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UNIT 4:. THERMO 1. UNIT 4: THERMODYNAMICS I Text: Chapter 6 Thermodynamics is Thermochemistry is TERMS: “ the system ” = object being studied (or focused on) “ the surroundings ” = everything else in the universe. t he study of energy and its interconversions.

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UNIT 4:

THERMO 1


UNIT 4: THERMODYNAMICS I Text: Chapter 6

Thermodynamics is

Thermochemistry is

TERMS:

“the system” = object being studied (or focused on)

“the surroundings” = everything else in the universe

the study of energy and its interconversions

the study of energy changes associated

with physical & chemical changes


H2O

DEMO: NaOH(s) -------> Na+(aq) + OH-(aq)

pellets dissolved in water

heat released by “the system”

(NaOH and the water)

heat absorbed by “the surroundings”

(glass beaker, your hand, air in room)

“the system”

released energy into

“the surroundings”

=

  • [energy lost by system = energy gained by surroundings]


What IS energy? that which is necessary

SI unit for ENERGY is the JOULE.

The English unit is the CALORIE (1 cal = 4.184J)

ENERGY is NOT measured directlylike mass or volume.

We can onlymeasure ________________ in terms of ___________

(or the capacity)

to do work “w” or to produce heat “q”

changes in energy

work and heat


ENERGY:

1) potential = “stored” energy due to

________

or ____________

2) kinetic = energy of “motion”

KEparticle =

Forms:

9.8m/s2 gravity

position

PE = m•g•h

composition

: stored in bonds

between particles

mv2

2

  • mechanical

  • nuclear

  • electrical

  • thermal(heat)

  • electromagnetic(light)

  • sound(sonar)

  • chemical


random

particle motion in matter

THERMAL(HEAT) ENERGY reflects the

and is affected by:

1) TEMPERATURE: as T _____, heat energy _____

2) QUANTITY of the substance: as MASS _____,

heat energy ____

Heat involves

due to

(from -----> )

the transfer of energy between two objects

temperature differences

warmer object cooler object


  • PHYSICAL CHANGE: ex: warm hand melts ice

  • H2O(s) ---> H2O(l)endothermic for the ice (“the system”)

  • CHEMICAL CHANGE:

  • atoms (or ions) “rearranging” to make products

  • breaking bonds _______ energy (_____thermic)

  • 2) making bonds ________ energy (____thermic)

endo

absorbs

releases

exo


2

COMBUSTION of methane is EXOTHERMIC because

energy needed to energy needed to

break bondsmake bonds

(absorbed) (released)

CH4(g) + O2(g) ------->

CO2(g) + H2O(g)

CH4(g) + O2(g) -----> CO2(g) + H2O(g) + energy(heat & light)

2

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

higher PE due to

weaker bonds

(more reactive, less “stable”)

  • PE

heat released

2

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

lower PE due to stronger bonds

(less reactive, more “stable”)

2

2

higher PE = lower PE +energy


SYNTHESIS of nitrogen monoxide is ENDOTHERMIC because

energy needed to energy needed to

break bondsmake bonds

(absorbed) (released)

NO(g)

N2(g) + O2(g) ---------->

energy + N2(g) + O2(g) -----> NO(g)

  • 2

  • - - - - - - - - - - - - -

  • higher PE, weaker bonds

  • (less “stable”)

  • PE

  • heat absorbed

  • - - - - - - - - - - - - - - - - -

  • lower PE, stronger bonds

  • (more “stable”)

2

  • energy+lower PE =higher PE


1st Law of Thermodynamics:

the energy of the universe is constant

 The E (internal energy) is defined as

Esystem = PEsystem + KEsystem

Ecan be changed by a “flow” of heat(q) or work(w) or both!!

DE = q + w

“heating water”

will have a positive q

(adding heat to the system)

“condensing a vapor”

will have a negative q

(subtracting heat from the system)

the sum of the

kinetic & potential energies of all particles in the system.

+q

-q

  • ∙∙


We focus on

“the system”.

heat

added

to

subtracted

from

  • +q

  • -q

  • system

added

to

subtracted

from

  • -w

  • +w

work

  • surroundings


1st Law of Thermodynamics: really means energy is neither created nor destroyed in ordinary chemical and physical changes.

