Energy & Chemical Change

Energy & Chemical Change

Thermochemistry The study of energy or heat changes during a chemical reaction Energy (E ) The capacity to do work or transfer heat Two types: KE & PE Kinetic Energy (KE) – (includes thermal energy) Energy of motion KE = ½mv 2

Thermochemistry Energy (cont. ) KE and PE are interconvertible
Potential Energy (PE) – (stored energy) Energy possessed as a result of position or chemical composition (chemical energy) Exists in attractions and repulsions - gravity, positive and negative charges, springs KE and PE are interconvertible

rock has potential energy due to it’s position but no KE
PE decreasing while KE increasing Energy associated with position can be converted to energy of motion traded PE for KE

Chemical Energy (PE) PE possessed by chemicals
Stored in chemical bonds Breaking bonds requires energy Forming bonds releases energy Chemical energy can be converted into heat – energy of molecular motion (KE)

Manifestations of Energy

Your Turn! Which of the following represents a decrease in the potential energy of the system? A book is raised six feet above the floor A ball rolls downhill Two electrons come close together A spring is stretched completely Two atomic nuclei approach each other

Your Turn! Which of the following represents an increase in the potential energy of the system? Na  Na+ + e– A periodic table falls off the wall Ca2+ + 2e–  Ca A firecracker explodes Gasoline is burned and creates CO2 and H2O

1st Law of Thermodynamics
Law of Conservation of Energy - Energy can neither be created nor destroyed Can only be converted from one form to another Total energy of universe is constant Total Energy Potential Energy Kinetic Energy = +

Units of Energy A calorie (cal)
Energy needed to raise the temperature of 1 g of H2O by 1 °C 1 cal = J (exactly) 1 kcal = 1000 cal 1 kcal = kJ A nutritional (food) Calorie (Cal) note capital C 1 Cal = 1000 cal = 1 kcal kcal = energy needed to raise 1000 g of water 1°C

Units of Energy Joule (J) calorie (cal) SI unit of energy 1000 J = 1kJ
Energy needed to raise the temperature of 1 g of H2O by 1 °C = J (exactly) 1 kcal = 1000 cal = kJ A nutritional (food) Calorie (Cal) note capital C 1 Cal = 1000 cal = 1 kcal = kJ

A bag of chocolate candy has 220 Cal. How much energy is this in kJ?

A bag of chocolate candy has 220 Cal. How much energy is this in kJ?
Answer: a

Temperature vs. Heat Temperature is a measure of the amount of thermal energy within a sample of matter. Proportional to the average kinetic energy of the particles A higher average kinetic energy means Higher temperature Faster moving molecules Heat (q) is the exchange of thermal energy between two objects. Heat exchange occurs when two objects have a difference in temperature. Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature. Thermal equilibrium

Heat Transfer Hot and cold objects placed in contact
Molecules in hot object move faster KE transfers from hotter object to colder object A decrease in average KE of hotter object An increase in average KE of colder object Over time Average KEs of both objects become the same Temperature of both become the same Thermal equilibrium is established

Kinetic Theory: Liquids and Solids
Atoms and molecules in liquids and solids are constantly moving Particles of solids jiggle and vibrate in place Distributions of KEs of particles in gas, liquid and solid are the same at same temperatures At same temperature, gas, liquid, and solid have Same average kinetic energy But very different potential energy

Your Turn! Which statement about kinetic energy (KE) is true?
Atoms and molecules in gases, liquids and solids possess KE since they are in constant motion. At the same temperature, gases, liquids and solids all have different KE distributions. Molecules in gases are in constant motion, while molecules in liquids and solids are not. Molecules in gases and liquids are in constant motion, while molecules in solids are not. As the temperature increases, molecules move more slowly.

System and Surroundings
Material or process we are interested in studying Example: reaction in a beaker Surroundings Everything else with which system can exchange energy Example: solvent in which chemicals are dissolved Boundary Separation between the system and surroundings Visible Example: Walls of beaker Invisible Example: Line separating warm and cold fronts What we study is the exchange of energy between the system and the surroundings.

Three Types of Systems Open System Closed System Open to atmosphere
Can gain or lose mass and energy across boundary Most reactions are done in open systems Closed System Not open to the atmosphere Energy can cross boundary, but mass cannot Open system Closed system

Three Types of Systems (cont.)
Isolated System No energy or matter can cross the boundary Energy and mass are constant Example: Thermos bottle Isolated system

Adiabatic Process Adiabatic Process
Process that occurs in an isolated system Process where neither energy nor matter crosses the system/surrounding boundary a + diabatos = not passable

Your Turn! A closed system can __________ include the surroundings
absorb energy and mass not change its temperature not absorb or lose energy and mass absorb or lose energy, but not mass

The surroundings gain the exact amount of energy lost by the system
Comparing the Amount of Energy in the System and Surroundings during Transfer Conservation of energy means that the amount of energy gained or lost by the system has to be equal to the amount of energy lost or gained by the surroundings. The surroundings gain the exact amount of energy lost by the system

Heat (q) exothermic – system liberates heat. q = – #
the energy transferred between objects of different temperatures exothermic – system liberates heat. q = – # endothermic – system absorbs heat. q = + #

Heat (q) exothermic endothermic q = – # q = + #

Heat (q) If the only type of energy transferred between two objects is heat energy, all of heat lost by one object is gained by another – Law of Conservation of Energy We cannot measure heat directly – only temperature changes – we use t to calculate changes in heat energy Heat (q) gained or lost by an object is directly proportional to temperature change (t) the object undergoes Adding heat, increases temperature Removing heat, decreases temperature

Heat (q) We measure changes in temperature to quantify the amount of heat transferred q = C × t q = heat C = heat capacity t = temperature change C = proportionality constant

Heat Capacity (C ) Amount of heat (q) required to raise temperature of an object by 1 °C Heat Exchanged = Heat Capacity × t q = C × t Units for heat capacity are J/°C or kJ°/C Heat Capacity is an extensive property Depends on two factors: Sample size or amount (mass) Doubling amount doubles heat capacity Identity of substance Water vs. iron It takes more heat to raise the temperature of one gram of water by 1o C than to cause the same temperature change in one gram of iron. The reason is that the heat capacity (C) of water is higher than that of iron.

Learning Check: Heat Capacity
A cup of water is used in an experiment. Its heat capacity is known to be 720 J/˚C. How much heat will it absorb if the experimental temperature changed from 19.2 ˚C to 23.5 ˚C? q = 3.1 × 103 J

If it requires J to raise the temperature of g of water by 1.00 C, calculate the heat capacity of 1.00 g of water. 4.18 J/°C

Your Turn! What is the heat capacity of 300. g of an object if it requires J to raise the temperature of the object by 2.00˚ C? 4.18 J/˚C 418 J/˚C 837 J/˚C 1.26 × 103 J/˚C 2.51 × 103 J/°C 1255 J/°C

Calorimetry units for specific heat are: J/g·C
calorimetry – the science of measuring heat changes specific heat – the heat required to raise 1 g of substance by 1 C units for specific heat are: J/g·C q = (mass) (specific heat) (ΔT) heat g sp. ht. ΔT = Tfinal - Tinitial

Specific Heat (s) Unique to each substance
Large specific heat means substance releases large amount of heat as it cools Think of how much heat your body absorbs on a hot day! Would iron be able to absorb the same amount of heat and still be at body temperature?

