Transcript Energy and the First Law - Winona State University

Chem. 412 – Phys. Chem. I
Laws of Thermodynamics
Oth Law
1st Law
2nd Law
3rd Law
Heat Flow
Energy
Entropy
Zero K
Heat & Work
Energy Distribution
Internal Energy
Enthalpy
Reaction Spontaneity
Sign Convention
 ΔU = Internal Energy.
 q = heat flow; transfer of energy between two objects.
 w = work; product of force applied to an object over a
distance.
Surroundings
System
Work – Basic Formulation
Work Done = Force x (Distance Moved)
Differentially =>
dw = F • dS
Work – Gas Expansions and Compressions – I (F15)
Work is area under curve of P-V Diagram.
 dw = F • dS = - P • dV
Work – Gas Expansions and Compressions – II (F15)
Work is area under curve of P-V Diagram.
 dw = F • dS = - P • dV
Work – Gas Expansions and Compressions – III (F15)
Work is area under curve of P-V Diagram.
Reversible versus Irreversible Work
Work – Gas Expansions and Compressions – IV (F15)
Work is area under curve of P-V Diagram.
Reversible versus Irreversible Work
Work – Gas Expansions and Compressions – V (F15)
Reversible versus Irreversible Work
Work – Gas Expansions and Compressions – I (F13)
Work is area under curve of P-V Diagram.
 dw = F • dS = - P • dV
Work – Gas Expansions and Compressions – II (F13)
Work is area under curve of P-V Diagram.
 dw = F • dS = - P • dV
Work – Gas Expansions and Compressions – III (F13)
Work is area under curve of P-V Diagram.
Reversible versus Irreversible Work
Work – Gas Expansions and Compressions – IV (F13)
Work is area under curve of P-V Diagram.
Reversible versus Irreversible Work
Work – Gas Expansions and Compressions – V (F13)
Reversible versus Irreversible Work
Tutorial Problem on ‘Work’
Consider one mole of an ideal gas kept in a right-circular cylinder with a movable
piston cap at constant temperature. The cap is attached to an external engine that is
capable of moving the piston in either directions.
(a) Initially, the cylinder has a volume of 1.0 m3 at a pressure of 10. Pa. The
cap was moved to give a final pressure of 1.0 Pa. In this case, the piston
cap was moved at an infinitely slow rate to achieve this final pressure,
thereby following the ideal gas law. Calculate the work involved in this
process. This work is referred to as reversible work, wrev . [ 23 J ]
(b) The initial and final volume/pressure was kept the same as in part (a) but
the piston cap was suddenly released to get to the final pressure value.
Calculate the work involved in this process. This work is referred to as
irreversible work, wirrev . [ 9 J ]
(c) Draw one P-V diagram for parts (a) and (b).
(d) Repeat parts (a), (b), and (c) by reversing the initial and final
volume/pressure conditions; that is, compression instead of expansion. [ 23
J, 90 J ]
(e) Compare/Contrast/Discuss the above results.
First Law of Thermodynamics
dU  q  w
U  q  w
First Law of Thermodynamics
 U = Internal Energy = Heat Flow under Constant Volume
 H = Enthalpy = Heat Flow under Constant Pressure
 H = U + PV
Two Heat Capacities from q
dU  CV  dT
C
q
dT
dH  CP  dT
Calorimetry
Bomb Calorimetry (Constant Volume
Calorimetry)
• Reaction carried out under
constant volume.
• Use a bomb calorimeter.
• Usually study combustion.
qv  U
qrxn  Ccal  T
Calorimetry
Constant Pressure (Solution) Calorimetry
• Atmospheric pressure is constant!
qP  H
qrxn  qsoln  specific heat of solution
 grams of solution  T
qrxn  S  msoln  T

Calorimetry
Constant Pressure Calorimetry
qrxn  S  ( total mass of solution)  T
H rxn
qrxn

mol *
* moles of species of interest
Calorimetry Examples
CyberChem video
1. In an experiment similar to the procedure set out for Part
(A) of the Calorimetry experiment, 1.500 g of Mg(s) was
combined with 125.0 mL of 1.0 M HCl. The initial
temperature was 25.0oC and the final temperature was
72.3oC. Calculate: (a) the heat involved in the reaction
and (b) the enthalpy of reaction in terms of the number of
moles of Mg(s) used. Ans: (a) –25.0 kJ (b) –406 kJ/mol
2. 50.0 mL of 1.0 M HCl at 25.0oC were mixed with 50.0
mL of 1.0 M NaOH also at 25.0oC in a styrofoam cup
calorimeter. After the mixing process, the thermometer
reading was at 31.9oC. Calculate the energy involved in
the reaction and the enthalpy per moles of hydrogen ions
used. Ans: -2.9 kJ , -58 kJ/mol [heat of neutralization
for strong acid/base reactions]
Calorimetry Examples: Hints
50.0 mL of 1.0 M HCl at 25.0oC were mixed with 50.0 mL of 1.0 M NaOH also at 25.0oC in a styrofoam cup
calorimeter. After the mixing process, the thermometer reading was at 31.9 oC. Calculate the energy involved in the
reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ , -58 kJ/mol [heat of neutralization for
strong acid/base reactions]
Comparison of Heat Capacities: CP vs. CV
 H 
CP  

