Energy & Chemical Reactions

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Chapter 6 Energy Chemical Reactions
General Chemistry I T. Ara
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A. Energy

• Thermodynamics science of heat, work the
  transformations of one to the other
• Thermochemistry the study of energy changes that
  occur during physical processes chemical
  reactions
• Energy the capacity to do work
• All energy can be classified as either kinetic or
  potential

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2. Kinetic Potential Energy

• Kinetic Energy energy that something has because
  it is moving
• Macroscale mechanical energy objects in motion
  (people, cars, baseballs, etc.)
• Nanoscale thermal energy nanoscale objects in
  motion (atoms, molecules, ions, etc.)

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2. Kinetic Potential Energy

• Potential Energy energy that something has as a
  result of its position and some force that is
  capable of changing that position
• Macroscale gravitational energy (E mgh)
• Nanoscale electrostatic energy (coulombic
  attraction between anions cations) chemical
  potential energy (attractions among atomic nuclei
  and electrons in molecules)

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2. Kinetic Potential Energy

Gravitational potential energy being converted
into kinetic energy.
The chemical potential energy stored in food
molecules being released (converted into thermal
energy).
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a) Units of Energy

• The SI Unit for energy is the joule (J).
• The joule is an SI derived unit.
• 1 joule 1 kg?m2
• s2
• It is frequently more practical to use a larger
  unit the kilojoule (kJ)
• 1 kJ 103 J

(Energy Force x Distance)
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a) Units of Energy

• Another common unit of energy is the calorie.
• 1 cal 4.184 J
• Energy changes related to most chemical processes
  are reported in kilocalories (kcal) or kilojoules
  (kJ).
• 1 kcal 1000 cal 4.184 kJ 4184 J

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a) Units of Energy

• The label on this Australian packet of Equal
  indicates that it supplies 16 kJ of nutritional
  energy. How many American Calories
  (kilocalories) are in this packet of Equal?

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a) Units of Energy

• The label on this Australian packet of Equal
  indicates that it supplies 16 kJ of nutritional
  energy. How many American Calories
  (kilocalories) are in this packet of Equal?
• 1 kcal 4.184 kJ
• 16 kJ x 1 kcal 3.82409
• 4.184 kJ
• 3.8 Calories

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3. Conservation of Energy

• The first law of thermodynamics states that
  energy can be neither created nor destroyed the
  total energy of the universe is constant.
• We will be studying the transfer of energy from
  one form or one place to another.
• How is energy
• transferred?
• Work Heat

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a) Work

• Doing work (or working on an object) is one
  process that transfers energy to an object.
• For example
• lifting an object against the force of gravity
  increases the potential energy of the object
• throwing a baseball increases the kinetic
  energy of an object

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b) Heat

• Transferring heat to an object is another way of
  transferring energy.
• Heating, or raising the temperature of, an object
  increases the thermal energy of that object.
• Thermal energy is kinetic energy on the
  nanoscale.
• Matter consists of nanoscale particles that are
  constantly in motion.
• Higher temperature ? faster motion of particles ?
  more thermal energy

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b) Heat

• Heating refers to the energy transfer that occurs
  whenever two samples of matter at different
  temperatures are brought into contact.
• Energy always transfers from the hotter sample to
  the cooler sample until both are at the same
  temperature a state called thermal equilibrium.

The hot steel bar transfers energy to the cool
water until the two come to thermal equilibrium.
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c) Defining the System

• When analyzing the thermodynamics of a process,
  it is crucial to keep track of ALL energy
  transfers taking place work and heat.
• In doing so, it is helpful to define the sample
  of matter you are investigating as the system
  (eg. a reaction, an object, etc.)
• Anything that can exchange energy with the system
  is defined as the surroundings.

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i) System vs. Surroundings

• The total amount of energy must remain constant,
  so any energy transferred out of the system must
  be transferred into the surroundings, and vice
  versa.

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ii) Thermodynamic Changes

• We will refer to ?E (change in energy) when
  keeping track of energy transfers.
• ?Esystem Efinal - Einitial
• -The sign of ?E indicates the direction of
  transfer.

