THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins
description
Transcript of THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins
![Page 1: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/1.jpg)
THERMODYNAMICS.Elements of Physical Chemistry. By P. Atkins
Concerned with the study of transformation of energy: Heat work
![Page 2: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/2.jpg)
CONSERVATION OF ENERGY – states that:
Energy can neither be created nor destroyed in chemical reactions. It can only be converted from one form to the other.
UNIVERSE System – part of world have special interest in… Surroundings – where we make our observations
![Page 3: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/3.jpg)
→ →
Open system Closed system Isolated system
Example:
↔ matter↔ energy ↔ energy not matter matter × Energy ×
![Page 4: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/4.jpg)
If system is themally isolated called Adiabatic system eg: water in vacuum flask.
![Page 5: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/5.jpg)
WORK and HEAT
Work – transfer of energy to change height of the weight in surrounding eg: work to run a piston by a gas.
Heat – transfer of energy is a result of temperature difference between system and surrounding eg:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) - heat given off.
If heat released to surroundings – exothermic.If heat absorbed by surroundings – endothermic.
![Page 6: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/6.jpg)
Example: Gasoline, 2, 2, 4 trimethylpentane
CH3C(CH3)2CH2CH(CH3)CH3 + 25/2 O2 → 8CO2(g) +H2O(l)
5401 kJ of heat is released (exothermic)
Where does heat come from?From internal energy, U of gasoline. Can represent chemical reaction:
Uinitial = Ufinal + energy that leaves system (exothermic)Or
Ui = Uf – energy that enters system (endothermic)
![Page 7: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/7.jpg)
Hence, FIRST LAW of THERMODYNAMICS (applied to a closed system)
The change in internal energy of a closed system is the energy that enters or leaves the system through boundaries as heat or work. i.e.
∆U = q +w ∆U = Uf – Ui
q – heat applied to system W – work done on system When energy leaves the system, ∆U = -ve i.e. decrease internal
energy When energy enter the system, ∆U = +ve i.e. added to internal
energy
![Page 8: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/8.jpg)
Different types of energies:
1. Kinetic energy = ½ mv2 (chemical reaction) kinetic energy
(KE) k T (thermal energy) where k = Boltzmann constant
2. Potential energy (PE) = mgh – energy stored in bonds
Now, U = KE + PE
![Page 9: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/9.jpg)
3. Work (W)w = force × distance moved in direction of force
i.e. w = mg × h = kg × m s-2 × m = kg m2 s-2
(m) (g) (h)
1 kg m2 s-2 = 1 Joule
- Consider work – work against an opposing force, eg: external pressure, pex. Consider a piston
![Page 10: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/10.jpg)
Piston
Pex
pressure (P)
pex
A = area of piston
P
h
h is distance moved
![Page 11: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/11.jpg)
w = distance × opposing forcew = h × (pex × A) = pex × hA
Work done on system = pex × ∆V∆V – change in volume (Vf – Vi)
\ Work done by system = -pex × ∆V Since U is decreased
![Page 12: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/12.jpg)
Example:
C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K 1 atm
(1 atm = 101325 Pa), -2220 kJ = q
What is the work done by the system?For an ideal gas;
pV = nRT (p = pex)n – no. of molesR – gas constantT = temperatureV – volumep = pressure
![Page 13: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/13.jpg)
\ V= nRT/p or Vi = niRT/pex
6 moles of gas:Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3
3 moles of gas:Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3
work done = -pex × (Vf – Vi) = -101325 (0.0734 – 0.1467) = +7432 J
![Page 14: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/14.jpg)
NB: work done = - pex (nfRT/pex – niRT/pex)
= (nf – ni) RT
Work done = -∆ngasRT
i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J
Can also calculate ∆U∆U = q +w q = - 2220 kJ
w = 7432.7 J = 7.43 kJ
\ ∆U = - 2220 + 7.43 = - 2212.6 kJ
![Page 15: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/15.jpg)
NB:qp ∆U why?
