Transcript CO 2 (g) - Miami East Schools

Thermochemistry
Chapter 5
Introduction
• All chemical and physical changes involve
energy
• Thermodynamics is the study of energy
and its transformations
• Thermochemistry is the study of the
energy changes involved in chemical
reactions
5.1 The Nature of Energy
Key Concepts
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Thermodynamics
Thermochemistry
Kinetic energy
Potential energy
System
Surroundings
Work
Force
Heat
5.1 The Nature of Energy
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Three fundamental forms of energy
Kinetic energy
Potential energy
Mass energy
E = mc2
discussed elsewhere
Kinetic Energy
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Any object in motion possesses KE or Ek
KE = ½ mv2
Examples
A cow of mass 2000 kg, walking at a speed of 0.5 m/s
has….
KE = 0.5 x 2000 kg x (0.5 m/s)2
= 250 Joules KE 1 J = 1 kg∙m2/s2
How fast would a 2.0 g fly have to fly to have the same
KE?
= 500 m/s
=1118 mph
Potential Energy
• Potential energy is “stored” energy
• Energy possessed by an object by virtue
of its position relative to other objects
• Gravity imparts PE
PE = mgh
• Electrostatic and chemical energy are
forms of PE
Energy Content
• Total energy of an object is the sum of KE
and all forms of PE
• E tot = KE + Σ PE
• Example:
A biker at the top of a hill has PE relative to the
bottom of the hill
A biker riding down the hill has both KE & PE
As a biker reaches the bottom of the hill, all PE
has been converted to KE
Conservation of Energy
Conservation of Mechanical Energy
0.80
Energy (J)
0.70
0.60
PE
0.50
KE
0.40
ME
0.30
Poly. (PE)
0.20
Poly. (KE)
0.10
Linear (ME)
0.00
-0.100.00
0.10
0.20
0.30
Time (s)
0.40
0.50
Units of Energy
• Joule 1 J = 1 N∙m = 1kg∙m2/s2
• 1 cal = 4.184 J
• calorie
The energy required to increase the temperature of
1 g of water by 1°C
• Nutritional calories …
• Are actually kilocalories
• 1 Cal = 1000 cal = 1 kcal
System & Surroundings
• Thermodynamic studies are conducted within a
defined system
• System: portion of the universe defined for
study
• Surroundings: everything else
• Open system can exchange energy with
surroundings
• Close system cannot exchange energy and
matter with surroundings self-contained
– self-contained
– Isolated from surroundings
Transferring Energy: Work and
Heat
• What is work?
• Force: a push or pull exerted on an object
• Work is the energy used to cause an
object to move
• Work is the product of Force x distance
• w = Fd
Units?
• Energy is transferred as work
Heat
• Energy is also transferred as heat.
• What is heat?
• Heat is energy transferred from an object
of higher temperature to an object of lower
temperature
• Heat is not temperature!
Energy
• So energy can be defined as…
• Capacity to do work or to transfer heat
• ΔE = w + q
Sample Problem
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A bowler lifts a 5.4-kg bowling ball from
ground level to a height of 1.6 m, and
then drops it back to the ground
What happens to PE as it is raised?
How much work was done to raise the
ball?
What is the speed of the ball at impact?
What is the weight of the ball in
newtons?
More sample problems
• Calculate the KE in J for
• An Ar atom moving with a speed of 650
m/s
• A mole of Ar atoms moving at the speed of
650 m/s
5.2
Key Concepts
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Internal energy
First law of thermodynamics
Endothermic
Exothermic
State functions
5.2 The First Law of Thermodynamics
• Energy is conserved
• Energy is neither created nor destroyed,
but can be transformed.