Using Specific Heat Heat Exchanged = (Specific Heat  mass)  t
q = m  s  t Units = J = g  J/(g °C)  °C Substances with high specific heats resist changes in temperature when heat is applied Water has unusually high specific heat Important to body (~60% water) Used to cushion temperature changes Why coastal temperatures are different from inland temperatures If a substance has a high specific heat, s is large. This means delta t will be small for the same amount of heat absorbed or given off.

Using Specific Heat How much heat is required to raise g H2O from 18.2 C to 84.3 C ? q = m x s x Dt q = g x J/g.oC x (84.3oC – 18.2oC) q = 4,682 J

Learning Check: Specific Heat
Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a g sample by 2.53 °C. The symbol q is used to mean “quantity of heat”.

Your Turn! The specific heat of copper metal is J/(g ˚C). How many J of heat are necessary to raise the temperature of a 1.42 kg block of copper from 25.0 ˚C to 88.5 ˚C? 547 J 1.37 × 104 J 3.47 × 104 J 34.7 J 4.74 × 104 J

Direction of Heat Flow Heat is the energy transferred between two objects Heat lost by one object has the same magnitude as heat gained by the other object Sign of q indicates direction of heat flow Heat is gained, q is positive (+) Heat is lost, q is negative (–) q1 = –q2 Example: A piece of warm iron is placed into beaker of cool water. Iron loses 10.0 J of heat, water gains J of heat qiron = –10.0 J qwater = J

Measuring Heat Flow Two methods for measuring heat changes
1. Constant pressure calorimetry (coffee cup calorimetry) 2. Constant volume calorimetry (bomb calorimetry)

Constant pressure calorimetry (coffee cup calorimetry)
g Al (s) was heated to 97.2 C then dumped into g H2O at 21.3 C. The temperature of the H2O rose to 24.2 C. What is the specific heat of Al (s) ?

Heat lost by the system is equal to the heat gained by the surroundings Heat lost by the Al (s) is equal to the heat gained by the H2O () -qmet = qw = g x J/g.oC x 2.9oC -qmet = qw = 988 J -qmet = 988 J qmet = -988 J

g Al (s) was heated to 97.2 C then dumped into g H2O at 21.3 C. The temperature of the H2O rose to 24.2 C. What is the specific heat of Al (s) ? -qmet = 988 J qmet = -988 J = g x s x (24.2C-97.2C) qmet = -988 J = g x s x (-73C) s = J/g·C

Constant volume calorimetry (bomb calorimetry)
put in material to burn pump in lots of oxygen, O2 put “bomb” in a container of water combust it and measure the temperature increase

1.236 g C3H8 and excess O2 was placed into a bomb calorimeter and submerged in H2O then combusted. The temperature rose from C to C. The heat capacity, Cbomb = 5.03 kJ/C. How much heat in kJ, kJ/g and kJ/mole did the combustion reaction liberate ? Note the units for heat capacity are J/C (or kJ/C)

Heat liberated by the system = heat gained by the surroundings Heat liberated by the combustion reaction is equal to the heat gained by the bomb calorimeter apparatus -qrxn = qbomb -qrxn = C x Dt = 5.03 kJ/C x (33.49C – 21.13C) qrxn = –62.2 kJ qrxn = kJ

qrxn = kJ qrxn/g = kJ/1.236g = kJ/g qrxn/mol = kJ/g x 44.10g/1 mol = 2, kJ/mol

Using Heat Capacity A ball bearing at ˚C is dropped into a cup containing 250. g of water. The water warms from 25.0 to 37.3 ˚C. What is the heat capacity of the ball bearing in J/˚C? Heat capacity of the cup of water = 1046 J / ˚C qlost by ball bearing = –qgained by water 1. Determine the temperature change of water t water = (37.3 ˚ C – 25.0 ˚C) = 12.3 ˚C 2. Determine how much heat is gained by water qwater = Cwater  twater = 1046 J/˚C  12.3 ˚C = ×103 J

Using Heat Capacity (cont)
A ball bearing at ˚C is dropped into a cup containing 250. g of water. The water warms from 25.0 to 37.3 ˚C. What is the heat capacity of the ball bearing in J/˚C? C of the cup of water = 1046 J /˚ C 3. Determine how much heat ball bearing lost qball bearing = – qwater = –12.87 × 103 J 4. Determine T change of ball bearing t ball bearing = (37.3 ˚C – ˚C) = –222.7 ˚ C 5. Calculate C of ball bearing = 57.8 J/˚ C

Specific Heat Calculation
How much heat energy must you lose from a 250. mL cup of coffee for the temperature to fall from 65.0 ˚C to 37.0 ˚C? (Assume density of coffee = 1.00 g/mL, scoffee = swater = 4.18 J/g˚C) q = s  m  t t = – 65.0 ˚ C = – 28.0 ˚C q = 4.18 J/g˚C  250. mL  1.00 g/mL  (– 28.0 ˚C) q = (–29.3  103 J) = –29.3 kJ

Using Specific Heat If a 38.6 g piece of gold absorbs 297 J of heat, what will the final temperature of the gold be if the initial temperature is 24.5 ˚C? The specific heat of gold is J/g˚C. Need to find tfinal t = tf – ti First use q = s  m  t to calculate t Next calculate tfinal 59.6 °C = tf – 24.5 °C tf = 59.6 °C °C = °C = ˚ C

Your Turn! What is the heat capacity of a container if 100. g of water (s = 4.18 J/g °C) at 100. ˚C are added to 100. g of water at 25.0 ˚C in the container and the final temperature is 61.0 ˚C? 35 J/˚C 4.12 × 103 J/˚C 21 J/˚C 4.53 × 103 J/˚C 50. J/˚C

Your Turn! - Solution What is the heat capacity of a container if 100. g of water (s = 4.18 J/g ˚C) at 100. ˚C are added to 100. g of water at 25.0 ˚C in the container and the final temperature is 61.0 ˚C? qlost by hot water = m × t × s =(100. g)(61.0 °C – 100.˚C)(4.18 J/g˚C) = –1.63  104 J qgained by cold water = (100. g)(61.0°C – 25.0˚C)(4.18J/g ˚C) = 1.50  104 J qlost by system = 1.50104 J + (–1.63104 J) = –1.3103 J qcontainer = –q lost by system = +1.3  103 J = 36 J/°C

Chemical Bonds and Energy
Consist of attractive forces that bind Atoms together to form molecules Ions together in ionic compounds Give rise to a compound’s potential energy Chemical energy Potential energy stored in chemical bonds Chemical reactions Involve breaking and making of chemical bonds Involves the absorption or release of energy

Chemical Reactions Formation of Bonds Breaking of Bonds
Atoms that are attracted to each other are moved closer together Decreases the PE of the reacting system Releases energy DE = Efinal – Einitial = negative Breaking of Bonds Atoms that are attracted to each other are forced apart Increases the PE of the reacting system Requires energy DE = Efinal – Einitial = positive

Exothermic Reaction Example:
A reaction where products have less chemical energy than reactants The reaction releases heat to the surroundings Some chemical (PE) energy is converted to kinetic energy Heat leaves the system; q is negative ( – ) Heat energy is a product Reaction gets warmer, the temperature increases Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat

Endothermic Reaction A reaction where products have more chemical energy than reactants The reaction absorbs heat from its surroundings Some kinetic energy is converted into chemical (PE) energy Heat is added to the system; q is positive (+) Heat energy is a reactant Reaction becomes colder, temperature decreases Example: Photosynthesis 6CO2(g) + 6H2O(g) + solar energy  C6H12O6(s) + 6O2(g)

