Ppt19, Thermochemistry

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.

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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)

Quick “Quiz”(PS8a, Q1)

True or false (correct if false):(ii) When a chemical reaction takes place, energy

is released.

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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).

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”

(relative concept)

• Things that are “hot” have a high T They have a high average KE per particle

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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x means “change in x”

• x = xf – xi “final minus initial”

• If T goes from 35 to 45 ºC, then: T = 45 – 35 = +10. ºC

• Did T increase or decrease?

• A positive delta means an increase in the variable

It increased

• If T goes from 45 to 35 ºC, then: T = 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)

Euniv= 0 Esys + Esurr = 0Esys = - Esurr (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.

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• At constant P, the difference between E and H is the amount of work (w) that is done on (or by) the system: H = E – w (P const.)

For typical chemical processes, w << E, and H E Since E = q + w, q =E – w …. = H! (IF P is constant)

More about “work” later

The change in Hsys (Hsys) 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:

Hsys qP (subscript “p” means “at const. P”)

If Hsys > 0, heat flows INTO the system (to make H increase)

“ENDOTHERMIC”

If Hsys < 0, heat flows OUT of the system (to make H decrease)

“EXOTHERMIC”

• Note: if P is NOT constant during some process, the amount of heat (flow) is NOT equal to the change in H (Hsys)! “H” is “enthalpy”, NOT heat! The change in H just happens to equal q when a process is carried out at constant P.

PS8a, Q5! Also Q6 on PS8b

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1st Law Applied, and convention for q

• qsys = -qsurr Key equation (&concept)

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• 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

• 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

Hsys = qsys AND qsys = - qsurr

Hsys = - qsurr Key equation

Hsys 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), Hsys ↔ PEsys

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• Thus, if Hsys < 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)!

Figure 6.2 (Zumdahl):

Exothermic Process

Hsys = - qsurr

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Endothermic Processes Generally convert KE of surroundings into PE in system

Hsys = - qsurr

• If endothermic (H > 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”!!!

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Concepts in Q’s 1-4, & 6 on PS8b have now been addressed. Try them!

Hsys is not determined by the surroundings—it is “assessed” by it!

• NOTE: The fact that the value of H equals - qsurr should not be interpreted to mean that the value is determined by the surroundings—it is not!!

• The value of Hsys 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 (Hsys’s) by measuring (changes in) temperature of the “surroundings” (Tsurr’s).

• Use:Hsys = - qsurr A property of the surroundings; can be

determined if Csurr is known via:

qsurr = Cs, surr x msurr x Tsurr

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)

• 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 TA

“It takes energy to raise a substance’s T” How much energy?

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Now try Q5 on PS8b!

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Thus, qsys = -qsurr reduces to:

Csys x msys x Tsys -(Csurr x msurr x Tsurr)

(no chemical or physical change in system)

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 T by:

• q = C x m x T in both the sys and surr

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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/(gC)

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Q8, Q9, and Q10 on PS8b use similar ideas/skills. Try them now!

Thus, qsys = -qsurr reduces to:

Hsys -(Csurr x msurr x Tsurr)

(IS a chemical or physical change in system)

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 (Hsys) converted to KE

qsurr is related to T

NOTE: qsys is NOT equal to Csys x msys x Tsys here!! The process dictates Hsys—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):

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1) The amount of H associated with a process depends on the amount of the process that occurs

2) The H for a chemical equation is not the same as the Hsys associated with an actual chemical reaction.

3) “Stoichiometry with energy” idea

• 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!

Follow up from PS8a Q2 (diff rxn eqn)

CaO(s) + 3 C(s) CaC2(s) + CO(g); H = 465 kJ

1) If 5 mol CO is formed:

How many moles of C react?

What is the H of the rxn?

2) If 8 mol C is to react:

# of moles of CaO needed?

Amt of energy absorbed?

3 mol C5 mol CO x 15 mol C

1 mol CO

465 kJ5 mol CO x 2325 kJ

1 mol CO

1 mol CaO8 mol C x 8 / 3 mol CaO

3 mol C

465 kJ8 mol C x 1240 kJ

3 mol C

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Stoichiometry with energy! (Example)


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); H = 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?