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ii) Thermodynamic Changes

• We will refer to ?E (change in energy) when
  keeping track of energy transfers.
• ?Esystem Efinal - Einitial
• -The sign of ?E indicates the direction of
  transfer.
• ?Esystem gt 0 Energy transferred into the
  system.
• - Final E greater than initial E
• ?Esystem lt 0 Energy transferred out of the
  system.
• - Final E lower than initial E

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ii) Thermodynamic Changes

• In many simple cases
• ?Esystem qsystem wsystem
• qsystem heat transferred to the system
• wsystem work done on the system

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ii) Thermodynamic Changes

• In many simple cases
• ?Esystem qsystem wsystem
• qsystem heat transferred to the system
• wsystem work done on the system
• q gt 0 heat absorbed by the system
• q lt 0 heat given off by the system

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ii) Thermodynamic Changes

• In many simple cases
• ?Esystem qsystem wsystem
• qsystem heat transferred to the system
• wsystem work done on the system
• q gt 0 heat absorbed by the system
• q lt 0 heat given off by the system
• w gt 0 work done on the system
• w lt 0 work done by the system

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ii) Thermodynamic Changes

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What is the change in internal energy of a system
that does 7.02 kJ of work and absorbs 888 J of
heat?
?E q w
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What is the change in internal energy of a system
that does 7.02 kJ of work and absorbs 888 J of
heat?

• ?E q w
• q 888 J 0.888 kJ
• w -7.02 kJ
• ?E q w
• ?E 0.888 7.02
• ?E -6.132 kJ -6.13 kJ

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iii) Internal Energy

• The internal energy of a system is defined as the
  sum of the individual energies of all of the
  nanoscale particles in a sample of matter.
• Magnitude of internal energy depends on
• Temperature
• Type of particles/molecules
• Number of particles

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iii) Internal Energy

• As a hot cup of coffee cools, its internal energy
  decreases.
• The energy is transferred, in the form of heat,
  to the surrounding room.
• The rooms internal energy increases.
• If the coffee cools from 90 C to 60 C (DT -30
  C), will the temperature in the room increase by
  30 C?

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B. Heat Transfer

• Heat Capacity (C) quantity of energy required to
  increase the temperature of a sample by one
  degree
• C q/?T
• The magnitude of the heat capacity depends on
• Mass of the sample
• Composition of the sample

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1. Heat Capacity
Calculate the heat capacity of an aluminum block
that must absorb 629 J of heat from its
surroundings in order for its temperature to rise
from 22 ?C to 145 ?C. Heat Capacity (C) C
q/?T
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1. Heat Capacity

• Calculate the heat capacity of an aluminum block
  that must absorb 629 J of heat from its
  surroundings in order for its temperature to rise
  from 22 ?C to 145 ?C.
• Heat Capacity (C) C q/?T
• C 629 J / (145 - 22 ?C)
• 629 J / 123 ?C
• 5.11 J/ ?C

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1. Heat Capacity

• a) When comparing the heat capacities of
  different substances with different masses, it is
  more useful to compare specific heat capacities.
• Specific Heat (c) quantity of energy needed to
  increase the temperature of one gram of a
  substance by one degree Celsius
• Molar Heat Capacity (cm) related to specific
  heat, but for one mole of substance

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1. Heat Capacity

Metals have low specific heats.
Water has high specific heat. (4.184 J/g?C
or 1.000 cal/g?C)
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1. Heat Capacity

• When dealing with specific heat capacities
  (c)
• Given c q/m?T
• Derive q cm?T
• ?T Tfinal Tinitial q/cm
• m q/c?T
• q thermal heat
• c specific heat
• m mass
• ?T change in temperature

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If a 10.0 g sample of each substance below has
250 J applied to it, which substance will have
the greatest increase in temperature? iron (c
0.46 J/g?C) water (c 4.184 J/g?C) copper
(c 0.39 J/g?C) aluminum (c 0.92 J/g?C)
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• If a 10.0 g sample of each substance below has
  250 J applied to it, which substance will have
  the greatest increase in temperature?
• iron (c 0.46 J/g?C)
• water (c 4.184 J/g?C)
• copper (c 0.39 J/g?C)
• aluminum (c 0.92 J/g?C)
• Copper! Lower heat capacity means less energy
  required to change the temperature larger
  temperature change.