Bond Strength Measured by how much energy is needed to break a bond or how much energy is released when a bond is formed. A larger amount of energy equals a stronger bond Weak bonds require less energy to break than do strong bonds A key to understanding reaction energies Example: If a reaction has Weak bonds in the reactants and Stronger bonds in the products Heat released (the system loses energy as it becomes more stable)

Why Fuels Release Heat Methane and oxygen have weaker bonds
Water and carbon dioxide have stronger bonds Heat is given off

Enthalpy (DH) of Reaction
Enthalpy - heat content of a system when measured at constant pressure DH = Hfinal – Hinitial DH = Hpdts – Hrcts It is easier to measure the difference in enthalpy between reactants and products than it is to measure the enthalpy of either of the two. Enthalpy, DH, of reaction aka heat of reaction

Enthalpy Change (H) H is a state function H = Hfinal – Hinitial
H = Hproducts – Hreactants Significance of sign of H Endothermic reaction System absorbs energy from surroundings H positive Exothermic reaction System loses energy to surroundings H negative

Enthalpy Changes in Chemical Reactions
The amount of heat a reaction produces or absorbs depends on the number of moles of reactants we combine. Endothermic Reactants + heat  products Exothermic Reactants  products + heat For heats of reaction to have meaning we must describe the system completely Amounts & concentrations of both reactants & products, temperature, and pressure – all of these can influence heats of reaction.

The Standard State – indicated by o sign
To compare heats of reaction chemists have defined a set of standard conditions. The standard state is the state of a material under a defined set of conditions. 1 bar pressure (~ 1 atm) 25°C 1 M solute concentration The standard heat of reaction, DHo is the value of DH for a reaction occurring under standard conditions and involving the actual numbers of moles specified by the coefficients of the equation – usually involving 1 mole.

H in Chemical Reactions
Standard Conditions for H 's 25 °C and 1 atm and 1 mole Standard Heat of Reaction (H ° ) Enthalpy change for reaction at 1 atm and 25 °C Example: N2(g) H2(g)  2 NH3(g) 1.000 mol mol mol When N2 and H2 react to form NH3 at 25 °C and 1 atm kJ released H = –92.38 kJ

Thermochemical Equation
Write H immediately after equation N2(g) + 3H2(g)  2NH3(g) H° = –92.38 kJ When 1 mol N2 and 3 mol H2 react to form 2 mol NH3 at 25oC and 1 bar, the reaction releases kJ of energy Must give physical states of products and reactants H different for different states CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) H ° rxn = – kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H ° rxn = – kJ Difference is equal to the energy to vaporize water

Thermochemical Equation
Write H immediately after equation N2(g) + 3H2(g)  2NH3(g) H°= –92.38 kJ Assumes coefficients is the number of moles 92.38 kJ released when 2 moles of NH3 formed If 10 mole of NH3 formed 5N2(g) + 15H2(g)  10NH3(g) H° = –461.9 kJ H° = (5 × –92.38 kJ) = – kJ Can have fractional coefficients Fraction of mole, NOT fraction of molecule ½N2(g) + 3/2H2(g)  NH3(g) H°rxn = –46.19 kJ

State Matters! Needs to be Specified!
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g) ΔH °rxn= –2043 kJ C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l ) ΔH °rxn = –2219 kJ Note: there is difference in energy because states do not match If H2O(l ) → H2O(g) ΔH °vap = 44 kJ/mol 4H2O(l ) → 4H2O(g) ΔH °vap = 176 kJ/mol Or –2219 kJ kJ = –2043 kJ If we burn propane, we produce 2043 kJ of energy and make gas phase products. But the standard enthalpy of combustion for propane is –2219 kJ. Why?

Learning Check: –1,256 kJ Consider the following reaction:
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) ΔE ° = –2511 kJ The reactants (acetylene and oxygen) have 2511 kJ more energy than products. How many kJ are released for 1 mol C2H2? –1,256 kJ

Learning Check: Given the equation below, how many kJ are required for 44 g CO2 (MM = g/mol) to react with H2O? 6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g) ΔH˚reaction = 2816 kJ If 100. kJ are provided, what mass of CO2 can be converted to glucose? 470 kJ 9.4 g

Your Turn! Based on the reaction
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g) H˚reaction = – 434 kJ/mol CH4 What energy change occurs when 1.2 moles of methane reacts? –3.6 × 102 kJ 5.2 × 102 kJ –4.3 × 102 kJ 3.6 × 102 kJ –5.2 × 102 kJ H = –434 kJ/mol × 1.2 mol H = –520.8 kJ = 5.2 × 102 kJ

Your Turn! Rxn. 1: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH °rxn= –2043 kJ Rxn. 2: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l ) ΔH °rxn = –2219 kJ Why does rxn. 2 release more heat than rxn. 1? It shouldn’t, and the DH values should be equal More reactants must have been used in rxn 2 In rxn 1 some of the heat of the reaction is used up converting liquid water to gas, so less heat can be given off. Liquids always have lower temperatures than gases

Reversing Thermochemical Equations
Consider CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H°reaction = – kJ We can reverse the thermochemical equation When we do so we change the sign of H CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) H°reaction = kJ

Multiple Paths; Same H °
We can often get from reactants to products by several different paths Intermediate A Products Reactants Intermediate B Enthalpy is a state function. This important fact permits us to calculate heats of reaction for reactions that we cannot actually carry out in the laboratory. We simply need to find a path to get from reactant to product along which enthalpy values are known. Should get same H ° for each path Enthalpy is a state function and is path independent

Example: Multiple Paths; Same H°
Path a: Single step C(s) + O2(g)  CO2(g) H° rxn = –393.5 kJ Path b: Two step Step 1: C(s) + ½O2(g)  CO(g) H° rxn = –110.5 kJ Step 2: CO(g) + ½O2(g)  CO2(g) H° rxn = –283.0 kJ Net Rxn: C(s) + O2(g)  CO2(g) H° rxn = –393.5 kJ Chemically and thermochemically, identical results True for exothermic and endothermic reactions

Example: Multiple Paths; Same Hrxn
Path a: N2(g) + 2O2(g)  2NO2(g) H° rxn = 68 kJ Path b: Step 1: N2(g) + O2(g)  2NO(g) H° rxn = kJ Step 2: 2NO(g) + O2(g)  2NO2(g) H° rxn = –112 kJ Net rxn: N2(g) + 2O2(g)  2NO2(g) H° rxn = 68 kJ Hess’s Law of Heat Summation For any reaction that can be written into steps, value of H °rxn for reactions = sum of H °rxn values of each individual step

Enthalpy Diagrams Graphical description of Hess’ Law
Vertical axis = enthalpy scale Horizontal line =various states of reactions Higher up = larger enthalpy Lower down = smaller enthalpy Gives energy description of alternative pathways

Enthalpy Diagrams Use to measure Hrxn Arrow down Hrxn = negative
Arrow up Hrxn = positive Calculate cycle One step process = sum of two step process Example: H2O2(l )  H2O(l ) + ½O2(g) –286 kJ = –188 kJ + Hrxn Hrxn = –286 kJ – (–188 kJ ) Hrxn = –98 kJ

Summarizing Hess’s Law
Hess’s Law of Heat Summation Going from reactants to products Enthalpy change is the same whether reaction takes place in one step or in many steps Chief Use Calculation of H °rxn for reaction that can’t be measured directly Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction

Rules for Manipulating Thermochemical Equations
When equation is reversed, sign of H° rxn must also be reversed. If all coefficients of equation are multiplied or divided by same factor, value of H° rxn must likewise be multiplied or divided by that factor Formulas canceled from both sides of equation must be for substance in the same physical state

