Chemistry

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 135.

Unit 4: THERMOCHEMISTRY AND NUCLEAR CHEMISTRY Chapter 6: Thermochemistry 6.1: The Nature of Energy and Types of Energy Energy (E): - the ability to do work or produce heat. Different Types of Energy:

1. Radiant Energy: - solar energy from the sun. 2. Thermal Energy: - energy associated with the random motion of atoms and molecules. 3. Chemical Energy: - sometimes refer to as Chemical Potential Energy. It is the energy stored in the

chemical bonds, and release during chemical change.

4. Potential Energy: - energy of an object due to its position. First Law of Thermodynamics: - states that energy cannot be created or destroyed. It can only be

converted from one form to another. Therefore, energy in the universe is a constant.

- also known as the Law of Conservation of Energy (ΣEinitial = ΣEfinal). 6.2: Energy Changes in Chemical Reactions Heat (q): - the transfer of energy between two objects (internal versus surroundings) due to the difference

in temperature. Work (w): - when force is applied over a displacement in the same direction (w = F × d). - work performed can be equated to energy if no heat is produced (E = w). This is known as the

Work Energy Theorem.

System: - a part of the entire universe as defined by the problem. Surrounding: - the part of the universe outside the defined system. Open System: - a system where mass and energy can interchange freely with its surrounding. Closed System: - a system where only energy can interchange freely with its surrounding but mass not

allowed to enter or escaped the system. Isolated System: - a system mass and energy cannot interchange freely with its surrounding.

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 136. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Exothermic Process (ΔE < 0): - when energy flows “out” of the system into the surrounding. (Surrounding gets Warmer.)

Endothermic Process (ΔE > 0): - when energy flows into the system from the surrounding.

(Surrounding gets Colder.) 6.3: Introduction of Thermodynamics Thermodynamics: - the study of the inter action of heat and other kinds of energy. State of a System: - the values of all relevant macroscopic properties like composition, energy,

temperature, pressure and volume. State Function: - also refer to as State Property of a system at its present conditions. - energy is a state function because of its independence of pathway, whereas work and heat

are not state properties. Pathway: - the specific conditions that dictates how energy is divided as work and heat.

- the total energy transferred (ΔE) is independent of the pathway, but the amounts of work and heat involved depends on the pathway.

Po te

nt ia

l E ne

rg y

Reaction Pathway

Reactant(s)

Product (s)

ΔE < 0 (Δq < 0 when w = 0)

Surrounding

Potential Energy Diagram for Exothermic Process

Energy Output

Po te

nt ia

l E ne

rg y

Reaction Pathway

Reactant(s)

Product (s)

ΔE > 0 (Δq > 0 when w = 0)

Surrounding

Potential Energy Diagram for Endothermic Process

Energy Input

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 137.

Internal Energy (E): - total energy from work and heat within a system.

Work as Compression and Expansion - During expansion on the system, w < 0 because the system is pushing out and work is done by the

system (energy output to the surrounding). - During compression on the system, w > 0 because the system is being pressed by the surround and

work is done on the system (energy input by the surrounding).

w = F × Δd (Pressure = Force per unit of Area, A F

P = or F = PA)

w = (PA) × Δd (Substitute PA as Force; A × Δd = Volume – 3 dimensions) w = −P ΔV (During Expansion V ↑, and w ↓. ∴ Negative is added to PΔV)

6.4: Enthalpy of Chemical Reactions

Enthalpy (H): - the amount of internal energy at a specific pressure and volume (H = E + PV).

ΔE = q − PΔV (ΔE = ΔH − PΔV Rearrange formula for enthalpy) ΔH − PΔV = q − PΔV (Equate ΔE and simplify by cancelling −PΔV on both sides)

ΔH = q (Change in Enthalpy is Change in Heat of a system at constant pressure and little change in volume.)

ΔE = q + w

ΔE = Change in System’s Internal Energy q = heat (q > 0 endothermic; q < 0 exothermic) w = work (w > 0 work done on the system; w < 0 work done by the system)

w = −P ΔV (1 L • atm = 101.3 J)

Change in Enthalpy in a Chemical Reaction

ΔH = q = Hproducts − Hreactants

ΔH > 0 Endothermic Reaction ΔH < 0 Exothermic Reaction

Po te

nt ia

l E ne

rg y

Reaction Pathway

Reactant(s)

Product (s)

ΔH < 0 (Hproducts < Hreactants)

Potential Energy Diagram for Exothermic Chemical Reactions / Physical Processes

ΔH = nΔHrxn

ΔH = Change in Enthalpy n = moles ΔHrxn = Molar Enthalpy of Reaction (kJ/mol)

Po te

nt ia

l E ne

rg y

Reaction Pathway

Reactant(s)

Product (s)

ΔH > 0 (Hproducts > Hreactants)

Potential Energy Diagram for Endothermic Chemical Reactions / Physical Processes

Assignment 6.1 to 6.3 pg. 198 #1 to 3, 6 to 11

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

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Writing ΔH Notations with Chemical Equations / Physical Process: a. Endothermic Reactions / Processes

Reactant(s) + Heat → Product(s) OR Reactant(s) → Product(s) ΔH = +____ kJ Example: Water is vaporized from its liquid state.

H2O (l) + 40.7 kJ → H2O (g) or H2O (l) → H2O (g) ΔH = +40.7 kJ b. Exothermic Reactions / Processes

Reactant(s) → Product(s) + Heat OR Reactant(s) → Product(s) ΔH = −____ kJ Example: Methane undergoes combustion at constant pressure.

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) + 802.5 kJ or CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) ΔH = −803 kJ Example 1: It takes 116.2 kJ to form 84.0 L of NO2 (g) from its elements at 1.00 atm and 25.0°C. Determine

the molar heat of enthalpy for the formation of NO2 (g). Express the answer in proper ΔH notation. Example 2: Given that 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g) + 5317 kJ, calculate the change in

enthalpy when 28.2 g of butane is burned.