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1. Heat Capacity

• How much heat (in joules) does it take to raise
  the temperature of 225 g of water from 25.0 to
  100.0 ?C? (c 4.184 J/g?C)

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1. Heat Capacity

• How much heat (in joules) does it take to raise
  the temperature of 225 g of water from 25.0 to
  100.0 ?C? (c 4.184 J/g?C)
• q cm?T
• q (4.184 J/g?C)(225 g)(100.0 25.0 ?C)
• q 7.06x104 J

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1. Heat Capacity
What will be the final temperature of a 5.00 g
silver ring at 37.0 ?C that gives off 25.0 J of
heat to its surroundings (c 0.235 J/g
?C)?
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1. Heat Capacity

• What will be the final temperature of a 5.00 g
  silver ring at 37.0 ?C that gives off 25.0 J of
  heat to its surroundings
• (c 0.235 J/g ?C)?
• ?T Tfinal Tinitial q/cm
• Tfinal 37.0 ?C -25.0 J / (0.235 J/g
  ?C)(5.00 g)
• Tfinal 37.0 ?C -21.3 ?C
• Tfinal 37.0 ?C - 21.3 ?C
• Tfinal 15.7 ?C

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1. Heat Capacity
148 J of heat are transferred to a a piece of
glass (c 0.84 J/g?C), raising the temperature
from 25.0 ?C to 49.4 ?C. What is the mass of the
glass?
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1. Heat Capacity

• 148 J of heat are transferred to a a piece of
  glass (c 0.84 J/g?C), raising the temperature
  from 25.0 ?C to 49.4 ?C. What is the mass of the
  glass?
• m q/c?T
• m (148 J)/(0.84 J/g?C)(24.4 ?C)
• m 7.2 g

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2. Phase Changes

• We just saw that energy transfers ALWAYS
  accompany temperature changes.
• Energy transfers also accompany physical and
  chemical changes, even when there is no change in
  temperature.
• eg. Energy is always transferred into or out of a
  system during a phase change.

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a) Melting/Freezing

• During a phase change
• The temperature of the sample remains constant
• Energy must be continually transferred into or
  out of the sample to facilitate the
  transformation
• Melting phase change from solid to liquid
  energy is transferred into the sample
• Freezing phase change from liquid to solid
  energy is transferred out of the sample

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a) Melting/Freezing

• Heat of Fusion quantity of thermal energy that
  must be transferred to a solid as it melts
  (qfusion - qfreezing)

Water Heat of fusion 333 J/g at 0
?C. Specific Heat (l) 1.00 cal/g?C Specific
Heat (s) depends on T ? 0.5 cal/g?C near 0 ?C
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a) Melting/Freezing

• Adding energy to a sample can cause temperature
  changes and/or phase changes.

• The energy required to melt a 1-gram ice cube at
  0 C would be enough to heat a 1-gram block of
  iron at 0 C to 738 C (red hot) or to melt a
  0.5-g ice cube and warm the resulting water to 80
  C.

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b) Vaporization/Condensation

• Vaporization phase change from liquid to gas
  energy is transferred into the sample
• Condensation phase change from gas to liquid
  energy transferred out of the sample
• Heat of Vaporization energy that must be
  transferred to convert a liquid to a gas
• qvaporization -qcondensation

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b) Vaporization/Condensation

• Heating the flask causes the water to boil
  liquid turning to a vapor.
• When the heating is stopped, energy is
  transferred to the surroundings as the steam
  condenses back into liquid water.

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How much heat is required to melt 125 g of ice at
0?C?
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How much heat is required to melt 125 g of ice at
0?C?

• Water Heat of fusion 333 J/g at 0 ?C
• (125 g)(333 J/g) 41625 J
• 4.16x104 J
• 41.6 kJ

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You have 1.00 g of ice at 0.0 ?C. How much
energy would it take to melt all of the ice, warm
the water from 0.0 ? C to 100.0 ? C, and boil the
water to vapor (gas) at 100.0 ? C?
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You have 1.00 g of ice at 0.0 ?C. How much
energy would it take to melt all of the ice, warm
the water from 0.0 ? C to 100.0 ? C, and boil the
water to vapor (gas) at 100.0 ? C?
Phase 1 (Melting Ice) Heat of Fusion 333 J/g
at 0 ?C q (1.00 g)(333 J/g) 333 J Phase 2
(Warming Water) c 4.184 J/g ?C q (4.184
J/g?C)(1.00 g)(100.0 ?C) q 418.4 J Phase 3
(Vaporizing Water) Heat of Vap. 2260 J/g at
100 ?C q (1.00 g)(2260 J/g) 2260 J q
333 418.4 2260 3011 J 3.011 kJ
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3. Endothermic vs. Exothermic

