generalchem_ls_22.pdf

8/10/2019 GeneralChem_LS_22.pdf

1/24

Phase Diagram

BITSPilani, Pilani Campus

Map showing ranges of Tand p over which variousphases are stable.

Phase Boundaries-Melting, Vaporization andBoiling Curves

Triple Point, Critical Point

Phase Rule F = C P + 2


2/24

Clapeyron equation

The slopes of the lines of two phase equilibrium in the p-Tphase diagrams are related to the entropy change and volumechange.

Gm(T,p) = Gm

(T,p)

For equilibrium of phases and at T and p. Suppose p ischanged to p + dp, and T to T + dT, and equilibrium is to bemaintained, then require dGm

(T,p) = dGm(T,p)

SmdT + Vmdp = SmdT+ Vmdp

Rearranging, dp/dT = (SmSm)/(VmVm= transS/transV

UsingtransS = transH/T, can rewrite the equation as

dp/dT = transH/T(transV)



3/24

Clausius Clapeyron equation

For solid-gas equilibrium or liquid-gas equilibrium awayfrom the critical point.

VmgVml VmgRT/p

dp/dT vap

H/TVm

g vap

Hp/RT2

or d(ln p)/dT = vapH/RT2 which is the Clausius-Clapeyronequation. Over small ranges of temperature,this may be integrated assuming constant vapH yielding

ln (p2/p1) = (vapH/R)(1/T21/T1)



4/24

Chemical potential and composition

For a perfect gas, Gm

(T,p) = Gm

0(T) + RT ln(p/p0), whereGm

0(T) is the standard molar Gibbs energy at the given Tand p0= 1 bar.

In a perfect gas mixture, i(T,p,x) = i0(T) + RT ln(pi/p

0)where piis the partial pressure of constituent i, and i

0= Gmi0

For any solutions which are very dilute in the solutes

i(T,p,x) = i0(T) + RT ln[i], where [i] is the molar

concentration.



5/24

Chemical Reactions

As with physical processes, the heat transferaccompanying chemical reactions at a giventemperature are also measured usually underconditions of constant volume or constant pressure.

At constant volume, qV= (U)VU

At constant pressure, qp= ()p H

H = U + (pV) U + gas

RT where gas

is thedifference (products reactants) in the sum of thestoichiometric coefficients of gas phase species.



6/24

Standard states

Reaction enthalpies are usually reported for a set of socalled standard conditions at a given temperature. Thestandard state of a substance is the pure substance in thespecified physical state of aggregation, ie., gas, liquid orsolid, and the particular crystalline modification eg.,

graphite/diamond, rhombic/ monoclinic sulfur, at the giventemperature and a pressure of 1 bar.



7/24

Standard reaction enthalpies

For example, for the reaction

CH2=CH2(g) + H2(g) CH3-CH3(g), the standard enthalpychange = 137 kJ at 298.15 K.

Similarly, for 2H2(g) + O2(g) 2H2O(l), 298= 572 kJ.

Combustion Reaction:

CH4(g) + 2O2(g)CO2(g) + 2H2O(l), = -890 kJ

Standard enthalpy of combustion C298(CH4,g) is 890kJ/mole



8/24

Hess law

The standard enthalpy of a reaction is the sum of thestandard enthalpies of the reactions into which the overallreaction may be divided. Example:

C3H6(g) + H2(g) C3H8(g) r1= 124 kJ

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)r2= 2220 kJ

H2O(l) H2(g) + O2(g) r3= 286 kJ

C3H6(g) + (9/2) O2(g) 3CO2(g) + 3H2O(l)

r= r1+ r2 + r3= 2058 kJ



9/24

Standard enthalpy of formation

The standard enthalpy of formation of a substance is definedas the standard enthalpy of the reaction leading to theformation of one mole of the substance from its elements intheir reference states, ie., the element in its most stable stateunder the prevailing conditions.

Standard enthalpies of reaction can be obtained fromstandard enthalpies of formation using

r= ifi(products) ifi(reactants)

H2(g) + O2(g) H2O(l) f298(H2O(l))= 286 kJ/mole

C(s,graphite) + O2(g) CO2(g) f298(CO2(g)) = 393.5kJ/mole



10/24

Enthalpy of Reaction

For C6

H6

(l) + 15/2 O2

(g) 6CO2

(g) + 3H2

O(l)

r= 6 f(CO2(g)) + 3 f(H2O(l))

f(C6H6(l)) 15/2 f(O2(g))

= 6x(393.5) + 3x(285.8) (49) (0) = 3268 kJThe enthalpy of combustion of benzene liquid is therefore3268 kJ/mole



11/24

Reaction Gibbs Energy

For the general reaction aA + bB cC + dD, at givenT and p, at some arbitrary composition, the ReactionGibbs Energy is given as rG = cC+ dDaAbB


Note slopes.

rG < 0 Spontaneous

rG > 0 Spontaneous At equilibrium,

(G/n)T,peq = rG eq = 0


12/24

rGand composition

Since J

= J

+ RT ln [J]

for each constituent,

rG = d(D+ RT ln [D]) + c(C+ RT ln [C])

a(A+ RT ln [A]) b(B+ RT ln [B})

rG =

rG+ RT ln {([C]c[D]d)/([A]a[B]b)}

rG = rG+ RT ln Q

where rGis the standard Reaction Gibbs Energy and Q iscalled the reaction quotient.

