math for heat capacity
TRANSCRIPT
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Or in general, 'T V
U UU U dV dTV T
For infinitesimal changes,
T V
U UdU dV dTV T
VV
U CT
T
T
UV
Some terms are familiar:
Math for Heat Capacity
Heat capacity at constant volume
Internal pressure at constant temp
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Heat Capacity
If dV = 0, then dU = CV(T)dT
However, CV(T) is approximately constant over small temperature changes and above room temperature
so integrate both sides: U = CVT = qv+ w
Since constant volume, w= o and qV=CVT
T V
U UdU dV dTV T
CV(T)T (V)
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Internal Energy (5)Many useful, general relationships are derived frommanipulations of partial derivatives, but I will(mercifully) spare you more.Suffice it to say that U is best used for processestaking place at constant volume, with only PV work:Then dU = dqV and U= U2 – U1 = qV
The increase in internal energy of a system in a rigidcontainer is thus equal to the heat qV supplied to it.
We would prefer a different state function forconstant pressure processes: enthalpy.
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Enthalpy Defined
At constant V and P, q = U = H
U = q + w
q= qP = U - w, w = -PV
qP = U + PV H
Enthalpy, H U + PV
At Constant P, H = U + PV
dH dU d PV dU PdV VdP
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Comparing H and Uat constant P
1. Reactions that do not involve gasesV 0 and H U
H = U + PV
2. Reactions in which ngas = 0V 0 and H U
3. Reactions in which ngas 0 V 0 and H U
A B C D
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ConcepTest #1
Which of the following reactions has the largest difference between H and U?
A. NH3 (g) + HCl (g) NH4Cl (s)
B. CO (g) + Cl2 (g) COCl2 (g)
C. ZnS (s) + 3/2 O2 (g) ZnO (s) + SO2 (g)
D. ZnO (s) + CO (g) Zn (s) + CO2 (g)
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Comparing H and Uat constant P (2)
1. Reactions that do not involve gasesV 0 and H U
H = U + PV
2. Reactions in which ngas = 0V 0 and H U
3. Reactions for which ngas 0
H U + ngas RT Ideal Gases
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Heat Capacity atConstant Volume or Pressure
CP = dqP/dT = (H/T)P = (U/T)P + (PV/T)P
Partial derivative of enthalpy withrespect to T at constant P
CV = dqV/dT = (U/T)V
Partial derivative of internal energy with respect to T at constant V
Ideal Gas: CP = CV + nR
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Heat Capacity, C
Molar Heat Capacity (J K-1 mol-1 )Heat needed to raise T of 1 mole by 1 K
q = CmnT
Heat Capacity (J K-1 )Heat needed to raise T of system by 1 K
q = CT
Specific Heat Capacity (J K-1 kg-1 ) Heat needed to raise T of 1 kg by 1 K
q = CSmT
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CalorimetryMeasure H and U for Reactions
Isolate sample, bomb, and water
bath from surroundings
Initiate reaction
Heat released causes temperature rise
in system
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Calorimetry ProblemFor the complete combustion of 1 mole of ethanol in a bomb calorimeter, 1364.5 kJ mol–1is released at 25 °C. Determine cU and cH.
cU = qV = –1364.5 kJ
cH = cU + nRT = –1364.5 –2.5 = –1367.0 kJ/mol
C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l)c
n = -1 At 25 C, RT =8.314 J/ K 298.15 K = 2.5 kJ
Now for cH
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Endothermic & Exothermic Processes
H Negative Negative amount of heat absorbed(i.e. heat released by the system)
Exothermic Process
H Positive Positive amount of heat absorbed by the system
Endothermic Process
H = Hfinal - Hinitial
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ConcepTest #2
A. H2O (s) H2O (l) H = +
B. CH4 + 2 O2 CO2 + 2 H2O H = –C. H2O (g) H2O (l) H = + D. 2 H (g) H2 (g) H = –
For which of the following reactions is the indicated sign of H incorrect?
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Thermochemical Equations
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)
(a combustion reaction) cH = – 890 kJ
Phases must be specified
H is an extensive property
Sign of H changes when reaction is reversed
Reaction must be balanced
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ConcepTest #3
2 CO2 (g) + 4 H2O (l) 2 CH4 (g) + 4 O2 (g)
A. -890 kJ B. -1780 kJ C. +890 kJD. +1780 kJ
What is H for this reaction?
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Calculation of rxnH
I. Hess’s LawH of an overall process is the sum of the Hs for the individual steps
II. Use of Standard Enthalpies of Formation
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I. Hess’s Law
Calculate H for the reaction:
H2O (solid at 0 °C) H2O (gas at 100 °C)
H2O (s, 0 °C) H2O (l, 0 °C) fusH = 6.0 kJ/mol
H2O (l, 0 °C) H2O (l, 100 °C) CT = 7.5 kJ/mol
H2O (l, 100 °C) H2O (g, 100 °C) vapH = 40.6 kJ/mol
H2O (s, 0 °C) H2O (g, 100 °C) rxnH = 54.1 kJ/mol
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II. Standard Enthalpy of Formation
The standard enthalpy of formation ( fH°)of a compound is the enthalpy change for the formation of one mole of compound from the elements in their standard state.
Designated by superscript o: H°
For example, CO2:
C (graphite) + O2 (g) CO2 (g)
rxnH° = -393.5 kJ/mol Table 2.7
f H° CO2 (g) = -393.5 kJ/mol