Chapter 10 Energy

10.3 Exothermic and Endothermic Processes 10.4 Thermodynamics 10.1 The Nature of Energy 10.2 Temperature and Heat 10.3 Exothermic and Endothermic Processes 10.4 Thermodynamics 10.5 Measuring Energy Changes 10.6 Thermochemistry (Enthalpy) 10.7 Hess’s Law 10.8 Quality Versus Quantity of Energy 10.9 Energy and Our World 10.10 Energy as a Driving Force Copyright © Cengage Learning. All rights reserved

Ability to do work or produce heat. Energy Ability to do work or produce heat. That which is needed to oppose natural attractions. Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed. The total energy content of the universe is constant. Copyright © Cengage Learning. All rights reserved

Potential energy – energy due to position or composition. Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Telekinesis Video Part 1 Telekinesis Video Part 2

Energy Transformations Types of Energy 1. Kinetic energy 2. Potential energy 3. Chemical energy 4. Heat energy 5. Electric energy 6. Radiant energy

Initial Position In the initial position, ball A has a higher potential energy than ball B.

Final Position After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.

Both of the above are at 70oF. Which has the most heat? Temperature = average kinetic energy. Heat= total energy. Both of the above are at 70oF. Which has the most heat?

A measure of the random motions of the components of a substance. Temperature A measure of the random motions of the components of a substance. Copyright © Cengage Learning. All rights reserved

Heat A flow of energy between two objects due to a temperature difference between the objects. Heat is the way in which thermal energy is transferred from a hot object to a colder object. Copyright © Cengage Learning. All rights reserved

Energy Changes Accompanying the Burning of a Match Copyright © Cengage Learning. All rights reserved

Energy Changes and Chemical Reactions 4 C3H5N3O9 6 N2 + 10 H2O + 12 CO2 + O2 + 5720kJ Nitroglycerine Exothermic

Endothermic Process: Heat flow is into a system. Absorb energy from the surroundings. q = positive(+) Exothermic Process: Energy flows out of the system. q = negative (-) Energy gained by the surroundings must be equal to the energy lost by the system.

Concept Check Is the freezing of water an endothermic or exothermic process? Explain. Exothermic process because you must remove energy in order to slow the molecules down to form a solid. Copyright © Cengage Learning. All rights reserved

Concept Check Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Exo Endo Exothermic (heat energy leaves your hand and moves to the ice) Endothermic (heat energy flows into the ice) Endothermic (heat energy flows into the water to boil it) Exothermic (heat energy leaves to condense the water from a gas to a liquid) Endothermic (heat energy flows into the ice cream to melt it) Copyright © Cengage Learning. All rights reserved

Concept Check For each of the following, define a system and its surroundings and give the direction of energy transfer. Methane is burning in a Bunsen burner in a laboratory. Water drops, sitting on your skin after swimming, evaporate. System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic) System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic) System=burner, surroundings= lab, energy to system System=drops, surroundings= air, energy to surroundings Copyright © Cengage Learning. All rights reserved

Hydrogen gas and oxygen gas react violently to form water. Concept Check Hydrogen gas and oxygen gas react violently to form water. Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? Explain. Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted. Copyright © Cengage Learning. All rights reserved

Study of energy Law of conservation of energy is often called the first law of thermodynamics. The energy of the universe is constant.

The common energy units for heat are the calorie and the joule. calorie – the amount of energy (heat) required to raise the temperature of one gram of water 1oC. Joule – 1 calorie = 4.184 joules

Example Convert 60.1 cal to joules.

