1 Topic 5. Chemical Energetics Created by S. Colgan; Modified by K. Slater Resources: 36,Sta.

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Topic 5. Chemical Energetics

Created by S. Colgan; Modified by K. SlaterResources: http://lincoln.pps.k12.or.us/lscheffler/Energetics.ppt#269,36,Standard Enthalpy Changes

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IB Topic 5: Energetics5.1: Exothermic and Endothermic Reactions

5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction

(ΔHo). 5.1.2 State that combustion and neutralization

are exothermic processes. 5.1.3 Apply the relationship between temperature

change, enthalpy change and the classification of a reaction as endothermic or exothermic.

5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction.

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Heat and TemperatureHeat and Temperature

HeatHeat is energy that is transferred from one is energy that is transferred from one object to another due to a difference in object to another due to a difference in temperaturetemperature

TemperatureTemperature is a measure of the average is a measure of the average kinetic energy of a bodykinetic energy of a body

Heat is always transferred from objects at a Heat is always transferred from objects at a higher temperature to those at a lower higher temperature to those at a lower temperaturetemperature

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5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (ΔHo).

Exothermic ReactionExothermic Reaction: A process that releases heat to its surroundings. Products have less energy than the reactants

Endothermic ReactionEndothermic Reaction : A process that absorbs heat from the surroundings. Products have more energy than the reactants.

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Standard Enthalpy Change of Reaction (Standard Enthalpy Change of Reaction (∆∆H):H): The heat energy exchanged with the surroundings when a reaction happens under standard conditions (NOT STP… see below).

Since the enthalpy change for any given reaction will vary with the conditions, esp. concentration of chemicals, ΔH are measured under standard conditions:

pressure = 101.3 kPa temperature = 25ºC = 298 K Concentrations of 1 mol dm-3 The most thermodynamically stable allotrope (which in the

case of carbon is graphite)

Only ΔH can be measured, not H for the initial or final state of a system.

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Pseudonyms (other names) for H Heat of Reaction: Hrxn heat produced in a chemical reaction

Heat of Combustion: Hcomb heat produced by a combustion reaction

Heat of Neutralization: heat produced in a neutralization reaction (when an acid and base are mixed to get water, pH = 7)

Heat of solution: Hsol heat produced by when something dissolves

Heat of Fusion: Hfus heat produced when something melts

Heat of Vaporization: Hvap heat produced when something evaporates

Heat of Sublimation: Hsub heat produced when something sublimes

Heat of formation: Hf change in enthalpy that accompanies the formation of 1 mole of compound from it’s elements (this has special uses in chemistry…)


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5.1.2 State that combustion and neutralization are exothermic processes.

CombustionExothermic reactionGeneral Combustion Reaction Formula:

Compound (usually hydrocarbon) + O2 CO2 + H2O + energy

CH4 + 2O2 CO2 + 2H2O + 890kJ∆H = -890kJ

NeutralizationExothermic reactionAcid + Base Salt + Water + energyHCl + NaOH NaCl + H2O + 57.3 kJ

∆H = -57.3kJ

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5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction as endothermic or exothermic.

ExothermicHeat flows out of the

systemSurroundings heat upHeat change (ΔH) < 0

(negative) C8H18+ 12½O2 8CO2 +

9H2O ΔH = -5512 kJ mol-1

H2 + ½O2 H2O ΔH = -286 kJ mol-1

EndothermicHeat flows into the

systemSurroundings cool down Heat change (ΔH) > 0

(positive)

H2O(s) H2O(l) ΔH = +6.01 kJ mol-1

½N2 + O2 NO2

ΔH = +33.9 kJ mol-1

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Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

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energy + H2O (s) H2O (l)

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Exothermic ReactionsProducts more stable than reactants

(lower energy).ΔH = Hproducts – Hreactants

Since the products have less energy than the reactants, the ΔH value is negative.

Endothermic ReactionsProducts less stable than reactants

(higher energy)ΔH = Hproducts – Hreactants

Since the products have more energy than the reactants, the ΔH value is positive.