Example: Calculate Horxn for C (s, graphite)  C (s, diamond)
Given C (s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ C (s, dia) + O2(g)  CO2(g) H°rxn = –396 kJ To get desired equation, must reverse second equation and add resulting equations C(s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ CO2(g)  C(s, dia) + O2(g) H°rxn = –(–396 kJ) C(s, gr) + O2(g) + CO2(g)  C(s, dia) + O2(g) +CO2(g) H° = –394 kJ kJ = + 2 kJ –1[ ]

Learning Check Calculate H °rxn for 2 C (s, gr) + H2(g)  C2H2(g)
Given the following: C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l ) H °rxn = – kJ C(s, gr) + O2(g)  CO2(g) H °rxn = –393.5 kJ H2(g) + ½O2(g)  H2O(l ) H °rxn = –285.8 kJ

Example: Calculate Horxn for 2C(s, gr) + H2(g)  C2H2(g)
2CO2(g) + H2O(l )  C2H2(g) + 5/2O2(g) H°rxn = – (– kJ) = kJ 2C(s, gr) + 2O2(g)  2CO2(g) H°rxn =(2–393.5 kJ) = –787.0 kJ H2(g) + ½O2(g)  H2O(l ) H°rxn = –285.8 kJ 2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g)  C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l ) 2C(s, gr) + H2(g)  C2H2(g) H°rxn = kJ

Your Turn! Which of the following is a statement of Hess's Law?
H for a reaction in the forward direction is equal to H for the reaction in the reverse direction. H for a reaction depends on the physical states of the reactants and products. If a reaction takes place in steps, H for the reaction will be the sum of Hs for the individual steps. If you multiply a reaction by a number, you multiply H by the same number. H for a reaction in the forward direction is equal in magnitude and opposite in sign to H for the reaction in the reverse direction.

Your Turn! Given the following data:
C2H2(g) + O2(g)  2CO2(g) + H2O(l ) Hrxn= –1300. kJ C(s) + O2(g)  CO2(g) Hrxn = –394 kJ H2(g) + O2(g)  H2O(l ) Hrxn = –286 kJ Calculate Hrxn for the reaction 2C(s) + H2(g)  C2H2(g) 226 kJ –1980 kJ –620 kJ –226 kJ 620 kJ Reverse 1st eqn. Add 2 × 2nd eqn. Add 3rd eqn. Hrxn = kJ + 2(–394 kJ) + (–286 kJ)

Tabulated H° Values Large databases of thermochemical equations have been compiled to make it possible to calculate H° values for Hess’ law calculations. A major problem is the vast number of reactions A better way is needed We can use Standard Heats of Formation

Standard Heats of Formation
Standard Enthalpy of Formation, Hf° the ΔH (heat) that accompanies the formation of 1 mole of anything from it’s free, uncombined elements at standard conditions Constitute a better way to use Hess’ Law without the need to manipulate thermochemical equations. standard conditions – 25 C and 1 atm pressure units for are kJ/mole coefficients represent number of moles product must be formed from free, uncombined elements

Standard State Most stable form and physical state of element at 1 atm (1 bar) and 25 °C (298 K) Element Standard state O O2(g) C C (s, gr) H H2(g) Al Al(s) Ne Ne(g) Note: All Hf° of elements in their standard states = 0 Forming element from itself. See Appendix in back of textbook and Table 6.2

Uses of Standard Enthalpy (Heat) of Formation, Hf°
1. From definition of Hf°, we can write balanced equations directly – bypassing thermochemical equations Hf° of C2H5OH(l ) 2C(s, gr) + 3H2(g) + ½O2(g)  C2H5OH(l ) Hf° = – kJ/mol Hf° of Fe2O3(s) 2Fe(s) + 3/2O2(g)  Fe2O3(s) Hf° = –822.2 kJ/mol Each compound is formed from its standard state elements using numbers for each compound derived from a table.

Your Turn! Which reaction corresponds to the standard enthalpy of formation of NaHCO3(s), Hf ° = – kJ/mol? A. Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr)  NaHCO3(s) B. Na+(g) + H+(g) + 3O2–(g) + C4+(g)  NaHCO3(s) C. Na+(aq) + H+(aq) + 3O2–(aq) + C4+(aq)  NaHCO3(s) D. NaHCO3(s)  Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr) E. Na+(aq) + HCO3–(aq)  NaHCO3(s)

Your Turn! Which reaction corresponds to the standard enthalpy of formation of C4H3Br2NO2(l)? 4C(s, gr) + 3H(g) + 2Br(s) + N(g) + 2O(g)  C4H3Br2NO2(l) B. 8C(s, gr) + 3H2(g) + 2Br2(g) + N2(g) + 2O2 (g)  2C4H3Br2NO2(l) C. 4C(s, gr) + 3/2H2(g) + Br2(l) + ½N2(g) + O2(g)  C4H3Br2NO2(l) D. 4C(s, gr) + 3/2H2(g) + Br2(s) + ½N2(g) + ½O2(g)  C4H3Br2NO2(l)

Your Turn! The standard enthalpy of formation of sulfur dioxide is kJ. What is DHf for the formation of g of sulfur dioxide in its standard state from its elements in their standard states? A kJ B kJ C. -4,759 kJ D kJ E kJ = ̶ kJ

Write the reaction describing the of NaCl (s).

Using Hf° 1. write and balance the reaction
standard heat of reaction – the ΔH (heat) that accompanies ANY reaction under standard conditions (units are kJ) = [(sum products) – (sum reactants)] 1. write and balance the reaction 2. use table of thermodynamic data of to calculate

Example: Calculate Horxn Using Hf°
Calculate H°rxn using Hf° data for the reaction SO3(g)  SO2(g) + ½O2(g) Multiply each Hf° (in kJ/mol) by the number of moles in the equation Add the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each product Subtract the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each reactant H°rxn has units of kJ H°f has units of kJ/mol H°rxn = 99 kJ

Determine the for the combustion of ammonia
4 NH3 (g) + O2 (g) N2 (g) H2O (g) 3 2 6 (kJ) 4 NH3 (g) 4 x (-46.19) 3 O2 (g) 3 x 0 2 N2 (g) 2 x 0 6 H2O (g) 6 x (-241.8) = pdts – rcts = [( )-( )] = kJ

Ethanol is used as an additive in many fuels today
Ethanol is used as an additive in many fuels today. What is ΔHºrxn (kJ) for the combustion of ethanol? 2 C2H5OH (l ) + 6 O2 (g) → 4 CO2 (g) + 6 H2O (l ) ΔHºf (kJ/ mol) C2H5OH (l) –277.6 CO2 (g) –393.5 H2O (g) –241.8 H2O (l) –285.8 – 401.7 –2469 + 2734 – 2734

Ethanol is used as an additive in many fuels today
Ethanol is used as an additive in many fuels today. What is ΔHºrxn (kJ) for the combustion of ethanol? 2 C2H5OH (l ) + 6 O2 (g) → 4 CO2 (g) + 6 H2O (l ) ΔHºf (kJ/ mol) C2H5OH (l) –277.6 CO2 (g) –393.5 H2O (g) –241.8 H2O (l) –285.8 – 401.7 –2469 + 2734 – 2734 Answer: e

Determine the for the following reaction at 84 °C and 0.79 atm.
2 N2O (g) + O2 (g) NO2 (g) 3 4 (kJ) 2 N2O (g) 2 x (81.57) 3 O2 (g) 3 x 0 4 NO2 (g) 4 x (33.8) = pdts – rcts = [(135.2)-( )] = kJ

Practice Problems

Your Turn! Which is a unit of energy? Pascal Newton Joule Watt Ampere

Your Turn! = 41.87 Cal Convert 175.2 kJ into nutritional Calories

Your Turn! 13.8 kcal Convert 57.9 kJ into calories 13800 Cal 13.8 cal
Pay attention to the difference between Calories (which equals a kcal) and calories with a small “c”.

