intoduction to physical chemistry. the energies of chemistry is the essential point of physical...
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Intoduction to Physical Chemistry
The energies of chemistry is the essential point of physical chemistry.
Physical chemistry is based on measurements.
Although certain techniques can measure energy change like measuring heat of
combustion, usually the measurement is made on concentration change which
is related to energetic driving force.
Experiments may be measured at constant pressure or constant volume
• Total energy change ΔH (enthalpy) ΔU (energy)~E• Disordered energy TΔS (entropy change) TΔS (entropy change)• Ordered Energy ΔG ΔA
ΔH= ΔA+ TΔS ΔH= ΔG+ TΔS ΔG= ΔH- TΔS ΔA= ΔH- TΔS
This is the driving force for chemical change if P=constant
This is the driving force for chemical change if V=constant
P=constantV≠constant
P ≠ constantV=constant
ΔG*
ΔGA
BG
RTK e
A
KB
G
RTRT
k eNh
d ARate k A
dt
K
A Bk
A B
BasicityOr
Acidity
NucleophilicityOr
ElectrophilicityMeasure ofreactivity
Measure ofstability
S H
R RTK e e
S H
R RTRT
k e eNh
lnS
RRT
eNh
H
R
1
T
H
R
ln K
1
T
S
R
• The driving force for chemical change is free energy change for the
reversible equilibrium ΔG and for reaction kinetics ΔG*
• As a result of the driving force the concentration will change and the
concentration change we can calculate the magnitude of the driving
force.
No matter how important equilibrium constants are, acid strengths are not measured by their Keq value. The Keq value represents the full equilibrium constant as in [3.17] or in [3.22]:
H2O + HA H3O(+) + A (-)Keq
[3.22a]
Keq=[H3O(+)] [A(-)]
[H 2O] [HA]
[3.22b]
but acid strength is measured by a pseudo equilibrium constant, Ka, from which the water concentration is omitted [3.23]:
HA H(+) + A(-)Ka
[3.23a]
Ka=[H(+)] [A (-)]
[HA]
[3.23b]
Clearly, the two constants are related to each other according to [3.24]:
Ka = Keq [H2O]
[3.24]
For historical reasons, acid strength has been expressed as the negative log of base 10 of Ka. This quantity is denoted as pKa (the power of Ka)
pKa = -log Ka
[3.25]
Now if we wish to relate G to pKa, we must take -log of equation [3.24] and rearrange it. The result of this elementary manipulation is given in [3.26]
G = 2.303RT {pKa + log [H2O]}
[3.26]
Knowing that RT = 0.002 x 300 = 0.6 and [H2O] = 55.56 M, thus, log [H2O] = 1.745, we obtain [3.26] in a numerical form.
G = 1.3818 {pKa + 1.745}
[3.27]
Note that G = 0 does not occur at pKa = 0. Instead if G = 0 then
pKa = - log [H2O] = -1.745
[3.28]
weak acid
strong acid
Go << 0
Go
Go >> 0
Keq << 1; Ka << [H2O] pKa >> -log[H2O]
Keq = 1; Ka = [H2O] = 55.56 pKa = -log[H2O] = -1.745
Keq >> 1; Ka >> [H2O] pKa << -log[H2O]
The relationships between structure and acidity are summarized below. (i) The acidity increases with the electronegativity of the atom that carries the H:
C H N H O H F H
[3.29]
(ii) Acidity increases as we descend a vertical column of the periodic table:
HF HCl HBr HI [3.30]
(iii) Acidity decreases with increasing hybridization:
sp sp2 sp3
[3.31]
(iv) Acidity increases with the inductive, i.e., more electron withdrawing, effect of the substituent:
FCH2–COOH ClCH2–COOH BrCH2–COOH
[3.32]
(v) Acidity increases with the number of electron withdrawing groups (EWG):
F3C–COOH F2HC–COOH FH2C–COOH H3C–COOH
[3.33]
(vi) Acidity increases with increasing resonance in the conjugate base:
Even though, pKa is defined as a measure of acid strength it can also be used to measure the strength of its conjugate base.