Only equal if no work is done i.e. ∆V = 0
i.e. qv = ∆U
![Page 16: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/16.jpg)
Example: energy diagram
C3H8 + 5 O2 (Ui)
3CO2 + 4H2O(l) (Uf)
U
U
progress of reaction
reaction path
![Page 17: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/17.jpg)
Since work done by system = pex∆V
System at equilibrium when pex = pint (mechanical equilibrium)
Change either pressure to get reversible work i.e.
pex > pint or pint > pex at constant temperature by an infinitesimal change in either parameter
![Page 18: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/18.jpg)
For an infinitesimal change in volume, dV Work done on system = pdV
For ideal gas, pV = nRT
\p = nRT/ V\ work = p dV = nRT dV/ V
= nRT ln (Vf/Vi) because dx/x = ln x Work done by system= -nRT ln(Vf/Vi)
Vf
ViVf
Vi
![Page 19: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/19.jpg)
Enthalpy, H
Most reactions take place in an open vessel at constant pressure, pex. Volume can change during the reaction
i.e. V 0 (expansion work). Definition: H = qp i.e. heat supplied to the system at
constant pressure.
![Page 20: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/20.jpg)
Properties of enthalpy
Enthalpy is the sum of internal energy and the product of pV of that substance.
i.e H = U + pV (p = pex)
Some properties of H
![Page 21: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/21.jpg)
Hi = Ui + pVi
Hf = Uf + pVf
\Hf – Hi = Uf – Ui +p(Vf – Vi)or
H = U + p V
![Page 22: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/22.jpg)
Since work done = - pex V
H = (- pex V + q) +p V(pex= p)
\ H = ( -p V + q) + p V = q
\ H = qp
![Page 23: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/23.jpg)
suppose p and V are not constant?
• H = U + ( pV) expands to: • H = U + pi V + Vi P + (P) (V)
• i.e. H under all conditions.
• When p = 0 get back
H = U + pi V U + p V• When V = 0: H = U + Vi p
![Page 24: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/24.jpg)
Enthalpy is a state function.
lattitude
A
Bpath 1
path 2
- does not depend on the path taken
![Page 25: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/25.jpg)
NB: work and heat depend on the path taken and are written as lower case w and q. Hence, w and q are path functions. The state functions are written with upper case.
eg: U, H, T and p (IUPAC convention).
![Page 26: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/26.jpg)
Standard States
By IUPAC conventions as the pure form of the substance at 1 bar pressure (1 bar = 100,000 Pa).
What about temperature? By convention define temperature as 298 K but could be at any
temperature.
![Page 27: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/27.jpg)
Example:
C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l)
at 1 bar pressure, qp = - 2220 kJmol-1.
Since substances are in the pure form then can write
H = - 2220 kJ mol-1 at 298 K
represents the standard state.
![Page 28: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/28.jpg)
H2(g) → H(g) + H(g), H diss = +436kJmol-1
H2O(l) → H2O(g), H vap = +44.0 kJmol-1
Calculate U for the following reaction:
CH4(l) + 2 O2(g) → CO2(g) + 2H2O(l), H = - 881.1kJmol-1
![Page 29: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/29.jpg)
H = U + (pV) = U + pi V + Vi p + p V
NB: p = 1 bar, i.e. p = 0 \ H = U + pi V
Since -pi V = - nRT,
U = H - nRT
![Page 30: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/30.jpg)
calculation
\ U = - 881.1 – ((1 – 2)(8.314) 298)/ 1000 = - 881.1 – (-1)(8.314)(0.298) = - 2182.99 kJ mol-1
![Page 31: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/31.jpg)
STANDARD ENTHALPY OF FORMATION, Hf
Defined as standard enthalpy of reaction when substance is formed from its elements in their reference state.
Reference state is the most stable form of element at 1 bar atmosphere at a given temperature eg.
At 298 K Carbon = Cgraphite
Hydrogen = H2(g)
Mercury = Hg(l)
Oxygen = O2(g)
Nitrogen = N2(g)
![Page 32: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/32.jpg)
NB: Hf of element = 0 in reference state
Can apply these to thermochemical calculationseg. Can compare thermodynamic stability of substances in their standard state.
From tables of Hf can calculate H f rxn for any reaction.
![Page 33: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/33.jpg)
Eg. C3H8 (g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Calculate Hrxn given that:
Hf of C3H8(g) = - 103.9 kJ mol-1
Hf of O2(g) = 0 (reference state)
Hf of CO2(g) = - 393.5 kJ mol-1
Hf of H2O(l) = - 285.8 kJ mol-1
Hrxn = n H (products)- n H(reactants)
![Page 34: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/34.jpg)
Hf(products) = 3 (- 393.5) + 4 (- 285.8)= - 1180.5 -1143.2 = - 2323.7 kJ mol-1
Hf(reactants) = - 103.9 + 5 0 = - 103.9 kJ mol-1
Hrxn = - 2323.7 – (- 103.9) = - 2219.8 kJ mol-1
= - 2220 kJ mol-1
![Page 35: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/35.jpg)
Answer same as before. Eq. is valid. Suppose: solid → gas (sublimation) Process is: solid → liquid → gas Hsub = Hmelt + Hvap
Ie. H ( indirect route) = . H ( direct route)
![Page 36: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/36.jpg)
Hess’ Law
- the standard enthalpy of a reaction is the sum of the standard enthalpies of the reaction into which the overall reaction may be divided. Eg.
