Both carbon and hydrogen are estimated by the same method. A known weight of the organic substance is burnt in excess of oxygen and the carbon and hydrogen present in it are oxidized to carbon dioxide and water, respectively.
The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the organic substance is calculated. The apparatus employed for the purpose consists of three units
(1) oxygen supply
(2) combustion tube
(3) absorption apparatus.
(1) Oxygen supply: To remove the moisture from oxygen it is allowed to bubble through sulphuric acid and then passed through aV-tube containing soda lime to remove CO₂. The oxygen gas free from moisture and carbondioxide enters the combustion tube.
(2) Combustion tube: A hard glass tube open at both ends is used for the combustion of the organic substance. It contains (i) an oxidized copper gauze to prevent the backward diffusion of the products of combustion (ii) a porcelain
boat containing a known weight of the organic substance (iii) coarse copper oxide on either side and (iv) an oxidized copper gauze placed towards the end of the combustion tube. The combustion tube is heated by a gas burner.
(3) Absorption Apparatus: The combustion products containing moisture and carbondioxide are then passed through the absorption apparatus which consists of (i) a weighed U'-tube packed with pumice soaked- in cone. H₂SO₄ to absorb water (ii) a set of bulbs containing a strong solution of KOH to absorb CO₂ and finally (iii) a guard tube filled with anhydrous CaCI₂ to prevent the entry of moisture from atmosphere. Procedure: The combustion tube is heated strongly to dry its content. It is then cooled slightly and connected to the absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced. The tube is again heated strongly till the substance in the boat is burnt away. This takes about 2 hours. Finally, a strong current of oxygen is passed through the combustion tube to sweap away any traces of carbon dioxide or moisture which may be left in it. The If-tube and the potash bulbs are then detached and the increase in weight of each of them is determined.
Calculation:
Weight of the organic substance taken= w g
Increase in weight of H₂O = xg
Increase in weight of CO₂ =yg
18 g of H20 contain 2g of hydrogen
\(\therefore \) x g of H2O contain\(\left( \frac { 2 }{ 18 } \times \frac { x }{ w } \right) \) g of hydrogen
Percentage of hydrogen = \(\left( \frac { 2 }{ 18 } \times \frac { x }{ w } \times 100 \right) \)%
44g of CO₂ contains 12g of carbon
\(\therefore \) y g of CO₂ contain \(\left( \frac { 2 }{ 44 } \times \frac { y }{ w } \right) \) g of carbon 
Percentage of carbon= \(\left( \frac { 2 }{ 44 } \times \frac { y }{ w } \times 100 \right) \)%
Note:
1. If the organic substance under investigation also contain N, it will produce oxides of nitrogen on combustion. A spiral of copper is introduced at the combustion tube, to reduce the oxides of nitrogen to nitrogen which escapes unabsorbed.
2. If the compound contains halogen a well, a spiral of silver is also introduced in the combustion tube. It converts halogen into dilver halide.
3. In case if the substance also contains sulphur, the copper oxide in the combustion tube is replaced by lead chromate. The SO₂ formed during combustion is thus converted to lead sulphate and prevented from passing into the absorption unit.

S_N1stands for unimolecular nucleophilic substitution
(i) Hence rate ofthe reaction = k(alkyl halide. This S_N1 reaction follows first order kinetics and occurs in two steps.
(ii) We understand ~l reaction-mechanism by taking a reaction between tertiary butyl bromide with aqueous KOH.
\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -Br\overset { OH(aq) }{ \underset { -Br }{ \longrightarrow } } { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -OH\)
Tert-Butyl bromide                      Tert-Butyl alcohol
This reaction takes place in two steps as shown below
Step - 1 Formation of carbocation :
(i) The polar C - Br bond breaks forming a carbocation and bromide ion. This step is slow and hence it is the rate determining step.

(ii) The carbocation has 2 equivalent lobes of the vacant 2p orbital, so it can react equally rapidly from either face.
Step - 2 :
(i) The nucleophile immediately reacts with the carbocation. This step is fast and hence does not affect the rate of the reactions.

