IF = 96500 C = charge of one mole of e^- charge of 6.022 x 10^-23 electron

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

 \(Rate=\frac { -1 }{ 2 } \frac { d[NH_3] }{ dt } \)
\(\frac { -d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { H}_{ 2 } \right] }{ dt } \)
\(=\frac { 1 }{ 2 } { k_1\left[ { NH}_{ 3 } \right] } = { k_2 \left[ { NH}_{ 3 } \right] }=\frac { 1 }{ 3 } { k_3\left[ { NH}_{ 3 } \right] } \)
[3/2] k₁ = 3k₂ = k₃
1.5k₁ = 3k₂ = k₃

(i) Equivalent conductance is defined as the conductance of 'V' m³  of electrolytic solution containing one gram equivalent of the electrolyte in a conductivity cell in which the electrodes are one meter apart
(ii) \(\Lambda=\frac{\kappa \times 10^{-3}}{N} \mathrm{Sm}^{2} \mathrm{~g} \mathrm{eq}{ }^{-1}\)

(i) AlCl₃ is obtained by heating a mixture of alumina and coke in a current of chlorine.
\(2 \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{C}+6 \mathrm{Cl}_{2} \rightarrow 4 \mathrm{AlCl}_{3}+3 \mathrm{CO}_{2}\)
(ii) On industrial scale it is prepared by chlorinating aluminium around 1000 K
\(2 \mathrm{Al}+3 \mathrm{Cl}_{2} \stackrel{1000 \mathrm{~K}}{\longrightarrow} 2 \mathrm{AlCl}_{3}\)

1. Photochemical reaction between H₂ and Cl₂ ;  \(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \stackrel{h v}{\longrightarrow} 2 \mathrm{HCl}_{(\mathrm{g})}\)
2. Decomposition of N₂O on hot platinum surface \(\mathrm{N}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}\)

\({ { \Lambda }_{ m }^{ o } }\) is called the limiting molar conductivity. i.e., the molar conductance of a solution at infinite dilution.

a) Primary and secondary nitroalkanes, having α-H, also show an equilibrium mixture of two tautomers namely nitro - and aci - form

b) Difference:

S.No	Nitro form	Aci - form
1.	Less acidic in nature.	More acidic
2.	Dissolves in NaOH slowly	Dissolves in NaOH instantly
3.	Decolourises FeCl3 solution	With FeCl3 gives reddish brown colour
4.	Electrical conductivity is low	Electrical conductivity is high

 

K = 1.54 \(\times\) 10^-3s^-1; t_1/2 = ?
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 1.54\times { 10 }^{ -3 } } \)
= 450 s

Arrhenius equation is,
\(k=Ae^\left ({ \frac { -Ea }{ RT } } \right )\)
Here,
A \(\rightarrow\) Frequency factor
E_a \(\rightarrow\) Activation energy of the reaction
R \(\rightarrow\) Gas constant
T \(\rightarrow\) Absolute temperature (in K)

CO acts as a strong reducing agent.
Example: 3CO + Fe₂O₃ \(\longrightarrow \)  2Fe + 3CO₂
It reduces metallic. oxides into metals.

Oxides of Carbon	Structure	Parameters
CO		Three electron pairs are shared between carbon and oxygen. The C-O bond distance is 1.128\(\overset{o}{A}\).
CO2		Equal bond distance for the both C-O bonds. Two C-O sigma bond, It has 3c-4e bond.

Catenation is an ability of an element to form chain of atoms.
The conditions for catenation.
(a) The valency of element is greater than or equal to two.
(b) Element should have an ability to bond with itself
(c) The self bond must be as strong as Its bond with other elements
(d) Kinetic inertness of catenated compound towards other molecules.
(e) Carbon possesses all the above properties and forms a wide range of compounds with itself and with other elements such as H, O, N, S and halogens.

In p-block elements, the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour
(i) Small size of the first member.
(ii) High ionisation enthalpy and high electronegativity.
(iii) Absence of d-orbitals in their valance shell.

When boric acid or borate salt is heated with ethyl alcohol in presence of conc. sulphuric acid, an ester, triethyl borate is formed. The vapour of this ester burns with a green edged flame and this reaction is used to identify the presence of borate.
H₃BO₃ + 2C₂H₅OH \(\frac{\text { Conc. }}{ H_2SO_4 } \longrightarrow \) 2B(OC₂H₅)₃ + 3H₂O

Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline cannot be prepared by this method because the arylhalides do not undergo nucleophilic substitution with the anion formed by phthalimide.

