Transcript Intro to Thermochemistry

Ch. 6 & 8.8 lecture
notes
THERMOCHEMISTRY!!!
Video: Exploding Gummy Bear!
I’m a wimp, so you’re not getting the live version! 
Please don’t try this at home! 
Terms to know
•System:
•Surroundings:
•Heat:
•Work:
•Energy:
Rvw: Endothermic vs.
Exothermic Reactions
st
1
Law of Thermodynamics
Nrg of universe is constant
◦ Nrg is conserved; cannot be created nor destroyed
Internal nrg (E) = total nrg of the system.
◦ can be changed by flow of work (w), heat (q), or both: ΔE
=q+w
q > 0 =
q < 0 =
W > 0 =
W < 0 =
State functions
Internal nrg (E) is a state function
State functions depend only on initial and final
state of the system
 Independent of path taken to get from start to finish.
 Example:
P-V work
P-V work
Work in terms of PV:
◦ For expanding gas:
◦ For compressed gas:
Enthalpy (H)
Relates nrg & P-V work; H = E + PV
At constant pressure (where only P-V work is allowed),
ΔH = nrg flow as heat = qp
 How is this so??? Derive the equation!
Enthalpy of rxns (ΔH )
ΔH = qp = Hfinal – Hinitial = Hprods – Hrxts
Characteristics of enthalpy changes (ΔH )
Enthalpy is an extensive property
ΔHforward = -ΔH reverse
ΔH depends on state of rxts and prods
 In terms of potential nrg, ….
 Assume everything is at standard conditions (25 oC, 1 atm) unless stated
otherwise
Practice problem #1
2H2O2 (l)  2H2O (g) + O2 (g)
ΔHrxn = -196 kJ
What is ΔH if 5.00 g of H2O2 is decomposed?
Calorimetry
Measures heat flow by
measuring . . .

Uses a calorimeter (see picture)
Constant-pressure calorimetry:



Patm remains constant
Used to determine changes in
enthalpy (heats of rxn) for rxns
occuring in solution
ΔH = qp
Coffee Cup Calorimeter
ON THE OUTSIDE…
ON THE INSIDE…
Constant-volume calorimetry:
A “bomb” calorimeter



No work is done. Why?
Weighed rxts placed inside rigid steel container (the
“bomb”) and ignited
ΔH = q + w = qV
Heat capacity
Heat capacity (C): nrg needed to raise temp of a
body by….. Units = ????


C = heat absorbed
increase in temp
Molar heat capacity: heat capacity of one
of substance; Units = ????
Specific heat capacity (c): heat capacity of one
of substance; Units = ????
Some specific heats
Substance
Lead
Gold
Silver
Copper
Iron
Carbon (graphite)
Granite
Olive Oil
Ethyl alcohol
Water, (liquid)
Specific Heat at 25oC (J/goC)
0.128
0.129
0.235
0.387
0.4498
0.711
0.803
2.0
2.45
4.1796
Notice anything about the first 5 substances?
Heat transfer
Heat transfer
Q = mcΔT = ΔH if at constant P
If a hot object touches a cooler object…
 Tf
Ti for hot object
 Tf
Ti for cooler object
 And Tf of hot object
Tf of cooler object
All heat lost by one object is gained by the other,
so…
 Q1 =
 Tip to avoid dealing with negative…
◦
Practice problem #2
A 150.0 g sample of a metal at 75.0 oC is added to 150.0 g
of water at 15.0 oC. The temperature of the water rises to
18.3 oC. Calculate the specific heat of the metal, assuming
that all of the heat lost by the metal is gained by the water.
(specific heat of water = 4.18 J/g oC)
Phase changes & enthalpy
Phase changes & enthalpy
How do you calculate
enthalpy for segments BC
and DE?
How do you calculate
enthalpy for segments AB,
CD and EF?
Phase change constants for
water
Specific heat (c) for ice = 2.09 J/g•oC
Specific heat (c) for water = 4.18
J/g•oC
Specific heat (c) for vapor = 1.84
J/g•oC
ΔHfus = Heat of fusion = 6.01 kJ/mol
ΔHvap
= Heat of vaporization = 40.67 kJ/mol
Phase change calorimetry
Practice Problems
Do problems on Calorimetry wkhst
Pre-assessment: Lewis
Structures
On your white board, draw the Lewis
structures for the following molecules:
◦H2O
◦CO2
◦HCN
Bond nrg & enthalpy of rxns
Reactants break bonds and new bonds form
to make products
Breaking bonds is….
Forming bonds is….
Bond enthalpy (D):
As bond strength increases, bond enthalpy….
Use bond enthalpies to calculate ΔH without
having to know ΔHf ’s for all species
More on bond enthalpies
Need to know Lewis structures for compounds. Just seeing “N2H4”
doesn’t help us calculate the bond enthalpy for it. We need to know
HOW it bonds:
So, there are
N-H bonds
and
N-N bond
ΔHrxn = Σ (D of bonds broken) - Σ (D of bonds formed)
NOTE: Sum of RXTS – Sum of PRODUCTS!!!
Multiply bond enthalpy by number of bonds
If ΔHrxn > 0, nrg to break bonds is
released when new bonds form
Vice-versa if ΔHrxn < 0.
nrg
Bond enthalpy, bond length
and bond strength
Bond
length (Å)
Bond
enthalpy
(kJ/mol)
C—C
C=C
CC
1.54
1.34
1.20
348
614
839
Interpreting the chart
As bond length , bond enthalpy
As bond length , # of bonds
As # of bonds , bond enthalpy
As bond strength , bond enthalpy
Therefore:
 As bond length , bond strength
 As # of bonds , bond strength
Bond enthalpies for
practice probs
Bond
Bond enthalpy (kJ/mol)
C—C
347
C=C
614
CC
839
H—H
432
N—N
160
N—H
391
Bond
Bond enthalpy (kJ/mol)
NN
Cl — Cl
C—H
H — Cl
941
239
413
427
C — Cl
339
Practice problem #3
Calculate ΔH for each of the following:
rxn
1. Cl2 + CH4  CH3Cl + HCl
2. N2H4  N2 + 2 H2
3. C2H4 + HCl  C2H3Cl + H2
Practice Problem #4
The standard enthalpy of formation of NH3 is -46
kJ/mol. Use this information as well as the balanced
equation below to estimate the N-H bond energy.
Compare your result with the value in previous table.
N2 (g) + 3H2 (g) → 2NH3 (g)