ΔH = 116.2 kJ V = 84.0 L P = 1.00 atm T = 25.0°C = 298.15 K R = 0.0821 K mol L atm ••

n = ? ΔHrxn = ?

n = RT PV

= ( )

( ) )K 15.298(0.0821 L 0.84)mta 00.1(

K mol L mta

• •

= 3.431636791 mol

ΔH = nΔHrxn ΔHrxn = n HΔ

= mol 13.43163679

kJ 2.116 ΔHrxn = 33.9 kJ/mol

N2 (g) + 2 O2 (g) → 2 NO2 (g) ΔH = 67.8 kJ (2 mol of NO2 in Eq) or ½ N2 (g) + O2 (g) → NO2 (g) ΔH = 33.9 kJ

ΔHrxn = mol 2

kJ 5317− = −2658.5 kJ/mol

n = /molg 58.14 g 2.28

= 0.4850361197 mol C4H10

ΔH = ?

(There are 2 moles of C4H10 in the chemical equation for 5317 kJ.)

ΔH = nΔHrxn = (0. 4850361197 mol)(−2658.5 kJ/mol)

ΔH = −1.29 × 10 3 kJ = −1.29 MJ (1 MJ = 1000 kJ)

Assignment 6.4 pg. 199 #21 to 28

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 139.

6.5 & 12.6: Calorimetry and Phase Changes Energy involved in Physical Change (Temperature or Phase Change): Heating Curve: - a graph of temeparture versus time as a substance is heated from a solid phase to a

gaseous phase. - when a substance is undergoing a phase change, its temperature remains at a constant

(the plateau on the heating curve) until all molecules aquired enough energy to overcome the intermoelcular forces nexessary. This is commonly refered to as the potential change of a subsatnce.

- when a substance is undergoing temperature change within a particular phase, it is refered to as kinetic change (because temeperature is also refered to as the average kinetic energy of a substance).

Molar Enthalpy of Fusion (ΔHfus): - the amount of heat needed to melt one mole of substance from solid

to liquid at its melting point (in kJ/mol). Molar Enthalpy of Vaporization (ΔHvap): - the amount of heat needed to evaporate one mole of substance

from liquid to gas at its boiling point (in kJ/mol) Molar Enthalpy of Sublimation (ΔHsub): - the amount of heat needed to sublime one mole of substance

from solid to gas (in kJ/mol). - because sublimation involves two phase change in one step, the

molar enthalpy of sublimation is the sum or the molar ethalphies of fusion and vapourization.

0°C

100°C

Time

T em

pe ra

tu re

Heating Curve of Water

Water

Ice

Steam

Phase Change Liquid / Gas

at Boiling Point

Phase Change Solid / Liquid

at Melting Point

ΔHsub = ΔHfus + ΔHvap

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 140. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Specific Heat (s): - the amount of heat (J or kJ) needed to change (1 g or 1 kg) of substance by 1°C or 1 K. - the stronger the intermolecular forces, the higher the specific heat capacity.

Heat Capacity (C): - the amount of heat (J or kJ) needed to a given amount of substance by 1°C or 1 K.

- usually used in a calorimeter (see section 6.5).

Physical Thermodynamic Properties of Some Common Substances (at 1.00 atm and 298.15 K)

Substance Melting

Point (°C) Boiling

Point (°C) Specific Heat [kJ/(kg • °C)]

ΔHfus (kJ/mol)

ΔHvap (kJ/mol)

Ice H2O(s) 0 ---- 2.03 6.01 ---- Water H2O(l) ---- 100 4.184 ---- 40.79 Steam H2O(g) ---- ---- 1.99 ---- ---- Ammonia NH3 (g) −77.73 −33.34 2.06 5.66 23.33 Methanol CH3OH (l) −98 64.6 2.53 3.22 35.21 Ethanol C2H5OH (l) −114.1 78.3 2.46 7.61 39.3 Aluminum Al (s) 660 2519 0.900 10.79 294 Carbon (graphite) C (s) 3338 4489 0.720 117 ---- Copper Cu (s) 1085 2562 0.385 12.93 300.4 Iron Fe (s) 1538 2861 0.444 13.81 340 Mercury Hg (l) −39 357 0.139 23.4 59.0

Example 1: What is the change in enthalpy involved when 36.04 g of water boils from liquid to gas at 100°C?

Physical Potential Change

q = nΔHfus q = nΔHvap q = nΔHsub q = Heat Change (J or kJ) n = moles ΔHfus = Molar Enthalpy of Fusion (kJ/mol) ΔHvap = Molar Enthalpy of Vaporization (kJ/mol) ΔHsub = Molar Enthalpy of Sublimation (kJ/mol) ΔHsub = ΔHfus + ΔHvap

Physical Kinetic Change

q = msΔT q = CΔT C = ms

q = Heat Change (J or kJ) m = mass (g or kg) ΔT = Change in Temperature (in °C or K) s = Specific Heat [J/(g • °C) or kJ/(kg • °C) or J/(g • K) or kJ/(kg • K)]

C = Heat Capacity [J/°C or kJ/°C or J/K or kJ/K)]

Since this question involves phase change (vaporization) only, we need to use q = nΔHvap.

q = nΔHvap q = (2.000 mol)(40.79 kJ/mol) q = 81.58 kJ

ΔHvap = 40.79 kJ/mol

n = /molg 18.02 g 04.36

= 2.000 mol H2O

q = ?

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 141.

Example 2: How much energy is needed to heat 100.0 g of water from 20.0°C to 80.0°C? Example 3: What is the total energy needed to sublime 40.0 g of solid ammonia to gaseous ammonia? (ΔHfus = 5.66 kJ/mol; ΔHvap = 23.33 kJ/mol) Example 4: What is the total energy needed to heat 18.02 g of water at 80.0°C to steam at 115°C? Molar Enthalpy of Solution (ΔHsoln): - the amount of heat needed to dissolve 1 mole of substance.

- it can easily be found using a constant-pressure calorimeter. - for molecular solutes, it involves overcoming the intermolecular

forces of the solute and solvent. This is followed by the hydration process as the solute and solvent molecules come together.