H3C C
O
O
H
H3C C
O(-)
O
+ H(+)
pKa = 4.75
[3.36a]
CH3 CH2 OH
pKa = 16
CH3 CH2 O(-)+ H(+)
[3.36b]
Thus, acetic acid is a stronger acid than ethyl alcohol but the acetate ion is a weaker base than the ethoxide ion. Thus the larger the pKa value of the conjugate acid, the stronger the base will be. The following set of equations illustrates this point for three families of compounds.
pKa = +9.24
pKa = -1.74 pKa = +15.74
pKa = +38
pKa = +3.17
H4N(+)
H3O(+)
H3N + H (+)
H2O + H (+)
HF
H2N:(-) + H (+)
HO(-) + H (+)
F(-) + H (+)
basestrength
basestrength
acidstrength
acidstrength
Table 6.1 Rate equations and their characteristics for reactions of various order
d[A]dt
= k[A]-
d[A]dt
= k-
d[A]dt
= k[A]2-
d[A]dt
= k[A][B]-
0
1
2
1[B]0 - [A]0
[B][A]
[B]0[A]0
- lnln = kt
1[A]0
1[A]
- = kt
ln[A]0 - ln [A] = kt
[A]0 - [A] = kt
ln2k =
[A]02k =
1k[A]0
=
Ms
s-1
M-1s-1
Order Differentialrate equation
Integrated rate equation Half-life time
Units ofk
Higher order reactions (i.e., 3, 4, ….. n) are not included in Table 6.1. If further clarification is needed, students can consult textbooks of Physical Chemistry or Physical Organic Chemistry.
We can now recognize that SN2 and E2 reactions follow second order kinetics, as shown in [6.17] and [6.18]
C C
H
Cl
+ :OH (-) C C
H
OH
+ Cl(-)kSN2
[6.17a]
= kvSN2 SN2 [R-Cl][OH-]
[6.17b]
[6.18a]
vE2 = kE2[R-Cl][OH-]
[6.18b]
Note that the SN2 mechanism occurs with an inversion of configuration at the carbon carrying the chlorine. This only occurs if and only if the carbon is a stereocentre and we start with an optically pure enantiomer.
C C
H
Cl
C C + H2O + Cl (-) HO: (-) +kE2
C Ck'E2
C C
H
OH
k'SN2
C C
H
Cl
[6.25]
k'SN2 = kSN2 [OH(-)] and k'E2 = kE2 [OH(-)]where This mechanism can be written in a simplified fashion as:
A
X
Y
k1
k2
[6.26]
Concentration profiles for parallel first order reactions.
The rate of consumption of A to form both X and Y can be characterized by the following differential [6.27] and integrated [6.28] rate equations:
d[A]dt
- = k1[A] + k2[A] = (k1 +k2)[A] = k[A]
[6.27]
[A] = [A]0 e-kt
[6.28]
Equation [6.29] and [6.30] give the individual rates of product formation.
d[X]dt
= k1[A] = [A]0k1e-kt
[6.29]
d[Y]dt
= k2[A] = [A] 0k2e-kt
[6.30]
Taking the initial conditions to be
[X]0 = [Y]0 = 0
[6.31]
one may obtain the following integrated rate equations:
[6.32]
[6.33]
The rate of consumption of A to form both X and Y can be characterized by the following differential [6.27] and integrated [6.28] rate equations:
[6.27]
[A] = [A]0 e-kt
[6.28]
Equation [6.29] and [6.30] give the individual rates of product formation.
[6.29]
[6.30]
Taking the initial conditions to be
[X]0 = [Y]0 = 0
[6.31]
one may obtain the following integrated rate equations:
kk1
[X] = [A]0(1-e-kt)
[6.32]
kk2
[Y] = [A]0(1-e-kt)
[6.33]
From these two equations we may calculate the product ratio:
[X][Y] =
k2
k1
[6.34]
Of course, both k1 and k2 are pseudo-first-order rate constants and they incorporate the nucleophile concentration, which cancels in [6.34]. [6.34] demonstrates the relative importance of SN2/E2 mechanisms, which we stated in [6.20]. The SN1 [6.21] and E1 [6.22] mechanisms are not only parallel but consecutive multi-step mechanisms. Let us examine the kinetic methods we may use to study multi-step reactions.
Arrhenius’s relationship, which was presented earlier in [1.3], is given here in both its logarithmic and exponential forms in terms of the pre-exponential factor (A) and the energy of activation (Ea).
lnk = - + lnAEaR
1T
[6.60a]
k = Ae-Ea/RT
[6.60b]
Logarithmic (a) and nonlogarithmic (b) plots of the rate constant against reciprocal absolute temperature (a) and absolute temperature (b), respectively.
In the Arrhenius’s equation, the pre-exponential factor is assumed to be independent of T. In contrast, the transition state theory stipulates that the pre-exponential factor depends on T:
RTNh
k = e-G /RT
Nhe- /RTRT
= e S /R
[6.61]
Dividing both sides of [6.61] by T and taking the log to base 10 of both sides we obtain [6.62]:
logkT
= - H2.303R
1T
+ log RNh
+ S2.303R
[6.62a]
An example of the determination of the energy of activation (Ea) according to [6.60b], and the enthalpy of activation (H‡ ) as well as the entropy of activation (S‡ ) according to [6.62b], is shown for a typical case in Figure 6.10.
