Loading in 2 Seconds...

THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins

Loading in 2 Seconds...

- By
**river** - Follow User

- 175 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins' - river

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### THERMODYNAMICS.Elements of Physical Chemistry. By P. Atkins

Concerned with the study of transformation of energy:

Heat work

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

→ →

Example:

↔matter↔energy ↔ energy not mattermatter× Energy ×

Open system

Closed system

Isolated system

If system is themally isolated called Adiabatic system eg: water in vacuum flask.

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.

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)

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

Different types of energies:

- Kinetic energy = ½ mv2 (chemical reaction) kinetic energy

(KE) k T (thermal energy)where k = Boltzmann constant

- Potential energy (PE) = mgh – energy stored in bonds

Now,U = KE + PE

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

w = distance × opposing force

w = 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

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 moles

R – gas constant

T = temperature

V – volume

p = pressure

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

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

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

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)

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.

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

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

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).

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.

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.

H2(g)→ H(g) + H(g), H diss = +436kJmol-1

H2O(l)→ H2O(g), Hvap = +44.0 kJmol-1

Calculate U for the following reaction:

CH4(l) + 2 O2(g)→ CO2(g) + 2H2O(l), H = - 881.1kJmol-1

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

calculation

U = - 881.1 – ((1 – 2)(8.314) 298)/ 1000

= - 881.1 – (-1)(8.314)(0.298) = - 2182.99 kJ mol-1

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)

NB: Hf of element = 0 in reference state

- Can apply these to thermochemical calculations

eg. Can compare thermodynamic stability of substances in

their standard state.

- From tables of Hf can calculate H f rxn for any reaction.

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)

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

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)

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

From thermochemical data:

- C (g) +O2(g)→ CO 2(g)Hcomb=-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) , Hcomb= -393.5 –
- (-283.0) = - 110.5 kJ mol-1

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

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

Variation of H with temperature

Suppose do reaction at 400 K, need to know

Hf at 298 K for comparison with literature value. How?

- As temp.î HmÎ ie. Hm T
- Hm = Cp,m T where Cp,m is the molar
- heat capacity at constant pressure.

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.

qv = Cv ( T2 – T1) or CvT = 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)

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:

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.

An Ideal or Perfect Gas

- NB

For 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)

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.

Calculation

- eg. A system consisting of 1mole of perfectgas 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.

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 = 0

therefore q = -w and

q = -(-1239) = +1239 J

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: piViγ = pfVfγ and since:
- -p dV = Cv dT

Work done for adiabatic path = Cv (Tf- Ti)

- For n mol of gas: w = n Cv (Tf –Ti)
- Since piViγ= pfVfγ
- piViγ/Ti= pfVfγ/ 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.

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 = pfVfγ for both reversible and irreversible adiabatic ideal gas.

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.

Entropy

- S = qrev /T ( J K-1) at equilibrium
- Sisolated system> 0 spontaneous change
- Sisolated system< 0 non-spontaneous change
- Sisolated system= 0 equilibrium

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.

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.

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

Third Law of Thermodynamics

- eg. Standard molar entropy, SmThe entropy of a perfectly crystalline substance is zero at T = 0
- Sm/ J K-1 at 298 K
- ice 45
- water 70 NB. Increasing disorder
- water vapour 189
- For Chemical Reactions:
- Srxn = n S(products) - n S( reactants)
- eg. 2H2 (g) + O2( g)→ 2H2O( l ), H = - 572 kJ mol-1

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

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.

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)

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

For chemical Reactions:

- For chemical reactions
- G = n G (products) - n G (reactants)
- and
- Grxn = Hrxn - T Srxn

Relation between Grxn and position of equilibrium

- Consider the reaction: A = B
- Grxn = GB - GA
- If GA> GB , Grxn is – ve ( spontaneous rxn)
- At equilibrium, Grxn = 0.
- ie. Not all A is converted into B; stops at equilibrium point.

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

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 a3H2
- Generally for a reaction:
- aA + bB → cC + dD
- K = Qeqlb = ( acC adD / aaA abB ) eqlb = Equilibrium constant

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.

Can also show that:

- ln K = - G / RT
- K = e - G / RT
- eg
- H2 (g) + I2 (s)= 2HI (g) , Hf HI = + 1.7 kJ mol-1 at 298K; HfH2 =0 ; HfI 29(s)= 0

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. p2HI / pH2 p =0.25 ( rem. p = 1 bar; p 2 / p = p )
- p2HI = pH2 x 0.25 bar

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

ForK >> 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.

Effect of temperature on K

- Since Grxn = - RT ln K = Hrxn - TSrxn
- ln K = - Grxn / RT = - Hrxn/RT + Srxn/R
- ln K1 = - Grxn / RT1 = - Hrxn/RT1 + Srxn/R
- ln K2 = - Grxn / RT2 = - Hrxn/ RT2 + Srxn/ R
- ln K1 – ln K2 = - Hrxn / R ( 1/ T1 - 1/ T2 ) 0r
- ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation

Download Presentation

Connecting to Server..