t_1/2 α \(\frac{1}{[A_{0}]^{n-1}}\)...(1)
If [A₀ = 2[A₀]; then t_1/2 = 2t_1/2
2t_1/2 α \(\frac{1}{[2A_{0}]^{n-1}}\)...(2)
(2)/(1) = \(2= \frac{1}{[2A_{0}]^{n-1}} \times \frac{1}{[A_{0}]^{n-1}}\)
\(2= \frac {[2A_{0}]^{n-1}} {[A_{0}]^{n-1}}\)
\(2= \frac {1} {2}^{n-1}\)
2 = (2^-1)^n-1
2¹ = (2^-n+1)
n = 0

\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 100: [A] = 25
\(6.909=\frac { 2.303 }{ t } \log\frac { \left[ {100 } \right] }{ \left[ 25\right] } \)
\(t =\frac { 2.303 }{ 6.909 } \log(4)\)
\(t =\frac { 1 }{ 3 } \log(2^2)\)
\(= \left( \frac { 2 }{ 3 } \right) \log2\)

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

Yb²⁺ - 4f¹⁴ - no unpaired electrons - diamagnetic

Mn²⁺ ⇒ 3d⁵ contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)

i) 3A + 5B₂ ⟶ 4CD
Rate = k[A]³ [B₂]⁵
ii) X₂ + Y₂ ⟶ 2XY
Rate = k[X₂ ] [Y₂]

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 \(\mathrm{N}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}\)

Aspirin is o-acetyl salicylic acid

(i) Many micro organisms (bacteria, fungi and moulds) produce certain chemicals which inhibit the growth or metabolism of some other micro organism. Such chemicals are called antibiotics.
(ii) Eg: Penicillin from the mould penicillium notatum

i) Rickets - Vitamin - D (Cholecalciferol - (D₃) Ergocalciferol - (D₂)
ii) Scurvy (bleeding gums) - vitamin - C (Ascorbic acid)

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)

Electronic configuration of Ce⁴⁺ = [Xe] 4f⁰5d⁰6s⁰
Electronic configuration of Co²⁺ = [Ar]3d⁷

Potassium dichromate is prepared from chromate ore (FeCr₂O₄). The ore is concentrated by gravity separation. It is then mixed with excess sodium carbonate and lime and roasted in a reverberatory furnace.

The roasted mass is treated with water to separate soluble sodium chromate from insoluble iron oxide. The yellow solution of Sodium Chromate is treated with concentrated Sulphuric acid which converts Sodium Chromate into Sodium dichromate.

The saturated solution of sodium dichromate is mixed with KCl and filtered, the filtrate is cooled to obtain K₂Cr₂O₇ crystals.
\(\begin{array}{cc} \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} & \rightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl} \\ \begin{array}{c} \text { Sodium dichromate } \\ \text { (orange red) } \end{array} & \begin{array}{c} \text { Potassium dichromate } \\ \text { (orange red) } \end{array} \end{array}\)

(i) Preservatives are chemicals added to food which are capable of inhibiting, retarding or arresting the process of fermentation, acidification or other decomposition of food by the growth of micro organisms (or)
(ii) Chemical substances which are added to food that prevents the spoilage of food materials by destroying the food - spoiling micro-organisms are called food preservatives.
(iii) Eg: sodium benzoate, NaCl, Acetic acid Sodium metabisulphite, potassium meta bisulphite etc.

DNA	RNA
DNA is mainly present in nucleus, mitochondria and chloroplast	RNA is mainly present in cytoplasm, nucleolus and ribosomes
It contains deoxyribose sugar	It contains ribose sugar
Base pair A = T. G≡C	Base pair A = U. C≡G
Double stranded molecules	Single stranded molecules
It's life time is high	It is short lived
It is stable and not hydrolysed easily by alkalis	It is unstable and hydrolyzed easily by alkalis
It can replicate itself	It cannot replicate itself It is formed from DNA

Let [A₀] = 100;
When t = t_99.9%; [A] = (100-99.9) = 0.1
\(k=\frac { 2.303 }{ t } \log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log\left( \frac { 100 }{ 0.1 } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log1000\)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } (3)\)
\({ t }_{ 99.9\% }=\frac { 6.909 }{ K } \)
\({ t }_{ 99.9\% }=10\times \frac { 0.69 }{ K } \)
\({ t }_{ 99.9\% }={ 10 } t_{ 1/2 }\)