Thermodynamic “STATE” of a system is defined as a set of conditions which includes

Temperature

Pressure

Volume

Physical state (solid, liquid, gas)

Composition (identity of substances and # of moles)


These are considered properties of a system called

“state functions”

[heat (q) & work (w) are NOT state functions, but part of the pathway]

The value of “state function” depends only on STATE of that system not on HOW it got to that state (pathway)!!

A change in “STATE” describes the difference between 2 states

(independent of the pathway)

“FINAL” - “INITIAL” = change in state function

Ex: DT = Tfinal – Tinitial or DV = Vfinal– Vinitial


E(internal energy) includes all energywithin a substance:

•KE of particles

•attractive forces “between” p+s and e-s

(ionization energies)

•intermolecular forces between atoms, molecules or ions, etc

E = (Efinal – Einitial) = (Eproducts – Ereactants) = q + w

chemical change


The ONLY type of work involved in most chemical & physical changes is pressure-volume work. (work = force x distance)

P = force = f V = (L x W x H) so V = d3

aread2

P x V = f x d3= f x d = work

d2

Work done “on” or “by” a SYSTEM depends on

the external pressure and the volume.

DE = q + w

DE = q + -PDV

d

d

d


A. When gas expands-PDV = -P(V2>V1) = -w

B. When gas contracts-PDV = -P(V2<V1) = +w

C. In constant volume reactions, no PDV is done (because nothing moves through a distance)

Solids & liquids do NOTexpand/contract significantly with pressure so DV~0


Ex: 2NH4NO3(s) ------> 2N2(g) + 4H2O(g) + O2(g)

solid ALL gases

Dn = nproduct gases - nreactant gases

Dn = so DV is ____ w = -PDV

so expansion means -w !!

(work comingout of system)

7mol of gases

0 mol of gas

= 7 mol– 0 mol

+7 mol

(+)

= -P(+)


Ex: 2SO2(g) + O2(g) ------> 2SO3(g)

w = -PDV

  • 3 molgases

  • 2molgases

Dn= 2 - 3

  • = -1

  • = -P(-)

  • so contractionmeans +w!!

  • (work beingdone on system)


  • Ex: H2(g) + Cl2(g) ------> 2HCl(g)

  • 2 mol gases

  • 2 mol gases

  • Dn = 2 - 2

w = -PDV

= 0

= -P(0)

  • so constant volumemeans w = 0

  • (no work being done)


  • ENTHALPY, H “heat content” (from Day 2)

  • DE = q + w and therefore

  • DE = q+ -PDV

  • +PDV +PDV then

  • DH = DE + PDVsince then

  • DH = (q - PDV) + PDV therefore

  • DH = qp

  • Denthalpyequals heat gained or [email protected] “constant pressure”

  • “change of heat for a reaction” or “enthalpy change”:

  • DHreaction = Hproducts - Hreactants


Ex: When KOH(s) is dissolved in water, heat is released :

KOH(s) -----> K+(aq) + OH-(aq) DH = -43kJ/mol(exo!)

Problem: How much heat is released when 14.0g of KOH

is dissolved in water?

Enthalpyis an “extensive property”

(depends on amount of substance present).

  • H2O

(means: for every 1 mol KOH dissolved, 43 kJ of heat will be released)

  • “direction

  • of heat flow”

K- 39.10

O- 16.00

H – 1.01

56.11g/mol

-11kJ

  • 1mol_

  • 56.11g

x-43kJ=

1mol

14.0g x

-10.7289..

Ans: 11kJ released


For the reverse reaction,

the enthalpy is opposite in signbut equal in magnitude.

Ex: CH4(g)+ 2O2(g)---> CO2(g)+2H2O(g)DH = -890 kJ

CO2(g)+2H2O(g)--->CH4(g)+ 2O2(g)DH = +890 kJ

(exo)

(endo)


  • A “thermochemical equation” expresses the energy change, DH, for a reaction “as written”:

  • Ex: the combustion of ethanol:

  • C2H5OH(l)+ 3O2(g)----> 2CO2(g)+3H2O(g)DH = -1367 kJ

  • So DH = -1367 kJ= -1367 kJ= -1367 kJ= -1367 kJ

  • 1molC2H5OH 3molO22molCO2 3molH2O

  • How much heat is released when 275.0g CO2 are produced?