- for ionic solutes, it first involves overcoming the lattice energy. Finally, the ions and solvent molecules come together during hydration process.

Since this question involves temperature (kinetic) change only, we need to use q = mcΔT.

q = msΔT q = (100.0 g)(4.184 J/(g • °C))(60.0°C) = 25104 J q = 2.51 × 104 J = 25.1 kJ

s = 4.184 J/(g • °C) m = 100.0 g H2O ΔT = 80.0°C − 20.0°C = 60.0°C q = ?

For this question, we have two kinetic changes (water and steam) and one phase change (vaporization).

qtotal = mswΔTw + nΔHvap + mssΔTs (water) (vaporization) (steam)

qtotal = (0.01802 kg)(4.184 kJ/(kg • °C))(20.0°C) + (1.000 mol)(40.79 kJ/mol) + (0.01802 kg)(1.99 kJ/(kg • °C))(15°C)

qtotal = 1.5079136 kJ + 40.79 kJ + 0.537897 kJ qtotal = 42.8 kJ

m = 18.02 g H2O = 0.01802 kg H2O

n = /molg 18.02 g 02.18

= 1.000 mol H2O

swater = 4.184 kJ/(kg • °C) ΔTwater = 100.0°C − 80.0°C = 20.0°C ΔHvap = 40.79 kJ/mol ssteam = 1.99 kJ/(kg • °C) ΔTsteam = 115°C − 100°C = 15°C qtotal = ?

Since this question involves phase change (sublimation) only, we need to use q = nΔHsub.

q = nΔHsub q = (2.34741784 mol)(28.99 kJ/mol) q = 68.1 kJ

ΔHsub = ΔHfus + ΔHvap ΔHsub = 5.66 kJ/mol + 23.33 kJ/mol ΔHsub = 28.99 kJ/mol

n = /molg 17.04 g 0.40

= 2.34741784 mol NH3

q = ?

Enthalpy of Solution

ΔH = nΔHsoln ΔH = Change in Enthalpy n = moles ΔHsoln = Molar Enthalpy of Solution (kJ/mol)

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 142. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Example 5: A cold pack consists of 40.0 g of NH4NO3 is dissolved in water. How much energy is absorbed or released into its surrounding if the ΔHsoln is 26.2 kJ/mol?

Example 6: 12.9 kJ of heat is released when CaCl2 is dissolved in water, find the mass of CaCl2 dissolved

if the molar enthalpy of solution of CaCl2 is −82.8 kJ/mol. Heat of Dilution: - the amount of heat associated during the dilution process of a solution, - a solution that has a endothermic heat of solution (ΔHsoln > 0) will have an endothermic heat of dilution. - a solution that has a exothermic heat of solution (ΔHsoln < 0) will have an exothermic heat of dilution. (Examples: When NaOH (s) is dissolved to become NaOH (aq), it releases heat. During the dilution process,

the solution gets even warmer. This is because more intermolecular forces form between the added water molecules and the ions present. More intermolecular forces or bonds form mean more heat is released. Similarly, when concentrated H2SO4 (aq) is diluted, more intermolecular forces are made and the process releases a lot of heat. Hence, we always add concentrated to water slowly with constant stirring.)

Calorimetry: - uses the conservation of energy (Heat Gained = Heat Lost) to measure calories (old unit

of heat: 1 cal = 4.184 J). - physical calorimetry involves the mixing of two systems (one hotter than the other) to reach

some final temperature. - the key to do these problems is to identify which system is gaining heat and which one

is losing heat. Example 7: Hot water at 90.0°C is poured into 100. g of cold water at 10.0°C. The final temperature of the

mixture is 70.0°C. Determine the mass of the hot water.

Heat Lost = Heat Gained (hot water, kinetic) (cold water, kinetic) mhwsΔThw = mcwsΔTcw

mhw = hw

cwcw

T Tm

Δ Δ

= ( )( )

( )C0.20 C0.60g .100

o

o

mhw = 300. g

m cold water = 100. g s = 4.184 J/(g • °C) ΔT hot water = 90.0°C − 70.0°C = 20.0°C ΔT cold water = 70.0°C − 10.0°C = 60.0°C m hot water = ?

ΔHsoln = +26.2 kJ/mol (ΔHsoln > 0; endothermic) (Heat is absorbed from the surrounding)

n = /molg 80.06 g 0.40

= 0.499625281 mol NH4NO3

ΔH = ?

ΔH = nΔHsoln ΔH = (0.499625281 mol)(26.2 kJ/mol)

ΔH = 13.1 kJ absorbed

ΔHsoln = −82.8 kJ/mol (ΔHsoln < 0; exothermic) (Heat is released into the surrounding) ΔH = −12.9 kJ M = 110.98 g/mol CaCl2

n = ? m = ?

ΔH = nΔHsoln n = solnH

H Δ Δ

= /molJk 8.82 Jk 9.12

− −

n = 0.1557971014 mol m = nM = (0.1557971014 mol)(110.98 g/mol)

m = 17.3 g

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 143.

Example 8: A 1500. g of iron metal at 330.°C is dropped into a beaker of 1000. g of water at 25.0°C. What will be the final temperature?

Energy involved in Chemical Change (Chemical Reaction):

Molar Heat of Combustion (ΔHcomb): - the amount of heat released when one mole of reactant is burned with excess oxygen.

- the reaction is often exothermic and therefore ΔHcomb < 0. - we often use a constant-volume calorimeter (or bomb calorimeter)

to determine ΔHcomb due to its well-insulated design. It is calibrated for the heat capacity of the calorimeter, Ccal, before being use for to calculate ΔHcomb of other substances. The sample is measured and burned using an electrical ignition device. Water is commonly used to absorb the heat generated by the reaction. The temperature of the water increases, allowing us to find the amount of heat generated. By applying the law of conservation of energy, we can then calculate the ΔHcomb of the sample.

