**Unformatted text preview: **CHAPTER 6
Thermochemistry 1 Energy & Chemistry Burning peanuts supply sufficient energy to boil a
cup of water. 2 Burning sugar (sugar
reacts with KClO3, a strong
oxidizing agent) • These processes are examples of
CONVERSION OF ENERGY Energy & Chemistry
ENERGY is the capacity to do
work or transfer heat. HEAT is the form of energy that flows between 2 objects because
of their difference in temperature.
Other forms of energy —
• light
• electrical energy
• kinetic and potential energy 3 Potential & Kinetic
Energy
• Potential energy — energy that is associated with the object’s position.
• Kinetic energy — energy associated with motion 4 Potential & Kinetic
Energy
less stable
change in potential energy
EQUALS
kinetic energy more stable
A gravitational system. The state of lower potential energy is more
stable. 5 Potential & Kinetic
Energy 6 less stable
change in potential energy
EQUALS
kinetic energy more stable A system of two balls attached by a spring. The state of lower potential energy
is more stable. The potential energy gained by a stretched spring is converted to
kinetic energy when the moving balls are released. Potential & Kinetic
Energy
less stable
change in potential energy
EQUALS
kinetic energy more stable
A system of oppositely charged particles. Again, the state of lower potential
energy is more stable. 7 Potential & Kinetic
Energy 8 less stable
change in potential energy
EQUALS
kinetic energy more stable
A system of fuel and exhaust. A fuel is higher in chemical potential energy than
the exhaust. As the fuel burns, some of its potential energy is converted to the
kinetic energy of the moving car. Potential
Potential Energy
Energy
on
on the
the Atomic
Atomic
Scale
Scale
• Positive and
negative particles
(ions) attract one
another.
• As the particles
attract, they have a
lower potential
energy.
NaCl — composed of
Na+ and Cl- ions. 9 Potential & Kinetic
Energy Kinetic energy — energy of motion.
All atoms, molecules, ions have kinetic energy.
rotate
rotate
vibrate
vibrate translate
translate 10 Law
Law of
of Conservation
Conservation of
of
Energy
Energy 11 The first law of thermodynamics states that
there is CONSERVATION OF ENERGY in nature. The total energy is unchanged
in any process.
Energy can be converted from one form to
another but can neither be created nor
destroyed.
For example, if in a chemical reaction the PE of products is less
than that of reactants, the difference must be released as KE. PE
PE Energy
Energy Change
Change in
in
Chemical
Chemical Processes
Processes 12 Reactants
Reactants
Kinetic Kinetic Energy
Energy Products
Products Potential Energy of system dropped. Kinetic Energy
increased. Therefore, you often feel a T increase. 13 Internal
Internal Energy
Energy (E)
(E)
• Internal energy of an object
contains potential and kinetic
energies of all its particles.
• it depends on
- number of particles
- type of particles
- temperature
- state of matter Heat
Heat transfer
transfer
Heat is a form of energy that transfers
between two objects due to a
temperature difference. Heat transfers from hotter object to cooler
one until thermal equilibrium is
established. 14 Units
Units of
of Energy
Energy calorie (cal) 1 calorie = heat
required to raise
temperature of 1.0 g
of water by 1.0 oC.
1000 cal = 1 kcal 15 16 Units of Energy
The SI unit is called JOULE (“J”)
1 J = 1 kg · m2/s2 1 kJ = 1000 J
1 cal = 4.184 Joules 17 HEAT CAPACITY
Different objects can have different response to a
given amount of heat. Which can be heated more quickly? Specific
Specific Heat
Heat
The heat (q) absorbed or lost by an object is related to
a) mass of object;
b) change in temperature and;
c) the material that object is made of. Every material has a property called specific heat. Specific heat shows how much heat is needed
to increase the temperature of 1 gram of a substance
by 1 Kelvin. 18 19 heat transferred, J
Specific Heat = mass, g × T change,°C q
C
C == m∆T
∆T = Tfinal - Tinitial
Note that ∆T in K is the same as ∆T in °C. 20 The higher the specific heat capacity, the more slowly
the temperature of material rises in response to heating. 21 Specific heat is an
INTENSIVE property.
does NOT depend on
sample size. Sample Problem 1 Finding the Specific Heat from Quantity of Heat 22 PROBLEM: A sample of copper weighs 125 g. It takes 1.38x104 J to raise the
temperature of the copper sample from 250C to 3100C. What is
the specific heat of Cu, in J/g*K ? PLAN: Given the mass, heat and change in temperature, we can use
q = C x mass x T to find the answer. T in 0C is the same as in K.