  • C- 12.01

  • O- 2(16.00)

  • 44.01g/mol

  • 1mol_

  • 44.01g

275.0g x

  • x-1367kJ=

  • 2molCO2

-4271kJ

-4270.90..

4271kJ released


(lesser H) - (greater H)

So DH = (-) exo

heat coming “out of”or being

“subtracted from” the reactants


  • So DH = (+) endo

  • heat going “into” or being

  • “added to” the reactants


1 g/mL

spec. heat:

4.18J

goC

DT = Tfinal – Tinitial

initial

final

= 85.0oC - 75.0oC

= 10.0oC

  • (4.18J)

  • (goC)

= (500.g)

(10.0oC)

20900

q = 20900J

= 20.9kJ absorbed


DT =

(78.2J) =

(45.6g)

(c)

(13.3oC)

(c)

=

  • (78.2J)

0.12894

(45.6g)

(13.3oC)

c = 0.129J/goC


M x VL = mol

(1.0M)(0.050L)= 0.050molHCl

DHneut=kJ/mol

= 2.7kJ

0.050mol

need total mass

of solution100.g

DHneut= -54kJ/mol

exo

  • (4.18J)

  • (goC)

= (100.g)

6.5oC

2717

6.5oC= DT

  • q = 2700J = 2.7kJ


DT = Tfinal – Tinitial

Tf=x

(x – 80.0oC)

(x - 55.0oC)

(4.18J)

(goC)

-(50.0g)

(4.18J)

(goC)

= (20.0g)

-50.0x

= 20.0x

+ 4000oC)

- 1100oC)

+50.0x

+50.0x

+ 1100oC)

+ 1100oC)

= 70.0x

5100oC

72.857

Tfinal=

72.9oC


(x – 100.0oC)

(x – 23.0oC)

(0.444J)

(goC)

-(15.0g)

(4.18J)

(goC)

= (55.0g)

(x – 100.0oC)

(x – 23.0oC)

= 230.J

oC

-6.66J

oC

= 230.x

-6.66x

+ 666oC

– 5290oC

+6.66x

+ 5290oC

+ 5290oC)

+6.66x

= 237x

5956oC

25.130

Tfinal=

25.1oC


cm3

q = mcDT

  • q = DHvapm

need J/g

425oC

3

2

356.6oC

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

1

need g Hg:

25oC

-38.8oC

- - - - - - - - -

125cm3 x 13.5g =

cm3

1687.5

1690g


25oC(l)-->356.6oC(l)

  • q = m c DT

  • (331.6oC)

  • = (1690g)

(0.139J)

(goC)

  • 356.6oC

  • -25.000…oC

  • 77,896.15

q = 77,900J

  • DT = 331.6oC

356.6oC(l)-->356.6oC(g)

  • q = DHvapm

  • (1690g)

59.1kJ

mol

x _1mol_

200.59g

x1000J

1kJ

= 295J

g

  • 498,550

= 295J/g

q = 499,000J

294.63


356.6oC(g)-->425oC(g)

  • q = m c DT

  • 425oC

  • -356.6oC

  • = (1690g)

  • (0.102J)

  • (goC)

  • (68oC)

  • 68.4

q = 12,000J

  • 11,721.8

  • DT = 68oC

  • 77,900J

  • 499,000J

  • +12,000J

588,900J = 588.9kJ absorbed


We can only measure the

“change in enthalpy”

USING HEATS of REACTION:

HESS’S LAW statesthatthe enthalpy change for a reaction is the samewhether it occurs in one stepor a series of steps.

(Law of Heat Summation)

one step

reactants----------------------> products energy difference

---> ---> ---> ---> will besame!

series of steps

DHreaction = Hproducts - Hreactants


Example: oxidation of nitrogen to nitrogen dioxide:

One step:N2(g)+ 2O2(g)-----> 2NO2(g)DH = 68kJ (endo)!

Two steps: N2(g)+ O2(g)-----> 2NO(g)DH = 180kJ (endo)!

2NO(g)+ O2(g) -----> 2NO2(g)DH = -112kJ (exo)!

cancel

common terms, ___________________________________ __________________

then add

N2(g) + 2O2(g)-----> 2NO2(g)DH = 68kJ (endo)!


Rules for calculating enthalpy changes:

Take the reactions given, rearrange them and combine them so that they add up to the net reactionyou are asked to find.