Heat Lost = Heat Gained (iron, kinetic) (water, kinetic) mFesFeΔTFe = mwswΔTw (1500. g)(0.444 J/(g • °C))(330.°C − Tf) = (1000. g)(4.184 J/(g • °C))(Tf − 25.0°C) 219780 − 666Tf = 4184Tf − 104600 −666Tf − 4184Tf = −104600 − 219780 −4850Tf = −324380

Tf = 4850

324380 − −

Tf = 66.9°C

Tf = ? mFe = 1500. g sFe = 0.444 J/(g • °C) ΔTFe = 330.°C − Tf mwater = 1000. g swater = 4.184 J/(g • °C) ΔTwater = Tf − 25.0°C

Schematic of a Bomb Calorimeter

Chemical Combustion Calorimetry

Heat Lost = Heat Gained (Combustion Reaction) (water, kinetic)

nsampleΔHcomb = CcalΔT (if bomb calorimeter is used) or nsampleΔHcomb = mwswΔT (if the heat absorbed by the calorimeter itself is ignored)

Enthalpy of Combustion

ΔH = nΔHcomb ΔH = Change in Enthalpy n = moles ΔHcomb = Molar Heat of Combustion (kJ/mol)

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 144. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Example 9: Octane, C8H18(l) was burned completely to CO2 (g) and H2O (l) in a bomb calorimeter. The following are the observations of the experiment.

Mass of C8H18 (l) burned 32.65 g Initial Temperature of Calorimeter and Water 16.50°C Final Temperature of Calorimeter and Water 77.30°C Heat Capacity of Calorimeter 24.70 kJ/°C

a. Determine the experimental molar heat of combustion of C8H18 (l). b. The theoretical ΔHcomb for C8H18 (g) is −5470.1 kJ/mol, calculate the % error of this experiment.

Molar Heat of Reaction (ΔHrxn): - the amount of heat released when one mole of reactant undergoes

various chemical changes. - examples are ΔHcomb, ΔHneut (neutralization), ΔHion (ionization).

Constant-Pressure Calorimeter (or Styrofoam Calorimeter) - commonly used to determine ΔHneut, ΔHion,

ΔHfus, ΔHvap, ΔHrxn of non-combustion reaction. First, the sample’s mass is measured. Water is commonly used to absorb or provide the heat for the necessary change. The initial and final temperatures of the water are recorded, allowing us to find the amount of heat change. By applying the law of conservation of energy, we can then calculate the necessary molar enthalpy of change.

Heat Lost = Heat Gained (Combustion Reaction) (water, kinetic) noctaneΔHcomb = CcalΔT

ΔHcomb = octanen

TCcalΔ = ( )( ) ( )mol 2857517942.0

C80.60CkJ/ 70.24 o o

ΔHcomb = −5255 kJ/mol (negative sign is added because ΔHcomb is exothermic)

a. moctane = 32.65 g

noctane = /molg 114.26 g 65.32

= 0.2857517942 mol C8H18

Ccal = 24.70 kJ/°C ΔT = Tf − Ti = 77.30°C − 16.50°C ΔT = 60.80°C

ΔHcomb = ?

b. % error = lTheoretica

alExperiment lTheoretica − × 100%

% error = ( ) ( )

( )kJ/mol 1.5470 kJ/mol 5255kJ/mol 1.5470

− −−−

× 100% % error = 3.932%

(The small % error means the bomb calorimeter was a good heat insulator.)

Enthalpy of Chemical Reactions

ΔH = nΔHrxn ΔH = nΔHneut ΔH = nΔHion ΔH = nΔHcomb

ΔH = Change in Enthalpy n = moles ΔHrxn = Molar Heat of Reaction (kJ/mol) ΔHneut = Molar Heat of Neutralization (kJ/mol) ΔHion = Molar Heat of Ionization (kJ/mol) ΔHcomb = Molar Heat of Combustion (kJ/mol)

A Simple Styrofoam Calorimeter

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 145.

Example 10: When 250.0 mL of HNO3 (aq) at 0.300 mol/L is reacted with 250.0 mL of KOH (aq) at

0.300 mol/L, the temperature of the final mixture reached 28.60°C from 22.10°C. Determine the molar heat of neutralization between HNO3 (aq) and KOH (aq).

6.6: Standard Enthalpies of Formation and Reaction

Standard State: - standard conditions of 1 atm and 25°C. It is denote by a superscript “o”.

Standard Molar Enthalpy of Formation (ΔHof): - the amount of heat required / given off to make 1 mole of compound from its elemental components under standard conditions.

- the Molar Heat of Formation of ALL ELEMENTS is 0 kJ. - the state of the compound affects the magnitude of Hf.

(H2O (g) has ΔHof = −241.8 kJ/mol; H2O (l) has ΔHof = −285.8 kJ/mol)

(See Appendix 2 on pg. A2 to A5 in the Chang 9th ed. Chemistry textbook for a list of ΔH°f)

Assignment 12.6 pg. 421−423 #69, 71, 72, 78, 112 6.5 pg. 199− 200 #29 to 38

Non-Combustion Calorimetry

Heat Gained /Lost = Heat Lost / Gained (Non-Combustion Change) (water, kinetic)

nsampleΔHrxn = mwswΔT (non-combustion chemical change) nsampleΔHfus = mwswΔT (physical change - molar heat of fusion)

nsample ΔHvap = mwswΔT (physical change - molar heat of vapourization) msamplessampleΔTsample = mwswΔTw (physical change – specific heat)

The reaction is acid-base neutralization and will produce water as a result.

HNO3 (aq) + KOH (aq) → H2O (l) + KNO3 (aq)

Assuming KNO3 (aq) does not affect the specific heat of water, we would have 500.0 mL of water produced. Since we have equal moles of acid and base (0.300 mol/L × 0.2500 L = 0.0750 mol), the ΔHneut for HNO3 would be the same as KOH. Heat Lost = Heat Gained (Neutralization) (water, kinetic)

nacidΔHneut = mwswΔT

ΔHneut = acidn

Tsm ww Δ

ΔHneut = ( )( )( )

( )mol 0750.0 C50.6C gkkJ/( 184.4gk 5000.0 o o•

ΔHneut = −181 kJ/mol

(negative sign is added due to increased surrounding temperature - an exothermic reaction)

nacid = nbase = 0.0750 mol mw = 0.5000 kg (500.0 mL is produced) sw = 4.184 kJ/(kg • °C)

ΔT = 28.60°C − 22.10°C ΔT = 6.50°C

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 146. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Example 1: Find the standard molar enthalpy of formation for table salt given that its formation reaction,

2 Na (s) + Cl2 (g) → 2 NaCl (s) + 822 kJ, at standard conditions.

Example 2: What is the amount of heat absorbed / released when 100. g of CO2 (g) is produced from its elements (CO2 has ΔHof = −393.5 kJ/mol)?

