Transcript Ppt19(PS8)_Thermo_Hess

Ppt19, Thermochemistry
I.
Basic Ideas and Definitions
 KE(review), heat, temperature, potential energy,
thermal energy, enthalpy, etc.
II. Calorimetry—Obtaining energy changes by
measuring T changes.
III. Thermochemical Equations: Stoichiometry
with Energy!
IV. Hess’s Law (and related Ideas)
 Using energy changes of known reactions to
calculate energy changes of related ones
V. Standard Enthalpies of Formation
 Using tabulated values to calculate energy changes
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Quick “Quiz”
(PS8a, Q1)
True or false (correct if false):
(i) When a chemical bond is broken, energy is
released.
Answer: FALSE. It “takes” (absorbs) energy to
break any chemical bond! No exceptions!!
Breaking a bond is like “pulling apart two
magnets” or “lifting a book”!!
• PE of system increases; energy (either in the form of heat
or work) comes from outside of system (surroundings)
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Quick “Quiz”
(PS8a, Q1)
True or false (correct if false):
(ii) When a chemical reaction takes place,
energy is released.
Answer: FALSE. Some chemical reactions
absorb energy (called “endothermic”), and some
reactions release energy (“exothermic”)
• If only bond breaking occurs: endo
Cl2  2 Cl
• If only bond making occurs: exo
Cl + Cl  Cl-Cl
• If some bonds are broken and some (other ones) are
made, it could be endo or exo (it depends).
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Thermodynamics is the study of
energy changes
• Thermo = “heat” (a type of energy)
• Dynamic = “motion”  “changes”
• Two basic KINDS of energy:
 Kinetic (KE): energy of motion (of a
particle)
 Recall: For a sample of particles:
• KEavg(per particle)  TKelvin
 KEavg (x # particles) is called “thermal energy”
• T is NOT an energy, but it is proportional to one
kind of energy (thermal energy or avg. KE of particles)
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NOTE: T is not the same as “heat”!
• Heat is a type of energy; T is not an “energy”
• Things that are “hot” have a high T
 They have a high average KE per particle
(relative
concept)
 They do not “have” a lot of “heat” in them
• Heat is energy that transfers from a hotter
sample to a colder one
• Confusing because we “sense” heat flow but
brain interprets it as “temperature”:
 An object feels hot if heat transfers into our skin!
 An object feels cold if heat energy transfers out!
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Follow up: Difference between T
and heat
• You go to your car on a hot summer day
after the car has been sitting out for
several hours.
 Which is hotter, the metal belt buckle or the
cloth seat?
• They are the same temperature!!
• The belt buckle feels hotter to you because it
conducts heat well, so the amount of heat that
transfers into your skin each ms is much greater
than the amount of heat that transfers in from the
cloth!
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Potential Energy (PE): The
Second Type of Energy
• PE is energy of position
 Results from forces (e.g., book in gravitational field)
• Chemical potential energy results from forces
between atoms or molecules
 It takes energy to pull bonded atoms apart
 It takes energy to pull molecules in a liquid apart (to
turn into a gas)
 It takes energy to pull an electron away from a
nucleus
• When physical or chemical changes take place,
positions of atoms or molecules change relative
to one another  PE changes!
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Dx means “change in x”
• Dx = xf – xi “final minus initial”
• If T goes from 35 to 45 ºC, then:
 DT = 45 – 35 = +10. ºC
• Did T increase or decrease? It increased
• A positive delta means an increase in the variable
• If T goes from 45 to 35 ºC, then:
 DT = 35 – 45 = -10. ºC
• Did T increase or decrease? It decreased
• A negative delta means a decrease in the variable
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Go to PS8a
first
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1st Law: Energy is neither created nor
destroyed. Euniverse is a constant
• Energy can change forms
 E.g., from KE to PE or vice versa
• Energy can transfer from one “place” to
another
 Define a SYSTEM and a SURROUNDINGS
• Universe = system + surroundings (See Board)
 DEuniv= 0  DEsys + DEsurr = 0
DEsys = - DEsurr (minus  “opposite of”, not “negative!)