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Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0 6.4

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REVIEW

Endothermic Exothermic

Definition

Examples (2) N/A

Change in Temperature of the container

∆H value

Direction of heat flow

Stability of reactants

Stability of products

Bonding

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REVIEW

Endothermic Exothermic

Definition A process that absorbs heat from the surroundings

A process that releases heat into the surroundings

Examples (2) Combustion & Neutralization reactions

Change in Temperature

Decreases Increases

∆H value Positive Negative

Direction of heat flow From surroundings into system

From system into surroundings

Stability of reactants More stable Less stable

Stability of products Less stable More stable

Bonding Bond breaking Bond making

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IB Topic 5: Energetics5.2: Calculation of Enthalpy Changes

5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.

5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water

5.2.4 Evaluate the results of experiments to determine enthalpy changes.

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Factors Affecting Heat QuantitiesFactors Affecting Heat Quantities

The amount of heat contained by an object The amount of heat contained by an object depends primarily on three factors:depends primarily on three factors: The mass of materialThe mass of material The temperatureThe temperature The kind of material and its ability to The kind of material and its ability to

absorb or retain heat. absorb or retain heat.

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Specific heat: The amount of heat energy required to raise the temperature of 1 g of a substance 1 oC.

Units are: J g-1 oC-1

Specific heat of water (or aqueous solutions) Specific heat of water (or aqueous solutions) = 4.184 Joules = 1 cal = 4.184 Joules = 1 cal

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Heat QuantitiesHeat Quantities

The heat required to raise the temperature of 1.00 The heat required to raise the temperature of 1.00 g of water 1 g of water 1 ooC is known as a C is known as a caloriecalorie

Calorie (with a capital “C”)Calorie (with a capital “C”): dietary measurement of heat. Food has potential energy stored in the chemical bonds of food.

1 Cal = 1 kcal = 1000 cal The SI unit for heat is the The SI unit for heat is the joulejoule. It is based on . It is based on

the mechanical energy requirements. the mechanical energy requirements. 1.00 calorie = 4.184 Joules1.00 calorie = 4.184 Joules

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Heat Energy Changeq = m x c x ΔT

ΔH = q = heat (joules or calories)m = mass (g)

c = specific heat (J g-1 oC-1)The amount of heat required to raise the

temperature of 1 g of a substance 1 oC.

Specific heat of water = 4.184 Joules / Specific heat of water = 4.184 Joules /

ΔT = change in temperature

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How much heat in joules will be absorbed when 32.0 g of water is heated from 25.0 oC to 80.0 oC?

q = m x c x ΔTq = ?m = 32.0 gc = 4.18 J g-1 oC-1

ΔT = 80.0-25.0 = 55.0 oC

q = 32.0 x 4.18 x 55.0 = 7,360 J

When 435 J of heat is added to 3.4 g of olive oil at 21 oC, the temperature increases to 85 oC. What is the specific heat of olive oil?

q = m x c x ΔTq = 435 Jm = 3.4 gc = ?ΔT = 85-21 = 64 oC

435 = 3.4 x c x 64 = 2.0 J g-1 oC-1

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Heat Transfer Problem 1Heat Transfer Problem 1

Calculate the heat gained in an aluminum cooking pan Calculate the heat gained in an aluminum cooking pan whose mass is 400 grams, from 20whose mass is 400 grams, from 20ooC to 200C to 200ooC. The C. The specific heat of aluminum is 0.902 J gspecific heat of aluminum is 0.902 J g-1-1 ooCC-1-1..

Solution

q = mCT

= (400 g) (0.902 J g0.902 J g-1-1 ooCC-1-1)(200)(200ooC – 20C – 20ooCC))

= 64,944 J = 60,000J OR 6 x 104 J OR 60 kJ= 64,944 J = 60,000J OR 6 x 104 J OR 60 kJ

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How much heat in joules is required to raise the temperature of 250 g of mercury 52oC? The specific heat capacity of Hg is 0.14 J g-1 oC-1

A 1.55 g piece of stainless steel absorbs 141 J of heat when its temperature increases by 178 oC. What is the specific heat of the stainless steel?