SOLUTION: T = (310-25)0C = 285 0C
The same in K: T = 583 K – 298 K = 285 K
q = C x mass x T
C = q / (mass x T) = 1.38x104 J
(125 g x 285 K) = 0.387 J/gK Specific
Specific Heat
Heat
Capacity
Capacity
If 25.0 g of Al cool from
266 oC to 37 oC, how
many joules of heat
energy are lost by
the Al? heat = q = (C)(mass)(∆T)
C for aluminum = 0.896 J/g•K 23 24 Specific
Specific Heat
Heat
If 25.0 g of Al cool from 266 oC to 37 oC, how many joules of
heat energy are lost by the Al? heat = q = (C)(mass)(∆T)
where ∆T = Tfinal - Tinitial
q = (0.896 J/g•K)(25.0 g)(37 - 266)oC
q = - 5130 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al. 25 Specific
Specific Heat
Heat and
and
Molar
Molar Heat
Heat Capacity
Capacity
specific heat capacity
heat capacity per gram = J/g•°C or J/g•K molar heat capacity
heat capacity per mole = J/mol•°C or J/mol•K 26 Specific
Specific Heat
Heat
• • A bucket containing 11.5 kg of cold water at 12°C is taken into a house at a
warmer temperature and left inside until it has reached thermal equilibrium
with its new surroundings.
If water absorbed 502 kJ of energy from the surroundings, what is the final
temperature of water? Q = 502 kJ
m = 11.5 kg
c = 4.18 J g-1 °C-1
Tfinal = ? qq ==(C)(mass)(∆T)
(C)(mass)(∆T) q
∆T=
∆T= m C ∆ T = 502,000 J / (11,500 g × 4.18 J g-1°C-1 ) = 10.4 °C (to 3 s.f.)
So temperature of the water = 12°C + 10.4 °C = 22.4 °C
Final T of water ~= 22 °C Heat
Heat
Exchange
Exchange Heat transfer can be used to
find specific heat capacity,
C When two substances exchange heat,
heat lost by hot object = heat gained by cold
object
(conservation of energy) although they have opposite signs q1 = –q2 27 28 Heat
Heat
Transfer
Transfer
• Use heat transfer as a way
•
•
•
• to find specific heat
capacity, C
55.0 g Fe at 99.8 ˚C are
dropped into 225 g water
at 21.0 ˚C
Water and metal come to
23.1 ˚C
C of water = 4.18 J/g•K
What is the specific heat
capacity of the metal? Heat
Heat Transfer
Transfer
Because of conservation of energy,
q(Fe) = –q(H2O) (heat out of Fe = heat into H2O)
or q(Fe) + q(H2O) = 0
q(H2O) = (4.184 J/g•K) (225 g)(23.1 ˚C – 21.0 ˚C)
q(H2O) = 1977 J
q(Fe) = (CFe) (55.0 g)(23.1 ˚C – 99.8 ˚C)
q(Fe) = – 4219 • CFe
q(Fe) + q(H2O) = – 4219 CFe + 1977 = 0
CFe = 0.469 J/g•K 29 System and
Surroundings 30 • SYSTEM
– The object under study • SURROUNDINGS
– Everything outside the
system Universe =
System +
Surroundings Directionality of Heat Transfer 31 • Heat always transfers from hotter object to
cooler one. • EXOthermic: heat transfers from SYSTEM
to SURROUNDINGS. T(system) goes down
T(surr) goes up Directionality of Heat Transfer
• ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up
T (surr) goes down 32 Endothermic
Endothermic and
and
Exothermic
Exothermic Processes
Processes
ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. (system gains heat) EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. (system loses heat) 33 34 Endothermic
Endothermic or
or Exothermic
Exothermic ??
vaporization melting solid liquid freezing gas condensation Endothermic (requires heat)
Exothermic (releases heat) 35 FIRST
FIRST LAW
LAW OF
OF
THERMODYNAMICS
THERMODYNAMICS
heat transferred ∆E = q + w
change in system’s
internal energy work done
by (or to) the
system Energy is conserved! 36 Two different paths for the energy change of a system. 37 heat transfer in
(endothermic), +q heat transfer out
(exothermic), -q SYSTEM
SYSTEM ∆E = q + w
w transfer in
(+w) w transfer out
(-w) 38 The Sign Conventions for q and w ∆E = q + w
All signs are from the point of view of the SYSTEM
• For q: + means system gains heat;
- means system loses heat.
