Thermodynamics Honors Unit 5 Energy Basic Principles p

Thermodynamics Honors Unit 5

Energy: Basic Principles p Thermodynamics – the study of energy changes p Energy – the ability to do work or produce heat n Note: Work is force acting over a distance

Energy: Basic Principles p Kinetic Energy – energy of motion n KE = Potential Energy – energy due to position or composition p

Law of Conservation of Energy p A. k. a. first Law of Thermodynamics p Energy can be converted from one form to another but can’t be created or destroyed n This means the total energy of the universe is CONSTANT!

Heat vs. Temperature p Temperature – measure of the random motion of a substance n p Temperature is proportional to kinetic energy (it is a measure of the average kinetic energy in a substance) Heat (q) – flow of energy due to a temperature difference

Important Aspects of Thermal Energy & Temperature p Heat is NOT the same as temperature! p The more kinetic energy a substance has, the greater the temperature of its atoms and molecules. p The total thermal energy in an object is the sum of its individual energies of all the molecules. p For any given substance, its thermal energy depends not only on its composition but also on the amount of substance

System vs. Surroundings p A system is the part of the universe we are studying. The surroundings are everything else outside of the system. p

Direction of Heat Flow p Heat transfer occurs when two objects are at two different temperatures. p Eventually the two objects reach the same temperature n At this point, we say that the system has reached equilibrium.

Thermal Equilibrium p Heat transfer always occurs with heat flowing from the HOT object to the COLD object.

Thermal Equilibrium The quantity of heat lost by the hotter object and the quantity of heat gained by the cooler object are EQUAL.

Exothermic vs. Endothermic p Exothermic process heat is transferred from the system to the surroundings n Heat is lost from the system (temperature in system decreases) p Endothermic process heat transferred from the surroundings to the system n Heat is added to the system (temperature in system increases)

Exothermic Process

Endothermic Process

Units of Energy p Joule (J) is the SI unit of energy & heat One kilojoule (k. J) = 1000 joules (J) p calorie (cal) = heat required to raise the temperature of 1. 00 g of water by 1 °C 1 calorie = 4. 184 J

Units of Energy p Food is measured in Calories (also known as kilocalories) instead of calories 1 Cal = 1 kcal = 1000 calories

Units of Energy 3800 cal = _____ Cal = _____ J

Units of Energy The label on a cereal box indicates that 1 serving provides 250 Cal. What is the energy in k. J?

Heat Transfer Direction and sign of heat flow – MEMORIZE! ENDOTHERMIC: heat is added to the system & the temperature increases (+q) EXOTHERMIC: heat is lost from the system (added to the surroundings) & the temperature in the system decreases (-q)

Heat Capacity p The quantity of heat required to raise an object’s temperature by 1 °C (or by 1 Kelvin) p Heat capacity is an extensive property. n Which will take more heat to raise the temperature by 1 °C?

Specific Heat (Specific Heat Capacity) p Specific Heat (C) - The quantity of heat required to raise the temperature of one gram of a substance by 1 °C n Intensive property p Units: n J/(g°C) or J/(g°K) n cal/(g°C) or cal/(g°K)

Examples of Specific Heat At the beach, which gets hotter, the sand or the water? Higher specific heat means the substance takes longer to heat up & cool down!

Examples of Specific Heat p Specific heat (C)= the heat required to raise the temperature of 1 gram of a substance by 1 °C Cwater = 4. 184 J/(g °C) Csand = 0. 664 J/(g °C)

Calculating Changes in Thermal E q = m x C x Δt q = m. CΔt q = heat (cal or J) m = mass (g) C = specific heat capacity, J/(g °C) Δt = change in temperature, Tfinal – Tinitial (°C or K) ***All units must match up!!!***

Example q = m. CΔt How much heat in J is given off by a 75. 0 g sample of pure aluminum when it cools from 84. 0°C to 46. 7°C? The specific heat of aluminum is 0. 899 J/(g°C).

Example q = m. CΔt What is the specific heat of benzene if 3450 J of heat are added to a 150. 0 g sample of benzene and its temperature increases from 22. 5 °C to 35. 8 °C?

Example q = m. CΔt A 50. 0 g sample of water gives off 1. 025 k. J as it is cooled. If the initial temperature of the water was 85. 0 °C, what was the final temperature of the water? The specific heat of water is 4. 18 J/(g°C).

Calorimetry p Calorimetry: measurement of quantities of heat n A calorimeter is the device in which heat is measured.

Calorimetry p Assumptions: n Heat lost = -heat gained by the system n In a simple calorimeter, no heat is lost to the surroundings

Coffee Cup Calorimetry ***Use styrofoam instead of a beaker to keep heat in Steps: 1. Add hot solid metal to cool water 2. Water will heat up (T rises) as metal cools 3. Eventually, water & metal are at same T. qmetal + qwater = 0 qmetal = -qwater Heat lost by metal = -heat gained by the water

Calorimetry qmetal = -qwater Heat lost by metal = -heat gained by water Since q = m. CΔT, mm. CmΔTm = -[m. H 2 OCH 2 OΔTH 2 O]

Sample Problem A 358. 11 g piece of lead was heated in water to 94. 1 °C. It was removed from the water and placed into 100. m. L of water in a Styrofoam cup. The initial temperature of the water was 18. 7 °C and the final temperature of the lead and water was 26. 1 °C. What is the specific heat of lead according to this data?

Bomb Calorimeter p Constant volume “bomb” calorimeter n p Some heat from reaction warms water n p qwater = m. CH 2 OΔT Some heat from reaction warms bomb n p Burn sample in O 2 qbomb = CbombΔT qrxn + q. H 2 O + qbomb = 0

Energy & Changes of State p All changes of state involve energy changes (more in Unit 9) n Note that fusion = melting

State Functions p A property where the change from initial to final state does not depend on the path taken n Ex. ) The change in elevation from the top to bottom of a ski slope is independent of the path taken to go down from the slope

Enthalpy Changes for Chemical Rxns. p Heat of reaction n The heat absorbed or given off when a chemical reaction occurs at constant T (temp. ) and P (pressure)

Enthalpy p Enthalpy (H) n The heat content of a reaction (chemical energy) p ΔH = change in enthalpy n The amount of energy absorbed by or lost from a system as heat during a chemical process at constant P n ΔH = ΔHfinal - ΔHinitial

Properties of Enthalpy p Enthalpy is an extensive property n p Enthalpy is a state function n p It does depend on quantity Depends only on the final & initial values Every reaction has a unique enthalpy value since ΔH = Hproducts - Hreactants

Representation of Enthalpy as a Graph

Two Ways to Designate Thermochemical Equations Endothermic: a) H 2 (g) + I 2 (s) 2 HI (g) b) H 2 (g) + I 2 (s) + 53. 0 k. J 2 HI (g) k. J ΔH = 53. 0

Two Ways to Designate Thermochemical Equations Exothermic: a) ½ CH 4 (g) + O 2 (g) ½ CO 2 (g) + H 2 O (l) ΔH = -445. 2 k. J b) ½ CH 4(g) + O 2(g) ½ CO 2(g) + H 2 O(l) + 445. 2 k. J

Two Ways to Designate Thermochemical Equations Note the meaning of the sign in ΔH in the equations above!! Endothermic: ΔH = + Exothermic: ΔH = -

Two Ways to Designate Thermochemical Equations Note the important of designating the physical state or phase of matter. Why? ? Because this will change the heat of reaction! (ΔH)

Thermochemical Equations What do the coefficients stand for? How can they differ from the ones we have used before? Coefficients = the number of moles (as before) BUT We can use fractional coefficients now!

Thermochemical Equations What is the standard state? How do we designate conditions of temperature and pressure that are not at standard state? Standard state = 1 atm pressure & 25 °C ΔH° = ΔH at standard state Must show conditions over arrow if not at standard state!

Thermochemical Equations How can we find the enthalpy of reaction when we reverse it? Reverse the reaction, reverse the sign of ΔH! Example: CO (g) + ½ O 2 (g) CO 2 (g) ΔH = -283 k. J CO 2 (g) CO (g) + ½ O 2 (g) ΔH = +283 k. J

Example Given Rxn. #1, find the ΔH for Rxns. 2 & 3 Reaction #1 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) ΔH = +197. 8 k. J Reaction #2 SO 2 (g) + ½ O 2 (g) SO 3 (g) ΔH = Reaction #3 4 SO 3 (g) 4 SO 2 (g) + 2 O 2(g) ΔH =

ΔH as a Stoichiometric Quantity Given the reaction below, how much heat is produced when 15. 0 g of NO 2 are produced? 2 NO (g) + O 2 (g) 2 NO 2 (g) ΔH = -114. 1 k. J

ΔH as a Stoichiometric Quantity Given: ΔH = -283 k. J CO (g) + ½ O 2 (g) CO 2 (g) (a) Calculate the enthalpy of the above reaction when 3. 00 g of product are formed

ΔH as a Stoichiometric Quantity Given: ΔH = -283 k. J CO (g) + ½ O 2 (g) CO 2 (g) (b) If only 10. 0 grams of oxygen and an unlimited supply of CO are available to run this reaction, how much heat will be given off?

ΔH as a Stoichiometric Quantity Given: ΔH = -283 k. J CO (g) + ½ O 2 (g) CO 2 (g) (c) How many grams of carbon monoxide are necessary (assuming oxygen is unlimited) to produce 500 k. J of energy in this reaction?

ΔH as a Stoichiometric Quantity Given: ΔH = -283 k. J CO (g) + ½ O 2 (g) CO 2 (g) (d) Calculate the heat of decomposition of two moles of carbon dioxide.

Hess’s Law p The heat of a reaction (ΔH) is constant, whether the reaction is carried out directly in one step or indirectly through a number of steps. p The heat of a reaction (ΔH) can be determined as the sum of heats of reaction of several steps.

Hess’s Law: Example Consider the formation of water: H 2(g) + ½ O 2(g) H 2 O(g) + 241. 8 k. J (Exothermic Rxn ΔH = -241. 8 k. J)

Hess’s Law

Hess’s Law

Hess’s Law Σ ΔH along one path = Σ ΔH along another path Since ΔH is a state function!!

Hess’s Law Given: C(s) + O 2(g) CO 2(g) ΔH = -393. 5 k. J 2 CO(g) + O 2(g) 2 CO 2(g) ΔH = -577. 0 k. J Determine the heat of reaction for: C(s) + ½ O 2(g) CO(g)

Hess’s Law Given: C(s) + O 2(g) CO 2(g) ΔH = -393. 5 k. J C 2 H 4(g)+3 O 2(g) 2 CO 2(g)+2 H 2 O(l) ΔH= -1410. 9 k. J H 2(g) + ½ O 2(g) H 2 O(l) ΔH = -285. 8 k. J Determine the heat of reaction for: 2 C(s) + 2 H 2(g) C 2 H 4(g)

Standard Enthalpies of Formation NIST (National Institute for Standards and Technology) gives values for ΔHf° = standard molar heat of formation Definition: The heat content or enthalpy change when one mole of a compound is formed at 1. 0 atm pressure and 25 °C from its elements under the same conditions.

Examples of Formation Equations H 2(g) + ½ O 2(g) H 2 O(g) ΔHf°(H 2 O, g) = -241. 8 kj/mol C(s) + ½ O 2(g) CO(g) ΔHf°(CO, g) = -111 kj/mol ***Elements/reactants 1 mol of compound Notice units are per mole

Standard Enthalpy of Formation Values p Can look up values of in reference book or textbook p By definition, ΔHf° = 0 for elements in their standard states n Example: Cl 2 (g) H 2 (g) Ca (s)

Summation Equation p In general, when all enthalpies of formation are known: ΔHrxn° = ΣΔHf°(products) ΣΔHf°(reactants) Must multiply all Hf values by coefficient from balanced equation!!!

Summation Equation Example Use the summation equation to determine the enthalpy of the following reaction: 4 NH 3(g) + 5 O 2(g) 4 NO(g) + 6 H 2 O(g) ΔHreaction° = ΣΔHf°(products) ΣΔHf°(reactants)

Unit 5: Part II Thermodynamics: Spontaneity, Entropy and Free Energy

Spontaneous Change What is a spontaneous process? A process that occurs by itself without an outside force helping it. [ A spark to start a process is OK though]

Spontaneous Change Which of the following are spontaneous processes? 1. Snowman melting in the sun 2. Assembling a jigsaw puzzle 3. Rusting of an iron object in humid air 4. Recharging of a camera battery

Spontaneous Reactions and Energy Many spontaneous reactions are exothermic, but not all! Example: H 2 O (s) H 2 O (l) is spontaneous and an ENDOTHERMIC reaction! (∆ H = + 6. 0 k. J)

What other Factor Influences Spontaneity? The Randomness Factor! Nature tends to move spontaneously to a more random state.

Entropy: Disorder and Spontaneity What is entropy? A measure of the randomness (disorder) of a system—a STATE property (state function)! Reaction of K with water

The Second Law of Thermodynamics states: In a spontaneous process, there is a net increase of entropy (taking into account system and surroundings).

Spontaneous Processes result in more random states (more disorder). EXAMPLE: H 2 O (s) H 2 O (l) Water molecules are more disordered as a liquid than as a solid.

Sample Problem Predict which of the following processes have a positive change in entropy: (an increase in the randomness or disorder) a. Taking dry ice from a freezer and allowing it to warm from -80 o. C to room temperature b. dissolving blue food coloring in water c. freezing water into ice cubes

Entropy p Entropy is used to quantify randomness or disorder. p Like enthalpy, entropy is also a state function.

The Third Law of Thermodynamics states: A completely ordered pure crystalline solid has an entropy of zero at 0 K.

Standard Molar Entropies: ΔSo (1 mole, standard conditions): Tells you entropy at 25 o. C and 1 atm (standard state conditions) Units: J/mol K Note: Elements DO NOT have ΔSo = 0! (like they did with ΔHo)

Standard Molar Entropies

For a substance, Entropy generally increases as: 1. Phase change occurs from l g 2. # moles of gas increase from reactants to products 3. T increases (KE increases) s

For a reaction, entropy generally increases as: 1. Reactants (solids or liquids) Products (gases) 2. Total # moles of products > Total # moles of reactants 3. Total # moles of gaseous products > Total # moles of gaseous reactants 4. T is increasing.

Sign of ΔSo for a reaction means: +∆S Entropy increases; S prod > S react -∆S Entropy decreases; S react > S prod

Example Predict the sign of ΔS in each of the following reaction, and explain your prediction. p NH 3 (g) + HCl (g) NH 4 Cl (s) p 2 KCl. O 3 (s) 2 KCl (s) + 3 O 2 (g) p CO (g) + H 2 O (g) CO 2 (g) + H 2 (g)

Calculating ∆S for a Reaction ∆So = So (products) - So (reactants) Calculation is similar to ∆Ho (from Part I) Note units are JOULES not k. J as before!

Example: Calculate ΔSo for the following reaction using the tables in your reference book and the summation equation. 2 H 2 (g) + O 2 (g) 2 H 2 O (l)

Gibbs Free Energy and Free Energy Change The Gibbs (also known as Gibbs-Helmholtz) Equation shows relationship between Energy, Entropy and Spontaneity: ΔG = ΔH - T ΔS Change in Free Energy = Change in Enthalpy – (Temp. x Change in Entropy)

What is free energy? Free energy = AVAILABLE energy

The Relationship between ΔGreaction and Spontaneity 1. If ΔG = positive, reaction is NONSPONTANEOUS. 2. If ΔG = zero, reaction is at equilibrium (balanced). 3. If ΔG = negative, reaction is SPONTANEOUS.

Gibbs Free Energy, G Spontaneous Processes Must Have a Negative Free Energy! J. Willard Gibbs 1839 -1903

How are these factors and spontaneity related? Case # ΔH ΔS ΔG Result 1 - + - spontaneous at all T 2 - - + spontaneous toward low T HOWEVER nonspontaneous toward high T 3 + + + - nonspontaneous toward low T HOWEVER spontaneous toward high T 4 + - + nonspontaneous at all T EXOTHERMIC reactions with Increasing Entropy are Always spontaneous!

Example Predict if the reaction will be spontaneous or not. Use ΔH as given and your estimate of the sign of ΔS. a. C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O (g) ΔH = -2540 k. J b. Cl 2 (g) 2 Cl (g) ΔH is positive

Two methods of calculating ∆Go = ∆Ho - T∆So a) Determine ∆Horxn and ∆Sorxn and use Gibbs equation. b) Use tabulated values of free energies of formation, ∆Gfo. (we will not do this calculation, since it is similar to the ∆Ho one we did in Part I)

Standard Free Energy Change, ∆Go = ∆Ho - T∆So Note: k. J p The units for ΔSo generally are in J p The units for ΔHo are generally in p You must convert FIRST before beginning the problem! p T is in K (o. C + 273)

Example: Gibbs Equation Calculate ΔGo for the reaction below, and predict whether the reaction is spontaneous at 25 o. C. C (s)+ 2 H 2 (g) CH 4 (g) ΔSo = -80. 8 J/mol K ΔHo = -74. 8 k. J/mol T= 298 K

Standard Free Energy of Formation ΔGorxn. = ΣGof(products) - ΣGof(reactants) SAME SUMMATION EQUATION as ΔHo !!!!! (Use reference book for values)