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How much heat in joules is required to raise the temperature of 250 g of mercury 52 oC?

q = m x c x ΔTq = ?m = 250 gc = 0.14 J g-1 oC-1

ΔT = 52 oC

q = 250 x 0.14 x 52 = 1800 JOR 1.8 kJ

A 1.55 g piece of stainless steel absorbs 141 J of heat when its temperature increases by 178 oC. What is the specific heat of the stainless steel?

q = m x c x ΔTq = 141 Jm = 1.55 gc = ?ΔT = 178 oC

141 = 1.55 x c x 178 = 0.511 J g-1 oC-1

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Calorimeter: Reactions used to heat up an external source of water.

Temperature change of water, mass of material and mass of water are measured.

Use q = m x c x ΔT to solve for q then find the heat of reaction in kJ/mol of reacting substance.

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CalorimetryCalorimetry

Calorimetry involves the measurement of Calorimetry involves the measurement of heat changes that occur in chemical heat changes that occur in chemical processes or reactions. processes or reactions. Determines

the ΔH by measuring temp Δ's created from the rxn

The heat change that occurs when a The heat change that occurs when a substance absorbs or releases energy is substance absorbs or releases energy is really a function of three quantities:really a function of three quantities: The massThe mass The temperature changeThe temperature change The heat capacity of the materialThe heat capacity of the material

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Heat Capacity and Specific HeatHeat Capacity and Specific Heat

The ability of a substance to absorb or retain heat The ability of a substance to absorb or retain heat varies widely.varies widely.

The heat capacity depends on the nature of the The heat capacity depends on the nature of the material.material.

The The specific heatspecific heat of a material is the amount of of a material is the amount of heat required to raise the temperature of 1 gram of heat required to raise the temperature of 1 gram of a substance 1 a substance 1 ooC (or Kelvin)C (or Kelvin)

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SubstanceSubstance CCJ gJ g-1-1 K K-1-1 C C J molJ mol-1-1KK-1-1

Water (liquid)Water (liquid) 4.184 4.184 75.327 75.327

Water (steam)Water (steam) 2.080 2.080 37.47 37.47

Water (ice) Water (ice) 2.050 2.050 38.09 38.09

CopperCopper 0.385 0.385 24.47 24.47

AluminumAluminum 0.897 0.897 24.2 24.2

Ethanol Ethanol 2.44 2.44 112 112

Lead Lead 0.127 0.127 26.4 26.4

Specific Heat values for Specific Heat values for Some Common Some Common

SubstancesSubstances

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Heat Energy Changeq = -q

HUH?!?

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Heat Transfer Problem 2Heat Transfer Problem 2 What mass of water will be heated from 5.72What mass of water will be heated from 5.72ooC to 20.65C to 20.65ooC if C if

it is mixed in a calorimeter with 80.72 g of water that starts it is mixed in a calorimeter with 80.72 g of water that starts at 29.5at 29.5ooC? Assume that the loss of heat to the surroundings C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 is negligible. The specific heat of water is 4.184 J g-1 ooCC-1-1

Solution: Q (Cold) = -Q (hot) mcT= -mcT

(m)(4.184)(20.65 20.65 – 5.725.72) = – (80.72)(4.184)(20.65 20.65 – 29.529.5)

(m)(4.184)(14.9314.93) = – (80.72)(4.184)(–8.858.85) 62.46712 (m) = 2988.932448 m = 47.84 g Using proper sig. fig.s…

m = 47.8 g

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What mass of water will be heated from 25What mass of water will be heated from 25ooC to 38C to 38ooC C when 5.0 grams of 98when 5.0 grams of 98ooC copper is added to the water? C copper is added to the water? Assume that heat loss to the surroundings is negligible.Assume that heat loss to the surroundings is negligible.