• For w: + means work done on system;
- means work done by system. Sample Problem 2
PROBLEM: PLAN: Determining the Change in Internal Energy of a
System 39 When gasoline burns in a car engine, the heat released causes the
products CO2 and H2O to expand, which pushes the pistons outward.
Excess heat is removed by the car’s cooling system. If the expanding
gases do 451 J of work on the pistons and the system loses 325 J to
the surroundings as heat, calculate the change in energy (E) in J,
and in kJ. Define system and surroundings, assign signs to q and w and calculate E.
The answer should be converted from J to kJ and then to kcal. SOLUTION: Heat released by system, q < 0
Work done by system, w < 0 q = - 325 J
w = - 451 J E = q + w = -325 J + (-451 J) = -776 J
-776J x 1 kJ
10 J
3 = -0.776kJ 40 w=Fxd
F=PxA
w = P x (A x d)
w = - PV
W is negative
because work
done by system Sample Problem 3
PROBLEM: PLAN: PV Work Calculate the work associated with the expansion of a gas from 46 L
to 64 L at a constant external pressure of 15 atm. Use the formula w = -PV SOLUTION: V = 64 L – 46 L = 18 L
w = - PV = -15 atm x 18 L = -270 L•atm
Sign of w is negative, because the gas does work on
surroundings while expanding.
To convert from L•atm to J, a conversion factor can
be used: 1 L·atm = 101.3 J 101.3 J
270 L atm 27350 J
1 L atm 41 6.2.
6.2. ENTHALPY
ENTHALPY and
and
CALORIMETRY
CALORIMETRY
w = - PV q = ∆E – w = ∆E + P∆V Lets define enthalpy as Then, at constant pressure, H = E + PV
H = E + PV
H = qp 42 43 ENTHALPY
ENTHALPY
∆H = qp Change of enthalpy equals heat
transferred at constant Pressure. •Enthalpy is the “heat portion” of energy.
• Units of enthalpy: same as for energy (J, kJ) often written in kJ/mol, representing the enthalpy of
just 1 mol of system. •The sign of ∆H tells us if the process is
endothermic or exothermic. ENTHALPY
ENTHALPY ∆
∆H = H
H = Hfinal Hinitial
final H
initial 44 If ∆H is positive, Hfinal > Hinitial final
initial System gained heat
Process is ENDOTHERMIC
If ∆H is negative, Hfinal < Hinitial final
initial System has lost heat Process is EXOTHERMIC
In a chemical reaction, ∆H = Hfinal – Hinitial = Hproducts – Hreactants Enthalpy diagrams for exothermic and endothermic processes. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) CH4 + 2O2 heat out Enthalpy, H Enthalpy, H H < 0 H2O(l) Hfinal Exothermic process Hfinal H > 0 CO2 + 2H2O A H2O(g) H2O(g) Hinitial B 45 heat in Hinitial Endothermic process Sample Problem 4
PROBLEM: PLAN: 46
Drawing Enthalpy Diagrams and Determining the
Sign of H In each of the following cases, determine the sign of H, state
whether the reaction is exothermic or endothermic, and draw the
enthalpy diagram.
(a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ (b) 40.7kJ + H2O(l) H2O(g) Look at whether heat is written as a “reactant” or as a “product”. If heat
is on the right side (“product”), then the system releases heat, and the
reaction is exothermic. The opposite is true for an endothermic reaction. SOLUTION: (a) The reaction is exothermic. H2(g) + 1/2O2(g)
EXOTHERMIC (reactants)
H = -285.8kJ H2O(l) (products) (b) The reaction is endothermic.
H2O(g)
ENDOTHERMIC
H2O(l) (products)
H = +40.7kJ
(reactants) 47 ConstantPressure
Calorimetry
How do they
experimentally
measure heats of
reactions? Coffee-cup calorimeter. Sample Problem 5 Determining the Heat of a Reaction 48 PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at
25.000C and add 25.0 mL of 0.500 M HCl, also at 25.000C. After
stirring, the final temperature is 27.210C. Assuming the final
solution has the mass 75.0 g and the same specfic heat capacity as
water: c = 4.18 J/g*K,
a) calculate heat gained by solution (qsoln) in J;
b) calculate heat released by reaction (qrxn);
c) calculate the heat (enthalpy change) of reaction (Hrxn) in kJ/mol.