1) if an equation is reversed, then the sign on DH is reversed (bec. heat flow is opposite)

2) the magnitude of DH is directly proportional to the moles of reactants and products in the equation.

If the coefficients in the equation are multiplied by an integer, then DHis multiplied by the same integer.

Thus DHreaction= DH1+ DH2 + DH3…….


flip

2

2

DH = -34kJ

SiH4(g) ---> Si(s) + 2H2(g)

  • Si(s) + O2(g)--->SiO2(s)

DH = -911kJ

DH = -484kJ

  • 2H2(g) + O2(g)--->2H2O(g)

  • SiH4(g) + 2O2(g)---> SiO2(s) + 2H2O(g)

DH = -1429kJ


  • 2

  • flip

2

  • a. 2S(s)+ 3O2(g) --->2SO3(g)

  • DH = -790.4kJ

  • b. 2SO3(g) ---> O2(g) + 2SO2(g)

  • DH = +198.2kJ

  • 2S(s)+ 2O2(g) --->2SO2(g)

  • DH = -592.2kJ

2

2

  • DH = -296.1kJ

  • S(s)+ O2(g) --->SO2(g)


STANDARD ENTHALPY OF FORMATION

The standard enthalpy of formation DHofof a compound is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their “standard states”.

(also called standard molar enthalpy of formation or just

heat of formation)

The degree symbol on a thermodynamic function, as in DHofindicates that the corresponding process has been carried out under “standard conditions” (not STP!)

The THERMOCHEMICAL“STANDARD STATE” of a substance is its most stable state at standard pressure

(1atm) and at room temp (25oC, 298K) unless otherwise specified.


DHof refers to “reactants in their standard states” ---> “products in their standard states”

STANDARD STATE:

For a Compound:

1. the standard state of a gaseous compound is a pressure of exactly 1atm.

2. the standard state of a liquid or solid compound is the

pure liquid or solid.

3. the standard state of a solution is a concentration of

exactly 1 Molar.

For an Element:

the standard state is the form in which the element exists at 1atm and 25oC.

ex: O2(g)Br2(l)Fe(s)

Note: the enthalpy of formation DHof for any element in its standard state is zero!!!


The enthalpy change for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants

from the enthalpies of formation of the products.

DHoreaction= SnpDHof(products) - SnrDHof(reactants)

S (sigma) means “sum of”

np means moles of products

nrmeans moles of reactants

Elements are not included in the calculation because elements require no change in form.


USING THE TABLE FOR “HEAT OF FORMATIONS”DHof

Problem #1: Calculate the DHrxn

2NH3(g) + 3O2(g) + 2CH4(g) ------> 2HCN(g) + 6H2O(g)

DHof: ____kJ/mol___kJ/mol_____kJ/mol_____kJ/mol_____kJ/mol

135.1

-75

-46

0

-242

DHorxn =

products - reactants

2mol(135.1kJ)

mol

  • +6mol(-242kJ)

  • mol

2mol(-46kJ) +

mol

3mol(0kJ) +

mol

2mol(-75kJ)

mol

270.2kJ

-92kJ

  • =

+ -1452kJ

+ -150kJ

-242kJ

-1181.8kJ

-939.8

  • DHorxn =

-940.kJ


Problem #2: If DHrxn = 85kJ, find DHoffor NaHCO3

2NaHCO3(s) -------> Na2CO3(s) + CO2(g) + H2O(l)

DHof: _?__kJ/mol_______kJ/mol_____kJ/mol_____kJ/mol

baking soda

  • -394

  • -286

  • -1131

1mol cancels mol

  • DHorxn= [products] – [reactants]

  • (-286kJ)

  • (-394kJ)+

  • (-1131kJ) +

85kJ =

2(x)

85kJ =

  • (-1811kJ) -

  • 2x

-85kJ

  • +2x

-85kJ

  • +2x

  • 2x = -1896kJ

  • x = -948kJ

  • DHof= -948kJ/mol


C + E --> C

C + C--> C

E + E --> C only!

2 ways to find DHrxn for this synthesis

1) DHrxn= 2mol (-826kJ) = -1652kJ

(mol)

2) DHrxn = SnprodDHof-SnreactDHof

-

+

0kJ

(mol)

=

0kJ

(mol)

2mol(-826kJ)

(mol)


synthesis

D

DHorxn = +1652kJ

2Fe2O3(s) ---> 4Fe(s) + 3O2(g)

g-->mol

25.0gFe x 1mol x 1652kJ =

55.85g 4molFe

185kJ absorbed

184.87


If 1164 kJ is required to decompose BaO,

find DHofof BaO.

