Transcript Chapter 10 Energy
Chapter 10 Energy Energy: Types Objectives: 1)To understand the general properties of energy 2)To understand the concepts of temperature and heat 3)To consider the direction of energy flow as heat Energy: Types Energy: the ability to do work or produce heat TWO TYPES Potential energy (energy due to position or composition) PE = mgh Kinetic energy: energy is due to the motion of an object KE=1/2 mv2 Law of Conservation of Energy: energy can be converted from one form to another but can be neither created nor destroyed. Energy: Types Work: force acting over a distance. As energy is transferred some of it is converted to heat (due to friction). STATE FUNCTION: property of the system that changes independently of its pathway. Examples: ENERGY is a state function work and heat are not. Energy: Temperature and Heat Temperature: measure of the random motion of the components of a substance. What will happen to the temperature of the water on both sides? Insulated Box and a thin metal sheet Energy: Temperature and Heat Heat: flow of energy due to a temperature difference. T final = T hot (initital) + T cold (initial) 2 Exothermic and Endothermic System: part of the universe on which we wish to focus Surroundings: include everything else in the universe. Exothermic: a process that results in the evolution of heat. ENERGY FLOWS OUT Endothermic: process that absorbs energy from the surroundings ENERGY FLOWS IN. 10.4 Thermodynamics • Objective: To understand how energy flow affects internal energy 10.4 Thermodynamics • Thermodynamics: study of energy • First Law of Thermodynamics: The energy of the universe is constant (same as Law of Conservation of Energy) • Internal energy (E)= sum of kinetic and potential energy of the system – This can be changed by a flow of work, heat or both 10.4 Thermodynamics • Change in Internal energy DE=q+w q= heat w= work Thermodynamic quantities consists of 2 Parts: a number and a sign. The sign reflects the system’s point of view. 10.4 Thermodynamics • Change in Internal energy DE=q+w Surroundings Surroundings Energy System System DE<0 DE>0 q is -, exothermic, heat out q is +, endothermic, heat into system w is -, system does work on surroundings w is +, if surroundings do work on system 10.5 Energy Changes Objective: 1) To understand how heat is measured 10.5 Energy Changes Calorie: (metric system) the amount of energy (heat) required to raise the temperature of one gram of water by one Celsius degree. In food, kilocalorie. C The joule (SI unit) can be most conveniently defined in terms of the calorie. 1 calorie=4.184 joule 1 cal=4.184 J 10.5 Energy Changes Express 60.1 cal of energy in units of joules. 60.1 cal x 4.184 J = 251 J 1 cal Calculating energy requirements Determine the amount of energy(heat) in joules required to raise the temperature of 7.40 g water from 29.0o to 46.0o 4.184 J x 7.40 g goC x 17.0 C = 526 J 10.5 Energy Changes Another important factor is IDENTITY OF SUBSTANCE. Specific heat capacity: amount of energy required to change the temperature of one gram of a substance by one Celsius degree. Q=s x m x DT Q= energy (heat) required s=specific heat capacity m= mass of the sample in grams DT=change in temperature in Celsius degrees. Table 10.1 p. 297 10.5 Energy Changes A 1.6 g sample of a metal that has the appearance of gold requires 5.8 J of energy to change its temperature from 23 C to 41 C. Is the metal pure gold. 10.5 Energy Changes 5.8= s x 1.6 x (41-23) 5.8 =s s=0.20 J/g C 1.6 x 18 pure gold s=0.13 J/g C Section 10.6 Thermochemistry (Enthalpy) Objective: To consider the heat (enthalpy) of chemical reactions. Section 10.6 Thermochemistry (Enthalpy) Enthalpy: a special energy function (DH) Under constant pressure, the change in enthalpy is equal to the energy that flows as heat. DHp=heat Section 10.6 Thermochemistry (Enthalpy) When 1 mole of methane(CH4) is burned at constant pressure, 890kJ of energy is released as heat. Calculate DH for a process in which a 5.8 g sample of methane is burned at constant pressure. qp=DH= -890kJ/mol CH4 5.8 g CH4 x 1mol= 16.0 g Section 10.6 Thermochemistry (Enthalpy) When 1 mole of sulfur dioxide reacts with excess oxygen to form sulfur trioxide at constant pressure, 198.2kJ of energy is released as heat. Calculate DH for a process in which a 12.8 g sample of sulfur dioxide reacts with oxygen at constant pressure. 39.6kJ Section 10.6 Thermochemistry (Enthalpy) Calorimeter is a device used to determine the heat associated with a chemical reaction. Figure 10.6: A coffee-cup calorimeter. NCSSM Distance Learning T.I.G.E.R. - Chemistry page 4#thermo 10.7 Hess’s Law • Objective: To understand Hess’s law 10.7 Hess’s Law • Enthalpy is a state function • Independent of path soooo – The change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Hess’s Law) – Important: because it allows us to calculate heats of reaction that might be difficult or inconvenient to measure directly in a calorimeter. 10.7 Hess’s Law N2(g) + O2(g) 2NO2(g) DH=68kJ N2(g) + O2(g) 2NO (g) DH2=180kJ 2NO (g) + O2(g) 2NO2 DH3= -112kJ N2(g) + O2(g) 2NO2(g) DH=68kJ 10.7 Hess’s Law Characteristics of Enthalpy Changes 1) If a reaction is reversed, the sign of DH is also reversed. 2) The magnitude of DH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of DH is multiplied by the same integer. 10.7 Hess’s Law Given the following data 4CuO(s) 2Cu2O(s) + O2(g) DH=288kJ Cu2O(s) Cu(s) + CuO(s) DH=11 kJ Calculate DH for the following reaction 2Cu(s) + O2(g) 2CuO(s) Cu(s) + CuO(s) 2 Cu2O + O2 Cu2O DH= -(11kJ) 4CuO(s) DH=-(288kJ) 2(Cu(s) + CuO(s) Cu2O DH= -2(11kJ) 2 Cu2O + O2 4CuO(s) DH=-(288kJ) 2Cu(s) + O2(g) 2CuO(s) DH= -310kJ 10.10 Energy as a Driving Force Objective: To understand energy as a driving force for natural processes. 10.10 Energy as a Driving Force Energy spread: in a given process, concentrated energy is dispersed widely. This distributions happens every time an exothermic process occurs. Matter spread: molecules of a substance spread out and occupy a larger volume. 10.10 Energy as a Driving Force Entropy: natural tendency of the universe to become disordered. S (as things become more disordered), the value of S increases Second law of thermodynamics: The entropy of the universe is always increasing. Putting it all together! Gibbs Free Energy G=H-TDS Temperature in Kelvin Change in Free Energy DG=DH-TDS (constant temperature) Related to system If DG is negative (spontaneous reaction), DS (+) Calculations • Let’s predict the spontaneity of the melting of ice H2O(s) H2O(l) DH =6.03 x 103 J/mol and DS=22.1 J/K mol What is DG at -10o, 0o and 10oC? Which is spontaneous? DG=2.2x 102 DG=0 DG= -2.2 x 102 More calculations Br2(l) Br2(g) DH=31.0 kJ/mol and DS=93.0 J/K mol What is the normal boiling point? DG= DH- T DS 0= DH- T DS T= DH/D S=333K Figure 10.7: Energy sources used in the United States. Figure 10.8: The earth’s atmosphere. Figure 10.9: The atmospheric CO2 concentration over the past 1000 years. Figure 10.10: Comparing the entropies of ice and steam.