# Ch5

```Ch.5
Energy, E
work, w
1st Law Thermo
Calorimetry
∆H, Enthalpy
THERMOCHEMISTRY
Enthalphy, H, heat, q
heat of rxn;
enthalphyies of formation
Hess’ Law
∆HRXN
Specific heat
Thermochemistry
relationship bet chem rxns & E es due to heat
Energy, E capacity to do work or transfer heat
Work, w E or force that causes a  in direction or position of
an object w = F*d
Heat, q E to cause increase of temp of an object
hotter -----> colder
sys ----> surr exothermic, sys losses q
surr ----> sys endothermic, sys gains q
ENERGY
PE: potential E stored E, amt E sys has available
KE: kinetic E E in motion, Ek = 0.5 mv2
2 objects mass1 > mass2 @ same speed
1 object
v1 < v2
@ same mass
Internal E
which more Ek?
which more Ek?
total KE + PE of system
∆E = ∑Ef - ∑Ei = ∑Epdt - ∑Ereact
∆Esys = -∆Esurr
∆E always => system
surr
Transfer of E results in work &/or heat
NOW, think of atoms & molecules in random motion colliding!!!!!!
What kind
of Energies
would be
involved?
H2
O2
E UNITS
Joule, J EK = 0.5(2 Kg)(1 m/s)2 = 1 Kg-m2/s2 = 1 J
calorie, cal E needed to raise 1 g H2O by 1oC
1 cal = 4.184 J
1 Cal (food) = 1000 cal= 1 kcal
Transfer of E results in work &/or heat
System – Surroundings
Defined as ………….? What???
System: a defined region
Surroundings:
everything that will ∆ by influences of the system
OPEN: matter & E ex w/ surr
CLOSED: ex E, not matter w/ surr
∑PE(H2O + CO2) < ∑PE(O2 + CH4)
2 mol O2
1 mol CH4
system
∆PE
E released to surroundings
as Heat
E
2 mol H2O
1 mol CO2
∑PE(NO) > ∑PE(O2 + N2)
system
2 mol NO
∆PE
Heat absord from surroundings
E
1 mol N2
1 mol O2
Determine the sign of H in each process under 1 atm; eno or exo?
1. ice cube melts
2. 1 g butane gas burned to give CO2 & H2O
must predict if
heat absorbed
or released
1. ice is the sys, ice absorbs heat to melt, H “+”, ENDO
2. butane + O2 is the sys, combustion gives off heat, H “-”, EXO
Conservation of E
1st Law of Thermodynamics: total E of universe is constant
- E is neither created nor destroyed but es form
q: Heat, Internal H
E transfer bet sys & surr w/ T diff
w: work, other form E transfer
mechanical, electrical,
∆E = q + w
sum of E transfer as heat &/or work
∆E = q + w
+ + +
- + - + : sys gain E; w > q
+ - : sys lost E |w| > |q|
What is ∆E when a process in which 15.6 kJ of heat
and 1.4 kJ of work is done on the system?
∆E = q + w
15.6 + 1.4 kJ = 17.0 kJ
State Functions
Property of variable depends on current state;
not how that state was obtained
T, H, E, V, P use CAPITAL letters to indicate state fcts
∆ : state fcts depend on initial & final states
∆H
Enthalpy
Must measure q & w
2 types: electrical, PV
- movement of charged particles
- w of expanding gas
w = -P∆V
@ constant P
∆H = ∆E + P∆V
q + w
3 Chemical Systems
#1 no gas involved
s, l, ppt, aq phases have little or not V change; P∆V ≈ 0, then ∆H ≈ ∆E
#2 amt of gas no change
H2 (g) + I2 (g) -- 2 HI (g) P∆V = 0, then ∆H = ∆E
#3 amt of gas does change
N2 (g) + 3 H2 (g) -- 2 NH3 (g)
∆H
∆Hcomb
combines w/ O2
P∆V ≠ 0, then ∆H ≈ ∆E
∆E mostly transfer as Heat
Enthalpy Changes
∆Hf
cmpd formed
∆Hfus
subst melts
s -- l
∆Hvap
subst vaporizes
l -- g
PV Work
Calculate the work associated with the expansion
of a gas from 46 L to 64L @ 15 atm.