(ii) As shown above, the nucleophilic reagent OHcan attack carbocation from both the sides.
(iii) In the above example the substrate tert-butyl bromide is not optically active, hence the obtained product is optically inactive. If halo alkane substrate is optically active then, the product obtained will be optically inactive racemic mixture. As nucleophilic reagent OR- can attack carbocation from both the sides, to form equal proportion of dextro and levorotatory optically active isomers which results in optically inactive racemic mixture

Molar mass of Mg CO₃ = 84.32 g mol^-1
Percentage of Mg =\({24\over 84.32}\times 100=28.46\%\)
Percentage of C=\({12\over 84.32}\times 100=14.23\%\)
Percentage of O₃=\({48\over 84.32}\times 100=57.0\%\)
\(\underset{84.32\ g}{MgCO_3}\rightarrow MgO+\underset{44\ g}{CO_2 }\)
84.32 g of 100% pure MgCO₃ gives 44g of CO₂.
\(\therefore\) 100 x 10³ g of 100 % pure MgCO₃ gives =\({44\over 84.32}\times 100\times 10^3\)
= 52.182 x 10³g CO₂ 
100% pure MgCO₃ gives 52.182 x 10³ g CO₂
\(\therefore\) 90% pure MgCO₃ will give \({52.182\times 10^3\over 100}\times 90\) = 46963.8 g CO₂
= 46.96 Kg CO₂.

Experiment	\(\frac{v_1}{T_1}\)=constant
1	\(\frac{1.54}{293}\) = 0.0053
2	\(\frac{1.65}{313}\) = 0.0053
3	\(\frac{1.95}{373}\) = 0.0053
4	\(\frac{2.07}{393}\) = 0.0053

The average value of the constant is 0.0053.

(i) Heat evolved at constant volume, is measured in a bomb calorimeter.
(ii) Apparatus setup: The inner vessel (the bomb) and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws.
(iii) Experiment: A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess oxygen. The bomb is immersed in water, in the inner volume of the calorimeter. A stirrer is placed in the space between the wall of the calorimeter and the bomb, so that water can be stirred, uniformly. The reaction is started by striking the substance through electrical heating.
(iv) Calculation: A known amount of combustible substance is burnt in oxygen in the bomb. Heat evolved during the reaction is absorbed by the calorimeter as well as the water in which the bomb is immersed. The change in temperature is measured using a Beckman thermometer. Since the bomb is sealed its volume does not change and hence the heat measurements is equal to the heat of combustion at a constant volume (ΔU)_c
The amount of heat produced in the reaction (ΔU)_c is equal to the sum of the heat absorbed by the calorimeter and water.
Heat absorbed by the calorimeter q₁ = k.ΔT
where k is a calorimeter constant equal to m_c C_c (m_c is mass of the calorimeter and C_c is heat capacity of calorimeter)
Heat absorbed by the water q₂ = m_w C_w ΔT
where m_w is molar mass of water
C_w is molar heat capacity of water (4,184 kJ K^-1mol^-1)
Therefore ΔU_c = q₁ + q₂
=k.ΔT + m _w C_w ΔT
=(k+m _w Cw) ΔT
Calorimeter constant can be determined by burning a know.n mass of standard sample (benzoic acid) for which the heat of combustion is known (-3227 kJmol^-1)
The enthalpy of combustion at constant pressure of the substance is calculated from the equation (7.17)
\(\Delta { H }_{ C(pressure) }^{ 0 }=\Delta { U }_{ C(vol) }^{ o }+\Delta { n }_{ g }RT\)

When conc. H₂SO₄ absorbs H₂O from moist H₂ it produces so much of heat since its highly exothermic and so it may catch fire.

(i) Magnesium and calcium also play a vital role in biological functions. A typical adult human body contains about 25 g of magnesium and 1200 g of calcium. Magnesium plays an important· role in many biochemical reactions catalysed by enzymes. It is the co-factor of all enzymes that utilize ATP in phosphate transfer and energy release. It is also essential for DNA synthesis and is responsible for the stability and proper functioning of DNA.It is also used for balancing electrolytes in our body. Deficiency of magnesium results into convulsion and neuromuscular irritation.
(ii) Calcium is a major component of bones and teeth. It is also present in blood and its concentration is maintained by hormones (calcitonin and parathyroid hormone). Deficiency of calcium in blood causes it to take longer time to clot. It is also important for muscle contraction.
(iii) The main pigment that is responsible for photosynthesis, chlorophyll, contains magnesium which plays an important role in photosynthesis.