Your Turn! Which of the following is not a form of kinetic energy?
A pencil rolls across a desk A pencil is sharpened A pencil is heated A pencil rests on a desk A pencil falls to the floor

For a system that gains thermal energy and does work on the surroundings,
heat (q) and work (w) are both positive. heat (q) is positive and work (w) is negative. heat (q) is negative and work (w) is positive. heat (q) and work (w) are both negative. Answer: b

Calculate the internal energy, ΔE, for a system that does 422 J of work and loses 227 J of energy as heat. + 649 J – 649 J 0 J – 195 J + 195 J

Calculate the internal energy, ΔE, for a system that does 422 J of work and loses 227 J of energy as heat. + 649 J – 649 J 0 J – 195 J + 195 J Answer: b

A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the following is true? Heat lost by the metal > heat gained by water Heat gained by water > heat lost by the metal Heat lost by the metal > heat lost by the water Heat lost by the metal = heat gained by water More information is required

A cup of water is used in an experiment. Its heat capacity is known to be 720 J/˚C. How much heat will it absorb if the experimental temperature changed from 19.2 ˚C to 23.5 ˚C? q = 3.1 × 103 J

If it requires J to raise the temperature of g of water by 1.00 C, calculate the heat capacity of g of water. 4.18 J/°C

Your Turn! What is the heat capacity of 300. g of an object if it requires J to raise the temperature of the object by 2.00˚ C? 4.18 J/˚C 418 J/˚C 837 J/˚C 1.26 × 103 J/˚C 2.51 × 103 J/°C 1255 J/°C

Your Turn! A copper mug has a heat capacity of 77.5 J/˚C. After adding hot water to the mug, the temperature of the mug changed from 77.0 ˚F to 185 ˚F. How much heat did the mug absorb from the water? 4.65 × 103 J 1.29 J 8.37 × 103 J 5.97 × 103 J 1.43 × 104 J Note units: temps. must be in oC ti = 25.0 oC, tf = 85.0 oC, So Dt = 85.0 oC – 25.0 oC = 60.0 oC = 4.65 × 103 J

Learning Check 4.18 J/g °C 4.18 J/g °C
Calculate the specific heat of water if the heat capacity of 100. g of water is 418 J/°C. What is the specific heat of water if heat capacity of g of water is 4.18 J/°C? Thus, heat capacity is independent of amount of substance – an intensive property! 4.18 J/g °C 4.18 J/g °C

Your Turn! The specific heat of silver J g–1 °C–1. What is the heat capacity of a 100. g sample of silver? 0.235 J/°C 2.35 J/°C 23.5 J/°C 235 J/°C 2.35 × 103 J/°C

Your Turn! A cast iron skillet is moved from a hot oven to a sink full of water. Which of the following is false? The water heats The skillet cools The heat transfer for the skillet has a negative (–) sign The heat transfer for the skillet has the same sign as the heat transfer for the water

Example 6. 2: How much heat is absorbed by a copper penny with mass 3
Example 6.2: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from −8.0 °C to 37.0 °C? Sort Information Given: Find: T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g q, J q = m ∙ Cs ∙ DT Note: Cs is same as s Cs = J/g•ºC (Table 6.4) Strategize Conceptual Plan: Relationships: Cs m, DT q Follow the conceptual plan to solve the problem Solution: Check Check: the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise

Specific Heat Capacity
If 50 g samples of all substances in Table 6.4 absorbed 100 J of energy, which would show the greatest temperature change? DT = q /Cs so DT 1/Cs If 50 g samples of all substances in Table 6.4 cooled by 75 oC, which would release the most energy? Delta T = q/mCs ; Delta T is proportional to 1/Cs Answer Top Question – Glass Answer Bottom Question - Water

Example 6. 3: A 32. 5-g cube of aluminum initially at 45
Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? Cs, Al = J/g•ºC, Cs, H2O = 4.18 J/g•ºC Given: Find: mAl = 32.5 g, Tal = 45.8 °C, mH20 = g, TH2O = 15.4 °C Tfinal, °C q = m ∙ Cs ∙ DT Cs, Al = J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4) Conceptual Plan: Relationships: Cs, Al mAl, Cs, H2O mH2O DTAl = kDTH2O qAl = −qH2O Tfinal Solution: Check: the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures

A hot piece of metal weighing 350. 0 g is heated to 100. 0 °C
A hot piece of metal weighing g is heated to °C. It is then placed into a coffee cup calorimeter containing g of water at 22.4 °C. The water warms and the copper cools until the final temperature is 35.2 °C. Calculate the specific heat of the metal and identify the metal. metal: g, T1 = °C, T2 = 35.2 °C H2O: g, T1 = 22.4 °C, T2 = 35.2°C, Cs = 4.18 J/g °C FIND: Cs , metal, J/gºC Cs= 0.377m; presumably copper Set up m ∙ Cs ∙ delta T for the water and the metal. Solve for the specific heat of the metal. m, Cs, DT q

A 3. 54 g piece of aluminum is heated to 96
A 3.54 g piece of aluminum is heated to 96.2 ºC and allowed to cool to room temperature, 22.5 ºC. Calculate the heat (in kJ) associated with the cooling process. The specific heat of aluminum is J/ g · K. – 236 + 236 – 0.236 – 0.638 q = m c delta T The metal (system) loses heat so the sign is negative

A 3. 54 g piece of aluminum is heated to 96
A 3.54 g piece of aluminum is heated to 96.2 ºC and allowed to cool to room temperature, 22.5 ºC. Calculate the heat (in kJ) associated with the cooling process. The specific heat of aluminum is J/ g · K. – 236 + 236 – 0.236 – 0.638 Answer: d

A 1. 5 g iron nail is heated to 95
A 1.5 g iron nail is heated to 95.0 °C and placed into a beaker of water. Calculate the heat gained by the water if the final equilibrium temperature is 57.8 ° C. The specific heat capacity of iron = J/ g ° C, and the specific heat capacity of water = 4.18 J/ g ° C. 6.0 J 25 J –25 J 39 J – 39 J m c delta T for iron = m c delta T for water

A 1. 5 g iron nail is heated to 95
A 1.5 g iron nail is heated to 95.0 °C and placed into a beaker of water. Calculate the heat gained by the water if the final equilibrium temperature is 57.8 ° C. The specific heat capacity of iron = J/ g ° C, and the specific heat capacity of water = 4.18 J/ g ° C. 6.0 J 25 J –25 J 39 J – 39 J Answer: b

Identical amounts of heat are applied to 50 g blocks of lead, silver, and copper, all at an initial temperature of 25 °C. Which block will have the largest increase in temperature? Lead Silver Copper None; all will be at the same temperature.

Identical amounts of heat are applied to 50 g blocks of lead, silver, and copper, all at an initial temperature of 25 °C. Which block will have the largest increase in temperature? Lead Silver Copper None; all will be at the same temperature. Answer: c q = m c delta T. Which has the smaller value for c? c and delta T are inversely proportional given the same amounts of heat and masses. Lead should be the correct answer.