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SubstanceSubstance C (C (J gJ g-1-1 K K-1-1)) C C (J mol(J mol-1-1KK-1-1))

CopperCopper 0.385 0.385 24.47 24.47

AluminumAluminum 0.897 0.897 24.2 24.2

Lead Lead 0.127 0.127 26.4 26.4

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What mass of water that will be heated from 25What mass of water that will be heated from 25ooC to C to 3838ooC when 5.0 grams of 98C when 5.0 grams of 98ooC copper is added to the C copper is added to the water? Assume that the loss of heat to the water? Assume that the loss of heat to the surroundings is negligiblesurroundings is negligible

Solution: Q (Cold: H2O) = -Q (hot Cu) mcT= -mcT Let T = final temperature

(x) (4.184)(38 - 25oC) = - (5.0g) (0.385)(38 - 98)

(x)(4.184)(13) = - (5.0 g)(0.385)(-60) 54.392x = 115.5 x = 2.123474 g With proper sig. figs. x = 2.1 g

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What is the final temperature when 50 grams of water at What is the final temperature when 50 grams of water at 2020ooC is added to 80 grams water at 60C is added to 80 grams water at 60ooC? Assume that C? Assume that the loss of heat to the surroundings is negligible. The the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 specific heat of water is 4.184 J g-1 ooCC-1-1

Solution: Q (Cold) = -Q (hot) mcT= -mcT Let T = final temperature

(50 g) (4.184 J g-1 oC-1)(T – 20oC) = –(80 g) (4.184 J g-1 oC-1)(T – 60oC)

(50 g)(T – 20oC) = –(80 g) (T – 60oC) 50T – 1,000 = 4,800 – 80T 130T = 5,800 T = 44.6 oC

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Frayer Model

HeatEnthalpyEndothermicExothermic

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Using Thermochemical Equations

Mg(s) + ½ O2(g) MgO(s) + heat Equation 1

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Equation A ∆H = -A

MgO(s) + 2HCl(aq) MgCl2(aq) +H2O(l) Equation B ∆H = -B

H2(g) + ½ O2(g) H2O(l) Equation C ∆H = +C

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.

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Mg(s) + ½ O2(g) MgO(s) + heat Equation 1

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Equation A ∆H = -A

MgO(s) + 2HCl(aq) MgCl2(aq) +H2O(l) Equation B ∆H = -B

H2(g) + ½ O2(g) H2O(l) Equation C ∆H = +C

The final ∆H = -A + B + C


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Using Thermochemical EquationsCalcium oxide combines with water to produce calcium hydroxide and heat

(exothermic reaction). CaO(s) + H2O(l) Ca(OH)2(s) + 65.2 kJ OR CaO(s) + H2O(l) Ca(OH)2(s) H = -65.2 kJ

These H values assume 1 mole of each compound (based on the coefficients)

How many kJ of heat are produced when 7.23 g of CaO react?1) Write out and balance equation: already balanced2) Determine the number of moles: 7.23 g/56.01 gmol-1 = 0.129 mol3) Multiply: 0.129 mol CaO x 65.2 kJ mol-1

* Notice that mol / mol cancel out and you’re left with kJ4) Solve = 8.41 kJ


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Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide.

How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?

1) Write out and balance the equation2) Determine the number of moles of NaHCO3(s)3) Set up the ratio4) Solve for x5) State, with justification, whether the reaction is endothermic

or exothermic.


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Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide.

2NaHCO3(s) + 129 kJ Na2CO3(s) + H2O(g) + CO2(g) OR 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) H = 129 kJ

How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?1) Balance equation2) Moles NaHCO3(s) = 2.24 mol3) Ratio: x kJ = 129 kJ

2.24 NaHCO3(s) 2 mol 4) Solve for x = 144 kJ5) The reaction is endothermic because energy is being absorbed /

the H is positive.


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Using Experimental Data

In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup calorimeter. At the start, the solutions and the calorimeter are all at 25.0 oC. During the reaction, the highest temperature observed is 32.0 oC. Calculate the heat (in kJ) released during this reaction. Assume the densities of the solutions are 1.00 g mL-1.