(Given: DHrxnFind: DHof)

2

2

DHrxn = +1164kJ

BaO(s) Ba (s) + O2(g)

now balance

now reverse to make it synthesis

DHrxn = -1164kJ

2Ba(s) + O2(g) 2BaO(s)

2

2

Now how can we make it a “formation” equation? Think of Def!

Ba(s) + 1 O2(g)BaO(s)

DHof= -582kJ/mol

2

1 mol


REVIEW FOR THERMO 1 TEST

Is work done BY or ON the system?

a) 2AB(g) + C2(g)------> 2ABC(g)

2mol

3mol

22.4L

22.4L

22.4L

22.4L

22.4L

compression

w = -PDV

DV = Vf - Vi

= -P(-1)

= 2 - 3

DV = -1

= +w

added to system

work done ON the system


b) 2C(s) + O2(g)------> 2CO(g)

1mol

2mol

22.4L

expansion

DV = Vf - Vi

w = -PDV

= -P(+1)

= 2 - 1

DV = +1

= -w

subtracted from system

work done BY system


2) If a system expands from 2.5L to 10.0L at a constant pressure of 5.0atm, find work done in Joules. Know this!! 1L•atm = 101.3J

DV = 10.0L – 2.5L

= 7.5L

w = -PDV

w = -(5.0atm)

(7.5L)

(101.3J)

(1L•atm)

-3798.75

w = -3800J

work done BY system, expansion


3) Write a “FORMATION” equation for NaNO3

write elements’formulas and state of matter symbols

balance the equation (this is the reaction equation)

÷ moles of product to get 1 moleof product (Def of “formation” eq)

REACTION EQUATION:

____________________________________________________

FORMATIONEQUATION:

____________________________________________________

Look up heat of formation of NaNO3 : DHof = __________________

2

2

3

Na(s) + N2(g) + O2(g)------> NaNO3(s)

Na(s) + 1N2(g) + 3O2(g)------> NaNO3(s)

2 2

-467kJ/mol


Problem: Find heat released when 1.0g N2 reacts in excess Na and O2.

Since we will have to use the “heat of the reaction” not

“heat of formation”: DHorxn = __________________

So the thermochemical equation is:

__________ = _________ = __________ = ___________

Now we can solve the problem:

1.0gN2 x __________ x ______ =

2mol

(-467kJ)=

mol

-934kJ

-934kJ

1molN2

-934kJ

2molNaNO3

-934kJ

3molO2

-934kJ

2molNa

1molN2

-934kJ

1molN2

33kJ released

28.02g

-33.33


4) “Standard State” DHorxnDHof

means T = ____________

P = _____ = 760.torr

lithium ( )

bromine ( )

fluorine ( )

AgNO3(aq) = __________

25oC = 298K

1atm

s

l

g

1 Molar


State Functions:

(capital letters)

properties of a system that are

independent of the pathway

V

P

T

H

E

volume

pressure

temperature

enthalpy (heat content)

internal energy

q & w  state functions


DHof

= kJ

mol

if (-), then STABLE compound

The greater the (-) value,

the more stable the compound,

the stronger the bonds

if (+), then UNSTABLE compound


MORE STABLE

MORE STABLE

DHrxn = (-) exo

DHrxn = (+) endo

heat coming out of

(subtracted from)

heat going into

(added to)

DHreaction = Hproducts - Hreactants


CALORIMETRY PROBLEMS

If 1) metal & water OR

2) hot water & cold water

-q = +q

metal calorim.

H2O

hot water cold water

-mcDT = +mcDT

If 1) chemical reaction (s) + (aq) OR

2) dissolving (s) in (aq)

q = mcDT

total

DHneut

kJ/mol

Solve for q (Joules), J-->kJ

Convert g-->mol,

 kJ by mol

DHsol’n

Is heat going into system (+) or

out of system (-) ?


  • DHvap

BPt

L-->G

G

MPt

  • DHfus

L

S-->L

S

c = “specific heat”

J/goC

Convert if need be:

  • J/moloC ---> J/goC

  • mol ---> g


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