Compound	Molecular formula	Empirical Formula
Fructose	C6 H12O6	C H2O
Caffeine	C8 H10 N4 O2	C4 H5 N2O

The electronic configuration of Sodium (Z = 11) Is²2s²2p⁶3s¹.
Magnesium (Z = 12) 1s²2s²2p⁶3s²
Magnesium atom has a smaller radius and higher nuclear charge than a sodium atom, thus more energy will be required to remove the electron from the same orbital (3s), making the first ionisation energy of magnesium higher than that of sodium.
However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing 1 electron, sodium attains the stable noble gas configuration of neon (1s²2s²2p⁶). On the other hand, magnesium, after losing 1 electron still has one electron in the 3s-orbital(1s²2s²2p⁶3s¹). In order to attain the stable noble gas configuration, Thus, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

s-character & electronegativity,
\(\therefore\)  sp hybrid orbitals are more electronegative among the three.

(i) Preparation of alkene by dehydration of alcohol:
When an alcohol is heated at 430-440 K with excess of concentrated sulphuric acid, a molecule of water from alcohol is removed and an alkene is formed. is reaction is called elimination reaction.
\({ C }_{ 2 }{ H }_{ 5 }OH\overset { Conc.{ H }_{ 2 }{ SO }_{ 4 } }{ \underset { 430-440K }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }\)
Ethanol                          Ethene
Ethene can also be prepared in laboratory by catalytic dehydration of alcohol.
\({ C }_{ 2 }{ H }_{ 5 }OH\overset { { A1 }_{ 2 }{ O }_{ 3 } }{ \underset { 623K-723K }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }\)
Ethanol                         Ethene
(ii) Preparation of ethene by Kolbe's electrolytic method:
When an aqueous solution of potassium succinate is electrolyzed between two platinum electrodes, ethene is produced at the anode.
\(\overset { { CH }_{ 2 }-COOK }{ \underset { { CH }_{ 2 }-COOK }{ | } } \overset { Electrolysis }{ \longrightarrow } \overset { { CH }_{ 2 }-{ COO }^{ - } }{ \underset { { CH }_{ 2 }-{ COO }^{ - } }{ | } +2 } { K }^{ + }\)
Potassium Succinate
At anode
\(\overset { { CH }_{ 2 }CO{ O }^{ - } }{ \underset { { CH }_{ 2 }CO{ O }^{ - } }{ | } \rightarrow } \overset { { CH }_{ 2 } }{ \underset { { CH }_{ 2 } }{ || } } +{ CO }_{ 2(g) }+{ 2e }^{ - }\)
                   Ethene
(iii) When alkene is treated with halogens like chlorine or bromine, addition takes place rapidly and forms 1,2- dihalo alkane (or) vicinal dihalide.
\({ CH }_{ 2 }={ CH }_{ 2 }\overset { { \\ CI }_{ 2 } }{ \longrightarrow } \overset { { CH }_{ 2 }- }{ \underset { CI }{ | } } \overset { { CH }_{ 2 } }{ \underset { CI }{ | } } \)
                         1,2 dichloroethane     

The chloro fluoro derivatives of methane and ethane are referred by trade name Freons. These Chloro Fluoro Carbon compounds are stable, non-toxic, noncorrosive and noninflammable, easily liquefiable and are used in refrigerators, air- conditioners and ln the production of plastic foams. CFC's are the exhaust of supersonic air craft's and jumbo jets in the upper atmosphere. They slowly pass from troposphere to stratosphere. ,They stay for very longer period of 50 - 100 years. In the presence of uv radiation, CFC's break up into chlorine free radical
CF₂Cl₂ \(\xrightarrow{hv}\) CF₂ CI +Cl.
CFCl₃ \(\xrightarrow{hv}\)CFCI₂+ Cl
Cl^o+O₃ ⟶ C10+O₂
Clo^o+O ⟶ CI+O₂
Chlorine radical is regenerated in the. course of reaction. Due to this continuous attack of Cl⁰ thinning of ozone layer takes place which leads to formation of ozone hole. It is estimated that for every reactive chlorine atom generated in the stratosphere 1,00,000 molecules of ozone are depleted.