• the amount of E that leaves the system equals the amount of
E that enters the surroundings [and vice versa]
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What is Enthalpy?
(Better question: what is the change in enthalpy?)
• Internal Energy (Esys) = sum of all KE & PE of particles
• Enthalpy (H) is a property of a system whose change
is similar to the change in E for typical chemical
processes
 The formal definition of H is not important. Its change is.
• At constant P, the difference between DE and DH is the
amount of work (w) that is done on (or by) the system:
 DH = DE – w
(P const.)
 For typical chemical processes, w << DE, and DH  DE
 Since DE = q + w, q = DE – w …. = DH! (IF P is constant)
 More about “work” later
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The change in Hsys (DHsys) equals the amount of
heat (flow) if a process occurs at constant P
• When a process occurs in the system at constant pressure,
the change in H equals the amount of heat transferred:
DHsys = qP
(subscript “p” means “at const. P”)
 If DHsys > 0, heat flows INTO the system (to make H increase)
 “ENDOTHERMIC”
 If DHsys < 0, heat flows OUT of the system (to make H decrease)
 “EXOTHERMIC”
PS8a, Q5! Also Q6 on PS8b
• Note: if P is NOT constant during some process, the amount of heat (flow) is NOT equal to
the change in H (DHsys)! “H” is “enthalpy”, NOT heat! The change in H just happens to
equal q when a process is carried out at constant P.
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1st Law Applied, and convention for q
• qsys = -qsurr  Key equation (&concept)
• qsys < 0 means heat flowed OUT of the
system (and into the surroundings) & qsurr > 0
• qsys > 0 means heat flowed INTO the
system (and out of the surroundings) & qsurr < 0
• E.g.: If 10 J flows from sys to surr:
qsys = -10 J and qsurr = +10 J
• E.g.: If 20 J flows into sys from surr:
qsys = +20 J and qsurr = -20 J
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DHsys often represents a conversion of PE
into KE or KE into PE
(PE in sys; KE into or from surroundings)
• For a chemical or physical process at
constant P (and T), DHsys ↔ DPEsys
DHsys = qsys AND qsys = - qsurr
 DHsys = - qsurr  Key equation
• Thus, if DHsys < 0 (exothermic), chemical PE in the
system ends up getting converted into KE in the
surroundings, and the energy transfer occurs as
heat (warming up the surroundings)!
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Figure 6.2 (Zumdahl):
Exothermic Process
DHsys = - qsurr
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Endothermic Processes Generally convert
KE of surroundings into PE in system
DHsys = - qsurr
• If endothermic (DH > 0), the rearrangement
in system requires energy to occur, and that
energy flows in from the surroundings (qsurr
< 0)
 [imagine the REVERSE of the process on prior slide]
• The “-” sign means “opposite of”, not “negative”!!!
Concepts in Q’s 1-4, & 6 on PS8b have now been addressed. Try them!
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DHsys is not determined by the
surroundings—it is “assessed” by it!
• NOTE:
The fact that the value of DH equals - qsurr
should not be interpreted to mean that the value
is determined by the surroundings—it is not!!
• The value of DHsys is determined by the
rearrangement (changes in position of atoms
/ molecules / ions) in the system.
 i.e., it is determined by the PROCESS in the system
 The surroundings is just a “reporter” of sorts
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II. Calorimetry
• Used to obtain changes in enthalpy
(DHsys’s) by measuring (changes in)
temperature of the “surroundings” (DTsurr’s).
• Use:
DHsys = - qsurr
A property of the surroundings; can be
determined if Csurr is known via:
qsurr = Cs, surr x msurr x DTsurr
The “surroundings” is usually “reduced” to a calorimeter, a liquid, a solid, etc.