• The total energy of the universe is a
constant
First Law of Thermodnamics
•If the energy of a system decreases, it must be
transferred to the surroundings
•If energy is gained by a system, it must be transferred
from the surroundings
Internal Energy
• Sum of all KE and PE of all components of
a system
• E = ΣKE + ΣPE
• Change in internal energy is …
• ∆E = Ef - Ei
• E cannot be known, but ∆E can be
measured
Internal Energy
E = the sum of all kinetic and potential energies of all
components of the system
Internal Energy
• Thermodynamic quantities must be
described with a number, a unit, and a
sign
• When the system has absorbed energy
from its surroundings…
• +∆E….. Ef > Ei, therefore ∆E is positive
• When the system loses energy to its
surroundings…
• -∆E ….. Ef < Ei, therefore ∆E is negative
Changing Internal Energy
• A system of H2 and O2
gas has more internal
energy than one
composed of H2O (l)
• E is lost to surroundings
when H2O (l)is formed
• E is gained when H2O (l)
is decomposed
Increasing Internal Energy
• Heat & work are
added to the system
• Ef > Ei
• Internal energy
increases
• ∆E = Ef -Ei
• ∆E > 0
• ∆E = q + w
Sample Problem
Relating ∆E to heat & work
• H2(g) & O2(g) ignite
• System loses 1150J of heat to
surroundings
• As gases expand, piston rises
• 480 J work done on the piston.
• What ∆E?
• ∆E = q + w
• ∆E = -1150J + (-)480J
• ∆E = -1630 J
• 1630 J was transferred from
system to the surroundings
Endothermic & Exothermic
Processes
• Endothermic….
• System absorbs heat
from surroundings
• Exothermic….
• System releases heat
into surroundings
Energy Diagram
Exothermic Reaction
State – specific characteristics of a
system (physical or chemical)
Examples:
Pressure
Temperature
Composition
State Function – a property of a system that
characterizes a state of a system and is
independent of how the system got to that
state.
Examples:
Pressure
Volume
Composition
www70.homepage.villanova.edu/kathleen.thrush/chapter_6_powerpoint_le.ppt
Thermodynamics
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
Energy it took each
hiker to get to the
top is different !!
www70.homepage.villanova.edu/kathleen.thrush/chapter_6_powerpoint_le.ppt
State Functions
• In the first law of thermodynamics
ΔE = q + w
• ΔE is a state function
• q and w are not
• See fig 5.8 p. 152
5.3 Enthalpy
• Symbol H
• Like internal energy, enthalpy cannot be
measured, but ∆H can
• ∆H equals the heat, qp, gained or lost by a
system under constant pressure
• Exothermic processes:
∆H < 0
• Endothermic processes:
∆H > 0
• ∆H = Hfinal – Hinitial = qp
5.4 Enthalpies of Reaction
Thermochemical Equations
• ∆H = Hfinal – H initial
• ∆Hrxn = Hproducts – Hreactants
• 2H2 (g) + O2 (g)  2H2O (g) ∆H = -486.3 kJ
• When 2 moles of H2 gas combusts to form
2 moles of H2O under conditions of
constant pressure, 486.3 kJ of heat are
released by the system.
• Coefficients indicate molar amounts
associated with ∆Hrxn
Enthalpy Diagrams
Guidelines for Thermochemical
Equations
1. Enthalpy is an extensive property
CH4(g) + O2(g)  CO2(g) + 2H2O(l) ∆H = -890 kJ
2. ∆H for reaction is equal in magnitude but
opposite in sign to ∆H for the reverse rxn
CO2(g) + 2H2O(l)  CH4(g) + O2(g) ∆H = +890 kJ
3. ∆H depends on the state of the reactants
& products
CH4(g) + O2(g)  CO2(g) + 2H2O(g) ∆H = -802 kJ
2H2O(l)  2H2O(g) ∆H = +88 kJ
5.5 Calorimetry
• Calorimetry is the measure of heat flow
involved in a chemical reaction
• Heat capacity: amount of heat energy required
to raise the temperature of an object by 1K
• Molar heat capacity: amount of heat energy
required to raise the temperature of 1 mole of a
pure substance by 1K
• Specific heat: amount of heat energy required
to raise the temperature of 1 g of a substance by
1K
• = heat capacity of 1 gram
Determining specific heat
• During a thermochemical process,
– Measure mass of substance
– Measure change in temperature
– Measure heat energy transferred
Specific heat = q/mΔT
Units: J/g∙K
q = mc(Tf-Ti)
Sample Problem 5.5
• How much heat is needed to warm 250g
of water from 22C to 98C. (Specific heat
of water is 4.18 J/gK)