Example 3: Iron (III) oxide, rust, is produced from its elements, iron and oxygen. What is the mass of rust produced when 1.20 MJ is released when iron is reacted with oxygen (ΔHof = −822.2 kJ/mol for Fe2O3)?

Example 4: Calculate the standard molar enthalpy of formation of silver (I) oxide when 91.2 g of Ag2O is

produced from its elements and 12.2 kJ of heat is released from the process. Standard Molar Enthalpy of Reaction (ΔHorxn): - the amount of heat involved when 1 mol of a particular

product is produced or 1 mol of a particular reactant is consumed under standard conditions of 1 atm and 25°C.

- it is equal to the difference between of all enthalpies of products and all enthalpies of reactants.

- if the reaction is a combustion, it is called the molar heat of combustion.

Standard Enthalpy of Formation (Chemical)

ΔH = nΔHof ΔH = Change in Enthalpy n = moles ΔHof = Standard Molar Enthalpy of Formation (kJ/mol)

ΔH = −822 kJ n = 2 mol of NaCl

ΔHof = ?

ΔH = nΔHof

ΔHof = n HΔ

= mol2

kJ 822− ΔHof = −411 kJ/mol

n = /molg 44.01 g .100

= 2.272210861 mol CO2

ΔHof = −393.5 kJ/mol

ΔH = ?

ΔH = nΔHof ΔH = (2.272210861 mol)(−393.5 kJ/mol)

ΔH = −894 kJ (894 kJ is released)

ΔH = −1.20 MJ = −1.20 × 103 kJ (exothermic) ΔHof = −822.2 kJ/mol M = 159.7 g/mol Fe2O3

n = ? m = ?

ΔH = nΔHof

n = o fH

H Δ Δ

= /molJk 822.2

Jk 1020.1 3

− ×−

= 1.459498905 mol

m = nM = (1.459498905 mol)(159.7 g/mol) m = 233 g

n = /molg 231.74 g 2.91

= 0.3935444895 mol Ag2O

ΔH = −12.2 kJ (exothermic)

ΔHof = ?

ΔH = nΔHof

ΔHof = n HΔ

= mol3935444895.0

kJ 2.12−

ΔHof = −31.0 kJ/mol

(Note: In hydrocarbon combustion, assume all products are gaseous

unless otherwise stated.)

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 147.

Example 5: Propane is a clean burning fossil fuel that is widely used in outdoor barbecue.

a. Calculate the standard molar enthalpy of combustion for propane. (ΔHof C3H8 = −103.9 kJ/mol; ΔHof CO2 = −393.5 kJ/mol; ΔHof H2O (g) = −241.8 kJ/mol)

b. Draw its potential energy diagram. c. How much energy will be absorbed or released when 15.0 g or propane is burned?

Example 6: Find the amount of heat released when 34.9 g of butane gas is burned at standard conditions. (ΔHof C4H10 = −124.7 kJ/mol; ΔHof CO2 = −393.5 kJ/mol; ΔHof H2O (g) = −241.8 kJ/mol)

b. Potential Energy Diagram

Direct Method to determine Standard Enthalpy of Reaction

ΔHorxn = ΣHoproducts − ΣHoreactants

ΔHorxn = Change in Enthalpy of Reaction ΣHoproducts = Sum of Heat of Products (from all nΔHof of products) ΣHoreactants = Sum of Heat of Reactants (from all nΔHof of reactants)

a. We have to first write out a balance equation for the combustion of propane.

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) ΔHof : −103.9 kJ/mol 0 kJ/mol −393.5 kJ/mol −241.8 kJ/mol

ΔHorxn = ΣHoproducts − ΣHoreactants ΔHorxn = [3 mol (−393.5 kJ/mol) + 4 mol (−241.8 kJ/mol)] − [1 mol (−103.9 kJ/mol) + 5 mol (0 kJ/mol)] ΔHorxn = [−2147.7 kJ] − [−103.9 kJ] = −2043.8 kJ

ΔHorxn = −2044 kJ/mol of C3H8 burned

c. From part a., for every 1 mol of C3H8 burned, 2043.8 kJ is released.

n = /molg 44.10 g 0.15

= 0.3401360544 mol C3H8

ΔHorxn = −2043.8 kJ/mol ΔH = ?

ΔH = nΔHorxn ΔH = (0. 3401360544 mol)(−2043.8 kJ/mol)

ΔH = −695 kJ (695 kJ is released)

ΔH = nΔHrxn ΔH = (0.6002751978 mol)(−2658.3 kJ/mol) = −1595.7 kJ

ΔH = −1.60 MJ (1.60 MJ is released)

n = /molg 58.14 g 9.34

= 0.6002751978 mol C4H10

ΔHrxn = −2658.3 kJ/mol ΔH = ?

We have to first write out a balance equation for the combustion of butane.

2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g) (We have to divide all coefficients by 2 because we are calculating ΔHrxn per mol of butane burned.)

