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Dec 23, 2018

Van't Hoff Equation

Van't Hoff Equation - Effect on temperature on Equilibrium Constant:

Equilibrium Constant KP of a reaction is constant at a given temperature but when temperature is changed, the value of equilibrium constant also changed. The quantitative relation, relation, Known as Van't Hoff equation is derived by using Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation is 

ΔG0 = ΔH0 + T[d(ΔG0)/dT]P 

Zero superscript is indicating the stranded values.
or, - (ΔH0/T2 ) = -(ΔG0/T2 ) + 1/T[d(ΔG0)/dT]P 
or, - (ΔH0/T2 )  [d/dTG0/T)]P 
Again Van't Hoff isotherm is, - RT lnKP = ΔG0
or, - R lnKP = ΔG0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dTG0/T)]P
Comparing the above two equation we have,
(dlnKP/dT) = (ΔH0/T2)
This is the differential form of Van't Hoff reaction equation.
Greater the value of ΔH0, the faster the equilibrium constant (KP) changes with temperature (T).
Separating the variables and integrating,
dlnKP  = (ΔH0/R) (dT/T2)
(Assuming that ΔH0 is independent of temperature)
or, lnKP  = - (ΔH0/R)(1/T) + C (integrating constant)
The integration constant can be evaluated and identified as ΔS0/R using the relation ΔG0 = ΔH0 - TΔS0.
Thus the Van't hoff Equation is 

lnKP  = - (ΔH0/R)(1/T) + ΔS0/




For a reaction 2A + B ⇆ 2C, ΔG0(500K) = 2 KJ mol-1 
Find the KP at 500K for the reaction A + ½B  C .

ΔG0(500K) for the reaction A + ½B  C  is 2 KJ mol-1/2 = 1 KJ mol-1/2
The relation is ΔG0 = - RT lnKP 
or, KJ mol-1  = - 8.31 × 10-3 KJ mol-1 K-1 × 500K lnKP 
or, lnKP  = 1/(8.31 × 0.5) = 0.2406

∴ KP = 1.27

Plot of lnKP vs 1/T :

For exothermic reaction, ΔH0 = (-) ve.
Examples are the formation of ammonia from Hand N2.
Van't Hoff Equation - Effect on temperature on Equilibrium Constant
Plot of lnKP vs 1/T
For endothermic reaction, ΔH0 = (+) ve.
Examples are the dissociation of HI into Hand I2.
For the reaction, ΔH0 = 0, lnKP is independent of T. Provided ΔS0 does not change much with T.
Since for idel system, H is not a function of P and ΔH0 = ΔH
and Van't Hoff equation is dlnKP/dT = ΔH/RTand the integrated equation is,
lnKP = (ΔH/R)(1/T) + ΔS/R
However S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS0 and ΔG0.
If we consider the integrated Van't Hoff equation at two temperature then, it becomes,

ln(KP2/KP1) = (ΔH/R){(T2 -T1)/T1T2)}

Where KP1 and KP2 are the equilibrium constants of the reaction at two different temperature T1 and T2 respectively.
Thus determination of KP1 and KP2 at two temperature helps to calculate the value of ΔH of the reaction.
The above reaction is called Van't Hoff reaction isobar since P remains constant during the change of temperature.



Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. Under What condition do you expect a linear relationship between logk and 1/T ?

Again KP = KC (RT)ΔƔ
or, lnKP = lnKC + ΔƔ lnR + ΔƔlnT
Differentiating with respect to T,
dlnKP/dT = dlnKC/dT + ΔƔ/T
But dlnKP/dT = ΔH0/RT2
Hence, dlnKC/dT = ΔH0/RT2 - ΔƔ/T = (ΔH0 - ΔƔRT)/RT2
or, dlnKC/dT = ΔU0/RT2
ΔU0 is standerd heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.
The integrated form of the equation at two temperature is,

ln(KC2/KC1) = (ΔU0/R){(T2 -T1)/T1T2)}
For ideal system ΔU0 = ΔU.
Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.
Two important assumptions are:
(i) The reacting system is assumed to behave ideally and 
(ii) ΔH is taken independent of temperature for small range of temperature change.
Due to assumption involved ΔH and ΔU do not produce precise value of these reactions.
As ΔG is independent of pressure for ideal system a and b also independent of pressure.
Hence ΔG0 = - RT lnKP 
or, [dlnKP/dT]T = 0



Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.

