)(4.184 J/g o C)(4.2 o C) = J c?c? = 0.31 J/g o C