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 associatedwith physical & chemical changes

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)

H2O

“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 directly like mass or volume.

We can only measure ________________ in terms of ___________

 

(or the capacity) to do work “w” or to produce heat “q”

changes in energywork and heat

ENERGY:

1) potential = “stored” energy due to  

________  or ____________

2) kinetic = energy of “motion” KEparticle =

Forms:

position PE = m•g•h

9.8m/s2 gravity

composition : stored in bonds between particles

mv2

2thermal(heat) electromagnetic(light)sound(sonar)chemical

mechanicalnuclearelectrical

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 -----> )

random particle motion in matter

↑ ↑

↑↑

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 1) breaking bonds _______ energy (_____thermic) 2) making bonds ________ energy (____thermic)

absorbs

releases

endo

exo

COMBUSTION of methane is EXOTHERMIC becauseenergy needed to energy needed to break bonds make bonds

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

  

 

 

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

 

PE

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

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

heat released

higher PE due toweaker bonds(more reactive, less “stable”)

lower PE due to stronger bonds (less reactive, more “stable”)

2 2

higher PE = lower PE + energy

2

2

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

SYNTHESIS of nitrogen monoxide is ENDOTHERMIC becauseenergy needed to energy needed to break bonds make bonds

(absorbed) (released) 

NO(g)

 

 

 

 

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

 

PE heat absorbed

higher PE, weaker bonds(less “stable”)

lower PE, stronger bonds (more “stable”)

energy + lower PE = higher PE

2- - - - - - - - - - - - -

2

the sum of the kinetic & potential energies of all particles in the system.

1st Law of Thermodynamics: the energy of the universe is constant The E (internal energy) is defined as  

Esystem = PEsystem + KEsystem

 

E can 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)

+q

-q∙∙∙∙ ∙∙∙ ∙

∙∙

 

We focus on “the system”.system

surroundings

heat

work

+q

+w

-q

-w

added

to

added

to

subtracted from

subtracted from

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

TemperaturePressureVolumePhysical 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 energy within 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

area d2

 

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) = -wB. When gas contracts -PDV = -P(V2<V1) = +wC. In constant volume reactions, no PDV is done

(because nothing moves through a distance)

Solids & liquids do NOT expand/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 coming out of system)

0 mol of gas 7 mol of gases

= 7 mol – 0 mol

+7 mol (+) = -P(+)

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

w = -PDV

3 mol gases 2 mol gases

Dn = 2 - 3= -1

= -P(-)

so contraction means +w !! (work being done on system)

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

2 mol gases 2 mol gases

Dn = 2 - 2 = 0 w = -PDV = -P(0)

so constant volume means 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 + PDV since then

DH = (q - PDV) + PDV therefore

DH = qp

Denthalpy equals heat gained or lost @ “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?

Enthalpy is an “extensive property” (depends on amount of substance present).

H2O

K- 39.10O- 16.00H – 1.01 56.11g/mol

14.0g x 1mol_56.11g

x -43kJ = 1mol

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

-10.7289..

-11kJ“direction of heat flow”

Ans: 11kJ released

For the reverse reaction, the enthalpy is opposite in sign but 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 3molO2 2molCO2 3molH2O  

How much heat is released when 275.0g CO2 are produced?C- 12.01O- 2(16.00) 44.01g/mol 275.0g x 1mol_

44.01gx -1367kJ = 2molCO2

-4270.90..

-4271kJ

4271kJ released

(lesser H) - (greater H)

heat coming “out of” or being “subtracted from” the reactants

So DH = (-) exo

(greater H) - (lesser H)So DH = (+) endo

heat going “into” or being “added to” the reactants

initial finalDT = Tfinal – Tinitial

= 85.0oC - 75.0oC

= 10.0oC

spec. heat: 4.18J goC

= (500.g) (4.18J) (goC)

(10.0oC)

20900q = 20900J = 20.9kJ absorbed

1 g/mL

DT =

(78.2J) = (45.6g) (c) (13.3oC)

0.12894

(c) (78.2J)(45.6g) (13.3oC)