w = -P∆V
w = -(15 atm)(18 L) = -270 L-atm
NOTE: “PV” work
- P in P∆V always refers to external P
- P that causes compression or
resists expansion
A balloon is inflated by heating the air inside. The vol changes from
4.00*106 L to 4.5*106 L by the addition of 1.3*108 J of heat. Find ∆E,
assuming const P = 1.0 atm
1 L-atm = 101.3 J
∆V = 5.0*105 L
P = 1.0 atm
∆E = q + w
w = -P∆V
w = -(1.0 atm)(5.0*105 L) = -5.0*105 L-atm
(-5.0*105 L-atm)(101.3 J / 1 L-atm) = -5.1*107 J
∆E = q + w = (1.3*108 J) + (-5.1*107 J) = 8*107 J
More E added by heating than gas expanding,
net increase in q, ∆E “+”
REVIEW
PE - KE
Enthalpy
sys - surr
J - cal 1st Law
PV
STATE Fcts
CALORIMETRY
Heats of Reaction
Measure of Heat flow, released or absorded, @ const P & V
Not as simple as: ∆Hfinal - ∆Hinitial
Solar-heated homes use rocks to store heat. An increase of 120C
in temp of 50.0 Kg of rocks, will absorb what quantity of heat?
Assume Cs = 0.82 J/Kg-K. What T would result in a release
of 450 kJ?
Heat Capacity, C
T when object absorbs heat
+q or -q?
gains loss
endo exo
Specific Heat, Cs
C of 1 g of subst
heat trans
q
C 

 J
gK
s g * (Tf - Ti ) m * T
q = Cs*m*T
How much Heat is transferred when 720 g of antifreeze cools 25.5 oC?
Cs = 2.42 J/g-K
q = Cs * mass * ∆T
∆ T = -25.5 ?
oC?
K or
 K = oC
THN IK!!!!
T?
q = (2.42 J/g-K) * (720 g) * (-25.5K)
= -44400 J or -44.4 kJ
HESS’S LAW
Heat Summation
Hess’ states:
overall H is sum of individual steps
Rxn are multi-step processes
Calculate H from tabulated values
REACTS ======> PDTS
Figure 05.22
THN IK!!!!
What effect H if
- reverse rxn?
Figure 05.22
Calculate HRXN for
Ca(s) + 0.5 O2(g) + CO2 (g) --------> CaCO3(s)
HRXN = ?
given the following steps:
Ca (s) + 0.5 O2 (g) -----> CaO (s)
Hof = - 635.1 kJ
CaCO3 (s) -----> CaO (s) + CO2 (g) Hof = + 178.3 kJ
Note: to obtain overall rxn ==> (1st rxn) + (-2nd rxn)
Ca (s) + 0.5 O2 (g) -----> CaO (s)
CaO (s) + CO2 (g) -----> CaCO3 (s)
Hof = - 635.1 kJ
Hof = - 178.3 kJ
Ca(s) + 0.5 O2(g) + CO2 (g) ------> CaCO3(s)
“o”??
Ho?
Horxn = - 813.4 kJ
Enthalpies of
Formation
STANDARD STATES
Set of specific conditions
- gas: 1 atm, ideal behavior
- aq solution: 1 M (mol/L)
- pure subst: most stable form @ 1 atm & Temp
T usually 25oC
- forms 1 mole cmpd; kJ/mol
Use superscript “o” indicates Std States
H
0
f
H
0
rxn
formation of subst from
its elemental parts
Individual ∆Hf 0 values from book table, appendix C, pg 1100
NOTE: look at state
H = - 635.5 kJ
Ca (s) + 0.5 O2 (g) -----> CaO (s)
0
f
CaO
0.5 O
H  0.0
Ca (s) H  0.0
2 (g)
(s)
H  - 635.5
0
f
0
f
H (Pdts) - H (Reacts)  H
O
f
O
f
-635.5 kJ - (0.0 + 0.0)kJ = -635.5 kJ
O
rxn
Solar-heated homes use rocks to store heat. An increase of 120C
in temp of 50.0 Kg of rocks, will absorb what quantity of heat?