• Processes involving energy transfers can be
  described as endothermic or exothermic.
• endothermic energy must be transferred into the
  system to maintain a constant temperature
• exothermic energy must be transferred out of the
  system to maintain a constant temperature

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3. Endothermic vs. Exothermic

• Melting and Vaporization are both endothermic
  processes (q gt 0)
• Heat must be transferred into the system to
  facilitate these phase changes
• Heat is absorbed by the system
• Freezing and Condensation are both exothermic
  processes (q lt 0)
• Heat must be transferred out of the system to
  facilitate these phase changes
• Heat is given off by the system

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C. Enthalpy (?H)

• The quantity of thermal energy transferred into a
  system at constant pressure (qp) is called the
  enthalpy change (?H).
• Enthalpy is an extensive property used to
  quantify the heat evolved or absorbed during a
  physical or chemical change.
• Extensive Property dependent on the quantity of
  substance in the sample (the enthalpy of 2 mol of
  A is exactly twice the enthalpy of 1 mol of A)

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C. Enthalpy

• Enthalpy is a state function a property whose
  value is always the same if a system is in the
  same state.
• A change in a state function does not depend on
  the path by which the system changes from one
  state to another.

This allows us to apply laboratory measurements
to real-life situations. eg. The enthalpy change
of a chemical reaction is the same whether it
occurs in a lab or in your body.
eg. Altitude is a state function.
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1. Enthalpy of Chemical Reactions

• We have already investigated energy transfers
  during physical transformations, what about
  chemical transformations reactions?
• ?H Hproducts Hreactants
• An endothermic reaction (?H gt 0) is a reaction in
  which heat is absorbed from the surroundings.
• An exothermic reaction (?H lt 0) is a reaction in
  which heat is given off to the surroundings.

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1. Enthalpy of Chemical Reactions

• The combustion of hydrogen in the presence of
  oxygen is highly exothermic. This means that the
  energy of the reactants is higher than the energy
  of the products. The extra energy is released
  during the reaction.
• 2 H2 (g) O2 (g) ?
• 2 H2O (g)

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2. Thermochemical Equations

• Thermochemical Equation a balanced chemical
  equation (including phase labels) with the molar
  enthalpy of reaction written directly after the
  equation
• N2 (g) 3 H2 (g) ? 2 NH3 (g) ?H -91.8 kJ
• a) Molar Interpretation When 1 mol of nitrogen
  gas reacts with 3 mol of hydrogen gas to form 2
  mol of ammonia gas, 91.8 kJ of energy is given
  off.

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b) Physical States of Matter

• Physical states are important in thermochemical
  equations.
• Phase changes are accompanied by enthalpy
  changes, so ?H depends on the phases of the
  reactants and products.
• H2 (g) ½ O2 (g) ? H2O (g) ?H? - 241.8 kJ
• H2 (g) ½ O2 (g) ? H2O (l) ?H? - 285.8 kJ

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c) Quantity of Matter

• Thermochemical equations obey the rules of
  stoichiometry the larger the magnitude of the
  reaction, the more energy is transferred.
• H2 (g) ½ O2 (g) ? H2O (g) ?H? - 241.8 kJ
• 2 H2 (g) O2 (g) ? 2 H2O (g) ?H? - 483.6 kJ

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d) Manipulating Thermochemical Equations

• When a thermochemical equation is multiplied by
  any factor, the value of ?H for the new equation
  is obtained by multiplying the ?H in the original
  equation by that same factor.
• When a thermochemical equation is reversed, the
  value of ?H is reversed in sign.