C4H10 (g) + 2

13 O2 (g) → 4 CO2 (g) + 5 H2O (g)

ΔHf : −124.7 kJ/mol 0 kJ/mol −393.5 kJ/mol −241.8 kJ/mol

ΔHorxn = ΣHoproducts − ΣHoreactants

ΔHorxn = [4 mol (−393.5 kJ/mol) + 5 mol (−241.8 kJ/mol)] − [1 mol (−124.7 kJ/mol) + 2

13 mol (0 kJ/mol)]

ΔHorxn = [−2783 kJ] − [−124.7 kJ] ΔHrxn = −2658.3 kJ/mol of C4H10 burned

Po te

nt ia

l E ne

rg y

Reaction Pathway

C3H8 (g) + 5 O2 (g)

3 CO2 (g) + 4 H2O (g)

ΔHorxn = −2044 kJ

−2147.7 kJ

−103.9 kJ

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 148. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Example 7: When 10.02 g of liquid heptane is burned in the reaction vessel of a calorimeter, 1.50 L of water around the vessel increased its temperature from 20.0°C to 85.0°C. Ignoring the metallic material of the calorimeter,

a. determine the experimental standard molar enthalpy of combustion heptane. b. find the theoretical standard molar enthalpy of combustion of heptane.

(ΔH°f heptane = −224.2 kJ/mol) c. explain why the experimental ΔH°rxn is different than its theoretical counterpart.

c. Some of the possible reasons why experimental ΔH°rxn (−4.08 MJ) is different than the theoretical ΔH°rxn (−4.46 MJ)

Some of the heat released by the reaction is absorbed by the metal calorimeter itself. Thus, the temperature gained by the water is not an exact reflection of the energy lost by the combustion.

The calorimeter is not a closed system. Heat might escape into the surrounding. Even if the system is closed, the buildup of gases from the reaction would increase pressure

and volume. Hence, some of the energy produced from the reaction is used to do work by the system. Thereby, lowering the heat available to warm the water.

a. We use the conservation of heat to calculate experimental ΔHrxn. nΔH°rxn = mwswΔT

ΔH°rxn = n

Tsm ww Δ = ( ) ( )( )( )

( )mol 0999700688.0 C0.65C gkkJ/ 184.4gk 50.1 oo•

ΔH°rxn = 4080.621377 kJ/mol (released) Experimental ΔH°rxn = −4.08 MJ/mol of C7H16 burned

n = /molg 100.23 g 02.10

n = 0.0999700688 mol C7H16

mwater = 1.50 kg (1 kg = 1 L of water) ΔT = 85.0°C − 20.0°C = 65.0°C swater = 4.184 kJ / (kg • °C)

ΔH°rxn = ?

b. To find theoretical ΔH°rxn for the combustion of heptane, we have to use the direct method.

C7H16 (l) + 11 O2 (g) → 7 CO2 (g) + 8 H2O (g) ΔHf : −224.2 kJ/mol 0 kJ/mol −393.5 kJ/mol −241.8 kJ/mol

ΔH°rxn = ΣH°products − ΣH°reactants ΔH°rxn = [7 mol (−393.5 kJ/mol) + 8 mol (−241.8 kJ/mol)] − [1 mol (−224.8 kJ/mol)] ΔH°rxn = [−4688.9 kJ] − [−224.8 kJ] = −4464.1 kJ

Theoretical ΔH°rxn = −4.46 MJ/mol of C7H16 burned

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 149.

Example 8: HCOOH (g) were completely burned to CO2 (g) and H2O (l) in a calorimeter. The following are the observation of the experiment.

Mass of HCOOH(g) burned 9.22 g Initial Temperature of Calorimeter and Water 21.5°C Final Temperature of Calorimeter and Water 37.3°C Heat Capacity of Calorimeter and Water 3.20 kJ/°C

a. Determine the experimental molar enthalpy of formation of HCOOH (l) assuming standard conditions. (ΔHof CO2 = −393.5 kJ/mol; ΔHof H2O (l) = −285.8 kJ/mol)

b. If the theoretical ΔH°f for HCOOH (g) is −363 kJ/mol, calculate the % error of this experiment.

a. We use the conservation of heat to calculate experimental ΔHrxn. nΔH°rxn = CcalΔT

ΔH°rxn = n

TCcalΔ = ( )( ) ( )mol 2003041495.0

C8.15CkJ/ 20.3 oo

ΔH°rxn = 252.4161388 kJ/mol (released) Experimental ΔH°rxn = −252 kJ/mol of HCOOH burned

n = /molg 46.03 g 22.9

n = 0.2003041495 mol HCOOH

Ccal = 3.20 kJ/°C ΔT = 37.3°C − 21.5°C = 15.8°C

ΔH°rxn = ?

Next, we use the direct method to find the ΔH°f of HCOOH.

HCOOH (l) + ½ O2 (g) → CO2 (g) + H2O (l) ΔH°f : ? kJ/mol 0 kJ/mol −393.5 kJ/mol −285.8 kJ/mol

ΔH°rxn = ΣH°products − ΣH°reactants −252.4161388 kJ = [1 mol (−393.5 kJ/mol) + 1 mol (−285.8 kJ/mol)] − [1 mol (ΔH°f)] −252.4161388 kJ = [−679.3 kJ] − [1 mol (ΔH°f)] ΔH°f = −679.3 kJ + 252.4161388 kJ = −426.8838612 kJ

Experimental ΔH°f of HCOOH = −427 kJ/mol

b. % error = %100 alTheroretic

alExperiment lTheoretica ×

−

% error = ( ) ( )

( ) %100

kJ 363 kJ 427 kJ 363

× −

−−− % error = 17.6%

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 150. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Hess’s Law: - the indirect method of obtaining overall ΔH°rxn of a net reaction by the addition of ΔH°rxn of a series of reactions.

- when adding reactions, compare the reactants and products of the overall net reaction with the intermediate (step) reactions given. Decide on the intermediate reactions that need to be reversed and / or multiply by a coefficient, such that when added, the intermediate products will cancel out perfectly yielding the overall net reaction.

- if a particular reaction needs to be reversed (flipped), the sign of the ΔH for that reaction will also need to be reversed.

- if a coefficient is used to multiply a particular reaction, the ΔH for that reaction will also have to multiply by that same coefficient.