More generally, use activity in place of concentration

For gases commonly use partial pressure pJin place of [J]



13/24

Equilibrium Constant

At equilibrium, r

G = 0, or

rG= RT ln Qeq = RT {([C]c[D]d)/([A]a[B]b)}eq

rG= RT ln K

The equilibrium constant is a function of the temperaturealone.



14/24

Spontaneity

Reaction feasible if rG> 1, and not so if rG>> 0, ie., K 0, then rG< 0 at allT

(b) If rH< 0 (exothermic) and rS< 0, then rG< 0for T < rH/ rS

(c) If rH> 0 (endothermic) and rS> 0, then rG< 0for T > rH/ rS

(d) If rH> 0 (endothermic) and rS< 0, then rG< 0 atno temperature

Of course, it is rG which will ultimately decide thefeasibility, with rG = RT ln(Q/K)



15/24

Effect of T on K and rG0

From r

G0= RT ln K, one has

d(lnK)/dT = d/dT (rG0/RT)

From ((G/T)/T) = G/T2S/T = -H/T2

d(lnK)/dT = rH0/RT2

which is calledvantHoff equation. If assumed independentof T, then

ln (K2/K1) = (rH0/R) (1/T1- 1/T2)

Note le Chatelier principle



16/24

Chemical Kinetics

How rapidly reactants are consumed and products formed

How reaction rates respond to changes in conditions suchas temperature and pressure, or the presence of a catalyst

Practical importance of predicting speed of reaction and

approach to equilibrium.

Leads to an understanding of reaction mechanism, theanalysis of a reaction into a sequence of elementary steps



17/24

Reaction Rate


Raw Data:

Concentrations of reactants/products at series of times afterreaction started.Information, especially structural, on

any intermediateInstantaneous rateis proportional tothe slope of the graph ofconcentration versus time

Consider the reaction N2(g) + 3H2(g) 2NH3(g)Rate = d[N2]/dt = 1/3 d[H2]/dt = d[NH3]/dt

If at some time rate of formation of NH3is 1.2 mmol l-1s-1,

then rate of consumption of N2is 0.6 mmol l-1s-1


18/24

Differential Rate Law

Equation that represents the rate of reaction in terms ofthe molar concentrations of the species in the overallreaction (including, possibly the products).

Often found to be proportional to the molar concentrationsof the reactants raised to simple powers.

If the rate of aA + bB Products is of the form, Rate =k[A]m[B]n, reaction said to be of order m inA, and order n inB, and order m+n overall. k is called the rate constant,

characteristic of reactionFor example, for H2+ I22HI, d[H2]/dt = k[H2][I2]

Reaction is first order in H2, first order in I2and second orderoverall



19/24

Rate Laws

NO2+ CO NO + CO2, Under certain conditions,d[CO]/dt = k[NO2]2so that the reaction is second order in

NO2, zero order in CO, and order two overall

Rate Law may be more complicated and it may not always

be possible to specify order. For eg., forH2+ Br22HBr

d[HBr]/dt = k[H2][Br2]3/2/{[Br2] + k[HBr]}

When k[HBr]


20/24

Determination of rate law

Measurement of any convenient physical property, such aspressure, density, absorbance, conductivity etc., or chemicalanalysis may be used to follow the rate

Method of Initial Rates:

Instantaneous rate measured at beginning of reaction forseveral different initial concentrations of reactants.

Often used in conjunction with Isolation Method.

If initial rate r0= k[A]0n

, thenln r0= ln k+ n ln[A]0so that a plot of lnr0versus ln[A]0hasslope n



21/24

Example


[I]0/M 1.0x10-5 2.0x10-5 4.0x10-5 6.0x10-5

[Ar]0/M

Rate/M s-1

Rate/M s-1

Rate/M s-1

Rate/M s-1

1.0x10-3 8.70x10-4 3.48x10-3 1.39x10-2 3.13x10-2

5.0x10-3 4.35x10-3 1.74x10-2 6.96x10-2 1.57x10-1

1.0x10-2 8.69x10-3 3.47x10-2 1.38x10-1 3.13x10-1

2I(g) + Ar(g) I2(g) + Ar(g)


22/24

Example


Slope = 2 Slope = 1

-(1/2)d[I]/dt = k[I]2[Ar]By plotting the intercepts of the first graph versuslog[Ar]0, determine k = 8.6x10

9 l2mol-2s-1


23/24

Integrated Rate Laws

Expression for the concentration of a species as a functionof time, which is what is experimentally measured typically.

First order reaction, say A P, with d[A]/dt = k[A]

Integrate from t = 0 when [A] = [A]0to desired time t to find

ln[A] ln[A]0= kt, so that a plot of ln[A] vs. t is linear withslope = -k.

[A] = [A]0ekt, and [P] = [A]0(1 ekt)

Similarly, for a reaction A Products with d[A]/dt = k[A]2,

one obtains

1/[A] = 1/[A]0+ kt so that a plot of 1/[A] versus t is a straightline of slope k



24/24

Half Life

Alternatively, for the second order reaction above,

[A] = [A]0/(1 + [A]0kt), and

[P] = kt([A]02)/(1 + kt[A]0)

The half-life t1/2of a reactant is the time taken for itsconcentration to fall to half the initial value

For a first order reaction, the half life of the reactant onwhose concentration the reaction rate depends is given by

t1/2= (ln 2)/k, seen to be independent of [A]0.

For a reaction, say A P, with -d[A]/dt = k[A]2, the half life isfound to be t1/2= 1/k[A]0

generalchem_ls_22.pdf

Documents