50.0 g of water at 22 °C is mixed with 125 g of water initially at 36 ° C. What is the final temperature of the water after mixing, assuming no heat is lost to the surroundings? 29 °C 42 ° C 33 ° C 31 ° C 36 ° C

50.0 g of water at 22 °C is mixed with 125 g of water initially at 36 ° C. What is the final temperature of the water after mixing, assuming no heat is lost to the surroundings? 29 °C 42 ° C 33 ° C 31 ° C 36 ° C Answer: c Use m c delta T. c is the same in both instances. Use T(final) – T(initial) for delta T.

the units are correct, the sign is as expected for combustion
When g of sugar is burned in a bomb calorimeter, the temperature rises from °C to °C. If Ccal = 4.90 kJ/°C, find DE for burning 1 mole (MM C12H22O11 = g/mol) Given: Find: 1.010 g C12H22O11, T1 = °C, T2 = °C, Ccal = 4.90 kJ/°C DErxn, kJ/mol x 10−3 mol C12H22O11, DT = 3.41°C, Ccal = 4.90 kJ/°C DErxn, kJ/mol Conceptual Plan: Relationships: Ccal, DT qcal qrxn qrxn, mol DE qcal = Ccal x DT = −qrxn MM C12H22O11 = g/mol Solution: Check: the units are correct, the sign is as expected for combustion

When 1.22 g of HC7H5O2 (MM ) is burned in a bomb calorimeter, the temperature rises from °C to °C. If DE for the reaction is −3.23 x 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? mol, DE qrxn qcal qcal, DT Ccal qcal = Ccal x DT = −qrxn MM HC7H5O2 = g/mol

Calorimeter Problem When g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from ˚C to ˚C. a) How many Calories are in olive oil, per gram? The heat capacity of the calorimeter is kJ/˚C. t = ˚C – ˚C = ˚C qabsorbed by calorimeter = Ct = kJ/°C × ˚C = kJ Calculate heat absorbed by calorimeter Convert this to heat released by combustion of olive oil Convert heat to Cal/gram qreleased by oil = – qcalorimeter = – kJ –8.740 Cal/g oil

Calorimeter Problem (cont)
Olive oil is almost pure glyceryl trioleate, C57H104O6. The equation for its combustion is C57H104O6(l) + 80O2(g)  57CO2(g) + 52H2O What is E for the combustion of one mole of glyceryl trioleate (MM = g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate. E = qV = –3.238 × 104 kJ/mol oil

Your Turn! A bomb calorimeter has a heat capacity of 2.47 kJ/K. When a 3.74×10–3 mol sample of ethylene was burned in this calorimeter, the temperature increased by 2.14 K. Calculate the energy of combustion for one mole of ethylene. –5.29 kJ/mol 5.29 kJ/mol –148 kJ/mol –1410 kJ/mol 1410 kJ/mol qcal = Ct = 2.47 kJ/K × 2.14 K = kJ qethylene = – qcal = – kJ

Coffee Cup Calorimetry
NaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH. The initial t = 25.5 °C and final t = 32.2 °C. What is H in kJ/mole of HCl? Assume for these solutions s = J g–1°C–1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1. NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(aq) qabsorbed by solution = mass  s  t massHCl = 50.0 mL  1.02 g/mL = 51.0 g massNaOH = 50.0 mL  1.04 g/mL = 52.0 g massfinal solution = 51.0 g g = g t = (32.2 – 25.5) °C = 6.7 °C 1. Determine how much heat is absorbed by the calorimeter.

qcal = g  J g–1 °C–1  6.7 °C = 2890 J Rounds to qcal = 2.9  103 J = 2.9 kJ qrxn = –qcalorimeter = –2.9 kJ = mol HCl Heat evolved per mol HCl = 2. Heat evolved by reaction = -58 kJ/mol

When 50.0 mL of M H2SO4 is added to 25.0 mL of 2.00 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution increases to 33.9 °C. Calculate H in kJ/mole of limiting reactant. Assume: specific heat of the solution is J/g°C, density is 1.00 g/mL, and the calorimeter absorbs a negligible amount of heat. Write balanced equation 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(aq) Determine heat absorbed by calorimeter masssoln = (25.0 mL mL) × 1.00 g/mL = 75.0 g qsoln = 75.0 g × (33.9 – 25.0)°C × J/g°C = 2.8×103 J

Determine Limiting Reagent
= mol H2SO4 present = mol NaOH needed Determine moles of H2SO4 present Determine moles of NaOH needed Determine moles of NaOH present = mol NaOH present NaOH is limiting = –56 kJ/mol

Your Turn! A g sample of solid is transferred from boiling water (t = 99.8 ˚C) to 152 g water at 22.5 ˚C in a coffee cup. The temperature of the water rose to 24.3 ˚C. Calculate the specific heat of the solid. –1.1 × 103 J g–1 ˚C–1 1.1 × 103 J g–1 ˚C–1 1.0 J g–1 ˚C–1 0.35 J g–1 ˚C–1 0.25 J g–1 ˚C–1 q = m × s × t Determine how much heat is absorbed by the calorimeter. Calculate the specific heat of the solid. = 1.1 × 103 J qsample = – qwater = – 1.1 × 103 J

How much heat is evolved in the complete combustion of 13
How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g) ? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = −2044 kJ MM = g/mol Given: Find: 13.2 kg C3H8, q, kJ/mol 1 kg = 1000 g, 1 mol C3H8 = −2044 kJ, Molar Mass = g/mol Conceptual Plan: Relationships: mol kJ kg g Solution: Check: the sign is correct and the number is reasonable because the amount of C3H8 is much more than 1 mole

C(s) + O2(g) → CO2 (g) DH = −395.4 kJ/mol C
How much heat is evolved when a g diamond is burned? (DHcombustion = −395.4 kJ/mol C) C(s) + O2(g) → CO2 (g) DH = −395.4 kJ/mol C kj is per mole. How many moles are there?

What is DHrxn/mol Mg for the reaction: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if g Mg completely reacts in mL of solution and changes the temperature from 25.6 °C to 32.8 °C? Given: Find: 6.499 x 10−3 mol Mg, 1.00 x 102 g, DT = 7.2 °C,,Cs = 4.18 J/g∙°C DH, kJ/mol 0.158 g Mg, mL, T1 = 25.6 °C, T2 = 32.8 °C, assume d = 1.00 g/mL, Cs = 4.18 J/g∙°C H, kJ/mol Conceptual Plan: Relationships: Cs, DT, m qsol’n qrxn qrxn, mol DH qsol’n = m x Cs, sol’n x DT = −qrxn, MM Mg= g/mol Solution: Check: the sign is correct and the value is reasonable because the reaction is exothermic

In a coffee cup calorimetry experiment, 50. 0 mL of 0
In a coffee cup calorimetry experiment, 50.0 mL of M NaOH is added to 50.0 mL of M HCl. The temperature of the solution increased by 14.9 °C. Calculate ΔH for the reaction. Use 1.00 g/ mL as the density of the solution and 4.18 J/ g ° C as the specific heat capacity. 1.25 kJ/ mol 6.23 kJ/ mol 31.0 kJ/ mol 1250 kJ/ mol 6230 kJ/ mol

In a coffee cup calorimetry experiment, 50. 0 mL of 0
In a coffee cup calorimetry experiment, 50.0 mL of M NaOH is added to 50.0 mL of M HCl. The temperature of the solution increased by 14.9 °C. Calculate ΔH for the reaction. Use 1.00 g/ mL as the density of the solution and 4.18 J/ g ° C as the specific heat capacity. 1.25 kJ/ mol 6.23 kJ/ mol 31.0 kJ/ mol 1250 kJ/ mol 6230 kJ/ mol Answer: d

Your Turn! Which of the following is an endothermic process?
Na+ + e–  Na wood burning CH4(g)  C(g) + 4H(g) a bomb exploding water condensing Attraction of charges releases heat Produces heat (i.e., heat released) Breaking bonds requires heat Produces heat (i.e., heat released) Heat must be released

Which of the following statements about enthalpy is FALSE?
The value of ΔH for a chemical reaction is the amount of heat absorbed or evolved in the reaction under conditions of constant pressure. An endothermic reaction has a positive ΔH and absorbs heat from the surroundings. An endothermic reaction feels cold to the touch. An exothermic reaction has a negative ΔH and gives off heat to the surroundings. An exothermic reaction feels warm to the touch. Statements (a), (b), and (c) are all false statements. Statements (a), (b), and (c) are all true statements.