The amount of substance being heated (number of grams). Energy (Heat) Required to Change the Temperature of a Substance Depends On: The amount of substance being heated (number of grams). The temperature change (number of degrees). The identity of the substance. Specific heat capacity is the energy required to change the temperature of a mass of one gram of a substance by one Celsius degree. Copyright © Cengage Learning. All rights reserved

Specific Heat Capacities of Some Common Substances Copyright © Cengage Learning. All rights reserved

To Calculate the Energy Required for a Reaction: Energy (heat) required, Q = s × m × ΔT Since q=ΔH (change of heat) ΔH = s x m x ΔT Q = energy (heat) required (J) s = specific heat capacity (J/°C·g) m = mass (g) ΔT = change in temperature (°C) Copyright © Cengage Learning. All rights reserved

Energy of a Snickers Bar There are 311 Cal in a Snickers Bar. If you added this much heat to a gallon of water at 25oC what would be the final temperature? 311 Cal = 311,000 cal 1 gallon of water weighs 3960.4 g Heat = SH x m x ΔT 311,000 cal = 1.000 cal/goC x 3960.4g x ?oC Solving for the change of temperature gives us 78.5oC Add this to the 25oC to get 104oC which is above the boiling point of water (100oC)

The mass of water and the temperature increase. Example Calculate the amount of heat energy (in joules) needed to raise the temperature of 6.25 g of water from 21.0°C to 39.0°C. Where are we going? We want to determine the amount of energy needed to increase the temperature of 6.25 g of water from 21.0°C to 39.0°C. What do we know? The mass of water and the temperature increase. Copyright © Cengage Learning. All rights reserved

What information do we need? Example Calculate the amount of heat energy (in joules) needed to raise the temperature of 6.25 g of water from 21.0°C to 39.0°C. What information do we need? We need the specific heat capacity of water. 4.184 J/g°C How do we get there? Copyright © Cengage Learning. All rights reserved

Exercise A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.) a) 0.052 g b) 4.6 g c) 19 g d) 590 g The correct answer is c. Q = s×m×ΔT m = Q/ s×ΔT m = {(142 cal)(4.184 J/1 cal)} / {(0.45 J/g°C)(92°C–23°C)} m = 19 g

(90o(100g)+ 10o(500g))/600gT = 23o Concept Check A 100.0 g sample of water at 90.°C is added to a 500.0 g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. The correct answer is c). The final temperature of the water is 23°C. - (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 23°C (90o(100g)+ 10o(500g))/600gT = 23o

The final temperature of the water is: a) Between 50°C and 90°C Concept Check You have a Styrofoam cup with 50.0 g of water at 10.C. You add a 50.0 g iron ball at 90.C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C The correct answer is c). The final temperature of the water is 18°C. - (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 18°C 50.0gFe(0.45SHFe)(90-XoC)=50.0gw(4.18SHw)(X-10oC)

Consider the reaction: S(s) + O2(g) → SO2(g) Example Consider the reaction: S(s) + O2(g) → SO2(g) ΔH = –296 kJ per mole of SO2 formed. Calculate the quantity of heat released when 2.10 g of sulfur is burned in oxygen at constant pressure. Where are we going? We want to determine ΔH for the reaction of 2.10 g of S with oxygen at constant pressure. What do we know? When 1 mol SO2 is formed, –296 kJ of energy is released. We have 2.10 g S. Copyright © Cengage Learning. All rights reserved

Consider the reaction: S(s) + O2(g) → SO2(g) Example Consider the reaction: S(s) + O2(g) → SO2(g) ΔH = –296 kJ per mole of SO2 formed. Calculate the quantity of heat released when 2.10 g of sulfur is burned in oxygen at constant pressure. How do we get there? Copyright © Cengage Learning. All rights reserved

Consider the combustion of propane: Exercise Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. –252 kJ (5.00 g C3H8)(1 mol / 44.094 g C3H8)(-2221 kJ / mol C3H8) ΔH = -252 kJ

Enthalpy, H is measured using a calorimeter. Calorimetry Enthalpy, H is measured using a calorimeter.