PLAN: a) qsoln can be calculated using the mass, c, and T.
b) qrxn can be found from conservation of energy.
c) we need to find the number of moles of product (water) produced,
which can be done from determining the limiting reactant. Then, the
heat qrxn is divided by the number of moles of water to give us the heat
of reaction per mole of water formed. Sample Problem 5 Determining the Heat of a Reaction continued
SOLUTION: a) qsoln = mass x specific heat x T
= 75.0 g x 4.18 J/g*0C x (27.21-25.00)0C
b) = 693 J qrxn + qsoln = 0
qrxn = - qsoln = - 693 J c) HCl(aq) + NaOH(aq) = - 0.693 kJ
NaCl(aq) + H2O(l) For NaOH: 0.500 M x 0.0500 L = 0.0250 mol NaOH For HCl: 0.500 M x 0.0250 L = 0.0125 mol HCl
HCl is the limiting reactant.
0.0125 mol of H2O will form during the rxn. Hrxn = (-0.693 kJ/0.0125 mol H2O) = -55.4 kJ/ mol 49 CALORIMETRY
CALORIMETRY 50 Measuring Heats of Reaction
Constant Volume
“Bomb” Calorimeter
• Burn combustible
sample.
• Measure heat evolved
in a reaction.
• Derive ∆H for
reaction. Fig. 6.6 51 Calorimetry
Some heat from reaction warms
water
qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms
“bomb”
qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb Measuring
Measuring Heats
Heats of
of Reaction
Reaction
CALORIMETRY
CALORIMETRY
Calculate heat of combustion of 1 mol
of octane.
C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O
• Burn 1.00 g of octane
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1200 g water
• Heat capacity of bomb = 837 J/K 52 Measuring
Measuring Heats
Heats of
of Reaction
Reaction
CALORIMETRY
CALORIMETRY
Step 1 Calculate heat transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2 Calculate heat transferred from reaction to bomb.
q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J
Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J
Heat of combustion of 1.00 g of octane = 48.0 kJ
Step 4 Heat of combustion per mole of octane =
= ( 48.0 kJ/g)(114 g/mol) = 5472 kJ/mol 53 Using
Using Enthalpy
Enthalpy
1) Enthalpy change (
1) Enthalpy change (
H
Hoo) depends on physical states ) depends on physical states of reactants and products.
of reactants and products.
H2(g) + 1/2 O2(g) --> H2O(g)
H2(g) + 1/2 O2(g) --> H2O(l) ∆H˚ = -242 kJ
∆H˚ = -286 kJ 54 Using
Using Enthalpy
Enthalpy
oo
2) If a reaction is reversed, H is also 2) 2) If a reaction is reversed, H is also 2) reversed.
reversed. H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ H2O(g) ---> H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ 55 Using
Using Enthalpy
Enthalpy
3) If the coefficients of a reaction are multiplied 3) 3) If the coefficients of a reaction are multiplied 3) by an integer, H
by an integer, Hoo is multiplied by that same is multiplied by that same integer.
integer.
H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 2 H2(g) + O2(g) --> 2 H2O(g) ∆H˚ = -484 kJ 56 57
Sample Problem 6
PROBLEM: Using Enthalpy Given the thermochemical equation:
N2(g) + O2(g) 2NO(g) H = 180 kJ find the enthalphy change for equations:
a) 2NO(g) N2(g) + O2(g)
b) 6NO(g) 3N2(g) + 3O2(g) H = ?
H = SOLUTION: a) Reverse the original equation; H switches sign.
2NO(g) N2(g) + O2(g)
H = 180 kJ
b) Triple all coefficients; H triples.
6NO(g) 3N2(g) + 3O2(g) H = 540 kJ Using
Using Enthalpy
Enthalpy
oo
4) The value of H given with a balanced 4) The 4) The value of H given with a balanced 4) The chemical equation assumes the moles of chemical equation assumes the moles of reactants and products are exactly equal to the reactants and products are exactly equal to the balancing coefficients.
balancing coefficients. H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 242 kJ released when 1 mol H2 reacts with ½ mol O2 58 Sample Problem 7
PROBLEM: Using the Heat of Reaction (H) Thermal decomposition of aluminum oxide can be represented by
Al2O3(s) 2Al(s) + 3/2O2(g) Hrxn = 1676 kJ How much heat is required to produce 12.5 g of Al?
