Date post: | 30-Dec-2015 |
Category: |
Documents |
Upload: | belinda-neal |
View: | 247 times |
Download: | 0 times |
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
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
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 (-) ?