• Determine the molar heat capacity of
water
Constant Pressure Calorimetry
• Calorimeter
• An insulated container in which aqueous
chemical reactions are conducted
• If the mass and specific heat of the solvent
(water) is known, q can be calculated if ΔT
of the water is measured
• qsoln = - qrxn and qrxn = - qsoln
• qrxn = - (msoln)(csoln )ΔT
Sample 5.6
• 50 mL of 1.0 M HCl mixed with 50 mL of
1.0 M NaOH.
• Given: Ti = 21.0C Tf = 27.5C; density of
soln = 1.0 g/mL; csoln = 4.18 J/gK
• Calculate enthalpy change for the
neutralization reaction
• msoln = V x d = 100mL x 1.0 g/mL = 100 g
• ΔTsoln = Tf - Ti = 27.5 – 21.0C = 6.5C
• = 6.5K
Sample 5.6
• qrxn = - qsoln
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= - (msoln)(csoln )ΔTsoln
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= - (100 g)(4.18J/gK)(6.5K)
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= - 2700 J
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= - 2.7 kJ
• How would you express this change on a
molar basis?
Sample 5.6
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At constant pressure ΔH = q
Express ΔH on a molar basis
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Molar quantities of reacants =
(0.050 L)(1 mol/L) = 0.050 mol
• ΔH = (-2.7 kJ/0.050 mol) = -54 kJ/mol
5.6 Hess’s Law
• If a rxn is carried out in steps, ΔHrxn = sum
of ΔH of each step.
• ΔHrxn = Σ ΔH1 + ΔH2 + ΔH3 + …step.
AD
ΔHrxn = -850 kJ
net rxn
AB
ΔH1 = -175 kJ
step 1
BC
ΔH2 = -400 kJ
step 2
C D
ΔH3 = -275kJ
step 3
Hess’s Law
Eq 1 CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
ΔH = -802 kJ
Eq 2 2H2O (g)  2H2O (l)
ΔH = -88 kJ
Add Eq 1 + Eq 2
CH4(g)+2O2(g)+2H2O (g)  CO2(g)+2H2O(g)+2H2O (l)
ΔH = -890 kJ
Net Eq
CH4(g) + 2O2(g)  CO2(g) + 2H2O (l)
ΔH = -890 kJ
Using Hess’s Law to Calculate ΔH
Sample 5.8
Given:
(1) C(s) + O2(g) → CO2(g)
(2) CO(g) + ½ O2(g) → CO2(g)
ΔH1 = -393.5 kJ
ΔH2 = -283.0 kJ
Determine the ΔH for the reaction
(3) C(s) + ½ O2 (g) → CO(g)
ΔH3 = ? kJ
Using Hess’s Law to Calculate ΔH
Sample 5.8
The basic idea: rearrange equations 1 and 2, if
necessary, in such a way that, when added,
produce equation 3
C(s )  O 2 ( g )  CO 2 ( g )
DH1  - 393.5 kJ
1
CO 2 ( g )  CO( g )  O 2 ( g ) - DH 2  283.0 kJ
2
_____________________________________________
1
C(s )  O 2 ( g )  CO( g )
2
DH 3  - 110.5 kJ
5.7 Enthalpies of Formation
• Thermodynamic quantities:
• ΔHº
Standard Enthalpy
Standard conditions: P = 1 atm, T = 298K
• ΔHvap
Enthalpy of vaporization
Vaporize liquids to gases
• ΔHf
Enthalpy of fusion
Melting solids to liquids
• ΔHc
• ΔHfº
Enthalpy of combustion
Enthalpy of formation
Enthalpy of Formation
• ΔHfº for the formation of 1 mol of a
substance from its elements
• 2 C (graphite) + 3H2 (g) + ½ O2 (g)  C2H5OH (l)
• ΔHfº = -277.7 kJ
• By definition, ΔHfº = 0 for any element
Calculation of DH using DHf°
We can use Hess’s law in this way:
DH =  nDHf°products –  mDHf° reactants
where n and m are the stoichiometric
coefficients.
Calculation of DH using DHf°
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
DH = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ
Energy in Foods
Most of the fuel in the
food we eat comes
from carbohydrates
and fats.
Energy in Fuels
The vast
majority of the
energy
consumed in
this country
comes from
fossil fuels.