This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (\(\triangle\)G^O) and equilibrium constant is
\(\Delta { G }^{ 0 }=-RTln\ K\)    ...(1)
We know that
\(\Delta { G }^{ 0 }=\Delta { H }^{ 0 }-T\Delta { S }^{ 0 }\)
Substituting (1) in equation (2)
\(-RTln\ K\ =\Delta { H }^{ 0 }-T\Delta { s }^{ 0 }\)
Rearranging
In \(K=\cfrac { -\Delta H^{ 0 } }{ RT } +\cfrac { { \Delta S }^{ 0 } }{ R } \)     ...(3)
Differentiating equation (3) with respect to temperature
\(\cfrac { d\left( In\quad K \right) }{ dT } =\cfrac { \Delta { H }^{ 0 } }{ { RT }^{ 2 } } \)       ...(4)
Equation 4 is known as differential form of Van't Hoff equation.
On integrating the equation 4, between T₁ and T₂ with their respective equilibrium constants K₁ and K_2.
\(\int _{ { k }_{ 1 } }^{ { K }_{ 2 } }{ d\left( In\ K \right) =\cfrac { \Delta { H }^{ 0 } }{ R } \int _{ { T }_{ 2 } }^{ { { T }_{ 2 } } }{ \cfrac { dT }{ { T }^{ 2 } } } } \)
\(\left[ In\quad K \right] _{ { K }_{ 1 } }^{ { K }_{ 2 } }=\cfrac { \Delta { H }^{ 0 } }{ R } \left[ -\cfrac { 1 }{ T } \right] ^{ { T }_{ 2 } }_{ { T }_{ 1 } }\)
\(In\quad { K }_{ 2 }-In\quad { K }_{ 1 }=\cfrac { \Delta { H }^{ 0 } }{ R } -\left[ \cfrac { 1 }{ { T }_{ 2 } } +\cfrac { 1 }{ { T }_{ 2 } } \right] \)
\(In\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)
\(log\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ 2.303R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)  ...(5)
Equation (5) is known as integrated form of Van't Hoff equation.

PCl₅ ⇌ PCl₃ + Cl₂
Given that [PCl₅]_initial = 1 mol; V = 1 dm³; K_C = 2

	PCl5	PCl3	Cl2
Initial no.of moles	1	-	-
No.of moles	x	-	-
No.of moles at equilibrium	1 - x	x	x
Equilibrium concentration	\({1-x\over 1}\)	\({x\over 1}\)	\({x\over 1}\)

\(K_c={[PCl_3][Cl_2]\over [PCl_5]}\)
\(2={x\times x\over (1-x)}\)
2 - 2x = x²|
x² + 2x - 2 = 0
Solution for a quadratic equation
\(a x^{2}+b x+c=0 \text { are }, x=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\)
a- = 1 b = 2 c = -2
\(x=\frac{-2-\sqrt{4-4 \times 1 \times-2}}{2 \times 1} \)
\(x=\frac{-2-\sqrt{12}}{2}=\frac{-2-\sqrt{4 \times 3}}{2}\)
\( x =\frac{-2-2 \sqrt{3}}{2} \)
\(=\frac{-2+2 \sqrt{3}}{2}, \frac{-2-2 \sqrt{3}}{2} \)
\(x =-1+\sqrt{3} ;-1-\sqrt{3} \)
\(=-1-\sqrt{3} \text { not possible } \)
Since x is + ve, 
x = -1 + 1.732
x = 0.732
Equilibrium concentration of
\( {\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=\frac{1-x}{1}=1-0.732=0.268 \mathrm{M}} \)
\({\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)
\({\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)

(i) O or F. Both O and F lie in 2^nd period. As we move from O to F the atomic size decreases. Due to smaller size of F nuclear charge increases.
Further, gain of one electron by
\(F\rightarrow F^-\)
ion has inert gas configuration, While tile gain of one electron by
\(O\rightarrow O^-\)
gives O^- ion which does not have stable inert gas configuration. Consequently, the energy released is much higher in going from
\(F\rightarrow F^-\)
than going from\(O\rightarrow O^-\). In other words electron gain enthalpy of F is much more negative than that of oxygen.
(ii) The negative electron gain enthalpy of CI \((e.g.\triangle H=-349\ kJ\ mol^{-1})\) is more than that of \(F(e.g.\triangle H=-328\ kJ\ mol^{-1})\) 
The reason for the deviation is due to the smaller size of F. Due to its small size, the electron repulsions in the relatively compact 2p-subshell are comparatively large and hence the attraction for incoming electron is less as in the case of Cl.

(i) System: A system is defined as any portion of matter (or universe) under thermodynamic.consideration, which is separated from the rest of the universe by real or imaginary boundaries.
(ii) Surroundings: Everything in the universe that is not the part of system and can interact with system is called as surroundings.
(iii) Boundary: Anything which separates the system from its surroundings is called boundary.