=

c = 0.129J/goC

M x VL = mol

6.5oC= DT

need total mass

of so

lution 100.g

(1.0M)(0.050L)= 0.050molHCl

DHneut =kJ/mol

= (100.g)(4.18J) (goC)

2717

q = 2700J = 2.7kJ

= 2.7kJ 0.050mol

DHneut= -54kJ/mol

exo

6.5oC

= (20.0g) (x - 55.0oC)(4.18J) (goC)

DT = Tfinal – Tinitial

Tf = x

-(50.0g) (4.18J) (goC)

(x – 80.0oC)

-50.0x + 4000oC)–

= 20.0x - 1100oC)–

+50.0x +50.0x + 1100oC)+ 1100oC)

= 70.0x 5100oC

Tfinal = 72.857

72.9oC

(4.18J) (goC)

-6.66J oC

= (55.0g) (x – 23.0oC)-(15.0g)(0.444J) (goC)

(x – 100.0oC)

= 230.J oC

= 237xTfinal =

25.13025.1oC

(x – 100.0oC)

– 5290oC-6.66x

+ 666oC = 230.x +6.66x

+6.66x

+ 5290oC+ 5290oC)5956oC

(x – 23.0oC)

cm3

-38.8oC

356.6oC

25oC

425oC

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

- - - - - - - - -

need g Hg:

125cm3 x 13.5g = cm3

1687.5 1690g

12 3

need J/g

q = mcDT

q = DHvapm

25oC(l)-->356.6oC(l) q = m c DT

= (1690g)(0.139J) (goC)356.6oC

-25.000…oC

DT = 331.6oC

(331.6oC)

77,896.15q = 77,900J

356.6oC(l)-->356.6oC(g) q = DHvapm

59.1kJ mol

x _1mol_ 200.59g

x 1000J 1kJ

= 295J/g 294.63

= 295J g

(1690g)

498,550q = 499,000J

q = m c DT356.6oC(g)-->425oC(g)

425oC-356.6oC

= (1690g) (68oC)

68.4DT = 68oC

(0.102J) (goC)

11,721.8q = 12,000J

77,900J499,000J+12,000J

588,900J = 588.9kJ absorbed

USING HEATS of REACTION: HESS’S LAW states that the enthalpy change for a reaction is the

same whether it occurs in one step or a series of steps.(Law of Heat Summation)  

one step

reactants ----------------------> products energy difference ---> ---> ---> ---> will be same! series of steps

 

We can only measure the “change in enthalpy”

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 reaction you 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 DH is multiplied by the same integer.   Thus DHreaction = DH1 + DH2 + DH3…….

flip

SiH4(g) ---> Si(s) + 2H2(g) DH = -34kJ

Si(s) + O2(g) ---> SiO2(s) DH = -911kJ

2H2(g) + O2(g) ---> 2H2O(g)DH = -484kJ

DH = -1429kJSiH4(g) + 2O2(g)---> SiO2(s) + 2H2O(g)

2 2

flip

b. 2SO3(g) ---> O2(g) + 2SO2(g)

DH = -790.4kJa. 2S(s) + 3O2(g) ---> 2SO3(g)

2

2S(s) + 2O2(g) ---> 2SO2(g)

DH = +198.2kJ

DH = -592.2kJ

S(s) + O2(g) ---> SO2(g)

2

DH = -296.1kJ

2 2

2

STANDARD ENTHALPY OF FORMATION

The standard enthalpy of formation DHof of 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 DHo

f indicates 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 DHo

f 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 = SnpDHo

f (products) - SnrDHof(reactants)

S (sigma) means “sum of” np means moles of products

nr means 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)  DHo

f : ____kJ/mol ___kJ/mol_____kJ/mol _____kJ/mol _____kJ/mol

 