C (g) + ½ O2(g) → CO (g) , Hcomb =? at 298K
![Page 37: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/37.jpg)
– From thermochemical data: C (g) +O2 (g) → CO 2(g) H
comb =-393.5 kJmol-1…………………………….(1)
CO (g) +1/2 O2 (g) →CO 2(g), Hcomb = -283.0 kJ mol-
1……………………. (2) Subtract 2 from 1 to give: C (g) + O2 (g) – CO (g) – 1/2 O2 (g) → CO2 (g) – CO2 (g) \ C (g) + ½ O 2 (g) → CO (g) , H
comb= -393.5 – (-283.0) = - 110.5 kJ mol-1
![Page 38: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/38.jpg)
Bond Energies
eg. C-H bond enthalpy in CH4
CH4 (g) → C (g) + 4 H (g) , at 298K. Need: Hf
of CH 4 (g) =- 75 kJ mol-1
Hf of H (g) = 218 kJ mol-1
Hf of C (g) = 713 kJ mol-1
![Page 39: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/39.jpg)
Hdiss = nHf (products) - nHf
( reactants)
= 713 + ( 4x 218) – (- 75) = 1660 kJ mol-1 Since have 4 bonds : C-H = 1660/4 = 415 kJ mol-1
![Page 40: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/40.jpg)
Variation of H with temperature
Suppose do reaction at 400 K, need to knowH
f at 298 K for comparison with literature value. How?
As temp.î HmÎ ie. H
m T
\ Hm = Cp,m T where Cp,m is the molar
heat capacity at constant pressure.
![Page 41: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/41.jpg)
Cp,m = Hm/ T = J mol-1/ K
= J K-1 mol-1 \ HT2 = HT1 + Cp ( T2 - T1) Kirchoff’s equation. and Cp = n Cp(products)- nCp(reactants)
For a wide temperature range: Cp ∫ dT between T1 and T2. Hence : qp = Cp( T2- T1) or H = Cp T and.
![Page 42: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/42.jpg)
qv = Cv ( T2 – T1) or Cv T = U ie. Cp = H / T ; Cv =U /T For small changes: Cp = dH / dT ; Cv = du / dT
For an ideal gas: H = U + p V For I mol: dH/dT = dU/dT + R \ Cp = Cv + R Cp / Cv = γ ( Greek gamma)
![Page 43: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/43.jpg)
Work done along isothermal paths
Reversible and Irreversible paths ie T =0 ( isothermal)
pV = nRT= constant
Boyle’s Law : piVi =pf Vf
Can be shown on plot:
![Page 44: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/44.jpg)
pV diagram
P
V
Pivi
Pfv
f
pV= nRT = constant
![Page 45: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/45.jpg)
Work done = -( nRT)∫ dV/V
= - nRT ln (Vf/Vi)
Equation is valid only if : piVi=pfVf and therefore: Vf/Vi = pi/pf and
Work done = -( nRT) ln (pi/pf) and follows the path shown.
![Page 46: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/46.jpg)
pV diagram
An irreversible path can be followed: Look at pV diagram again.
V
P
Isothermal reversible process (ie. at equlb. at every stage of the process)
Irreversible reaction
PiVi
PfVf
![Page 47: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/47.jpg)
An Ideal or Perfect Gas
NBFor an ideal gas, u = 0
Because: U KE + PE k T + PE (stored in bonds)
Ideal gas has no interaction between molecules (no bonds broken or formed)
![Page 48: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/48.jpg)
Therefore u = 0 at T = 0
Also H = 0 since (pV) = 0 ie no work done
This applies only for an ideal gas and NOT a chemical reaction.
![Page 49: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/49.jpg)
Calculation
eg. A system consisting of 1mole of perfect gas at 2 atm and 298 K is made to expand isothermally by suddenly reducing the pressure to 1 atm. Calculate the work done and the heat that flows in or out of the system.
