Name KEY Period CRHS Academic Chemistry Unit 15 Energy and Thermochemistry Notes Quiz Date Exam Date Lab Dates Notes, Homework, Exam Reviews and Their KEYS located on CRHS Academic Chemistry Website: https://cincochem.pbworks.com
Page 2 of 20 Unit 15 Notes 15.1 MATTER AND ENERGY Energy, & Heat Energy the capacity for doing work or supplying Heat. Heat (_q ) _Energy that transfers from one object to another because of a temperature difference between them. The standard international unit (SI unit) of heat is the Joule ( J ). Law of Conservation of Energy Energy is neither created nor destroyed, but it can be from one form to another. Types of Energy In this course we consider 4 types of energy 1. Kinetic Energy due to motion measured by temperature 2. Thermal Energy total of all kinetic energies within a given system the energy of a system responsible for temperature does not include bond energy (potential energy) 3. Potential Energy STORED in the bonds between atoms in a substance 4. Chemical Energy Energy released or absorbed due to the making and breaking of chemical bonds
Unit 15 Notes Page 3 of 20 Endothermic vs. Exothermic Processes All chemical reactions involve a loss or gain of heat. Exothermic Process - System releases heat to surroundings. Temperature _rises and an exothermic process feels warm. Endothermic Process System absorbs heat from surroundings. Temperature decreases and an endothermic process feels cool. Heat Transfer *Heat flows _from a warmer object _to_ a cooler object. Three Types of HEAT Transfer Type of heat transfer Drawing Example 1. Conduction Touching a stove -Occurs through direct contact of heated particles -Occurs best in solids 2. Convection -Heated particles move through a fluid (liquid or gas) Ocean/Air currents Boiling Water 3. Radiation -electromagnetic waves travel through empty space Sun Feeling heat near a fire
Page 4 of 20 Unit 15 Notes 15.2 HEAT CALCULATIONS FACT: Simply being alive generates heat, which is commonly measured in calories. calorie (cal) the quantity of heat needed to raise the temperature of 1 g of water 1 C Joules are the SI unit for heat. Below is the conversion factor. 1 calorie = 4.184 Joules 1 _Calorie = 1 _kilocalorie = 1000 kilocalorie Practice: Convert 444 calories to joules. 1860 J 444 cal x 4.184 J 1 cal = 1857.7 J = 1860 J Practice: Convert 8341 joules to calories 1994 J 8341 J x 1 cal = 1993.5 cal = 1994 cal 4.184 J Heat capacity (C ) and Specific Heat Capacity (Cp) Heat Capacity (C) (unit = J/ o C ) the quantity of heat needed to raise the temperature of an object 1 C. Heat capacity is an _Extensive physical property of a substance because it depends on the size of the object. From heat capacity, we can derive specific heat capacity by dividing heat capacity by the object s mass. Specific Heat Capacity (Cp) = Heat Capacity (C) mass
Unit 15 Notes Page 5 of 20 Specific heat capacity, (Cp) (unit = _J/g o C or cal/g o C ) the quantity of heat needed to raise the temperature of 1 g of substance 1 C at constant pressure. Every substance has a unique value. Specific heat capacity is an intensive property of matter because it is independent of size. Specific heat capacity is also referred to as specific heat. COMMON SPECIFIC HEATS Substance J/g-C cal/g-c water 4.18 1.00 grain alcohol 2.4 0.58 ice 2.1 0.50 steam 1.7 0.40 chloroform 0.96 0.23 aluminum 0.90 0.21 glass 0.50 0.12 iron 0.46 0.11 silver 0.24 0.057 mercury 0.14 0.033 Water Water (H2O) has a high specific heat capacity which means water can hold a large amount of heat, so water - Is slow to _heat up and cool down; and holds or stores heat more easily Questions: 1) In the days before central heating, why did pioneers go to sleep with hot water bottles, rather than hot iron pans? Water stores more heat, stays warmer longer, releases heat slower 2) Why do coastal cities have a more moderate climate than inland areas? Coastal water absorbs heat from sun during day and releases heat during night and breezes across water cool or warm land next to it. 3) Why do we pump water through radiators in automobiles? Water holds heat produced by engine and then is cooled in radiator and returned to engine so it can gain more heat to cool engine.