(i) In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200^oC and 20-40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes.
(ii) Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
\(2 \mathrm{H}_{2(\mathrm{~g})}+4 \mathrm{OH}_{(a q)}^{-} \rightarrow 4 \mathrm{H}_{2} \mathrm{O}_{(l)}+4 \mathrm{e}^{-}\)
Reduction occurs at the cathode:
\(\mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(t)}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}_{(\mathrm{aq})}^{-}\)
(iii) The overall reaction is \(2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(1)}\)
(iv) The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H₂-O₂ fuel cell. Other fuel cells like propane -O₂ and methane O₂ have also been developed.

(i) The half cell reactions are :
\(2Br^{-} \rightarrow Br_{2}+2e^{-}\)     \(E^{0}_{ox}=-1.09V\) ...(1)
\(2Fe^{3+}+2e^{-}\rightarrow2Fe^{2+}\)  \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=+0.771V\) ..(2)
(ii) Adding (1) of (2) :
\(2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{2+}+Br_{2}\)   \(E^{0}_{cell}=?\) ...(3)
\(E^{0}_{cell}=E^{0}_{ox}+E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}\)
= (-1.09 + 0.771)V
= -0.319V
(iii) E⁰_cell is – ve; \(\Delta G\) is +ve and the cell reaction is non spontaneous.
(iv) Hence Fe³⁺ cannot oxidises Br^- to Br₂.

First law:
The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.
m α Q \(\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}} \Rightarrow \mathrm{Q}=\mathrm{It}\right]\)
m α It
m = ZIt
Where Z = electro chemical equivalent of the substance
I = current
t = time of passage of current
Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.
m α Z
\(\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}\)
m = mass of the metal deposited
Z = electro chemical equivalent

(i) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.
(ii) The reaction between KMnO₄ and H₂SO₄ and oxalic acid is catalysed by MnSO₄ and increases the rate of oxidation of C₂O₄^2- by MnO₄^-.

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentration is called a zero order reaction.
Let us consider the following by hypothetical zero order reaction.
 A⟶ product 
The rate law can be written as,
Rate = k[A]⁰ 
\(\frac { -d\left[ A \right] }{ dt } =k(1)\ \ \ \therefore \left( { \left[ A \right] }^{ 0 }=1 \right) \)
\(\Rightarrow -d\left[ A \right] =kdt\)
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time 't',
\(-\int _{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }{ d\left[ A \right] } =k\int _{ 0 }^{ t }{ dt } \)
\(-{ \left( \left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
[A₀] - [A] = kt   
\(k=\frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \)
Straight line equation y = mx + c
ie., \(\left[ A \right] =-kt+\left[ { A }_{ 0 } \right] \)
⇒ y = c + mx
A plot [A] vs time gives a straight line with a slope of -k and y intercept of [A₀]

Generally on descending a group the ionisation energy decreases and hence the metallic character increases.

(i) Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
(ii) They are used for making water proofing clothes.
(iii) They are used as insulting material in electrical motor and other appliances.
(iv) They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

(i) In diborane two BH₂ units are linked by two bridged hydrogens.
(ii) It has eight B-H bonds.
(iii) Diborane has only 12 valance electrons.
(iv) The four terminal B-H- bonds is 2c - 2e bond (two centre - two electron bond.)
(v) Two three centred B - H - B bonds two electrons each. (3c - 2e).
(vi) In diborane, the boron is sp³ hybridised
(vii) Three of the four sp³ hydridised orbitals contains single electron and the fourth orbital is empty.

1. Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,
xA + yB ⇌ IC + mD
2. The reaction quotient Q for the above reaction is given below
\(Q=\cfrac { \left[ C \right] ^{ ' }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)  .....(1)
3. We have already learnt that,
ΔG = ΔG + RT In Q .........(2)
4. The Gibbs free energy can be related to the cell emf as follows.
[ஃ equation (1) and (2)]
ΔG = - nFE_cell; ΔG^o = - nFE^o_cell
5. Substitute these values and Q from (1) in the equation (2)
(2) ⇒\(-nF{ E }_{ cell }=-n{ FE }^{ o }_{ cell }+RT\quad In\cfrac { \left[ C \right] '\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) ....(3) 
6. Divide the whole equation (3) by (-nF)
(4) ⇒\({ E }_{ cell }={ E }_{ cell }-\cfrac { RT }{ nF } In\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) 
(or)
\({ E }_{ cell }={ E }_{ cell }-\cfrac { 2.303RT }{ nF } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) .....(4)
7. The above equation (4) is called the Nernst equation
8. At 25°C (298 K), the above equation (4) becomes,
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 2.303\times 8.314\times 298 }{ n(96500) } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)    
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 0.0591 }{ n } log\cfrac { { \left[ C \right] }^{ l }{ \left[ D \right] }^{ m } }{ { \left[ A \right] }^{ x }\left[ B \right] ^{ y } } \) ......(5) 
\(\left[\begin{array}{l} \text { Where } \\ \therefore \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ \mathrm{~T}=298 \mathrm{~K} \\ \mathrm{IF}=96500 \mathrm{C} \mathrm{mol}^{-1} \end{array}\right]\)

Kohlraush's law:
(i) At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
(ii) For a uni - univalent electrolyte such as NaCl, the Kohlraush's law is expressed as
\(\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{+}}+\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Cl}}^{-}\)
(iii) In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_x B_y, is given below.
\(\left(\Lambda_{m}^{0}\right)_{A_{x} B_{y}}=x\left(\lambda_{m}^{0}\right)_{A^{y+}}+y\left(\lambda_{m}^{0}\right)_{B^{x-}}\)
b) Calculation of molar conductance at infinite dilution for weak electrolytes experimentally.
(i) However, the same can be calculated using Kohlraush's Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.
\(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0} \) ..........(1)
\(\Lambda_{\mathrm{HCl}}^{0}=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0} \) .........(2)
\(\Lambda_{\mathrm{NaCl}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\) ...........(3)
(ii) Equation (1) + Equation (2) - Equation (3) gives,
\(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}\right)+\left(\Lambda_{\mathrm{HCl}}^{0}\right)-\left(\Lambda_{\mathrm{NaCl}}^{0}\right)=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}-}^{0} \)
\(=\Lambda_{\mathrm{CH}_{2} \mathrm{COOH}}^{0}\)

Given data: Time required for 99.9% completion is about 10 times its half life period. Prove that for a first order reaction the time required for 99.9% completion is about 10 times its half life period.
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution:
\({ t }_{ 99.9 }=\frac { 2.303 }{ k } \log\frac { 100 }{ 100-99.9 } \)
\(=\frac { 2.303 }{ k } \log\frac { 100 }{ 0.1 } \)
\(=\frac { 2.303 }{ k } \log1000\)
\({ t }_{ 99.9 }=\frac { 2.303\times 3 }{ k } \)
\({ t }_{ 99.9 }=\frac { 6.909 }{ k } ;\)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)  \(\therefore k=\frac { 0.693 }{ { t }_{ \frac { 1 }{ 2 } } } \)
\(\frac { { t }_{ 99.9\% } }{ { t }_{ \frac { 1 }{ 2 } } } =\frac { 6.909 }{ k } \times \frac { k }{ 0.693 }\)
\({ t }_{ 99.9 }=\frac { 6.909 }{ 0.693 } \times { t }_{ \frac { 1 }{ 2 } }\)
t_99.9 = 10 x t_1/2
Time required for 99.9% completion of the reaction = 10t_1/2

(i) Collision theory is based on the kinetic theory of gases. According to this theory, a chemical reaction occurs as a result of collisions between the reacting molecules.
(ii) Let us understand this theory by considering the following reaction.
A₂(g) + B₂(g) ⟶ 2AB(g)
(iii) If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
(iv) Rate ∝ number of molecules colliding per litre per second (collision rate).
(v) The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collison rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]
(vi) Where, Z is a constant
(vii) A fraction of effective collisions (f) is given by the following expression
\(f={ e }^{ \frac { { -E }_{ a } }{ RT } }\)
(xiv) This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collides with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.
(viii) The diagram illustrates the importance of proper alignment of molecules which leads to reaction.

(ix) The fraction of effective collisions (f) having proper orientation is given by the steric factor p.
⇒ Rate = p x f x collision rate
\(\Rightarrow Rate=p\times { e }^{ \frac { -Ea }{ RT } }\times Z\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(1)\)
As per the rate law,
Rate = \(k=\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(2)\)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)