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Using Experimental Data

In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup calorimeter. At the start, the solutions and the calorimeter are all at 25.0 oC. During the reaction, the highest temperature observed is 32.0 oC. Calculate the heat (in kJ) released during this reaction. Assume the densities of the solutions are 1.00 g mL-1.

Use q = m x c x ΔT

m = mass of solution = 50.0 mL x 1.00 g mL-1 = 50.0 gC = 4.18 j g-1 oC-1

ΔT = 32.0 – 25.0 = 7.0 oCq = 50.0 g x 4.18 J g-1 oC-1 x 7.0 oC = 1463 J = 1.5 kJ


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Using Thermodynamic Quantities (Standard Heats of Formation)

Heat of reaction can be found by: sum the heats of formation of all the products – sum of heats of formation of all the reactants

Hrxn = Hf products – Hf reactants

Hf = standard enthalpy of formation. Energy required to form a compound from its elements.

“standard” is a term used a lot in chemistry. It usually means that the values are experimentally determined and compared to an agreed upon reference value

Since the Hf is given per mole, we must multiply by coefficients


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Using Thermodynamic Quantities (Standard Heats of Formation)

Using the table of thermodynamic quantities, calculate the heat of reaction for 2SO2(g) + O2(g) 2SO3(g)

Heat of reaction = Hf products – Hf of reactantsHeat of products: Hf (SO3) = -395.2 kJ/mol x 2 = -790.4 kJ

Heat of reactants = Hf (SO2) + Hf (O2)(-296.9 kJ/mol x 2) + (0) = -593.8 KJ

Heat of reaction = -790.4 kJ – (-593.8 kJ) = -196.6 kJ


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N2O4 + 3 CO N2O + 3CO2

Hf products = 1(81 kJ mol-1) + 3(-393 kJ mol-1) = -1098 kJ/mol

Hf reactants = 1(-9.7 kJ mol-1)+ 3(-110 kJ mol-1) = -320.3 kJ mol-1

Hf products – Hf reactants = (-1098 kJ mol-1) – (-320.3 kJ mol-1)

= -778 kJ mol-1

Hrxn = -778 kJ mol-1

Therefore it is exothermic


Reactants Hf Products Hf

N2O4 9.7 kJ mol-1 N2O 81 kJ mol-1

CO -110 kJ mol-1 CO2 -393 kJ mol-1

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Ca(OH)2(s) + CO2 (g) H2O(g) + CaCO3 (s)

Hf products = 1(-241.8 kJ mol-1) + 1(-1206.9 kJ mol-1) = -1448.7 kJ/mol

Hf reactants = 1(-986.1 kJ mol-1)+ 1(-393.5 kJ mol-1) = -1379.6 kJ mol-1

Hf products – Hf reactants = (-1448.7 kJ mol-1) – (-1379.6 kJ mol-1)

= -69.1 kJ mol-1

Hrxn = -69.1 kJ mol-1

Therefore it is exothermic


Reactants Hf Products Hf

Ca(OH)2 -986.1 kJ mol-1 H2O -241.8 kJ mol-1

CO2 -393.5 kJ mol-1 CaCO3 -1206.9 kJ mol-1

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IB Topic 5: Energetics5.3 Hess’s Law & 5.4 Bond Enthalpies

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.

5.4.1 Define the term average bond enthalpy 5.4.2 Explain, in terms of average bond

enthalpies, why some reactions are exothermic and others are endothermic

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Hess’s Law: Reactions can be added together in order to determine heats of reactions that can’t be measured in the lab.