PLAN: Hrxn, which is equal to 1676 kJ, is given as heat of the reaction
assuming 1 mole of reactant decomposes. Therefore, this Hrxn is
per 1 mole of a reactant, and units should be 1676 kJ/mol.
We need to find how many moles of Al2O3 are reacting, and combine
it with Hrxn. SOLUTION: 59 12.5 g 1 mol Al 1 mol Al2O3 0.239 mol Al2O3
27.0 g Al
2 mol Al
1676 kJ
0.239 mol 401 kJ
1 mol 6.3.
6.3. Hess’s
Hess’s Law
Law 60 Consider the formation of water
H2(g) + 1/2 O2(g) > H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ 61 Hess’s
Hess’s Law
Law
Making liquid H2O from H2 + O2 involves two exothermic steps. H2 + O2 gas H2O vapor Liquid H2O Hess’s Law
Making H2O(l) from H2 involves two steps.
H2(g) + 1/2 O2(g) > H2O(g) + 242 kJ
H2O(g) > H2O(l) + 44 kJ H2(g) + 1/2 O2(g) > H2O(l) + 286 kJ
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more
others, the net ∆H is the sum of
the ∆H’s of the other rxns. 62 Hess’s Law 63 The change in enthalpy
(H) is the same whether
the reaction takes place in
one step or a series of
steps, and is the same no
matter which path is
followed. 64 Hess’s Law
& Energy Level
Diagrams
Forming CO2 can occur
in a single step or in a
two steps. ∆Htotal is the
same no matter which
path is followed. 65 ∆∆HH along
along one
one path
path == ∆∆HH along
along another
another path
path
• This equation is valid because ∆H is a
STATE FUNCTION
(depends only on the initial and final
states of the system, but does not
depend on the pathway).
• Other example of state function: change
of altitude during a trip to Denver.
• NOT a state function: mileage during a
trip to Denver. Using Hess’s Law
How to find H of a reaction using H’s of
some other reactions? •• Work backward from the required reaction, using Work backward from the required reaction, using “target” reactants and products to decide how to “target” reactants and products to decide how to manipulate other reactions.
manipulate other reactions.
•• Reverse available reactions and/or multiply them Reverse available reactions and/or multiply them by appropriate numeric factors, and then add them by appropriate numeric factors, and then add them together.
together. 66 Sample Problem 8 Using Hess’s Law to Calculate an Unknown H 67 PROBLEM: Given the following information:
Equation A: CO(g) + 1/2O2(g)
Equation B: N2(g) + O2(g) CO2(g) HA = -283.0 kJ
2NO(g) HB = 180.6 kJ calculate the unknown H of the following equation:
CO(g) + NO(g)
PLAN: CO2(g) + 1/2N2(g) H = ? Equations A and B have to be manipulated by reversal and/or multiplication by
factors in order to sum to the first, or target, equation. SOLUTION: Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g) CO(g) + NO(g) CO2(g) HA = -283.0 kJ 1/2N2(g) + 1/2O2(g) CO2(g) + 1/2N2(g) HB = -90.3 kJ
Hrxn = -373.3 kJ Sample Problem 9 Using Hess’s Law to Calculate an Unknown H 68 PROBLEM: Calculate the H for the following reaction: 3C (s) + 4H2 (g) C3H8 (g) Given the following information: C (s) + O2 (g) CO2 (g) H = 394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = 2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = 286 kJ
PLAN: Identify the target reactants and products in given equations, then manipulate
them by reversal and/or multiplication by factors.
SOLUTION: Step 1. Only reaction #1 has C(s), but we need 3 C(s). Therefore, we will
multiply it by 3. 3C (s) + 3O2 (g) 3CO2 (g) H = 1182 kJ 69 Sample Problem 9 3C (s) + 4H2 (g) C3H8 (g) 3 C (s) + O
3 2 (g) CO
3 2 (g) H = 394 kJ 1182 kJ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = 2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = 286 kJ Step 2. To get C3H8 on the product side of the reaction, we
need to reverse reaction #2. 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ Step 3: Only reaction #3 has H2(g) but we need 4H2(g).
Therefore, we will multiply by 4. 4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ 70 Sample Problem 9 Step 4: Add the equations together a...

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