Topic 5. Chemical Energetics

Topic 5. Chemical Energetics

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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.

http://upload.wikimedia.org/wikipedia/commons/d/d5/Archery_Target_80cm.svg

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Heat and Temperature• Heat is energy that is transferred from one

object to another due to a difference in temperature

• Temperature is a measure of the average kinetic energy of a body

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

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5.1.1 Define the terms exothermic reaction, endothermic reaction and

standard enthalpy change of reaction (ΔHo).

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

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


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5.1.1 Define the terms exothermic reaction, endothermic reaction and

standard enthalpy change of reaction (ΔHo).

Standard Enthalpy Change of Reaction (∆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…)

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


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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)

6.2

energy + H2O (s) H2O (l)

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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.

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

Hproducts < Hreactants

DH < 0Hproducts > Hreactants

DH > 0 6.4

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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.


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 15

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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 Quantities• The amount of heat contained by an object

depends primarily on three factors:– The mass of material– The temperature– The kind of material and its ability to

absorb or retain heat.

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Heat Quantities• The heat required to raise the temperature of

1.00 g of water 1 oC is known as a calorie• 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 joule. It is based on

the mechanical energy requirements. • 1.00 calorie = 4.184 Joules

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

changed.

Heat Energy Changeq = m x c x ΔT

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 /

ΔT = change in temperature


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changed.

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 1 Calculate the heat gained in an aluminum cooking pan

whose mass is 400 grams, from 20oC to 200oC. The specific heat of aluminum is 0.902 J g-1 oC-1.

Solution

Q = mCT = (400 g) (0.902 J g-1 oC-1)(200oC – 20oC)

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

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changed.

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

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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changed.

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 (Table

11.2)ΔT = 52 oC

q = 250 x 0.14 x 52 = 1800 J

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


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

reactions.

• 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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Calorimetry• Calorimetry involves the measurement of

heat changes that occur in chemical processes or reactions. Determines

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

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

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Heat Capacity and Specific Heat

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

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

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

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Substance CJ g-1 K-1 C J mol-1K-1

Water (liquid) 4.184 75.327

Water (steam) 2.080 37.47

Water (ice) 2.050 38.09

Copper 0.385 24.47

Aluminum 0.897 24.2

Ethanol 2.44 112

Lead 0.127 26.4

Specific Heat values for Some Common

Substances

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

reactions.

Heat Energy Changeq = -q

HUH?!?


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Heat Transfer Problem 2 What is the final temperature when 50 grams of water at

20oC is added to 80 grams water at 60oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 oC-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)(60oC- T)

(50 g)(T- 20oC) = (80 g)(60oC- T) 50T -1000 = 4800 – 80T 130T =5800 T = 44.6 oC

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5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes,

quantities of reactants and mass of water.

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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Using Thermochemical EquationsSodium 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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Using Thermochemical EquationsSodium 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

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




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

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 kJb) 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 kJb) 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 kJb) 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 reaction• Chemical bonds are broken• Atoms are rearranged• New chemical bonds are formed• These processes always involve

energy changes

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Energy Changes Breaking chemical bonds requires energy

Endothermic Forming new chemical bonds releases

energy Exothermic

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5.4.1 Define the term average bond enthalpy

Enthalpy changes of reactions are the result of bonds breaking and new bonds being formed. 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 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 enthalpy change for a chemical reaction.

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

Energy is released when chemical bonds form. When a chemical bond is formed its enthalpy change is expressed as a negative value

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

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Exothermic and Endothermic Processes

Exothermic processes release energy C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4H2O (g)

+ 2043 kJ

Endothermic processes absorb energy C(s) + H2O (g) +113 kJ CO(g) + H2 (g)

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Energy Changes in endothermic and exothermic processes

In an endothermic reaction there is more energy required to break bonds than is released when bonds are formed. The opposite is true in an exothermic 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.

average bond enthalpy reactants > average bond enthalpy products

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.

average bond enthalpy reactants < average 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

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


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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 Calculations

Calculate the enthalpy change for the reaction N2 + 3 H2 2 NH3

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Bonds broken1 N=N: = 9453 H-H: 3(435) = 1305 Total = 2250 kJBonds formed2x3 = 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



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

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Topic 5. Chemical Energetics

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