Vant Hoff reaction isotherm is ΔG = - RT lnKa + RT lnQa
But when the reaction attain equilibrium Qa = Ka
Thus, ΔG = 0.

Expression of Le -Chatelier Principle from Van't Hoff Equation:

Van't Hoff equations gives quantitative expression of Le-Chatelier Principle. From the equation,

lnKP = -(ΔH/R)(1/T) + C

It is evident that for endothermic reaction (ΔH〉0), increase of T increases the value of a of the reaction. But for exothermic reaction (ΔHㄑ 0), with rise in temperature, a is decreased. 
This change of KP also provides the calculation of quantitative change of equilibrium yield of products.
This above statement is in accordance with Le - Chatelier Principle which states that whenever stress is placed on any system in a state of equilibrium, the system is always react in a direction to reduce the applied stress. Thus when T is increased, the system in equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.   




Electron Affinity

Electron Affinity and its Variation:

Electron Affinity of an element is defined as the amount of energy released when an electron is added to a neutral gaseous atom in its lowest energy state that is, ground state to convert it into uni-negative gaseous ion.
In simple words, the electron affinity of an atom is defined as the energy liberated when a gaseous atom captures an electron.

A (neutral gaseous Atom) + e (electron) A⁻ (gaseous anion) + Electron Affinity (EA)

Evidently electron affinity is equal in magnitude to the ionisation energy of the species formed. The definition of electron affinity as the exothermicity of the electron capture reaction would lead to the negative sign according to our usual thermodynamic convention. But electron affinity is normally describes with positive sign. Negative electron affinities are known. These negative values indicate that the species concerned does not like to have any more electron.

Measurement of Electron Affinity:

Electron affinities are difficult to obtain. These are often obtained from indirect measurements, by analysis of Born - Haber energy cycles in which one step is electron capture or by the direct study of electron capture from heated filaments.

Calculate the electron affinity of chlorine from the Born -  Haber cycle, given the following date : lattice energy = - 774 kJ mol⁻¹ , Ionization Potential of Na = 495 kJ mol⁻¹, heat of sublimation of Na = 108 kJ mol⁻¹, energy for bond dissociation of chlorine (Cl₂) = 240 kJ mol⁻¹ and heat of formation of NaCl = 410 kJ mol⁻¹.

Born - Haber Cycle for formation of NaCl (S) is:
How to determine Electron Affinity
Born - Haber Cycle Applied to NaCl
From the above Born - Haber cycle we can written as:

- UNaCl - INa + ECl - SNa - 1/2DCl - ΔHf = 0
or,  ECl = UNaCl + INa + SNa + 1/2DCl + ΔHf 
∴ ECl = -774 + 495 + 108 + 120 + 410
= 359 kJ mol⁻¹

Factors Influencing the magnitude of Electron Affinity :

The magnitude of Electron Affinity (EA) is influenced by following factors such as,
1. Atomic Size. 2. Effective Nuclear Charge.  3. Electronic Configuration.

Atomic Size:

Larger the atomic radius lesser the tendency of the atom to attract the additional electron towards itself and lesser is the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom.
Thus EA values decreases with increases atomic atomic radius.

Effective Nuclear Charge:

Higher the magnitude of effective nuclear charge (Z*) towards the periphery of an atom greater is the tendency of the atom to attract the additional electron electron towards itself. Again greater the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom. As a result higher the energy released when extra electron is added to form an anion.
Thus the magnitude of Electron Affinity (EA) of he atom increases with increasing Z* value.

Electronic Configuration:

The magnitude of electron affinity depends on the electronic configuration of the atom. The elements having nS², nS² nP³, nS² nP⁶ valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration. 
Examples:
(a) Electron Affinity of Be and Mg are -0.60 eV and - 0.30 eV respectively
These elements posses filled S-Orbital in their valence shell which is stable hence Be and Mg do not prefer to except the additional electron in order to form anion. In such cases energy is required to form anion and consequently the electron affinity of Be and Mg posses negative sign with low magnitude.
From the above discussion it is clear that elements of Group - IIA posses low electron affinity vales with negative sign due to stable nS² valence shell electronic configuration. 

Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.