-46 0 -75 135.1 -242

DHorxn =

2mol(135.1kJ) mol

+ 6mol(-242kJ) mol

2mol(-46kJ) + mol

3mol(0kJ) + mol

2mol(-75kJ) mol

= 270.2kJ+ -1452kJ -92kJ + -150kJ

DHorxn = -939.8-940.kJ

-1181.8kJ -242kJ

products - reactants

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

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

f : _?__kJ/mol _______kJ/mol _____kJ/mol _____kJ/mol

baking soda

DHorxn = [products] – [reactants]

85kJ =

-1131 -394 -286

(-1131kJ) +

(-394kJ)+

(-286kJ)

2(x)

85kJ = (-1811kJ) -

2x+2x+2x -85kJ -85kJ

2x = -1896kJ

x = -948kJ

DHof = -948kJ/mol

1mol cancels mol

C + E --> CC + C --> CE + E --> C only!

2 ways to find DHrxn for this synthesis

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

f - SnreactDHof

2mol(-826kJ) (mol)

- 0kJ(mol)

0kJ(mol)

+=

2Fe2O3(s) ---> 4Fe(s) + 3O2(g) DHorxn = +1652kJ

g-->mol

25.0gFe x 1mol x 1652kJ = 55.85g 4molFe

184.87

185kJ absorbed

synthesis

D

If 1164 kJ is required to decompose BaO, find DHo

f of BaO.

BaO (s) Ba (s) + O2(g) DHrxn = +1164kJnow balance

2 2

now reverse to make it synthesis

2Ba(s) + O2(g) 2BaO(s) DHrxn = -1164kJ

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

2 2

Ba(s) + 1 O2(g) BaO(s)2DHo

f = -582kJ/mol1 mol

(Given: DHrxn Find: DHof )

REVIEW FOR THERMO 1 TEST

1) Is work done BY or ON the system?

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

 

 

 

 

3mol 2mol

22.4L22.4L22.4L 22.4L22.4L

compression

w = -PDV DV = Vf - Vi

= 2 - 3DV = -1

= -P(-1)

= +wadded to system

work done ON the system

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

2mol1mol

22.4L

expansion

= 2 - 1

DV = +1

DV = Vf - Viw = -PDV= -P(+1)

= -wsubtracted 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

w = -(5.0atm)

= 7.5L

(7.5L)

w = -PDV

(101.3J) (1L•atm)

-3798.75w = -3800J

work done BY system, expansion

3) Write a “FORMATION” equation for NaNO3

a) write elements’ formulas and state of matter symbols

b) balance the equation (this is the reaction equation)c) ÷ moles of product to get 1 mole of product (Def of “formation” eq)

REACTION EQUATION:

____________________________________________________FORMATION EQUATION:

____________________________________________________

Look up heat of formation of NaNO3 : DHof = __________________

Na(s) + N2(g) + O2(g) ------> NaNO3(s) 22 3

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 ______ =

(-467kJ) = mol

2mol -934kJ

-934kJ2molNa

-934kJ1molN2

-934kJ3molO2

-934kJ2molNaNO3

-934kJ1molN2

1molN2

28.02g

-33.33

33kJ released

 

4) “Standard State” DHorxn DHo

f

means T = ____________

P = _____ = 760.torr lithium ( ) bromine ( ) fluorine ( )

AgNO3(aq) = __________

25oC = 298K1atm

s

lg

1 Molar

State Functions: (capital letters)

properties of a system that are independent of the pathway

VPTHE

volume

pressure

temperature

enthalpy (heat content)

internal energy

q & w state functions

DHof = kJ

molif (-), then STABLE compound

if (+), then UNSTABLE compound

The greater the (-) value, the more stable the compound, the stronger the bonds

DHreaction = Hproducts - Hreactants

DHrxn = (-) exo DHrxn = (+) endo

heat coming out of (subtracted from)

heat going into (added to)

MORE STABLEMORE STABLE

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

Solve for q (Joules), J-->kJConvert g-->mol,

kJ by mol

DHneut

DHsol’n

kJ/mol

Is heat going into system (+) or out of system (-) ?

MPt

BPt

DHfus

DHvap

c = “specific heat” J/goC

SL

G

Convert if need be: J/moloC ---> J/goC

mol ---> g

S-->L

L-->G