![Page 50: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/50.jpg)
w = -pex V = pex(Vf -Vi)Vi = nRT/pi = 1 x 8.314 x (298)/202650
= 1.223 x 10-2 m3
Vf = 1 x 8.314 x 298/101325 = 2.445 x 10-2 m3
therefore, w = -pex (Vf- Vi) = -101325(2.445-1.223) x 10-2 = -1239 J
U = q + w; for a perfect gas U = 0therefore q = -w and
q = -(-1239) = +1239 J
![Page 51: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/51.jpg)
Work done along adiabatic path
ie q = 0 , no heat enters or leaves the system. Since U = q + w and q =0 U = w When a gas expands adiabatically, it cools. Can show that: pVγ = constant, where ( Cp/Cv =γ ) and: piVi
γ = pfVfγ and since:
-p dV = Cv dT
![Page 52: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/52.jpg)
Work done for adiabatic path = Cv (Tf- Ti) For n mol of gas: w = n Cv (Tf –Ti) Since piVi
γ = pfVf γ
piViγ/Ti
= pfVf γ/ Tf
\ Tf = Ti(Vi/Vf)γ-1
\ w =n Cv{ ( Ti( Vi/ Vf)γ-1 – Ti}
An adiabatic pathway is much steeper than pV = constant pathway.
![Page 53: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/53.jpg)
Summary
piVi = pfVf for both reversible and irreversible Isothermal processes. For ideal gas: For T =0, U = 0, and H=0 For reversible adiabatic ideal gas processes: q=0 , pVγ = constant and Work done = n Cv{ ( Ti( Vi/ Vf)γ-1 – Ti}
piViγ = pfVf
γ for both reversible and irreversible adiabatic ideal gas.
![Page 54: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/54.jpg)
2nd Law of Thermodynamics
Introduce entropy, S (state function) to explain spontaneous
change ie have a natural tendency to occur- the apparent driving force of spontaneous change is the tendency of energy and matter to become disordered. That is, S increases on
disordering.
2nd law – the entropy of the universe tends to increase.
![Page 55: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/55.jpg)
Entropy
S = qrev /T ( J K-1) at equilibrium
Sisolated system > 0 spontaneous change
Sisolated system < 0 non-spontaneous change
Sisolated system = 0 equilibrium
![Page 56: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/56.jpg)
Properties of S
If a perfect gas expands isothermally from Vi to Vf then since U = q + w = 0 \ q = -w ie qrev = -wrev and wrev = - nRT ln ( Vf/Vi) At eqlb., S =qrev/T = - qrev/T = nRln (Vf/Vi) ie S = n R ln (Vf/Vi)
Implies that S ≠ 0 ( strange!) Must consider the surroundings.
![Page 57: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/57.jpg)
Surroundings
Stotal = Ssystem + Ssurroundings
At constant temperature surroundings give heat to the system to maintain temperature.
\ surroundings is equal in magnitude to heat gained or loss but of opposite sign to make
S = 0 as required at eqlb.
![Page 58: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/58.jpg)
Rem: dq = Cv dT and dS = dqrev / T \ dS = Cv dT/ T and S = Cv ∫ dT /T between Ti and Tf
S = Cv ln ( Tf/ Ti ) When Tf/ Ti > 1 , S is +ve eg. L → G , S is +ve S → L , S is +ve and since qp = H Smelt = Hmelt / Tmelt and Svap = Hvap / Tvap
![Page 59: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/59.jpg)
Third Law of Thermodynamics
eg. Standard molar entropy, SmThe entropy of a perfectly
crystalline substance is zero at T = 0
S
m/ J K-1 at 298 K ice 45 water 70 NB. Increasing disorder water vapour 189 For Chemical Reactions: S
rxn = n S (products) - n S
( reactants)
eg. 2H2 (g) + O2( g) → 2H2O( l ), H = - 572 kJ mol-1
![Page 60: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/60.jpg)
Calculation
Ie surroundings take up + 572kJ mol-1 of heat
Srxn = 2S(H2Ol) - (2 S
(H2g ) + S (O2g) )
= - 327 JK-1 mol-1 ( strange!! for a spontaneous reaction; for this S is + ve. ).
Why? Must consider S of the surroundings also. S total = S system + S surroundings S surroundings = + 572kJ mol-1/ 298K = + 1.92 x 103JK-1 mol-1
\ S total =( - 327 JK-1mol-1) + 1.92 x 103 = 1.59 x 103 JK-1 mol-1 Hence for a spontaneous change, S > 0
![Page 61: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/61.jpg)
Free Energy, G
Is a state function. Energy to do useful work. Properties Since Stotal = Ssystem + Ssurroundings
Stotal = S - H/T at const. T&p Multiply by -T and rearrange to give: -TStotal = - T S + H and since G = - T Stotal
ie. G = H - T S
Hence for a spontaneous change: since S is + ve, G = -ve.