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d) Manipulating Thermochemical Equations
Given the equation 3 O2 (g) ? 2 O3 (g) ?H
285.4 kJ Calculate ?H for the following
equation 3/2 O2 (g) ? O3 (g) ?H ?
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d) Manipulating Thermochemical Equations

• Given the equation
• 3 O2 (g) ? 2 O3 (g) ?H 285.4 kJ
• Calculate ?H for the following equation
• 3/2 O2 (g) ? O3 (g) ?H ?
• ? H 285.4 kJ / 2 142.7 kJ

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d) Manipulating Thermochemical Equations

• Given the equation
• H2 (g) Cl2 (g) ? 2 HCl (g) ?H -184.6 kJ
• Calculate ?H for the following equation
• 2 HCl (g) ? H2 (g) Cl2 (g) ?H ?

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d) Manipulating Thermochemical Equations

• Given the equation
• H2 (g) Cl2 (g) ? 2 HCl (g) ?H -184.6 kJ
• Calculate ?H for the following equation
• 2 HCl (g) ? H2 (g) Cl2 (g) ?H ?
• ? H - (-184.6) 184.6 kJ

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d) Manipulating Thermochemical Equations

• Given the following equation
• N2 (g) 3 H2 (g) ? 2 NH3 (g) ?H -91.80 kJ
• Write the thermochemical equation for
• a) Formation of one mole of ammonia gas.

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d) Manipulating Thermochemical Equations

• Given the following equation
• N2 (g) 3 H2 (g) ? 2 NH3 (g) ?H -91.80 kJ
• Write the thermochemical equation for
• a) Formation of one mole of ammonia gas.
• 1/2 N2 (g) 3/2 H2 (g) ? NH3 (g) ?H -45.90 kJ

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d) Manipulating Thermochemical Equations

• Given the following equation
• N2 (g) 3 H2 (g) ? 2 NH3 (g) ?H -91.80 kJ
• Write the thermochemical equation for
• b) Decomposition of 6 mol of ammonia gas.

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d) Manipulating Thermochemical Equations

• Given the following equation
• N2 (g) 3 H2 (g) ? 2 NH3 (g) ?H -91.80 kJ
• Write the thermochemical equation for
• b) Decomposition of 6 mol of ammonia gas.
• 2 NH3 (g) ? 3 H2 (g) N2 (g) ?H 91.80 kJ
• 6 NH3 (g) ? 9 H2 (g) 3 N2 (g) ?H 275.4 kJ

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When 0.250 mol of solid CaO is mixed with 0.250
mol of liquid water, solid Ca(OH)2 is formed and
16.3 kJ of heat is released. Write a
thermochemical equation for the reaction
producing 1 mol of Ca(OH)2.
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When 0.250 mol of solid CaO is mixed with 0.250
mol of liquid water, solid Ca(OH)2 is formed and
16.3 kJ of heat is released. Write a
thermochemical equation for the reaction
producing 1 mol of Ca(OH)2.

• 0.250 CaO (s) 0.250 H2O (l) ? _____ Ca(OH)2 (s)
• 0.250 CaO (s) 0.250 H2O (l) ? 0.250 Ca(OH)2 (s)
• ?H -16.3 kJ
• 0.250 CaO (s) 0.250 H2O (l) ? 0.250 Ca(OH)2
  (s) x 4
• ?H -16.3 kJ x 4
• CaO (s) H2O (l) ? Ca(OH)2 (s)
• ?H -65.2 kJ

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What is the enthalpy change when 22.5 g of CH4
are burned in excess O2?CH4(g) 2 O2(g) ?
CO2(g) 2 H2O(l) ?H? -890 kJ
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What is the enthalpy change when 22.5 g of CH4
are burned in excess O2?CH4(g) 2 O2(g) ?
CO2(g) 2 H2O(l) ?H? -890 kJ

• CH4 12.011 4(1.008) 16.043 g/mol
• 22.5 g CH4 1.402 mol CH4
• 16.043 g/mol
• (1.402 mol CH4)(-890 kJ/mol) -1247.78 kJ
• -1.25x103 kJ

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D. Chemical Energy

• In an exothermic chemical reaction (?H lt 0),
  where does the energy come from?
• In an endothermic reaction (?H gt 0), where does
  the energy go?
• In a chemical reaction, chemical bonds are broken
  and new chemical bonds are formed.
• As nuclei and electrons move closer together or
  further apart, their energies change.