(Check out Hess’s Law Animation at http://intro.chem.okstate.edu/1314F00/Lecture/Chapter6/Hesslaw2.html) Example 9: Calculate ΔH°rxn for the reaction N2 (g) + 2 O2 (g) → 2 NO2 (g), when the following reactions are

given. N2 (g) + O2 (g) → 2 NO (g) ΔH°rxn = 180 kJ

2 NO2 (g) → 2 NO (g) + O2 (g) ΔH°rxn = 112 kJ Example 10: Determine the ΔH°rxn for the reaction S (s) + O2 (g) → SO2 (g), when the following reactions are

given.

S (s) + 2 3

O2 (g) → SO3 (g) ΔH°rxn = −395.2 kJ

2 SO2 (g) + O2 (g) → 2 SO3 (g) ΔH°rxn = −198.2 kJ

Note that 2 NO2 in the net reaction is on the product side, whereas 2 NO2 in the second reaction is on the reactant side. Hence, we need to reverse the second reaction and its sign of the ΔH°rxn.

N2 (g) + O2 (g) → 2 NO (g) ΔH°rxn = 180 kJ (Flipped) 2 NO (g) + O2 (g) → 2 NO2 (g) ΔH°rxn = −112 kJ

N2 (g) + 2 O2 (g) → 2 NO2 (g) ΔH°rxn = + 68 kJ

a. SO2 in the net reaction is on the product side, whereas 2 SO2 in the second reaction is on the reactant side. Hence, we need to reverse the second reaction and its sign of the ΔH°rxn.

b. There is only 1 SO2 in the net reaction, whereas there are 2 SO2 in the second reaction. Therefore the second reaction and its ΔH°rxn need to be multiply by the coefficient of ½.

S (s) + 2 3

O2 (g) → SO3 (g) ΔH°rxn = −395.2 kJ

(Flipped and × ½) ½ (2 SO3 (g) → 2 SO2 (g) + O2 (g)) ΔH°rxn = ½(+198.2 kJ)

S (s) + 2 3

O2 (g) → SO3 (g) ΔH°rxn = −395.2 kJ

SO3 (g) → SO2 (g) + ½ O2 (g) ΔH°rxn = 99.1 kJ

S (s) + O2 (g) → SO2 (g) ΔH°rxn = −296.1 kJ

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 151.

Example 3: Find the ΔH°rxn for the overall reaction of 2 N2 (g) + 5 O2 (g) → 2 N2O5 (g), when the following reactions are given.

H2 (g) + ½ O2 (g) → H2O (l) ΔH°rxn = −285.8 kJ N2O5 (g) + H2O (l) → 2 HNO3 (l) ΔH°rxn = − 76.6 kJ

½ N2 (g) + 2 3

O2 (g) + ½ H2 (g) → HNO3 (l) ΔH°rxn = −174.1 kJ

6.7: Present Sources of Energy and New Energy Sources Fossil Fuel: - hydrocarbon fuels that came from fossils of decayed organisms. 1. Natural Gas: - fossil fuel that consists of mainly small alkanes (80% methane, 10% ethane, 4%

propane, 2% butane, 4% nitrogen). - usually burns efficiently (complete combustion).

Complete Combustion: - where the products of combustion are carbon dioxide and water vapour only. -characterized by a blue flame.

Example: Propane burns completely. C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

Assignment 6.6 pg. 200−202 #39 to 42, 45 to 49, 51, 52, 54 to 56, 58, 60 to 64, 67, 68, 74, 76, 80, 81

a. 2 N2O5 in the net reaction is on the product side, whereas N2O5 in the second reaction is on the reactant side. Hence, we need to reverse the second reaction and its sign of the ΔH°rxn.

b. There are 2 N2O5 in the net reaction, whereas there is only 1 N2O5 in the second reaction. Therefore the second reaction and its ΔH°rxn need to be multiply by the coefficient of 2.

c. There are 2 N2 in the next reaction on the reactant side. Since ½ N2 is on the reactant side of the third reaction, we need to multiply the third reaction and its ΔH°rxn by the coefficient of 4.

d. In order for H2O to cancel from the first and second reactions, we have to multiple the first reaction by 2 and flipped. This is because H2O in the second reaction has also flipped and has been multiplied by 2.

(Flipped and × 2) 2 (H2O (l) → H2 (g) + ½ O2 (g)) ΔH°rxn = 2(+285.8 kJ) (Flipped and × 2) 2 (2 HNO3 (l) → N2O5 (g) + H2O (l)) ΔH°rxn = 2(+ 76.6 kJ)

(× 4) 4 (½ N2 (g) + 2 3

O2 (g) + ½ H2 (g) → HNO3 (l)) ΔH°rxn = 4(−174.1 kJ)

2 H2O (l) → 2 H2 (g) + O2 (g) ΔH°rxn = +571.6 kJ 4 HNO3 (l) → 2 N2O5 (g) + 2 H2O (l) ΔH°rxn = +153.2 kJ

2 N2 (g) + 6 O2 (g) + 2 H2 (g) → 4 HNO3 (l) ΔH°rxn = −696.4 kJ

2 N2 (g) + 5 O2 (g) → 2 N2O5 (g) ΔH°rxn = + 28.4 kJ

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 152. Copyrighted by Gabriel Tang B.Ed., B.Sc.

Incomplete Combustion: - where the main product of combustion is carbon monoxide, along with carbon dioxide and water vapour.

- happens when carbon particles started to form during combustion and deposited as soot as they cooled, or when there is insufficient oxygen.

- characterized by a yellow flame.

Example: Incomplete combustion of Propane. C3H8 (g) + 4 O2 (g) → 2 CO (g) + CO2 (g) + 4 H2O (g) 2. Petroleum (Crude Oil): - fossil fuels that consist mainly of heavier alkanes along with small amounts

of aromatic hydrocarbons, and organic compounds that contain sulfur, oxygen and nitrogen.

- gasoline is composed of 40% of crude oil, whereas natural gas is composed of only 10%.

Fractional Distillation: - a method of heating

crude oil in a tall column to separate its different components by their different boiling points.

- lighter alkanes in the natural gas will rise up to the top of the column because of their low boiling points.

- the heavier, fuel and lubricating oils will boil off at the bottom of the column due to their high boiling points.

Petroleum Refining: - a process to isolate different types of fuel from crude oil using fractional distillation or cracking.