![Page 62: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/62.jpg)
Free energy
ie. S > 0, G < 0 for spontaneous change ;
at equilibrium, G = 0.
Can show that : (dG)T,p = dwrev ( maximum work)
\ G = w (maximum)
![Page 63: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/63.jpg)
Properties of G
G = H - T S dG = dH – TdS – SdT H = U + pV \dH = dU + pdV + Vdp Hence: dG = dU + pdV + Vdp – TdS – SdT dG = - dw + dq + pdV + Vdp – TdS – SdT \ dG = Vdp - SdT
![Page 64: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/64.jpg)
For chemical Reactions:
For chemical reactions
G = n G (products) - n G (reactants) and
Grxn = H
rxn - T Srxn
![Page 65: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/65.jpg)
Relation between Grxn and position of
equilibrium
Consider the reaction: A = B
Grxn = G
B - GA
If GA> G
B , Grxn is – ve ( spontaneous rxn)
At equilibrium, Grxn = 0.
ie. Not all A is converted into B; stops at equilibrium point.
![Page 66: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/66.jpg)
Equilibrium diagram
![Page 67: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/67.jpg)
For non-spontaneous rxn. GB > GA, G is + ve
![Page 68: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/68.jpg)
Gas phase reactions
Consider the reaction in the gas phase: N2(g) + 3H2(g)→ 2NH3(g)
Q =( pNH3 / p)2 /( ( pN2/ p) (pH2/ p)3 ) where :
Q = rxn quotient ; p = partial pressure and p = standard pressure = 1 bar
Q is dimensionless because units of partial pressure cancelled by p .
At equilibrium: Qeqlb = K = (( pNH3 / p)2 / ( pN2/ p) (pH2/ p)3
)eqlb
![Page 69: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/69.jpg)
Activity ( effective concentration)
Define: aJ = pJ / p where a = activity or effective concentration. For a perfect gas: aJ = pJ / p For pure liquids and solids , aJ = 1 For solutions at low concentration: aJ = J mol dm-3
K = a2NH3 / aN2 a3
H2
Generally for a reaction: aA + bB → cC + dD
K = Qeqlb = ( acC ad
D / aaA ab
B ) eqlb = Equilibrium constant
![Page 70: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/70.jpg)
Relation of G with K
Can show that:
Grxn = Grxn + RT ln K
At eqlb., Grxn = 0
\ Grxn = - RT ln K
Hence can find K for any reaction from thermodynamic data.
![Page 71: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/71.jpg)
Can also show that:
ln K = - G / RT
\K = e - G / RT
eg H2 (g) + I2 (s) = 2HI (g) , H
f HI = + 1.7 kJ mol-1 at 298K; H
f H2 =0 ; Hf I 29(s)= 0
![Page 72: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/72.jpg)
calculation
Grxn = 2 x 1.7 = + 3.40 kJ mol-1
ln K = - 3.4 x 103 J mol-1 / 8.314J K-1 mol-1 x 298K = - 1.37 ie. K = e – 1.37 = 0.25 ie. p2 HI / pH2 p =0.25 ( rem. p = 1 bar; p 2 / p = p )
\ p2 HI = pH2 x 0.25 bar
![Page 73: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/73.jpg)
Example: relation between Kp and K
Consider the reaction: N2 (g) + 3H2 (g) = 2NH3 (g)
Kp =( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3
and K = [( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3] eqlb
\ Kp = K (p)2 in this case. ( Rem: (p)2 / (p)4 = p -2
![Page 74: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/74.jpg)
For K >> 1 ie products predominate at eqlb. ~ 103
K<< 1 ie reactants predominate at eqlb. ~ 10-3
K ~ 1 ie products and reactants in similar amounts.
![Page 75: THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins](https://reader035.fdocuments.in/reader035/viewer/2022081417/5681645f550346895dd6348b/html5/thumbnails/75.jpg)
Effect of temperature on K
Since \ Grxn = - RT ln K = H
rxn - TSrxn
ln K = - Grxn / RT = - H
rxn/RT + Srxn/R
\ ln K1 = - Grxn / RT1 = - H
rxn/RT1 + Srxn/R
ln K2 = - Grxn / RT2 = - H
rxn/ RT2 + Srxn/ R
ln K1 – ln K2 = - H
rxn / R ( 1/ T1 - 1/ T2 ) 0r
ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation