Use your textbook or other resources available to answer the following questions General Information: Thermochemistry Phase Change A change in the physical form/state but not a change in the chemical identity of the substance. PHASE CHANGES INVOLVE HEAT ENERGY!! Types of Phase Changes: Fusion (melting) solid to liquid Vaporization liquid to gas Sublimination solid to gas Freezing liquid to solid Condensation gas to liquid Deposition gas to solid To calculate heat (q) for a given temperature change: heat (q) = (specific heat) (mass) ( T) where T = T f T i To calculate heat (q) for phase change: heat (q) = (mass) ( H phase change ) Activity 1: On the heating curve of water below label the phase(s) that are present at each stage of the curve. Specific heat J/(g o C) H phase change Water(s) 2.03 H fus = +6.01 kj/mole Water(l) 4.179 H vap = +40.67 kj/mole T 150 o C To calculate heat for temperature change: heat = (specific heat) (mass) ( T) where T = T f T i 100 o C 50 o C 0 o C To calculate heat for phase change: heat = (mass) ( H phase change ) -50 o C Heat added Calculate the heat required to change one mole of ice at 45.0 o C to steam at 100 o C [use 3 sig.figs.] 1
To make meaningful observations and measurements of changes in energy (work and/or heat), we have to define the system and surroundings System The part of the universe we are studying Isolated System A system that exchanges neither energy nor matter with the surroundings Example: A Thermos bottle containing hot soup with the lid on tightly Closed System A system that exchanges energy but not matter with the surroundings Example: Cup of hot soup with a lid Open System A system that exchanges both energy and matter with the surroundings Example: An open cup of hot soup Surroundings Everything else Example: A chemical reaction in a beaker Within an isolated system, energy (in the form of heat) is not transferred to the surroundings, but it can be transferred between objects within the system. q hot + q cold = 0 By law!! q cold = -q hot By law!! So if q = mc T, then m cold c cold T cold = -m hot c ho t T hot Activity 2: Solve the following heat gained/heat lost problems. A 15.5 gram sample of a metal alloy is heated to 98.9 o C and then dropped into 25.0 grams of water in a perfect calorimeter (i.e. it is isolated). The temperature of the water rises from 22.5 o C to 25.7 o C. 1. Fill in the temperature diagram T = o C T = o C T = o C T = T f T i = o C o C = o C T = T f T i = o C o C = o C 2. Examine the application of the first law of thermodynamics to this system. 1st Law: Heat lost + Heat gained = 0 What substance lost heat: What substance gained heat: 3. Apply the first law of thermodynamics to solve for the specific heat of the metal alloy. 2
4. A 50. g sample of water at 30. C is added to 40. g of water at 90 C. What is the final temperature of the resulting mixture? The Enthalpy of Reaction: Calculations using Hess's Law, Heats of Formation, Bond Dissociation Energies and Calorimetry Enthalpy of reaction ( H rxn ) values have been determined experimentally for numerous reactions, and these H values may be used to calculate H values for other reactions involving the same chemical species. The reason this is possible is that enthalpy H is a state property so H is independent of path. (Similarly, the height of a mountain above sea level is independent of the path you follow to climb the mountain.) Because H is independent of path, we can determine the enthalpy of foods by burning them in a bomb calorimeter in the laboratory to produce the same products that are obtained by the complicated metabolic pathways in our body! There are four principle methods used to calculate H values for a reaction, both of which are based on the idea that H for a reaction is independent of the path used to go from reactants to products. Use of Hess's Law to Calculate H rxn Hess's Law The H for an overall reaction can be found indirectly by summing H values for any set of reactions which sum to the desired overall reaction. Hrxn = H rxn1 + H rxn2 + H rxn3 etc. Usually before reactions are added together, some of them must be reversed and/or multiplied by a factor n in order that they sum to the desired reaction. In this process the rules are: Whenever you multiply a reaction by n, H for the reaction is also multiplied by n. If you reverse a reaction, H changes sign. Standard Heat of Formation ( H f) The enthalpy change of a reaction for the formation of 1 mole of substance in its standard state formed from elements in their standard states Standard State: 1 atm at 25 C & 1 M solution concentration Pure elements in its most stable form under standard conditions, H f = 0 3
To calculate H rxn using standard heats of formation: aa + bb cc + dd H rxn = [chfc + dhfd ] - [ahfa + bhfb ] Therefore: H rxn = H fproducts H freactants Bond Dissociation Energy The amount of energy that must be supplied in order to break a chemical bond in an isolated molecule in the gaseous state Hrxn = BDEreactants BDEproducts Breaking a bond is always endothermic! Forming a bond is always exothermic! Must include # of bonds! Calorimetry The measurement of the quantity of heat transferred during a physical change or chemical process. It is an experimental method used to determine the enthalpy of a reaction (H rxn ) For enthalpy of a reaction, it is important to realize that the evolution of heat gained or lost during a reaction is per mole! Steps for Solving Calorimetry Problems: 1. Calculate heat using q = mc T 2. Use a conversion factor (may have to calculate some of the variables in the conversion factor) 3. H rxn = q X coefficient of compound # of moles of compound 4. Determine the sign (Consider the solution the surroundings and the reaction the system ) Activity 3: Solve the following using Hess s Law From the following heats of reaction ( H rxn ): 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) H = 196 kj 2 S(s) + 3 O 2 (g) 2 SO 3 (g) H = 790 kj Calculate the H rxn for the overall reaction: S(s) + O 2 (g) SO 2 (g) H =??? 4
Activity 4: Solve the following using Standard Heats of Formation Calculate the H rxn using standard heats of formation. 2C 2 H 5 OH (l) + 2CO 2 (g) C 6 H 12 O 6 (s) H rxn =?? Compound H f (kj/mol) C 2 H 5 OH -277.7 CO 2-393.5 C 6 H 12 O 6-1260 Activity 5: Solve the following problem using Bond Dissociation Energies. Calculate the heat of reaction ( Hrxn) for the following combustion reaction 2H 2 O (g) 2H 2 (g) + O 2 (g) Hrxn =?? Compound H BDE (kj/mol) H-H 436 H-O 460 O=O 502 O-O 138 5
Activity 6: Solve the following problems using Calorimetry. A 100.0 ml sample of 0.500 M HCl at 21.36 o C is added to 100.0mL of 0.500M NaOH also at 21.36 o C in a perfect calorimeter cup. After mixing, the temperature rises to 24.70 o C. Assume the specific heat of the reaction is 4.18 J/g C, the density of the solution is 1.00 g/ml and the calorimeter absorbs a negligible amount of heat. Determine the H rxn. 1. Examine the application of the first law of thermodynamics to this system. 1st Law: Heat gained + Heat lost = 0 What gained heat: What lost heat/energy: 2. How much heat (q) (in J) is absorbed by the solution? 3. What is the heat of this reaction in kj/mole of the HCl. Hint: You will have to write a reaction! 6