(Assume no heat is lost to the “rest” of the surroundings)
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Reminder (PS 8a, Q3)
“It takes energy to raise a substance’s T” How much energy?
• Cs is the specific heat (capacity) of a
substance
 amount of heat energy needed to raise 1 g of
a substance by 1 C
 A large(r) Cs means “hard(er) to change its T”
 (Other abbreviations: s, S.H., c)
• If the only thing that happens to a
substance (A) is that it changes T, then:
qA = Cs, A x mA x DTA
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Now try Q5 on PS8b!
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Helpful Question—Is a chemical or physical
process taking place, or “just” heat flow?
• If there is NO process in the system (or
surroundings):
 Heat flow is a result of different initial T’s in sys & surr
 T of both sys and surr changes because of heat flow
 q is related to DT by:
• q = C x m x DT in both the sys and surr
 Thus, qsys = -qsurr reduces to:
Csys x msys x DTsys = -(Csurr x msurr x DTsurr)
(no chemical or physical change in system)
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Example—Calorimetry
(case with heat transfer only—no phys or chm change)
#1 on Handout Sheet:
If a 40.1 g piece of iron at 652 °C is
dropped into a sample of 328 g of water at
32.4 °C, what will be the final temperature
after thermal equilibrium is established?
Assume that no heat is lost during the
process. CFe = 0.45 J/(gC)
Q8, Q9, and Q10 on PS8b
use similar ideas/skills. Try
them now!
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Helpful Question—continued
• If there IS a process in the system (but not in the
surroundings) and Tsys is kept constant (common):
 q flow is ultimately caused by the process (see next slide)

Because PE change in the system (DHsys) converted to KE
 qsurr is related to DT
 Thus, qsys = -qsurr reduces to:
DHsys = -(Csurr x msurr x DTsurr)
(IS a chemical or physical change in system)
NOTE: qsys is NOT equal to Csys x msys x DTsys here!! The process
dictates DHsys—surroundings responds to energy change in system
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III. Short but important interlude—
Meaning of Thermochemical Equations
• Before we work with the calorimetry eqn on the
prior slide, recall ideas from PS8a (Q2 & next slide):
1) The amount of DH associated with a process depends
on the amount of the process that occurs
2) The DH for a chemical equation is not the same as the
DHsys associated with an actual chemical reaction.
•
Just like the coefficient in a chemical equation is not the
same as the amount of moles of a substance that actually
“reacts” or “forms” during an actual chemical reaction!
3) “Stoichiometry with energy” idea
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Follow up from PS8a Q2 (diff rxn eqn)
CaO(s) + 3 C(s)  CaC2(s) + CO(g);
5 mol CO x
1) If 5 mol CO is formed:
DH = 465 kJ
3 mol C
= 15 mol C
1 mol CO
How many moles of C react?
What is the DH of the rxn? 5 mol CO x
2) If 8 mol C is to react:
8 mol C x
# of moles of CaO needed?
Amt of energy absorbed?
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465 kJ
= 2325 kJ
1 mol CO
1 mol CaO
= 8 / 3 mol CaO
3 mol C
8 mol C x
465 kJ
= 1240 kJ
3 mol C
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Stoichiometry with energy!
(Example)
#2 on Handout Sheet:
How much heat (in kJ) is evolved or absorbed in the
reaction of 233.0 g of carbon with enough CaO to
produce calcium carbide?
CaO(s) + 3 C(s)  CaC2(s) + CO(g); DH = 464.8 kJ
(b) Is the process exothermic or endothermic?
Reminder: If there’s a “process”, q flow is ultimately caused by that
process, with the amount being dependent on how much
process occurs)
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Another Example
# 3 on Handout Sheet:
85.8 kJ of energy is evolved (i.e., released) at constant
pressure when 3.56 g of P4 is burned according to:
P4(s) + 5 O2(g) → P4O10(s)
What is the ΔH for the (thermo)chemical equation?