Page 6 of 20 Unit 15 Notes Specific Heat Equation To calculate the heat required to change a substance s temperature (T) we use the following equation: q = mcpt Where: q = heat m = mass Cp = specific heat T = change in temp Practice 1) How much heat is absorbed by 500 g of water as its temperature is increased from 10C to 85C? q = m c deltat = 500 g x 4.184 J g o C x 75o C = 156750 J = 200000 J 200000 J delta T = 85 o C 10 o C = 75 o C 2) The temperature of a piece of copper with a mass of 95.4 g increases from 25.0 C to 48.0 C when the metal absorbs 849 J of heat. What is the specific heat of copper? q 849 J J = 0.3869 J m x delta T 95.4 g x 23.0 o C g o C g o C 0.387 J g o C
Unit 15 Notes Page 7 of 20 Calorimetry is the accurate and precise measurement of HEAT_ change for chemical processes. Calorimetry is based on the following principle. Heat LOST by a system = Heat GAINED by its surroundings q lost by system = q gained by calorimeter In order for the process to be accurate and precise it must be carried out in a device called a CALORIMETER. Constant volume (bomb) calorimetry: 1) Allows reaction to take place inside an enclosed container (bomb) which is surrounded by water 2) Measures the temperature change of the water. The bomb absorbs some energy so its heat capacity must be considered in calculations. 3) Uses the specific heat equation, q=mcprt, to calculate the heat gained or lost by the calorimeter.
Page 8 of 20 Unit 15 Notes 15.3 WHAT IS THERMOCHEMISTRY? THERMOCHEMISTRY is the study of the transfers of energy as heat that accompany chemical reactions and physical changes (changes in state). In Thermochemistry, we will look at how heat flows between a system and its surroundings. o o The system is the object or objects you are investigating. The surroundings are everything else. In thermochemical equations, heat flow is given from the point of view of the system. EXAMPLE 1: Ice melts System: water Surroundings: everything else Heat + H2O (s) H2O (l) ENDOTHERMIC Heat is a reactant Reverse the process Water freezes System: water Surroundings: everything else H2O (l) Heat + H2O (s) EXOTHERMIC Heat is a product EXAMPLE 2: Another example is a burning log that is releasing heat, but that energy was contained in the chemical bonds as potential energy of the log before it began to burn.
Unit 15 Notes Page 9 of 20 Heat Energy transferred during a reaction Every reaction involves a change in chemical potential energy. Reactants transform into products which either have GREATER or LESS potential energy than the reactants. So, in any reaction there is always either a ABSORBTION or RELEASE of energy. This change is called the Change in Enthalpy of Reaction. (In the past it was also called heat of reaction.) Change in Enthalpy of Reaction (H) = the CHANGE of heat for a reaction under constant pressure. Energy released or Energy absorbed H = H products - H reactants Heat lost to surroundings EXOTHERMIC Heat gained by system ENDOTHERMIC Heat flows out of system Heat flows into the system Heat expressed on the Product side of rxn Heat expressed on the Reactant side of rxn - ΔH + ΔH ΔH +ΔH Potential Energy of system Thermal energy of surroundings FEELS HOT Potential Energy of system Thermal energy of surroundings FEELS COLD Practice:
Page 10 of 20 Unit 15 Notes The graph shows the energy change during the reaction A + B C. 1) The products have ( MORE / LESS ) potential energy than the reactants. 2) The reaction is ( ENDOTHERMIC / EXOTHERMIC). So, Heat is _GAINED BY THE SYSTEM. Practice: On the graph below, draw the reaction progress of an exothermic reaction. Practice: On the graph below, draw the reaction progress of an endothermic reaction.