Assume Cs = 0.82 J/Kg-K. What T would result in a release
of 450 kJ?
What is the change in enthalpy for the reaction of sulfur dioxide
and oxygen to form sulfur trioxide. All in gas form.
Is this endo- or exo-thermic?
2 SO2 (g) + O2 (g) -----> 2 SO3 (g)
q = Cs*m*T
q = (0.82 J/g-K)*(5.0*104 g)*(12.0 K) = 4.9 * 105 J
T = q/[Cs*m]
T = (4.5*105 J)/[(0.82 J/g-K)*(5.0*104 g)] = 11O decrease
Find Hof per mole in tables (kJ/mol)
SO2 = -296.8
SO3 = -396.0 O2 = 0.0 free element
Sum Hf reactants using stoich coeff & also pdts
2 mol SO 2   296.8 kJ   1 mol O 2  0.0 kJ   - 593.6 kJ

1 mol

 1 mol 
  396.0 kJ 
2 mol SO 3 
  - 792.0 kJ
 1 mol 
H = Hf Pdts - Hf reacts
H = (-792.0 kJ) - (-593.6 kJ) = -198.4 kJ
Exothermic, -H
What if rxn were reversed?????
Write balanced eqn for the formation of 1 mol of NO2 gas from
nitrogen monoxide gas and oxygen gas. Calculate HOrxn
1 NO(g) + .5 O2(g) ---> NO2(g)
HOf : 90.3 kJ + .5(0) kJ ---> 33.2 kJ
(33.2 kJ) - (90.3 + 0)kJ = -57.1 kJ
Find the overall rxn, CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g),
from the given steps:
H2(g) + CO(g) ---> CH3OH(l)
CO(g) + H2O(l) ---> CO2(g) + H2(g)
Calculate Hrxn for each step and find the overall Hrxn
1 CH3OH(l) ---> 2 H2(g) + 1 CO(g)
Hf: -238.6
0
-110.5
Hrxn = 128.1 kJ
1 CO(g) + 1 H2O(l) ---> 1 CO2(g) + 1 H2(g)
Hf: -110.5 -285.8
-393.5
0
Hrxn = 2.8 kJ
1 CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g)
Hrxn = 130.9 kJ
Toss a ball upward.
a. Does KE increase or decrease
Decrease, KE converts to PE
b. As ball goes higher, want effect to PE
Increases
Define
a. System
b. Closed system
c. Not part of system
Region of study w/ E changes
exchange E not mass
surroundings
Explain
a. 1st Law
b. Internal E c. How internal E of closed system increase
E not created nor destroyed, changes form
Total E of system, KE + PE
System absorbs heat or work done on system
Calculate E of system, is endo- or exo- thermic
a. Balloon cooled, remove 0.655 kJ heat, shrinks, &
atmosphere does 382 J work on
q “-” w “+” E =-0.655 kJ + 0.382 kJ = -0.273 kJ EXO
b. 100 g metal bar gains 25oC, absorbs 322 J of heat. Vol is constant
q “+” w = 0 E = +322 J ENDO
c. Surroundings do 1.44 kJ work compressing gas in
perfectly insulated container
q = 0 (perfectly insulated) w “+” E = +1.44 kJ ENDO
Ca(OH)2(s) ----- CaO(s) + H2O(g)
Requires addition of 109 kJ of heat per mol of Ca(OH)2
a. Write balanced thermochemistry equation
Ca(OH)2(s) ----- CaO(s) + H2O(g) H = 109 kJ
b. Draw enthalpy diagram
CaO(s) + H2O(g)
H = 109 kJ
Ca(OH)2(s)
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