Q7 on PS8b uses “Stoichiometry
with energy” ideas/skills. Try it now!
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Return to Calorimetry
• Recall:
qsys = - qsurr
• If there IS a process in the system (but not in
the surroundings), This reduces to
DHsys = -(Csurr x msurr x DTsurr)
(chemical or physical change in system)
DHsys is “caused” by the process in the system (“stoichiometry with energy”),
but we can determine its value experimentally in a particular situation by
measuring the T change of the surroundings.
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Example—Calorimetry
(case with a physical or chemical change)
#4 on Handout Sheet:
Instant cold packs contain solid NH4NO3 and a pouch of
water. When the pack is squeezed, the pouch breaks and
the solid dissolves, lowering the temperature because of the
endothermic process:
NH4NO3(s)  NH4NO3(aq); DH = +25.7 kJ
What is the final temperature in a squeezed cold pack that
contains 50.0 g of NH4NO3 dissolved in 125 mL of water?
Assume the specific heat capacity of the dissolved NH4NO3 is negligible
compared to water, an initial temperature of 25.0 C, and no heat
transfer between the cold pack and the environment. dwater ~ 1.0 g/mL
Q11 and Q12 on PS8b use similar ideas/skills. Try them now!
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Exp 14 Part B
• Dissolve some solid in some water
• In an insulated cup
• System is the solid (plus the small amount of water
molecules that interact with the dissolved FUs of the solid)
– Process that occurs in sys is “dissolution”
• Assume (excess) H2O is the surroundings
• DHsys = - qsurr becomes: DHdiss = - qwater
• qwater = Cwater x mwater x DTwater (assume Twater=Tsol’n)
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Figure 6.9
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Reminder (DE vs DH)
(Slide 11, recopied here)
• Internal Energy (Esys) = sum of all KE & PE of particles
• Enthalpy (H) is a property of a system whose change
is similar to the change in E for typical chemical
processes
 The formal definition of H is not important. Its change is.
• At constant P, the difference between DE and DH is
the amount of work (w) done on (or by) the system:
 DH = DE – w
(P const.)
 For typical chemical processes, w << DE, and DH  DE
 Since DE = q + w, q = DE – w …. = DH! (IF P is constant)
 More about “work” later
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Return to Internal Energy (E)—Heat
and Work Matter here.
• There are two ways to increase the
(internal) energy of a system:
– Have heat (q) flow into it (qsys > 0)
– Have the surroundings do work (w) on it
(w > 0; chemists’ convention)
In equation form:
DEsys = q + w
(chemists)
• Of course, if heat flows out of the system,
or if the system does work, Esys decreases
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Return to Internal Energy (E)—Heat
and Work Matter here.
• Recall that qp = DH, so at constant P
DEsys = q + w becomes
DEsys = DHsys + w and thus
DHsys = DEsys - w (as noted earlier)
 If work is small, DH is
approximately equal to DE
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Figure 6.7 If the system expands against
an external pressure (i.e., piston moves
upward), DVsys is positive and “w” is negative
(system does work on surroundings).
w = -PDV
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Example
#5 on Handout Sheet:
Assume that a particular reaction
produces 244 kJ of heat and that 35 kJ of
PV (expansion/contraction) work is done on the
system. What are the values of DE and
DH for the system? For the surroundings?
Q13 & Q14 on PS8b use similar
ideas/skills. Try them now!
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IV. Hess’s Law
• Value of H (or E) for a system depends
ONLY on the state of the system (i.e., the P,
T, moles of substances, states of substances, etc.)
• It doesn’t matter how you got to that state
• Called a “state function”
• The change in H in going from State 1 to
State 2 does not depend on how you get
there (i.e., “path”).  “Hess’s Law”:
• DHoverall= DH1 + DH2 + DH3 + etc. (1,2,3 are
processes that “add up” to the overall)
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Example 1
3 H2 + N2  2 NH3 ; DH = ???
What is the DH for the above equation if we
know the following?