Unit 15 Notes Page 11 of 20 Thermochemical Equation A thermochemical equation is a chemical equation for a reaction that includes HEAT. *Thermochemical equations must be BALANCED.. *Treat the coefficients as number of moles, never as number of molecules *Thermochemical equations treat enthalpy change (_H ) just like any other REACTANT or PRODUCT Practice 1. 4Fe (s) + 3O2 (g) 2Fe2O3 (s) + 1625 kj Does this reaction release heat or absorb heat? RELEASES How much? 1625 kj What does kj mean? 1000 JOULES (measurement of ENERGY ) Endothermic or exothermic? EXOTHERMIC 2. C (s) + 2S (s) + 89.3 kj CS2 (l) Is heat released or absorbed in this thermochemical reaction? ADSORBED How much? 89.3 kj Endothermic or exothermic? ENDOTHERMIC Thermochemical reactions are normally written by designating the value of H, rather than by writing the energy as a reactant or product. 2H 2(g) = O 2(g) -> 2H 2O (g) H = -483.6 kj A negative H indicates an exothermic reaction, i.e. heat is a product. A positive H indicates an endothermic reaction, i.e. heat is a reactant. Chemical equations involving H are similar to _STOICHIOMETRY problems; they depend on the number of MOLES of reactants and products involved, for example: CaO(s) + H2O(l) Ca(OH)2 (s) + 65.2 kj
Page 12 of 20 Unit 15 Notes and 2 CaO(s) + 2 H2O(l) 2 Ca(OH)2 (s) + 130.4 kj Practice H2 (g) + F2 (g) 2HF(g) H = -536 kj Calculate the heat change (in kj) for the conversion of 10.1 g of H 2 gas to HF gas at constant pressure. 1 mole H2 563 kj 10. 1g x x = 2680 kj released 2. 02 g H2 1 mol H2 Practice 2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe (s) H = -851 kj Calculate the heat change (in kj) for the thermite reaction of 320 g of Fe 2O 3 into Al 2O 3 at constant pressure 1 mole Fe2O3 320 g Fe2O3 x 165. 7 g Fe2O3 x 852 kj = 1643 kj or 1600 kj released 1 mol Fe2O3 Practice K 2O (s) + H 2O (l) 2KOH (aq) H = 215 kj What is the heat change for the above reaction, at constant pressure if you begin with 282.6 g of K 2O? 1 mole K2O 282 g K2O x 71. 1 g K2O x 215 kj = 853 kj absorbed 1 mol K2O
Unit 15 Notes Page 13 of 20 15.4 DRIVING FORCES OF REACTIONS The majority of chemical reactions are EXOTHERMIC, where products are more stable. In nature, reactions tend to proceed in a direction that leads to lower energy state. However, some endothermic reactions do occur SPONTANEOUSLY! Question: Why do some reactions occur spontaneously that are endothermic? Answer: Two factors determine whether a reaction will occur spontaneously. 1. Change in ENTHALPY and 2. Change in RANDOMNESS of the particles in the system, which is known as ENTROPY Example of Endothermic Spontaneous Change: Melting Ice Ice will melt SPONTANEOUSLY at room temperature by absorbing heat from the surroundings even though this is endothermic process. The well-ordered arrangement of water molecule in ice crystals is lost, and the less-ordered liquid phase of higher energy is formed. There is tendency in nature to proceed in direction that INCREASES the randomness of a system. Entropy (S) is the measure of the RANDOMNESS of the particles, such as molecules, in a system. In general S(gases) > S(liquids) > S(solids) Entropy Change (S) can be measured for reactions and is the difference between the entropy of the products and the reactants.
Page 14 of 20 Unit 15 Notes Practice: Predict whether the value of S for each of the following reactions will greater than, less than, or equal to zero. 3H2 (g) + N2 (g) -> 2 NH3 (g) 2Mg (s) + O2 (g) -> 2 MgO (s) C6C12O6(s) + O2(g) -> 6 CO2(g) + 6 H2O(g) KNO3(s) - > K + 1(aq) + NO3-1 (aq) S is less than zero S is less than zero S is greater than zero S is greater than zero Free Energy Processes in nature are driven in two directions: toward least ENTHALPY and toward greatest ENTROPY A function called Gibbs Free Energy (G o ) has been defined that simultaneously assesses the tendency for enthalpy and entropy to change. Natural process proceed in the direction that lowers the Gibbs Free Energy. G o = H o - TS o The following table shows the four possible scenarios for Gibbs Free Energy. Remember that for a reaction to be spontaneous, the Gibbs Free energy should be NEGATIVE for the reaction. H S G - value (exothermic) + value (more random) always negative - value (exothermic) - value (less random) negative at lower temperatures + value (endothermic) + value (more random) negative at higher temperature + value (endothermic -value (less ransom) never negative
Unit 15 Notes Page 15 of 20 Practice: Predict whether the reactions will be spontaneous at 298K by calculating G o from the given enthalpies and entropies. 1. C 2H 2(g) + H 2O (g) -> C2H6 (g) H o = -136.9 kj/mol and -S o = -0.127 kj/(mol*k) G o = H o - TS o = -136.9 kj/mol (298K x -.127 kj/(mol*k) = -98.2 kj/mol (Spontaneous) 2. CH 4(g) + H 2O(g) -> CO(g) + 3 H 2(g) H o = +206.1 kj/mol and -S o = -0.215 kj/(mol*k) G o = H o - TS o = +206.1 kj/mol (298K x -.215 kj/(mol*k) = +270 kj/mol (Not Spontaneous)
Page 16 of 20 Unit 15 Notes