C(diamond) C(graphite)This reaction is too slow to be measured in the lab. Two reactions

can be used that can be measured in the lab:a) C(graph) + O2(g) CO2(g) H = -393.5 kJ

b) C(diam) + O2(g) CO2(g) H = -395.4 kJ

Since C(graphite) is a product, write equation a) in reverse to give:c) CO2(g) C(graph) + O2(g) H = 393.5 kJ

Now add equations b) and c) together:C(diam) + O2(g) + CO2(g) C(graph) + O2(g) + CO2(g)

H = -395.4 kJ + 393.5 kJ = -1.9 kJ

Final equation: C(diamond) C(graphite) H = - 1.9 kJ

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Given the following thermochemical equations, calculate the heat of reaction for:

C2H4(g) + H2O(l) C2H5OH(l)

a) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) H = -1367 kJ

b) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1411 kJ

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Given the following thermochemical equations, calculate the heat of reaction for

C2H4(g) + H2O(l) C2H5OH(l)

a) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) H = -1367 kJ

b) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1411 kJ

Since C2H5OH(l) is a product, write equation a) in reverse order:

c) 2CO2(g) + 2H2O(l) C2H5OH(l) + 3O2(g) H = 1367 kJ

Add equations b) & c) together, cancelling out substances on opposite sides of the arrow. Add the heat values to obtain the heat of reaction.

C2H4(g) + H2O(l) C2H5OH(l) H = -44 kJ

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C(s) + 2H2(g) CH4(g)

a) C(s) + O2(g) CO2(g) H = -393 kJ mol-1

b) H2(g) + ½ O2(g) H2O(l) H = -286 kJ mol-1

c) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ mol-1

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C(s) + 2H2(g) CH4(g)

C(s) + O2(g) CO2(g) H = -393 kJ

2(H2(g) + ½ O2(g) H2O(l)) H = 2(-286 kJ mol-1)

2H2(g) + O2(g) 2H2O(l)) H = -572 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ mol-1

CO2(g) + 2H2O(l) CH4(g) + 2O2(g) H = 890

C(s) + 2H2(g) CH4(g) H = -75 kJ

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5.4.1 Define the term average bond enthalpy

In a chemical reactionIn a chemical reaction Chemical bonds are brokenChemical bonds are broken Atoms are rearrangedAtoms are rearranged New chemical bonds are formedNew chemical bonds are formed These processes always involve These processes always involve

energy changesenergy changes

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Energy ChangesEnergy Changes

Enthalpy changes of reactions are the result of bonds breaking and new bonds being formed.

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Energy ChangesEnergy Changes

Breaking chemical bonds requires energyBreaking chemical bonds requires energy EndothermicEndothermic

Forming new chemical bonds releases Forming new chemical bonds releases energyenergy ExothermicExothermic

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5.4.1 Define the term average bond enthalpy

Remember… Breaking bonds requires energy Forming new bonds releases energy

Bond enthalpy is the energy required to break one mole of a certain type of bond in the gaseous state averaged across a variety of compounds.

FYI: Bond enthalpies for unlike atoms will be affected by surrounding bonds and will be slightly different in different compounds so average bond enthalpies are used.

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5.3.2 Using bond enthalpy to determine enthalpy change of a reaction

The average bond enthalpies for several types of The average bond enthalpies for several types of chemical bonds are shown in the table below:chemical bonds are shown in the table below:

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H = ∑ (energy required – ∑(energy released

to break bonds) when bonds are formed)

OR

H = ∑ (bond enthalpy – ∑(bond enthalpy

of reactants) of products)


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H =

∑ (bond enthalpy – ∑(bond enthalpy

of reactants) of products)


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Bond enthalpies can be used to calculate the Bond enthalpies can be used to calculate the enthalpy change for a chemical reactionenthalpy change for a chemical reaction..

Energy is required toEnergy is required to break chemical bondsbreak chemical bonds. . Therefore when a chemical bond is broken Therefore when a chemical bond is broken its its enthalpy changeenthalpy change carries a carries a positive signpositive sign..

