Thermochemistry Chapter 6

Energy Capacity to do work or to produce heat. 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

Energy 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. Copyright © Cengage Learning. All rights reserved

Initial Position In the initial position, ball A has a higher potential energy than ball B. Copyright © Cengage Learning. All rights reserved

Final Position After A has rolled down the hill, the potential energy lost by A (changed to kinetic energy) has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B (which is less than the P.E of A ) The temp. of the hill slightly increases. Copyright © Cengage Learning. All rights reserved

Transfer of Energy System – part of the universe on which we wish to focus attention. Surroundings – include everything else in the universe. Copyright © Cengage Learning. All rights reserved

System and Surrounding

Chemical Energy Endothermic Reaction: Heat flow is into a system. Absorb energy from the surroundings. Exothermic Reaction: Energy flows out of the system. Energy gained by the surroundings must be equal to the energy lost by the system. Copyright © Cengage Learning. All rights reserved

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

(system is underlined) CONCEPT CHECK! Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. (system is underlined) 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

Methane is burning in a Bunsen burner in a laboratory. 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) Copyright © Cengage Learning. All rights reserved

Hydrogen gas and oxygen gas react violently to form water. Explain. CONCEPT CHECK! Hydrogen gas and oxygen gas react violently to form water. Explain. Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? 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

Thermodynamics The study of energy and its interconversions is called thermodynamics. Law of conservation of energy is often called the first law of thermodynamics. Copyright © Cengage Learning. All rights reserved

Internal Energy Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system. To change the internal energy of a system: ΔE = q + w q represents heat w represents work Copyright © Cengage Learning. All rights reserved

Work vs Energy Flow To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Internal Energy (magnitude and direction) Thermodynamic quantities consist of two parts: Number gives the magnitude of the change. Sign indicates the direction of the flow. Copyright © Cengage Learning. All rights reserved

Internal Energy Sign reflects the system’s point of view. Endothermic Process: q is positive Exothermic Process: q is negative Copyright © Cengage Learning. All rights reserved

Internal Energy Sign reflects the system’s point of view. System does work on surroundings: w is negative Surroundings do work on the system: w is positive Copyright © Cengage Learning. All rights reserved

Work Work = P × A × Δh = PΔV P is pressure. A is area. Δh is the piston moving a distance. ΔV is the change in volume.

Work For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = –PΔV To convert between L·atm and Joules, use 1 L·atm = 101.3 J. Copyright © Cengage Learning. All rights reserved

They perform the same amount of work. EXERCISE! Which of the following performs more work? a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. They both perform the same amount of work. w = -PΔV a) w = -(2 atm)(4.0-1.0) = -6 L·atm b) w = -(3 atm)(3.0-1.0) = -6 L·atm Copyright © Cengage Learning. All rights reserved

Change in Enthalpy State function ΔH = q at constant pressure ΔH = Hproducts – Hreactants Energy is a state function; work and heat are not State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Copyright © Cengage Learning. All rights reserved

–252 kJ Consider the combustion of propane: EXERCISE! ΔH = –2221 kJ 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 Copyright © Cengage Learning. All rights reserved

Calorimetry Science of measuring heat Specific heat capacity: The energy required to raise the temperature of one gram of a substance by one degree Celsius. Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved

HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity?

specific heat(s) and heat capacity (C) The specific heat(s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released q = m x s x Dt q = C x Dt Dt = tfinal - tinitial

specific heat(s) and heat capacity (C) This Table is not from Zumdahl’s book

Calorimetry If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution. Copyright © Cengage Learning. All rights reserved

A Coffee–Cup Calorimeter Made of Two Styrofoam Cups RReview Expt.11RReview Expt.11 Copyright © Cengage Learning. All rights reserved

Constant-Volume Calorimetry q qrxn = - (qwater + qbomb) rxn = - (qwater + qbomb) Reaqwater = m x s x Dt ction at Constant V qwater = m x s x Dt qbomb = Cbomb x Dt Reaction at Constant V DH ~ q rxn DH ~ rxn No heat enters or leaves!

Calorimetry Energy released (heat) = s × m × ΔT s = specific heat capacity (J/°C·g) m = mass of solution (g) ΔT = change in temperature (°C) Copyright © Cengage Learning. All rights reserved

ENTHALPY ∆H = Hfinal - Hinitial Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆H where H = enthalpy and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = Hfinal - Hinitial

A 466-g sample of water is heated from 8.50°C to 74.60°C. Calculate the amount of heat absorbed (in kilojoules) by the water.