Which of the following statements about enthalpy is FALSE?
The value of ΔH for a chemical reaction is the amount of heat absorbed or evolved in the reaction under conditions of constant pressure. An endothermic reaction has a positive ΔH and absorbs heat from the surroundings. An endothermic reaction feels cold to the touch. An exothermic reaction has a negative ΔH and gives off heat to the surroundings. An exothermic reaction feels warm to the touch. Statements (a), (b), and (c) are all false statements. Statements (a), (b), and (c) are all true statements. Answer: e

If a 50. 00 mL sample of 0. 250 M HCl(aq) is added to a 50
If a mL sample of M HCl(aq) is added to a mL sample of NaOH(aq) in a coffee cup calorimeter, the temperature rises 1.71°C. What is DHrxn (kJ/mol NaOH)? Assume the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.18 J/g∙°C) HCl NaOH → NaCl H2O Answer: kJ/mol

Your Turn! What is your change in altitude if you climb up a 7,005 ft. mountain, drive a car from that spot down 2,003 ft, then parachute down another 11,508 ft. into a canyon? Draw a diagram that reflects your changes and your calculation. -6506 ft. 20516 ft. 16310 ft. ft. No change 7,005 ft. -2,003 ft. 5,002 ft. +7,005 ft. Start -11,508 ft. Alt. = -6,506 ft. -6,506 ft. Note: Change is negative because altitude decreased Final state is lower than initial state

Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
Given the following equations, Find DHrxn for Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ 2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ 2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ

Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ
Given the following equations, find DHrxn for Cu(s) + CuCl2(s)  2 CuCl(s) DH° = ? kJ Cu(s) + Cl2(g)  CuCl2(s) DH° = −206 kJ 2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ CuCl2(s)  Cu(s) + Cl2(g) DH° = +206 kJ 2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = kJ Cu(s) + CuCl2(s)  2 CuCl(s) DH° = kJ

Given the following equations, Find DHrxn for 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g)
2 NO(g) + O2(g)  2 NO2(g) DH° = −116 kJ 2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = −256 kJ N2(g) + O2(g)  2 NO(g) DH° = +183 kJ 2 NO2(g)  2 NO(g) + O2(g) DH° = +116 kJ 2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = −256 kJ 2 NO(g)  N2(g) + O2(g) DH° = −183 kJ X 3/2 3 NO2(g)  3 NO(g) O2(g) DH° = (+174 kJ) N2(g) O2(g) + 1 H2O(l)  2 HNO3(aq) DH° = (−128 kJ) 2 NO(g)  N2(g) + O2(g) DH° = (−183 kJ) 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = − 137 kJ X 1/2

Example: Hess’s Law Given the following information:
2 NO(g) + O2(g)  2 NO2(g) DH° = −116 kJ 2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = −256 kJ N2(g) + O2(g)  2 NO(g) DH° = +183 kJ Calculate the DH° for the reaction below: 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = ? [3 NO2(g)  3 NO(g) O2(g)] DH° = (+174 kJ) [1 N2(g) O2(g) + 1 H2O(l)  2 HNO3(aq)] DH° = (−128 kJ) [2 NO(g)  N2(g) + O2(g)] DH° = (−183 kJ) 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = − 137 kJ [2 NO2(g)  2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+116 kJ) [2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)] x 0.5 DH° = 0.5(−256 kJ) [2 NO(g)  N2(g) + O2(g)] DH° = −183 kJ

Use Hess’s Law to determine ΔH for the following target reaction
Use Hess’s Law to determine ΔH for the following target reaction. ½ N2 (g) + ½ O2 (g) → NO (g) ΔH = kJ NO (g) + ½ Cl2 (g) → NOCl (g) ΔH = –38.6 kJ 2 NOCl (g) → N2 (g) + O2 (g) + Cl2 (g) ΔH = ? –51.7 kJ 51.7 kJ –103.4 kJ 103.4 kJ 142.0 kJ

Use Hess’s Law to determine ΔH for the following target reaction
Use Hess’s Law to determine ΔH for the following target reaction. ½ N2 (g) + ½ O2 (g) → NO (g) ΔH = kJ NO (g) + ½ Cl2 (g) → NOCl (g) ΔH = –38.6 kJ 2 NOCl (g) → N2 (g) + O2 (g) + Cl2 (g) ΔH = ? –51.7 kJ 51.7 kJ –103.4 kJ 103.4 kJ 142.0 kJ Answer: c

Which of the following has a negative value of ΔH?
The combustion of gasoline The melting of snow The evaporation of ethanol The sublimation of CO2 at room temperature All of the above have positive ΔH values.

Which of the following has a negative value of ΔH?
The combustion of gasoline The melting of snow The evaporation of ethanol The sublimation of CO2 at room temperature All of the above have positive ΔH values. Answer: a

Standard Enthalpies of Formation

Writing Formation Reactions: Write the Formation Reaction for CO(g)
The formation reaction is the reaction between the elements in the compound, which are C and O. C + O  CO(g) The elements must be in their standard state. there are several forms of solid C, but the one with DHf° = 0 is graphite. oxygen’s standard state is the diatomic gas. C(s, graphite) + O2(g)  CO(g) The equation must be balanced, but the coefficient of the product compound must be 1. use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient. C(s, graphite) + ½ O2(g)  CO(g)

Write the formation reactions for the following:
CO2(g) Al2(SO4)3(s) C(s, graphite) + O2(g)  CO2(g) 2 Al(s) + 3/8 S8(s, rhombic) + 6 O2(g)  Al2(SO4)3(s)

Calculating Standard Enthalpy Change for a Reaction
Any reaction can be written as the sum of formation and decomposition reactions involving constitutive elements in their standard states. reactants  elements ΔH1 = - S ΔHof (reactants) elements  products ΔH2 = +S ΔHof (products) reactants  products ΔHorxn = ΔH1 + ΔH2 When employing decomposition reactions (ie. reactants  elements), we reverse the sign of the standard enthalpy of formation. C(s, graphite) + O2(g) → CO(g) ΔHfo = kJ/mol CO(g) → C(s, graphite) + O2(g) ΔHfo = kJ/mol

Calculating Standard Enthalpy Change for a Reaction
The DH° for the desired reaction is then the sum of the DHf° for the component reactions. DH°reaction = S n DHf°(products) − S n DHf°(reactants) S means sum n is the coefficient of the reaction component