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1. Bond Enthalpies

• Bond Enthalpy (Bond Energy) the enthalpy change
  that occurs when two bonded atoms in a gas-phase
  molecule are separated completely at constant
  pressure
• Bond breaking is always an endothermic process
  energy must be put in to break a chemical bond.
• Bond enthalpies are always positive.
• The potential energy of atoms are lower when they
  are bonded together.

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1. Bond Enthalpies

• Atoms that are chemically bonded together are
  thought of as being in a stable energy well.
  (More on this later.)

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1. Bond Enthalpies

• Bond breaking is always endothermic
• Cl2 (g) ? 2 Cl (g) ?H? 242 kJ
• Bond formation (the reverse reaction) is always
  exothermic
• 2 Cl (g) ? Cl2 (g) ?H? - 242 kJ
• Bond enthalpies are usually expressed per mole of
  bonds.
• eg. The bond enthalpy (bond strength) of Cl2 is
  242 kJ/mol.

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1. Bond Enthalpies

• Knowing the bond enthalpies for the bonds broken
  formed in a chemical reaction allows you to
  predict whether the reaction will be endothermic
  or exothermic.

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1. Bond Enthalpies

• Use the table of bond enthalpies to answer the
  questions.
• a) Which of the molecules in the table has
  the strongest bond? The weakest bond?
• b) Calculate the enthalpy change for the
  following reactions
• H-H Br-Br ? 2 H-Br
• H-H F-F ? 2 H-F
• c) Which is the most exothermic?

Bond Bond Enthalpy
(kJ/mol) _ H-H 436 H-Br
366 H-F 566 Br-Br 193 F-F
158
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1. Bond Enthalpies

• Strongest Bond?
• Highest Bond Enthalpy ? HF
• Weakest Bond?
• Lowest Bond Enthalpy ? F2

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1. Bond Enthalpies
H-H (g) Br-Br (g) ? 2 H-Br (g) ?H ?
80
1. Bond Enthalpies

• H-H (g) Br-Br (g) ? 2 H-Br (g) ?H ?
• Bonds Broken H-H ?H 436 kJ/mol
• Br-Br ?H 193 kJ/mol
• Bonds Formed H-Br ?H - 366 kJ/mol
• ?H 436 193 (2 x 366)
• ?H - 103 kJ/mol

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1. Bond Enthalpies
H-H (g) F-F (g) ? 2 H-F (g) ?H ?
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1. Bond Enthalpies
H-H (g) F-F (g) ? 2 H-F (g) ?H ? Bonds
Broken H-H ?H 436 kJ/mol F-F ?H
158 kJ/mol Bonds Formed H-F ?H - 566
kJ/mol ?H 436 158 (2 x 566) ?H - 538
kJ/mol
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1. Bond Enthalpies
H-H Cl-Cl ? 2 H-Cl ?H - 184 kJ/mol H-H
F-F ? 2 H-F ?H - 538 kJ/mol Which is the
most exothermic?
84
1. Bond Enthalpies
H-H Cl-Cl ? 2 H-Cl ?H - 184 kJ/mol H-H
F-F ? 2 H-F ?H - 538 kJ/mol Which is the
most exothermic? Formation of H-F
85
2. Calorimetry Measuring Enthalpy Changes

• Thermochemical equations and bond enthalpies can
  be used to calculate ?H for a chemical reaction.
• Thermal energy transfers can also be measured
  directly using a calorimeter.
• Calorimeter device for measuring the quantity of
  thermal energy transferred during a chemical
  reaction or some other process

86
2. Calorimetry Measuring Enthalpy Changes

• When at least one of the reactants or products is
  a gas, a bomb calorimeter is used because it is
  closed to the atmosphere at constant volume.
• System burning sample (reaction)
• Surroundings water bomb
• qreaction -(qbomb qwater)
• qreaction ?Ereaction

87
A bomb calorimeter has a heat capacity of 783
J/?C and contains 254 g of water. How much energy
is evolved by a reaction when the temperature of
the calorimeter goes from 23.73?C to 26.01?C?
88
A bomb calorimeter has a heat capacity of 783
J/?C and contains 254 g of water. How much energy
is evolved by a reaction when the temperature of
the calorimeter goes from 23.73?C to 26.01?C?