Cracking: - a chemical process whereby bigger alkanes are broken up into smaller ones using a catalyst

and heat. - since gasoline and natural gas only consists of 50% of crude oil, cracking is necessary to

convert heavier fuel to more common fuel used in today’s world.

Example: The Cracking of Hexadecane. C16H34 + 2 H2 ⎯⎯⎯⎯ →⎯ heat andcatalyst C8H18 + C8H18

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 153.

Reforming: - a chemical process where smaller alkanes are combined together and hydrogen is removed to form heavier alkanes or changed unbranched alkanes into branched alkanes.

- branched alkanes are easier to burn and has a higher octane value in gasoline. (isooctane or 2,2,4-trimethylpentane has the best octane rating – assigned as 100)

3. Coal: - a carbon-based mineral consists of very dense hydrocarbon ring compounds with high molar

masses. - leaves a lot of soot and burns incompletely. - usually contains 7% sulfur and when combusted with oxygen gives off SO2 and SO3, which is

the main source of air pollution and acid rain. Greenhouse Effect: - the emission of greenhouses gases that traps more of the sun’s radiant (heat) energy

in the atmosphere than it occurs naturally. Greenhouses Gases: - man-made and naturally occur gases that contribute to the Greenhouse Effect. 1. Carbon dioxide (CO2) is released to the atmosphere when solid waste, fossil fuels (oil, natural gas, and

coal), and wood and wood products are burned.

2. Methane (CH4) is emitted during the production and transport of coal, natural gas, and oil. Methane emissions also result from the decomposition of organic wastes in municipal solid waste landfills, and the raising of livestock.

3. Nitrous oxide (N2O) is emitted during agricultural and industrial activities, as well as during combustion of solid waste and fossil fuels.

Unit 4: Thermochemistry and Nuclear Chemistry Honour Chemistry

Page 154. Copyrighted by Gabriel Tang B.Ed., B.Sc.

4. Hydrofluorocarbons (HFCs), Perfluorocarbons (PFCs), and Sulfur Hexafluoride (SF6) are very powerful greenhouse gases that are not naturally occurring that are generated in a variety of industrial processes.

Each greenhouse gas differs in its ability to absorb heat in the atmosphere. HFCs and PFCs are the most heat-absorbent. Methane traps over 21 times more heat per molecule than carbon dioxide, and nitrous oxide absorbs 270 times more heat per molecule than carbon dioxide. Often, estimates of greenhouse gas emissions are presented in units of millions of metric tons of carbon equivalents (MMTCE), which weights each gas by its GWP value, or Global Warming Potential. (Information from US. EPA)

- Automobiles and Major Transportations account for 34% of CO2 emissions globally (Power Plants

contributes 33%; Major Industries and Home Heating contribute the remaining 33%). - Presently 89% of Energy Productions involve the burning of Fossil Fuels (Coal, Petroleum, Natural

Gas and Biomass). - Heat and Electricity generated from combustion of fossil fuel is at most 30% efficient.

(Data from University of Michigan: http://www.umich.edu/~gs265/society/greenhouse.htm) The Environmental Effect of Using Fossil Fuel: (Greenhouse Effect)

1. Global Warming: - the warming of global temperature due to an increased of greenhouse gases in the atmosphere.

2. Rise of Water Level: - low-lying islands and coastal area are endangered as polar icecaps melt due to the rise of temperature as a result of the greenhouse effect.

3. Unpredicted and Erratic Climate: - greenhouse effect is related to droughts and dry whether in many parts of the world.

4. Deforestation: - another cause of an increased in CO2 level in the atmosphere. As forests disappeared, there is a lack of plants to absorb carbon dioxide using photosynthesis.

- also causes mud and landslides, demineralization of the soil, lost animal habitats and extinction, destruction of entire ecosystems. Plants that may have important medicinal values can also be destroyed.

Alternate Energy Sources without the Emission of Greenhouse Gas 1. Solar Energy: - the most efficient energy source where energy from the sun is converted directly to

electricity through the use of photovoltaic cells (solar panels) or heat using high efficient insulated glass and an effective water heating system.

- technology exists but fairly expensive; requires many solar panels to generate adequate amount of electricity.

2. Wind Power: - the use of wind turbines to generate electricity.

- very efficient and extremely economical, but location specific and not very reliable when there is no wind.

- can disrupt migratory routes of birds (they get caught in the turbine), aesthetic problems for the various landscapes.

Honour Chemistry Unit 4: Thermochemistry and Nuclear Chemistry

Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 155.

3. Geothermal Power: - the use of underground steam to generate electricity. - very efficient and reliable, but location specific. - geothermal power is widely use in Iceland where it is sitting on the Atlantic ridge

and there are lots of hot springs. 4. Tidal Power: - the use of tidal current to generate electricity.

- very efficient and somewhat reliable, but location specific. - tidal power involves placing electric turbines at a narrow mouth of a channel where

normal tides can cause bigger current and quick rise in water levels. It is being used in the Bay of Fundy at Nova Scotia, Canada and Kvalsund at the Arctic tip of Norway.

- tidal power can sometimes disrupt migratory routes of marine species. 5. Hydroelectricity: - the use of dam and reservoir to turn electric turbines as water falls from a higher

level to the spillway (potential energy converted to kinetic energy to electricity). - very efficient and no emission of greenhouse gas. - location specific and very expensive to built. The reservoir flooding can destroy

ecological habitats and force migrations of people from towns and villages (Aswan Dam in Egypt and the Three Gorges Dam in China displaced thousands of people and submerged ancient cities). The presence of the dam can disrupt aquatic migratory routes as well.

- dams have a limited life span (the collection of silt and mud at the bottom of the dam has to be clear periodically to maintain the structural integrity of the dam). Dams can burst during earthquakes or poor maintenance. Flash flooding of towns along spillway is always a danger.

6. Hydrogen Fuel: - burning hydrogen to form water and generate heat and electricity.

- very efficient and zero pollution. - hydrogen is very explosive and technologies are still needed for supplying and storing

hydrogen safely in automobiles and homes.