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3. N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ ΔH1 = ΔH2 + ΔH3 = 68 kJ Copyright © Cengage Learning. All rights reserved

Characteristics of Enthalpy Changes If a reaction is reversed, the sign of ΔH is also reversed. The magnitude of ΔH 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 ΔH is multiplied by the same integer. Copyright © Cengage Learning. All rights reserved

Consider the following data: Example Consider the following data: Calculate ΔH for the reaction Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products. Multiply reactions to give the correct numbers of reactants and products. Copyright © Cengage Learning. All rights reserved

Reverse the two reactions: Example Reverse the two reactions: Desired reaction: Copyright © Cengage Learning. All rights reserved

Example Multiply reactions to give the correct numbers of reactants and products: 4( ) 4( ) 3( ) 3( ) Desired reaction: Copyright © Cengage Learning. All rights reserved

Final reactions: Desired reaction: ΔH = +1268 kJ Example Copyright © Cengage Learning. All rights reserved

-792/2 + 297 = -99 kJ Concept Check Calculate ΔH for the reaction: SO2 + ½O2 → SO3 ΔH = ? Given: (1) S + O2 → SO2 ΔH = –297 kJ (2) 2S + 3O2 → 2SO3 ΔH = –792 kJ a) –693 kJ b) 101 kJ c) 693 kJ d) –99 kJ The correct answer is d. We have to rearrange the equations so that they add up to the desired equation. First, reverse equation (1) to give : SO2 → S + O2 ΔH = +297 kJ Then add ½ equation (2) : S + 3/2O2 → SO3 ΔH = –396 kJ The two new eqns add up: SO2 + ½O2 → SO3 ΔH = –99 kJ Turn the first reaction around and divide the second reaction by 2. Do the same with the ΔH’s. -792/2 + 297 = -99 kJ

Carbon based molecules from decomposing plants and animals Fossil Fuels Carbon based molecules from decomposing plants and animals Energy source for United States Copyright © Cengage Learning. All rights reserved

Thick liquids composed of mainly hydrocarbons. Petroleum Thick liquids composed of mainly hydrocarbons. Hydrocarbon – compound composed of C and H. Copyright © Cengage Learning. All rights reserved

Gas composed of hydrocarbons. Natural Gas Gas composed of hydrocarbons. Copyright © Cengage Learning. All rights reserved

Coal Formed from the remains of plants under high pressure and heat over time. Copyright © Cengage Learning. All rights reserved

Effects of Carbon Dioxide on Climate Greenhouse Effect Copyright © Cengage Learning. All rights reserved

Effects of Carbon Dioxide on Climate Atmospheric CO2 Controlled by water cycle Could increase temperature by 10oC

Global Warming

Global Warming

Global Warming

Global Warming

New Energy Sources Solar Nuclear Biomass Wind Synthetic fuels

Consider a gas trapped as shown Natural processes occur in the direction that leads to an increase in the disorder of the universe. Example Consider a gas trapped as shown Copyright © Cengage Learning. All rights reserved

What happens when the valve is opened? Copyright © Cengage Learning. All rights reserved

Gives off energy when solid forms. Spreads out when dissolves. Two Driving Forces Gives off energy when solid forms. Spreads out when dissolves. Energy spread Matter spread Copyright © Cengage Learning. All rights reserved

In a given process concentrated energy is dispersed widely. Energy Spread In a given process concentrated energy is dispersed widely. This happens in every exothermic process. Copyright © Cengage Learning. All rights reserved

Molecules of a substance spread out to occupy a larger volume. Matter Spread Molecules of a substance spread out to occupy a larger volume. Processes are favored if they involve energy and matter spread. Copyright © Cengage Learning. All rights reserved

Entropy, S Function which keeps track of the tendency for the components of the universe to become disordered. Measure of disorder or randomness

Entropy, S What happens to the disorder in the universe as energy and matter spread? Copyright © Cengage Learning. All rights reserved

Second Law of Thermodynamics The entropy of the universe is always increasing. The heat death of the universe will occur when all particles of matter ultimately have the same average kinetic energy and exist in a state of maximum disorder. From this To this