2 H2 + N2 
N2H4 ; DH = 95.4 kJ
H2 + N2H4  2 NH3 ; DH = -187.6 kJ
Answer: The sum of these two, because the
sum of these two equals the overall process!
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Generalized “Procedure”
for Creating a set of equations that sum to the
equation of interest (“target”)
• See handout sheet examples and
boardwork
• Apply the following ideas (PS8a, Q2!):
1) If you reverse an equation, the sign of DH
opposite
becomes the __________.
2) If you multiply an equation through by a
number x, the DH becomes ____
x times the
original value.
Q’s 15-17 on PS8b use similar
ideas/skills. Try them now!
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“Special” Example
Find DH for
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) given:
C(s) + 2 H2(g) → CH4(g);
O2(g) → O2(g);
DH1 = -74.8 kJ
DH2 =
?
0
kJ
C(s) + O2(g) → CO2(g);
DH3 = -393.5 kJ
H2(g) + ½ O2(g) → H2O(l);
DH4 = -285.8 kJ
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b/c nothing
changed!
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“Special” Example
Find DH for
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
CH4(g) → C(s) + 2 H2(g) ;
DH1’ = - DH1
2 O2(g) → 2 O2(g);
DH2’ = -2DH2
------------------------------------------------------------------------------------------------------------------
DH3’ =
C(s) + O2(g) → CO2(g);
2 H2(g) + O2(g) → 2 H2O(l);
DH3
DH4’ = 2DH4
Elements!
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“Special” Example
Find DH for
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) given:
C(s) + 2 H2(g) → CH4(g);
O2(g) → O2(g);
DH1 =
DH2 =
C(s) + O2(g) → CO2(g);
DH3 =
H2(g) + ½ O2(g) → H2O(l);
DH4 =
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Q: Is there an “easy” way to calculate
the DHeqn without doing calorimetry?
• Yes and no
• Calorimetry does have to be done, but it can be
done ahead of time and not on the reaction of
interest.
 Obtain an effective value of “H” for one mole of
every substance (via calorimetry)
 Effective “H” is actually called the “standard
enthalpy of formation” of a substance X: DH°f(X)
• Use these “H” values of reactants and products
(substances) to calculate DHrxn for any chemical
equation!
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How to calculate DH°eqn
(from tabulated data)
For an equation:
a A + b B  c C + d D;
DH°eqn
DH°eqn = “H” of all products” - “H” of all reactants
“Hfinal”
“Hinitial”
= [c DH°f(C) + d DH°f(D)] - [a DH°f(A) + b DH°f(B)]
mol C x
kJ
mol C
From Tro:
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CH3Cl(g)
Find DH for :
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
Find DH for :
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
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How to calculate DH°eqn
EXAMPLE
Find DH for : CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
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What’s really going on with the prior
equation/procedure
• Imagine making ELEMENTS from
reactants and then turning ELEMENTS
into products. (Generalized “path” for any
reaction!!)
• Determine DH for making a substance
from its ELEMENTS (called DHformation)
• Tabulate these “DHf’s” for ALL
SUBSTANCES
• Use them to calculate DHrxn for any
chemical equation! Ppt19
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NOTES
• The DH°f of a substance is the:
– Enthalpy change associated with forming one mole of
a substance from its elements
– As such, the value for any element is zero*
• The DH°f for a substance depends on the
physical state of the substance in question
(because it takes or releases energy to change
a substance’s state)
* In its standard state
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Figure 6.8 Pathway for the Combustion
of Methane
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Figure 6.9 Schematic Diagram of
Energy Changes
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Figure 6.11 (Tro):
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Figure 6.10 A Pathway for the Combustion
of Ammonia
I showed on the board how to use data from the Table to find DH for this
process. Can generalize: DHoverall = Sum(n x DHf [P’s]) - Sum(n x DHf [R’s])
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