Atomic number and the electronic distribution of lithium and beryllium are:
Li    : 3  : 1S² 2S¹
Be   : 4  : 1S² 2S²
Lithium has an incompletely filled 2S sub-shell while beryllium has the sub shell filled. Thus lithium can receive an electron in its 2S sub-shell but for beryllium a still higher energy 2P level has to be made of. A filled shell or sub-shell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.
(b) The elements of Group - VA (group - 15) having nS² nP³ valence shell configuration also posses low electron affinity values.

Why electron affinity of Nitrogen is - 0.10 eV while that of Phosphorus is + 0.70 eV ?

Due to smaller size of Nitrogen atom when extra electron is added to the stable half - filled 2P orbital some amount of energy is required and hence electron affinity of nitrogen is negative. 
On the other hand, due to bigger size of P atom in compare to nitrogen less amount of energy released when the extra electron is added to the stable half - filled 3P orbitals and thus electron affinity of Phosphorus is expressed with positive sign.
(c) Electron Affinity data on noble gases are not available.
Complete lack of evidence for stable noble gas anion is sufficient prof of their low electron affinity. An electron, even if accepted by a noble gas atom, would have to go into an orbital of next quantum shell and presumably the nuclear charge not high enough to hold an electron placed so far out.

Periodic Variations:

In a Group:

In a group when move from top to bottom the size of atom generally increases with increasing atomic number and consequently the magnitude of electron affinity decreases in the same direction.

Exceptions: 

There are some exceptions to this general rule as is evident from the following examples:
(a) Although the elements of the second period of the periodic table are relatively smaller in size than those of the third period elements, but the electron affinity values of elements of second period is smaller than the electron affinity values of third period elements.
This unexpected behavior is explained by saying that the much smaller sizes of the second period elements give a very much higher value of charge densities for the respective negative ions. A high vales of electron density is opposes by the inter - electronic repulsion forces. For example, electron affinity of fluorine is lower than that of Chlorine . 
Thus the magnitude of electron affinity of the elements of group 17 decreases in the order :
Cl〉F〉Br〉I

Electron affinity of Cl is greater than that of F - Explain.

The halogen possess large positive electron affinity values indicating their strong tendency to form anions. This is easily understandable because their electronic configurations are only one electron short of the next noble gas element.
The electron affinity of chlorine is greater than that of F. This is probably due to very small size of F atom which results in a very high charge density and consequently strong repulsion between existing valence shell electrons and the entering electron (new coming electron). On the other hand Cl being bigger size , charge density is small and thus such repulsion is not strong enough. Hence the electron affinity Cl is greater than that of F.

In a Period:

In a period, when we move from left to right Z* value towards the new coming electron gradually increases with the increasing atomic number and hence the electron affinity gradually increases in the same direction.

Exceptions:

Exceptional cases may arise in case of the elements having stable nS², nS²nP³ and extraordinarily stable 1S², nS² nP⁶ valence shell configurations.
For example in case of the elements in 2nd period :
Be ,N and Ne the magnitude of electron affinity decreases with increasing atomic number.
A plot of electron affinities of elements up-to Chlorine against atomic number shown as:
What is Electron Affinity
Electron affinities as Function of Atomic Numbers






Slater's Rules

Slater's Rule for Effective Nuclear Charge:

Slater proposed some set of empirical rules to calculate the screening constant (σ) of various electrons present in different orbitals of an atom or an ion.
Once we get the value of screening constant it is easy enough to find the effective nuclear charge (Z*).

Screening Effect:

The Valence electron in a multi - electron atom is attracted by the nucleus and repelled by the electrons of inner-shells.
What is the Slater's rule ?
Shielding Effect
The combined effect of this attractive and repulsive force acting on the valence electron is that the Valence - electron experiences less attraction from the nucleus. This is known as the screening effect.