Pseudo first order reaction:
A second order reaction can be altered to a first order reaction by taking one of the reactants in large excess, such reaction is called pseudo first order reaction.
Example: Acid Hydrolysis of an ester,
\(C{ H }_{ 3 }COOC{ H }_{ 3(aq) }+{ H }_{ 2 }{ O }_{( l) }\overset { { H }^{ + } }{ \longrightarrow } C{ H }_{ 3 }COO{ H }_{ (aq) }+C{ H }_{ 3 }O{ H }_{ (aq) }\)
Rate = \(k\left[ { CH }_{ 3 }COOC{ H }_{ 3 } \right] \left[ { H }_{ 2 }O \right] \)
If the reaction is carried out with a large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., the concentration of water remains almost constant.
Now, we can define k [H₂O] = k'; Therefore the above rate equation becomes
Rate = \(k'\left[ { CH }_{ 3 }COOC{ H }_{ 3 } \right] \)
Thus it follows first order kinetics.

Rate law:
(i) Rate law or rate equation is an expression which relates the rate of a reaction with rate constant and the concentration of reactants.
(ii) For xA + yB → products
(iii) The rate law is r = k[A]^m[B]ⁿ
Rate constant:
(i) It is the rate of the reaction when the concentration of the reactants are taken unity.
In above rate law if[A] = [B] = 1, rate constant k = Rate

S.No	Lanthanoids	Actinoids
1.	Differentiating electron enters in 4f orbital	Differentiating electron enters in 5f orbital
2.	Binding energy of 4f orbitals are higher	Binding energy of 5f orbitals are lower
3.	They show less tendency to form complexes	They show greater tendency to form complexes
4.	Most of the lanthanoids are colourless	Most of the actinoids are coloured For Example: U3+ (red) U4+ (green).
5.	They do not form oxo cations	They do form oxo cations such as UO22+, NpO22++ etc.
6.	Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases	Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7

(i) Fe³⁺ - electronic configuration - [Ar] 3d⁵
(ii) It has exactly half-filled stable electronic configuration.
(iii) Fe²⁺ - electronic configuration -[Ar]3d⁶
(iv) It has only partially filled d-orbitals.
Hence Fe³⁺ is more stable than Fe²⁺.

A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following Cl₂ first order reaction,
A → product
Rate law can be expressed as
Rate = k[A]¹
Where, k is the first order rate constant
\(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^{1}  \)
\(\Rightarrow \frac{-\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]}=\mathrm{kdt}\)  .....(1)
Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A₀] to [A] at the later time.
\(\int_{\left[A_{0}\right]}^{[A]} \frac{-d[A]}{[A]}=k \int_{0}^{t} d t  \)
\((-\ln [A])_{\left[A_{0}\right]}^{[A]}=k(t)_{0}^{t}  \)
\(-\ln [A]-\left(-\ln \left[A_{0}\right]\right)=k(t-0) \)
\(-\ln [\mathrm{A}]+\ln \left[\mathrm{A}_{0}\right]=\mathrm{kt}  \)
\(\ln \left(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\right)=\mathrm{kt}\)  .....(2)
This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303.
\(2.303 \log \left(\frac{\left[A_{0}\right]}{[A]}\right)=k t  \)
\(k=\frac{2.303}{t} \log \left(\frac{\left[A_{0}\right]}{[A]}\right)\)   .....(3)

Structure of glucose: Glucose is an aldohexose. It is optically active with four asymmetric carbons. Its solution is dextrorotatory and hence it is also called as dextrose. The proposed structure of glucose is shown in the figure which was derived based on the following evidences

(i) Elemental analysis and molecular weight determination show that the molecular formula of glucose is C6H120 6'
(ii) On reduction with concentrated HI and red phosphorus at 373K, glucose gives a mixture of n hexane and 2, iodohexane indicating that the six carbon atoms are bonded linearly.

(iii) Glucose reacts with hydroxylamine to form oxime and with HCN to form: cyanohydrin. These reactions indicate the I presence of carbonyl group in glucose

(v) Glucose is oxidised to gluconic acid with ammonical silver nitrate (Tollen's reagent) and alkaline copper sulphate (Fehling's solution). Tollens reagent is I reduced to metallic silver and Fehlings I solution to cuprous oxide which appears as red precipitate. These reactions further: confirm the presence of an aldehyde group


(vi) Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol groups.

(vii) Glucoseis a stable compound and does not undergo dehydration easily. It indicates that not more than orie hydroxyl group is bonded to a single carbon atom. Thus the five hydroxyl groups are attached to five different carbon atoms and the sixth carbon is an aldehyde group.