• ?T 2.28 ?C
• qbomb (heat capacity)(?T) (783 J/?C)(2.28 ?C)
• qbomb 1785 J
• qwater cm?T (4.184 J/g?C)(254 g)(2.28 ?C)
• qwater 2423 J
• qreaction -(qbomb qwater) - (1785 2423)
  -4208 J
• qreaction -4.21x103 J 4.21 kJ

89
2. Calorimetry Measuring Enthalpy Changes

• When a reaction takes place in solution, it is
  easier to use a calorimeter that is open to the
  atmosphere.
• This allows for the direct measurement of the
  enthalpy change (?H q at constant pressure).
• A cheap and easy way to do this is with a
  coffee-cup calorimeter.

90
Coffee Cup Calorimeter

• ?Hreaction -qsolution
• The energy transferred to the solution comes from
  the reaction.
• The masses of other substances in the solution
  are so small compared to the mass of the solvent
  (water) that their heat capacities can usually be
  ignored (c ? 4.184 J/g?C)

91
Coffee Cup Calorimeter

• When a 13.0 g sample of NaOH (s) dissolves in
  400.0 g water in a coffee cup calorimeter, the
  temperature of the water changes from 22.6 to
  30.7 ?C. Calculate
• The heat transfer from system to
  surroundings.
• ?H for the process NaOH (s) ? Na (aq) OH-
  (aq)

92
Coffee Cup Calorimeter

• When a 13.0 g sample of NaOH (s) dissolves in
  400.0 g water in a coffee cup calorimeter, the
  temperature of the water changes from 22.6 to
  30.7 ?C. Calculate
• The heat transfer from system to
  surroundings.
• ?H for the process NaOH (s) ? Na (aq) OH-
  (aq)
• qsolution cm?T
• (4.184 J/g ?C)(400.0 13.0 g)(30.7 22.6
  ?C)
• (4.184 J/g ?C)(413.0 g)(8.1 ?C)
• 13996.735 1.4x104 J

93
Coffee Cup Calorimeter

• When a 13.0 g sample of NaOH (s) dissolves in
  400.0 g water in a coffee cup calorimeter, the
  temperature of the water changes from 22.6 to
  30.7 ?C. Calculate
• qsolution 13996.735 1.4x104 J
• ?H for the process NaOH (s) ? Na (aq) OH-
  (aq)
• ? H qreaction / mols NaOH
• qreaction - qsolution - 1.4x104 J
• 13.0 g NaOH x 1 mol 0.325 mol NaOH
• 39.9971 g

94
Coffee Cup Calorimeter

• When a 13.0 g sample of NaOH (s) dissolves in
  400.0 g water in a coffee cup calorimeter, the
  temperature of the water changes from 22.6 to
  30.7 ?C. Calculate
• qsolution 13996.735 1.4x104 J
• ?H for the process NaOH (s) ? Na (aq) OH-
  (aq)
• ? H (-13996.735)/(0.325 mol)
• ? H - 4.3x104 J
• ? H - 43 kJ

95
Coffee Cup Calorimeter
A 19.6-g sample of metal was heated to 61.67 ?C
and placed in 26.70 g of water in a coffee cup
calorimeter. The temperature of the water
increased from 25.00 ?C to 30.00 ?C. What is the
specific heat of the metal? - First, calculate
the magnitude of the heat transferred to the
water. Then, use qwater to calculate c for the
metal.
96
Coffee Cup Calorimeter

• A 19.6-g sample of metal was heated to 61.67 ?C
  and placed in 26.70 g of water in a coffee cup
  calorimeter. The temperature of the water
  increased from 25.00 ?C to 30.00 ?C. What is the
  specific heat of the metal?
• - First, calculate the magnitude of the heat
  transferred to the water. Then, use qwater to
  calculate c for the metal.
• qwater cm?T (4.184 J/g ?C)(26.7 g)(5.00
  ?C)
• 558.56 J
• qmetal - qwater

97
Coffee Cup Calorimeter
A 19.6-g sample of metal was heated to 61.67 ?C
and placed in 26.70 g of water in a coffee cup
calorimeter. The temperature of the water
increased from 25.00 ?C to 30.00 ?C. What is the
specific heat of the metal? qwater 558.56
J qmetal - qwater qmetal - 558.56 J
c(19.6 g)(30.00 61.67 ?C) c (- 558.56
J) / (19.6 g)(- 31.67 ?C) c 0.900 J/g?C