The carbon monoxide of the water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400^oC and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.
CO + H₂O ➝ CO₂ + H₂

(i) The decreasing order of bond angles is: NH₄⁺ > NH₃>NH₂^-
(ii) The N-atoms in all the three molecules are sp³ hybridised
(iii) The number of lone-pair of electrons present on nitrogen of NH₄⁺, NH₃ and NH₂^- are 0, 1 and 2.
(iv) Greater the number of lone pairs, greater is the repulsion and lesser is the bond angle.

CH₃ - CH = O, CH₃ -C = N and CH₃-I, the underlined carbon atoms are electrophilic centres as they will have partial positive charge due to polarity of the C-O or C-N or C-I bond.

S.No	Classical somg (london smog)	Photochemical smog (Los Angels smog)
1	It was first observed in London in Dec. 1952	It was first observed in Los Angels in 1950.
2	It occurs in cool, humid climate	It occurs in warm, dry and sunn climate
3	It consists of coal smoke and fog	It is formed by the combination of smoke dust and fog with air pollutants like N2 and hydrocarbons in presence of light.
4	It generally occurs in the morning and becomes worse when the sunshines	It forms when the sunshines and becomes worse in the afternoon.
5	This is mainly due to the induced oxidation of SO2 to SO3, which reacts with water yielding sulphuric acid aerosol.	This is mainly due to the induced oxidation of N2 to NO, NO2 and [O] + O2 to O3 NO and O3 are strong oxidising agents and can react with unburnt hydrocarbons in polluted air to form HCHO, Acrolein and PAN.
6	Chemically it is reducing in nature because of high concentration of SO2.	Chemically it is oxidising in nature because of high concentrations of oxidising agents like NO2 and O2
7	It is called reducing smog	It is called oxidising smog
8	Responsible for acid rain, causes poor visibility, affects air and road transport and causes bronchial irritation.	Causes irritation to nose, throat, eyes, skin and lungs, increases asthma, causes chest pain, uncomfortable breathing, PAN attacks young leaves, causes corrosion of metals, stones painted surfaces, etc.,

i) Raschig process ChIoro benzene is commercially prepared by passing a mixture of benzene vapour, air and HCI over heated cupric chlooride is reaction is called Raschig process.

ii) Dows Process :
\(\underset { Chlorobenzne }{ { C }_{ 6 }{ H }_{ 5 }Cl+NaOH } \overset { 350C }{ \underset { 300atm }{ \longrightarrow } } \underset { Phhenol }{ { C }_{ 6 }{ H }_{ 5 }OH+NaCl } \)
This reaction is known as Dow's Process
iii) Darzens process :

This reaction is known as Dow's process

Functional group :
A functional group is an'atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present.
Acetaldehyde - CHO.

i = \({observed\ property\over Theoritical\ property\ (calculated)}\)
Given ΔT_f = 0.680 K
m = 0.2 m
ΔT_f (observed) = 0.680 K
ΔT_f (calculated) = K_f m
= 1.86 K Kg mol^–1 × 0.2 mol Kg^–1
= 0.372 K
i = \({(\Delta T_f)\ observed\over (\Delta T_f)\ calculated}={0.680\ K\over 0.372\ K}=1.82\)

\(\underset { 2\ volumes }{ { 2H }_{ { 2 }_{ (g) } } } +\underset { 1\ volumes }{ { O }_{ { 2 }_{ (g) } } } \rightarrow \underset { 2\ volumes }{ { 2H }_{ 2 }{ O }_{ (g) } } \)
Thus 2 volumes of H₂ reacts with 1 volume of O₂ to produce 2 volumes of H₂O (g)
\(\therefore\) 10 volumes of H₂ would react with 5 volumes of O₂ to produce 10 volumes of H₂O(g).
Thus 10 volumes of H₂O will be produced.

Na₂O reacts with water and forms a base i.e, sodium hydroxide.
Na₂O + H₂O \(\rightarrow\)2NaOH
Cl₂ O₇ reacts with water and forms perchloric acid
Cl₂O₇ + H₂O \(\rightarrow\) 2HClO₄

Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. The mathematical form of Avogadro's hypothesis may be expressed as
V∝n, \(\frac { { V }_{ 1 } }{ { n }_{ 1 } } =\frac { { V }_{ 2 } }{ { n }_{ 2 } } =constant\)
Where V₁and n₁ are the volume and number of moles of a gas and V₂ and n₂ are a different set of values of volume and number of moles of the same gas at same temperature and pressure.

a) Principal quantum number defines energy and size of an orbital.
b) Azimuthal quantum number defines shape of an orbital
c) Magnetic quantum number defines spatial orientation (direction) of an orbital.