Slater's Rules to Calculate the Screening Constant (σ):

Rules for an electron in the nS, nP level:

(i) The first to do is to write out the electronic configuration of the atom or the ion in the following order and grouping.
(1S) (2S, 2P) (3S, 3P) (3d) (4S, 4P) (4d) (4f) (5S, 5P) etc.
It may be noted that so far as the screening effect is concerned the S and P electrons belonging to the same principal quantum shell have the same effect as advocated by Slater.
Na atom :
(1S)² (2S, 2P)⁸ (3S)¹
(ii) An electron in a certain (nS, nP) level is screened only by electrons in the same level and by the electrons of lower energy level.
Electrons lying above (nS, nP) level do not screen any (nS, nP) electron to any extent. Higher energy electrons have no screening effect on any lower energy electrons.
Examples:
Estimation of screening constant of the valence electron of the Sodium atom,
(1S)² (2S, 2P)⁸ (3S)¹ 
(The Valence electron will be excluded from our Calculation).
Estimation of screening constant of the 3d electron of the Sodium atom,
(1S)² (2S, 2P)⁸ (3S)¹
 (The one 2P electron and 3S electrons will be excluded from our Calculation).
(iii) Electrons of an (nS, nP) level shield a valence electrons in the same group by 0.35 each. This is also true for the electrons of the nd or nf that is for the electrons in the same group.
(iv) Electrons belonging to one lower quantum shell, that is (n-1) shell shield the valence electrons by 0.85 each.
(v) Electrons belonging to (n-2) or still lower quantum shell shield the valence electron by 1.0 each. This means the screening effect is complete.
Examples:
(a) For the valence electron of Na atom:
Screening Constant (σ 
= (2 × 1) + (8 × 0.85) + (0 × 0.35) 
8.8
(a) For the 2P electron of Na atom:
Screening Constant (σ ) 
= (2 × 0.85) + (7 × 0.35) 
4.15
Na+ Ion:
  (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (11 - 4.5) = 6.5
 Mg+2 : (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (12 - 4.5) = 7.5

 K+ (1s)2(2s, 2p)8(3s, 3p)8
Screening Constant 
σ  = (2×1.0)+(8×0.85)+(8×0.35)
= 11.6
Effective Nuclear Charge
Z⋆ = (19 - 11.6) = 7.4

Valence Electron of F ion:
  (1s)2(2s, 2p)7
Screening Constant 
σ  = (2×0.85)+(6×0.35)
= 3.8
Effective Nuclear Charge
Z⋆ = (9 - 3.8) = 5.2

F- ion (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (9 - 4.5) = 4.5
Estimate the screening constant for the outermost 4S electron of Vanadium.
Vanadium has atomic number 23 and the electronic configuration according to the Slater's Rule is :
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²
We have consider only one electron of the two 4S electrons.
∴Screening Constant (σ
(2 ×1.0) + (8×1.0) + (8×0.85) + (3×0.85) (1×0.35) 
19.7.

Rules for an electron in the nd, nf level:

The above rules are quite well for estimating the screening constant of S and P orbitals. However when a d or f electron being shielded the (iv) and (v) rule replaced by a new rules for estimation of screening constant.
The replaced rules are:
All electrons below the nd or nf level contribute 1.0 each towards the screening constant.
Example
Screening Constant for a 3d electron of Vanadium:
Electronic Configuration according to Slater's Rule is:
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²
We have consider only two electron of the three 3d electrons.
∴Screening Constant (σ
(2 ×1.0) + (8×1.0) + (8×1.0) + (3×0.35)
 = 18.70
Ion 
Electronic Configuration
 Screening 
Constant
(σ)
Effective 
Nuclear Charge
(Z) 
V(II) 
Ion
  (1s)2
(2s, 2p)8
(3s,3p)8
(3d)3
(2×1.0)+(8×1.0)+(8×1.0)+(3×0.35)
= 19.05
Z=
(23-19.05)
= 3.95
Cu(II) - Ion
(1s)2
(2s, 2p)8
(3s,3p)8
(3d)9
(2×1.0)+(8×1.0)+(8×1.0)+(9×0.35)
= 21.15
Z=
(29-21.15)
= 7.85
Find out the Screening Constant of the 4S and 3d electron of Chromium Atom.
Chromium has atomic number 24 and the electronic configuration according to the Slater's Rule is :
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)⁵ (4S)¹
∴ Screening Constant (σ) for the 4S electron is :
σ =(2×1.0) + (8×1.0) + (8×1.0) 
(3×0.85) (0×0.35) 
21.05.
And the Screening Constant (σ) for the 3d electron is :
σ =  (2×1.0) + (8×1.0) + 
(8×1.0) + (4×0.35) 
= 19.4.
Calculate the effective nuclear charge of the hydrogen atom.
Hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of single proton.
Thus, σ = 0 and Z* = 1.0 -0 = 1.0
Thus hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.