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Q7 on PS8b uses “Stoichiometry with energy” ideas/skills. Try it now!

Return to Calorimetry

• Recall:qsys = - qsurr

• If there IS a process in the system (but not in the surroundings), This reduces to

Hsys -(Csurr x msurr x Tsurr)

(chemical or physical change in system)

Hsys 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)


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); H = +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

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Q11 and Q12 on PS8b use similar ideas/skills. Try them now!

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

Hsys = - qsurr becomes: Hdiss = - qwater

• qwater = Cwater x mwater x Twater (assume Twater=Tsol’n)

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Figure 6.9

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Reminder (E vs H)(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.

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• At constant P, the difference between E and H is the amount of work (w) done on (or by) the system: H = E – w (P const.)

For typical chemical processes, w << E, and H E Since E = q + w, q =E – w …. = H! (IF P is constant)

More about “work” later

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: Esys = 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 = H, so at constant P

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Esys = q + w becomes

Esys = Hsys + w and thus

Hsys = Esys - w (as noted earlier)

If work is small, H is approximately equal to E

Figure 6.7 If the system expands against an external pressure (i.e., piston moves upward), Vsys is positive and “w” is negative (system does work on surroundings).

w = PV

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Example


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 E and H for the system? For the surroundings?

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Q13 & Q14 on PS8b use similar ideas/skills. Try them now!

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”:

• Hoverall= H1 + H2 + H3 + etc. (1,2,3 are

processes that “add up” to the overall)

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Example 1

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2 H2 + N2 N2H4 ; H = 95.4 kJ

H2 + N2H4 2 NH3 ; H = -187.6 kJ

What is the H for the above equation if we know the following?

3 H2 + N2 2 NH3 ; H = ???

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

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• Apply the following ideas (PS8a, Q2!):1) If you reverse an equation, the sign of H

becomes the __________.

2) If you multiply an equation through by a number x, the H becomes ____ times the original value.

opposite

x

Q’s 15-17 on PS8b use similar ideas/skills. Try them now!

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“Special” Example

Find H for

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) given:

C(s) + 2 H2(g) → CH4(g); H1 = -74.8 kJ

O2(g) → O2(g); H2 = kJ

C(s) + O2(g) → CO2(g); H3 = -393.5 kJ

H2(g) + ½ O2(g) → H2O(l); H4 = -285.8 kJ

? 0

b/c nothing changed!

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“Special” ExampleFind H for

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

CH4(g) → C(s) + 2 H2(g) ; H1’ = - H1

2 O2(g) → 2 O2(g); H2’ = -2H2

------------------------------------------------------------------------------------------------------------------

C(s) + O2(g) → CO2(g); H3’ = H3

2 H2(g) + O2(g) → 2 H2O(l); H4’ = 2H4

Elements!

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“Special” Example

Find H for

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) given:

C(s) + 2 H2(g) → CH4(g); H1 =

O2(g) → O2(g); H2 =

C(s) + O2(g) → CO2(g); H3 =

H2(g) + ½ O2(g) → H2O(l); H4 =

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Q: Is there an “easy” way to calculate the Heqn 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.

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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: H°f(X)

• Use these “H” values of reactants and products (substances) to calculate Hrxn for any chemical equation!

How to calculate H°eqn (from tabulated data)

For an equation:

a A + b B c C + d D; H°eqn

[c H°f(C) + d H°f(D)] “Hfinal” “Hinitial”

[a H°f(A) + b H°f(B)]

H°eqn “H” of all products” “H” of all reactants

mol CkJ

x mol C From Tro:

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Find H for : CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

Find H for : CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)

CH3Cl(g) -86.3

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How to calculate H°eqn EXAMPLE

Find H 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 H for making a substance from its ELEMENTS (called Hformation)

• Tabulate these “Hf’s” for ALL SUBSTANCES

• Use them to calculate Hrxn for any chemical equation!

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NOTES

• The H°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 H°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 H for this process. Can generalize: Hoverall Sum(n x Hf [P’s]) - Sum(n x Hf [R’s])

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Ppt19, Thermochemistry