Figure 6.10 Determination of energy of activation (Ea ---), and enthalpy of activation (H‡ ) for the reaction MeSPh +NaIO4 MeS(O)Ph + NaIO3
The Arrhenius-type energy of activation (a) and enthalpy of activation (H‡) are related to each other according to equation [6.63].
Ea = H‡ + RT
[6.63]
Since RT at room temperature is about of 0.6 kcal mol-1 or 2.51 kJ mol-1, the numerical values of Ea and H are usually close to each other. As discussed earlier, for parallel (i.e. competing) reaction mechanisms the product ratio at any time during the reaction is also the rate constant ratio (c.f. [6.20] and [6.24]). This may be used to calculate the differences in the activation parameters, H‡ and S‡, as shown in [6.64]:
RTNhRTNh
e
e
e
S1 /R
H2 /R
H1 /RT
H2 /RT
ek1
k2=
[X][Y]
=
[6.64b]
[X][Y]
= -logH1 - H2 S1 - S 2
2.303R2.303R1T
[6.64c]
In closing this section, we might add that for a reaction, in which the Transition State is more disordered than the Reactant State, the entropy change is expected to be positive. This is the case for SN1 and E1 reactions. In contrast to this if the disorder is reduced when the system reaches the Transition State, then the entropy change is expected to be negative. This is the case for SN2 and E2 mechanisms.
S > (for SN1 and E1)
[6.65a]
S < (for SN2 and E2)
[6.65b]
This expectation is illustrated in Table 6.2 for the following hydrolytic reactions following either the SN1 or SN2 mechanisms
R-L + H2O R-OH + LH
[6.66]
Table 6.2 Activation parameters for unimolecular (SN1) and bimolecular (SN2) hydrolysis of alkyl halide-type compounds.
Compounds ‡ S‡ Mechanism
(Kcal/mol) (cal/mol deg)
Me-Cl 25.3 -8.6 SN2
Me-Br 24.1 -6.7 SN2
iPr-Cl 24.9 -5.3 SN2
iPr-Br 24.4 -1.4 SN2
tBu-Cl 20.5 +3.4 SN1
tBu-SMe2(+) 31.6 +15.7 SN1
The numerical value of S‡ is also dependent on solvent polarity. The value of S‡ in a polar solvent is usually a smaller negative number than in an apolar solvent. In the case of an apolar
Figure 6.11 Three translational coordinates and 3 rotational coordinates of a general n-atomic molecule leave 3n - 6 internal coordinates. During a chemical reaction, all 3n - 6 internal coordinates change. Thus, the potential energy hypersurface may be regarded as a multivariable function.
E = f(x1, x2, x3, ............, x3n - 6 )
However, if we are lucky, we can single out 2 of the 3n - 6 coordinates that change more drastically than the others during the chemical reaction. Let us consider nucleophilic substitution, both SN2 and SN1.
C LNu:(-) + Nu C + :L(-)
r2 r2r1 r1
Let us consider the elimination reaction; both E2 and E1. In fact there is a third mechanism labeled as E1cb (elimination first order conjugate base). In the E1 mechanism, the leaving group (L) leaves first followed by proton loss, while in the E1cb the proton is transferred first to the base (Bronsted acid base reaction), followed by the loss of the leaving group, taking its bonding electron pair with it.
H C C L
H C C (+)
C C L(-)
C C
E1
E1cb
[6.70]
For all three mechanisms, the following pair of bond lengths may be singled out as those internal coordinates that change the most during the elimination reaction.
C C
H
L
rC-L
rC-H
[6.71]
The three reaction mechanisms correspond to three routes on the (PES). Of course the molecular structure, the nature of the leaving group, the basicity and the concentration of the nucleophile, as well as the polarity of the solvent, will influence which route will be the most energetically favorable.
There are rare molecular structures that favor E1cb mechanisms and some of examples are shown below.
C C
NR3
O2N
H
H
H
H(+)
HO(-)C C
NR3
O2N
HH
(+)
H2O +
H
(-) C C
H
O2N
H
H
Ph
OMe
H
NO2
Ph
OMeNO2 NO2
PhHO: (-)
H2O + (-) + (-):OMe
ClH HH
HO: (-)
ClH
H2O +
H(-)
+ Cl(-)
C C
CN
H Ar
NC
CNNC C C
CN
C ArCN
CN
C C
C
Ar
C
C
N
N
N
HO: (-)
H2O + (-)
N