CH4(g)+ 2 O2(g)→ CO2(g) + 2 H2O(g)
C(s, graphite) + 2 H2(g) → CH4(g) DHf°= − 74.6 kJ/mol CH4 CH4(g) → C(s, graphite) + 2 H2(g) DH° = kJ DH° = [(DHf° CO2(g) + 2∙DHf°H2O(g)) − (DHf° CH4(g) + 2∙DHf°O2(g))] Note that these equations all constitute enthalpies of formation of the reactants or products. There is no need to include oxygen since it is already in it’s standard state and ΔHf = 0. The enthalpy of reaction for the overall reaction must write equations that breaks down the reactants into their constitutive elements and then builds the products from their constitutive elements. Thus, the enthalpy of formation equation for methane must be reversed. This necessitates a change in sign. Note also the need to change the stoichiometry in the third step. This is to ensure cancellation of the H2 from the first step and includes doubling ΔH° for step 3. Do Example 6.11 on Page 277 DH° = [((−393.5 kJ)+ 2(−241.8 kJ) − ((−74.6 kJ)+ 2(0 kJ))] = −802.5 kJ C(s, graphite) + O2(g) → CO2(g) DHf°= −393.5 kJ/mol CO2 2 H2(g) + O2(g) → 2 H2O(g) DH° = −483.6 kJ H2(g) + ½ O2(g) → H2O(g) DHf°= −241.8 kJ/mol H2O CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) DH° = −802.5 kJ

Calculate the DHrxn for 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
1. Write formation reactions for each compound and determine the DHf° for each 2 C(s, gr) + H2(g) ® C2H2(g) DHf° = kJ/mol C(s, gr) + O2(g) ® CO2(g) DHf° = − kJ/mol H2(g) + ½ O2(g) ® H2O(l) DHf° = − kJ/mol 2. Arrange equations so they add up to desired reaction 2 C2H2(g) ® 4 C(s) + 2 H2(g) DH° = 2(−227.4) kJ -DHf° C2H2 4 C(s) + 4 O2(g) ® 4CO2(g) DH° = 4(−393.5) kJ DHf° CO2 2 H2(g) + O2(g) ® 2 H2O(l) DH° = 2(−285.8) kJ DHf° H2O 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DH = − kJ

DH°reaction = S n DHf°(products) − S n DHf°(reactants)
Calculate the DHrxn for 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DHf°C2H2 = kJ, DHf°H2O = −285.8 kJ, DHf°CO2 = −393.5 kJ DH°reaction = S n DHf°(products) − S n DHf°(reactants) DHrxn = [(4•DHf°CO2 + 2•DHf°H2O) – (2•DHf°C2H2 + 5•DHf°O2)] DHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))] DHrxn = − kJ

Calculate the DH° for decomposing 10
Calculate the DH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions (MMCaCO3 = g/mol) CaCO3(s) → CaO(s) + O2(g) DHf°’s DH°rxn per mol CaCO3 MMlimestone = g/mol, g CaCO3 mol CaCO3 kJ from above

The main engines of the space shuttle burn hydrogen to produce water
The main engines of the space shuttle burn hydrogen to produce water. How much heat (in kJ) is associated with this process if 1.32 × 105 kg of liquid H2 is burned? 2 H2 (l ) + O2 (l ) → 2 H2O (l ) ΔHºrxn = – kJ – 571.6 – 285.8 – 1.87 × 1010 – 3.74 × 1010 – 7.55 × 107

The main engines of the space shuttle burn hydrogen to produce water
The main engines of the space shuttle burn hydrogen to produce water. How much heat (in kJ) is associated with this process if 1.32 × 105 kg of liquid H2 is burned? 2 H2 (l ) + O2 (l ) → 2 H2O (l ) ΔHºrxn = – kJ – 571.6 – 285.8 – 1.87 × 1010 – 3.74 × 1010 – 7.55 × 107 Answer: c

Example: Other Calculations
Don’t always want to know H°rxn Can use Hess’s Law and H°rxn to calculate Hf° for compound where not known Example: Given the following data, what is the value of Hf°(C2H3O2–, aq)? Na+(aq) + C2H3O2–(aq) + 3H2O(l )  NaC2H3O2·3H2O(s) H°rxn = –19.7 kJ/mol Na+(aq) DHof = –239.7 kJ/mol NaC2H3O2•3H2O(s) DHof = kJ/mol H2O(l) DHof = kJ/mol

Example (cont.) H°rxn = Hf° (NaC2H3O2·3H2O, s) – Hf° (Na+, aq) – Hf° (C2H3O2–, aq) – 3Hf° (H2O, l ) Rearranging Hf°(C2H3O2–, aq) = Hf°(NaC2H3O2·3H2O, s) – Hf°(Na+, aq) – H°rxn – 3Hf° (H2O, l) Hf°(C2H3O2–, aq) = –710.4 kJ/mol – (–239.7kJ/mol) – (–19.7 kJ/mol) – 3(–285.9 kJ/mol) = kJ/mol

Learning Check Calculate H for this reaction using Hf° data.
2Fe(s) + 6H2O(l)  2Fe(OH)3(s) + 3H2(g) Hf° – – H°rxn = 2×Hf°(Fe(OH)3, s) + 3×Hf°(H2, g) – 2× Hf°(Fe, s) – 6×Hf°(H2O, l ) H°rxn = 2 mol× (– kJ/mol) + 3×0 – 2×0 – 6 mol× (–285.8 kJ/mol) H°rxn = –1393 kJ kJ H°rxn = kJ

Learning Check Calculate H°rxn for this reaction using Hf° data.
CO2(g) + 2H2O(l )  2O2(g) + CH4(g) Hf° – – – 74.8 H°rxn = 2×Hf°(O2, g) + Hf°(CH4, g) –Hf°(CO2, g) – 2× Hf°(H2O, l ) H°rxn = 2 × mol × (–74.8 kJ/mol) – 1 mol × (–393.5 kJ/mol) – 2 mol × (–285.8 kJ/mol) H°rxn = –74.8 kJ kJ kJ H°rxn = kJ

Your Turn! Calculate Hf° for FeO(s) using the information below. Hf° values are shown below each substance. Fe3O4(s) + CO(g)  3FeO(s) + CO2(g) Ho=21.9 kJ kJ kJ ?? – kJ A kJ B kJ C kJ D J E kJ

Your Turn! sol’n Important H°rxn = [3Hf° (FeO, s) + Hf°(O2, g)] – [Hf°(Fe3O4, s) + Hf°(CO, g)] +21.9 kJ = [3Hf° (FeO, s) kJ)] – [ kJ kJ)] +21.9 kJ = [3Hf° (FeO, s) kJ] kJ = 3Hf° (FeO, s) kJ = Hf° (FeO, s)

Learning Check Calculate H°rxn using Hf° for the reaction 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l ) H°rxn = [136 – ] kJ H°rxn = –1395 kJ

Check Using Hess’s Law 4 ×[NH3(g)  ½ N2(g) + 3/2 H2(g)] – 4 × Hf°(NH3, g) 7 ×[ O2(g)  O2(g) ] –7 × Hf°(O2, g) 4 ×[ O2(g) + ½ N2(g)  NO2(g)] 4 × Hf°(NO2, g) 6 ×[ H2(g) + ½ O2(g)  H2O(l ) ] 6 × Hf°(H2O, l) 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l ) Same as before

Energy & Chemical Change

Similar presentations

Presentation on theme: "Energy & Chemical Change"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Energy & Chemical Change

Similar presentations

Presentation on theme: "Energy & Chemical Change"— Presentation transcript:

Similar presentations

About project

Feedback