 T  P
H  f(P, T)
and
and
 U 
CV  

 T V
U  f(V, T)
 V  
 U  
CP  CV  
  P  
 
 T  P 
 V T 
CP  CV  n  R
(Ideal Gas)
Comparison of Heat Capacities: CP vs. CV
Comparison of Heat Capacities: CP vs. CV
(F08)
Example on State and Path Functions
20.0 grams of Argon gas was heated from 20.0oC to 80.0oC. Considering Argon
behaving as an ideal gas, calculate q, w, U, and H for the following processes:
(a) under Constant Volume, and
(b) under Constant Pressure.
Constant Volume
Constant Pressure
q
374 J
623 J
q
w
0
-249 J
w
U
374 J
374 J
U
H
623 J
623 J
H
CVm
12.5 J mol-1 K-1
20.8 J mol-1 K-1
CPm
Enthalpy of Phase Transitions
Constant P:
sg
s
g
H v  heat of vaporization  H g  H 
H f  heat of fusion  H   H s
 H 
CP  

 T  P
Constant P : H   CP  dT
Enthalpy of Phase Transitions
Enthalpy of Phase Transitions
35
Heat Flow at Const. P
30
25
20
15
10
5
0
-50
0
50
100
o
Temperature ( C)
150
Relating Urxn and Hrxn
H  U  PV
H  U   (PV )
For solids & liquids : ( PV ) is negligible ,  omit.
However for gases : (PV) is quite significan t.
For Ideal Gases : PV  nRT
 ( PV )   (nRT )

at constant T

 ( PV )  RT  (ng )
 H rxn  U rxn  RT  (ng )
Hess’s Law
• Hess’s law: if a reaction is carried out in a number of
steps, H for the overall reaction is the sum of H for
each individual step.
• For example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
2H2O(g)  2H2O(l)
H = -88 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -890 kJ
Hess’s Law
Given: (i)
(ii)
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
CO(g) + ½O2(g)  CO2(g)
Calculate the heat of reaction for:
∆H = -26.7 kJ/mol
∆H = -283.0 kJ/mol
2Fe(s) + 3/2 O2(g)  Fe2O3(s)
Ans: -822.3 kJ/mol
Enthalpies of Formation
If 1 mol of compound is formed from its constituent elements
(standard state), then the enthalpy change for the reaction is
called the enthalpy of formation, Hof .
• Standard conditions (standard state): Most stable form of
the substance at 1 atm and 25.00 oC (298.15 K).
• Standard enthalpy, Ho, is the enthalpy measured when
everything is in its standard state.
• Standard enthalpy of formation: 1 mol of compound is
formed from substances in their standard states.
• Standard enthalpy of formation of the most stable form of
an element is zero.
Enthalpies of Formation: Example
Example: Write the balanced reaction equation for the standard
enthalpy of formation of solid ammonium carbonate.
Enthalpies of Formation
Substance
Hof (kJ/mol)
C(s, graphite)
0
O(g)
247.5
O2(g)
0
N2(g)
0
Bucky Ball drawn by HyperChem
Enthalpies of Formation
Using Enthalpies of Formation to
Calculate Enthalpies of Reaction
• For a reaction
H rxn   nH f products    mH f reactants 
• Calculate/Compare heat of reactions for the combustion
of methanol gas and ethanol gas giving carbon dioxide
and water.
Temperature Dependence of CP and Hrxn
CP  a  b  T  c  T 2
H rxn 
 H
  ( H rxn ) 



T

P
f , prods
  H f , reacts
  ( H f , prods ) 
  ( H f , reacts ) 

 

T
T

P

P
 at constant P
d ( H rxn )
  C P ( prods)   C P ( reacts)  C p
dT

T2
T1
T2
d (H rxn )   CP  dT
T1
where : CP  a  b  T  c  T 2
Temperature Dependence of CP and Hrxn
Example: Find the heat of reaction for the following reaction
at 1000. K ( Horxn,1000 K ? )
NaCl(s)  Na(g) + ½ Cl2(g)
[ Given: Horxn,298K = 519.23 kJ ]
Species
CP / J K-1 mol-1
Na(g)
20.80
NaCl(s)
49.70
Cl2(g)
31.70 + 10.14x10-3 T – 2.72x10-7 T2
dU  q  w
Laws of Thermodynamics
w  P  dV
Oth Law
1st Law
2nd Law
3rd Law
Heat Flow
Energy
Entropy
Zero K
Heat & Work
Energy Distribution
qV  U
qP  H
CP  CV  n  R
H  U  PV
Internal Energy
Enthalpy
 H 
CP  

 T  P
Reaction Spontaneity
&
 U 
CV  

 T V
H rxn  U rxn  RT  (ng )