Strategy We know the quantity of water and the specific heat of water Strategy We know the quantity of water and the specific heat of water. With this information and the temperature rise, we can calculate the amount of heat absorbed (q). Solution Using Equation (6.12), we write Check The units g and °C cancel, and we are left with the desired unit kJ. Because heat is absorbed by the water from the surroundings, it has a positive sign.

q= q= ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J 17. How many kJ of heat must be removed from 1000 g of water (specific heat = 4.184 J /g oC) to lower the temperature from 18.0°C to 12.0°C?   A. 2.5 x 10–2 kJ B. 1.4 kJ C. 4.2 kJ D. 25 kJ q= ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J = 25 kJ q= q= ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J

CONCEPT CHECK! A 100.0 g sample of water at 90°C is added to a 100.0 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 The correct answer is b). Copyright © Cengage Learning. All rights reserved

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. 23°C 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 Copyright © Cengage Learning. All rights reserved

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 Copyright © Cengage Learning. All rights reserved

Qreaction for the combustion of the one-gram sample. Problem A 1.000 g sample of rocket fuel hydrazine, N2 H4 , is burned in a bomb calorimeter containing 1200 g of water. The temperature rises from 24.62 to 28.16 ⁰C. Taking C for the bomb to be 840 J/ ⁰C , calculate Qreaction for the combustion of the one-gram sample. Qreaction for the combustion of the one mole of hydrazine in the bomb calorimeter.

QAAaction = - (qwater + q calorimeter) qQreaction = - (qwater + q calorimeter) qwater = 1200 g x 4.184 J/g ⁰C x (28.16 – 24.62) ⁰C = 17773.6 J qcalorimeter = 840J/ ⁰C x (28.16-24.62) = 2973.6 J (a) Qreaction for 1 g N2H4 = 17773.6 J + 2973.6 J = 20747.2 J = -20.7 kJ (b) Qreaction for 1 mole of N2H4 = 20.7 kJ /gx 32 g/mole = -662.4 kJ water – 1200 g x 4.184 J/g ⁰C x (28.16 – 24.62) ⁰C Qreaction for 1 g N2H4 = 17773.6 J + 2973.6 J = 20747.2 J = -20.7 kJ Qreaction for 1 mole of N2H4 = 20.7 kJ /gx 32 g/mole =

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. Copyright © Cengage Learning. All rights reserved

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

The Principle of Hess’s Law Copyright © Cengage Learning. All rights reserved

To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 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

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

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:

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

Hess’s Law FoFor example, suppose you are given the following data: r example, suppose you are given the following data: Could you use these data to obtain the enthalpy change for the following reaction? you use these data to obtain the ealpy change for the following reaction? 3

Hess’s Law If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

Standard Enthalpy of Formation (ΔHf°) Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. Copyright © Cengage Learning. All rights reserved

Table 6.4 is not from Zumdahl’s book

A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H◦f of the compounds involved are: CH4: –75 kJ , CO2: -394 kJ, H2O: –286 kJ ∆H◦rxn = ∆H◦f products - ∆H◦f reactants = {[-394 + 2 x (-286)] –(-75)} = –891 kJ Answer: –891 kJ Copyright © Cengage Learning. All rights reserved

Petroleum, Natural Gas, and Coal Wood Hydro Nuclear Fossil Fuels Petroleum, Natural Gas, and Coal Wood Hydro Nuclear Copyright © Cengage Learning. All rights reserved

Energy Sources Used in the United States Copyright © Cengage Learning. All rights reserved

The Earth’s Atmosphere Transparent to visible light from the sun. Visible light strikes the Earth, and part of it is changed to infrared radiation. Infrared radiation from Earth’s surface is strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere. Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be. Copyright © Cengage Learning. All rights reserved

The Earth’s Atmosphere Copyright © Cengage Learning. All rights reserved

Other Energy Alternatives Oil shale Ethanol Methanol Seed oil Coal Conversion Hydrogen as a Fuel Other Energy Alternatives Oil shale Ethanol Methanol Seed oil Copyright © Cengage Learning. All rights reserved