Energy is released when chemical bonds Energy is released when chemical bonds formform. When a chemical bond is formed its . When a chemical bond is formed its enthalpy changeenthalpy change is expressed as a is expressed as a negative negative valuevalue

By combining the enthalpy required and the By combining the enthalpy required and the enthalpy released for the breaking and enthalpy released for the breaking and forming chemical bonds, one can calculate forming chemical bonds, one can calculate the enthalpy change for a chemical reactionthe enthalpy change for a chemical reaction

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Exothermic and Endothermic Exothermic and Endothermic ProcessesProcesses

Exothermic processes release energyExothermic processes release energy

CC33HH88 (g) + 5 O (g) + 5 O22 (g) (g) 3 CO 3 CO22 (g) + 4H (g) + 4H22O (g)O (g)

+ 2043 + 2043 kJkJ

Endothermic processes absorb energy Endothermic processes absorb energy

C(s) + HC(s) + H22O (g) O (g) +113 kJ+113 kJ CO(g) + H CO(g) + H22 (g) (g)

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Energy Changes in Energy Changes in endothermic and endothermic and exothermic processesexothermic processesIn an In an endothermic endothermic reaction there is reaction there is more energy more energy required to required to break bonds break bonds than is released than is released when bonds are when bonds are formed. formed.

The opposite is The opposite is true in an true in an exothermic exothermic reaction.reaction.

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5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic

If the amount of energy required to break the bonds in the reactants is greater than the amount of energy released when bonds are formed in the products, the reaction is endothermic.

bond enthalpy reactants > bond enthalpy products

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If the amount of energy required to break the bonds in the reactants is less than the amount of energy released when bonds are formed in the products, the reaction is exothermic.

bond enthalpy reactants < bond enthalpy products

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Use the following average bond enthalpies (kJ mol-1) to determine the heat of reaction for 3F2 + NH3 3HF + NF3

F-F = 158; N-H = 388; H-F = 562; N-F = 272

Energy in (kJ mol-1)3F-F is 3(158) + 3N-H is 3(388) = 1638 kJ

Energy out (kJ mol-1)3H-F is 3(562) + 3N-F is 3(272) = 2502 kJ

Since Energy in < Energy out, the reaction is exothermic

Heat of reaction is -864 kJ mol-1


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Using the Bond Enthalpy Table, determine the heat of reaction for:CO(g) + 2H2(g) CH3OH


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Energy in (kJ)1CO (triple bond) 1074 + 2H-H is 2(436) = 1946 kJ mol-1

Energy out (kJ)3C-H is 3(413) + 1C-O is 358 + 1 O-H is 464 = 2061 kJ mol-1

Hrxn = 1946 – 2061 = -115 kJ mol-1

Since Energy in < Energy out, reaction is exothermic


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Energy in (kJ)1CO (triple bond) 1074 + 2H-H is 2(436) = 1946 kJ mol-1

Energy out (kJ)3C-H is 3(413) + 1C-O is 358 + 1 O-H is 464 = 2061 kJ mol-1

Hrxn = 1946 – 2061 = -115 kJ mol-1

Since Energy in < Energy out, reaction is exothermic


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Bond Enthalpy Bond Enthalpy CalculationsCalculations

Calculate the enthalpy change for the Calculate the enthalpy change for the reaction Nreaction N22 + 3 H + 3 H22 2 NH 2 NH33

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Bonds broken

1 N=N: = 9453 H-H: 3(435) = 1305 Total = 2250 kJ

Bonds formed

2x3 = 6 N-H: 6 (390) = - 2340 kJ

Net enthalpy change

= + 2250 - 2340 = - 90 kJ

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In the 1960s NASA considered the relative merits of using hydrogen and oxygen compared with hydrogen and fluorine as rocket fuels. Assuming all the reactants and products are in the gaseous state, use bond enthalpies to calculate the enthalpy change of reaction (in kJ mol-1 of product) for both fuels. As mass is more important than amount in the choice of rocket fuels, which reaction would give more energy per kilogram of fuel?

Bond enthalpies (kJ mol-1):H-H: 435; O O: 496; H-O: 464 kJ; F-F: 158; H-F: 562


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Terms to Know

Endothermic Exothermic Temperature Heat Standard enthalpy change of reaction Standard enthalpy of formation Enthalpy of combustion Average bond enthalpy Hess’ Law Standard conditions

1 Topic 5. Chemical Energetics Created by S. Colgan; Modified by K. Slater Resources: 36,Sta.

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Transcript of 1 Topic 5. Chemical Energetics Created by S. Colgan; Modified by K. Slater Resources: 36,Sta.