math for heat capacity
TRANSCRIPT
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
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)
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.
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
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
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)
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
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
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
CalorimetryMeasure H and U for Reactions
Isolate sample, bomb, and water
bath from surroundings
Initiate reaction
Heat released causes temperature rise
in system
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
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
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?
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
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?
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
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
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