(i) Due to small size of Nitrogen atom.
(ii) Due to polar nature of N-H bond.
(iii) Due to inter molecular H- bonding which are stronger than London forces present in other hydrides. Other hydrides lack H - bonding

(i) Free energy is defined as G = H - TS. 'G' is a state function.
(ii) G- Extensive property; ΔG - intensive property. When mass remains constant between initial and final states of system.
(iii) 'G' has a single value for the thermodynamic state of the system.
(iv) G and ΔG values correspond to the system only.

(v) Gibbs free energy and the net work done by the system:
For any system at constant pressure and temperature
ΔG = ΔH - TΔS .....(1)
We know that,
ΔH = ΔU + PΔV
ΔG =ΔU + PΔV-TΔS
from first law of thermodynamics
ΔU = q +w
from second law of thermodynamics
Δ S=\(\frac{q}{T}\) Δ G=q+w+PΔ V-T\((\frac{q}{T}) \)
Δ G = w+PΔV
-ΔG = -w - PΔV ......(2)
But -PΔV represents the work done due to expansion against a constant external pressure.

Given:
Mg(s)+Br₂(I) ⟶ MgBr₂(s) ΔH_f⁰ = -524 KJ mol^-1
Sublimation:
Mg(s) ⟶ Mg(g), ΔH₁⁰ = +148 KJ mol^-1
Ionisation:
Mg(g) ⟶ Mg²⁺(g)+2e^- , ΔH₂⁰ =2187 KJ mol^-1
Vapourisation:
Br₂(I) ⟶ Br₂(g), ΔH₃⁰ = +31 KJ mol^-1
Dissociation:
Br₂(g) ⟶ 2Br(g), ΔH₄⁰ = +193 KJ mol^-1
Electron affinity:
Br(g) + e^- ⟶ Br^-(g), ΔH₅⁰ = -331 KJ mol^-1

ΔH_f = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + 2ΔH₅ + u
-524 = 148 + 2187 + 31 + 193 + (2 x -331) + u
-524 = 1897 + u
u = -524 - 1897
u = -2421 kJ mol^-1.

(k_H)_bonzene = 4.2 x 10^–5 mm Hg
Solubility of methane = ?
P = 750mm Hg           
P = 840 mm Hg
According to Henrys Law,
P = K_H . x_{in solution}.
750 mm Hg = 4.2 x 10^–5 mm Hg . x_{in solution}
\(\Rightarrow X_{insolution}={750\over 4.2\times 10^{-5}}\)
i.e, solubility = 178.5 x 10⁵
similarly at P = 840 mm Hg
solubility = \({840\over 4.2\times 10^{-5}}\)
= 200 x 10^-5.

(w) = 0.33 g
y = 0.397 g
\(\% \mathrm{P}=\frac{62}{222} \times \frac{\mathrm{y}}{\mathrm{w}} \times 100=\frac{62}{222} \times \frac{0.397}{0.33} \times 100=33.60 \%\)

Inductive effect (I):
(i) Inductive effect is defined as the change in the polarisation of a covalent bond due to the presence of adjacent bonds, atoms or groups in the molecule. This is a permanent phenomenon.
(ii) Let us explain the inductive effect by considering ethylchloride as example. The C-C bond in ethyl chloride is polar.
We know that chlorine is more electronegative than carbon, and hence it attracts the shared pair of electron between C-Cl in ethyl chloride towards itself. is develops a slight negative charge on Chlorine and a slight positive charge on carbon to which chlorine is attached.
To compensate it, the C₁ draws the shared pair of electron between itself and. C₂ . This polarisation effect is called inductive effect.
(iii) The magnitude of the charge separation decreases rapidly, as we move away from C₁ and is observed maximum for 2 carbons and almost insignicant after 4 bonds from the active group.
\(\overset { \delta }{ C } \overset { \delta + }{ { H }_{ 3 } } \longrightarrow \underset { 1 }{ C\overset { \delta + }{ { H }_{ 2 } } } \twoheadrightarrow \overset { \delta - }{ C } { l }_{ 2 }\)
It is important to note that the inductive effect does not transfer electrons from one atom to another but the displacement effect is permanent. The inductive effect represents the ability of a particular atom or a group to either withdraw or donate electron density to the attached carbon. Based on this ability the substituents are classified as +I groups and -I groups. Their ability to release or withdraw the electron through sigma covalent bond is called +I effect and -I effect respectively.
Highly electronegative atoms and atoms of groups which are carry a positive charge are electron withdrawing or -I group
Example : -F, -CI, -COOH, -NO₂, NH₂,
Higher the electronegativity of the substitutent, greater is the -I effect.
The order of the -I effect of some groups are given below :
NH³⁺ > NO₂ > CN > SO₃H > CHO > CO > COOH > COCI > CONH₂ > F > Cl > Br > I > OH > OR, NH₂ > C₆H₅ > H
Highly electropositive atoms and atoms are groups which carry a negative charge are electron donating or +I groups.
Example. Alkali metals, alkyl groups such as methyl, ethyl, negatively charged groups such as CH₃O^-, C₂H₅O^-, COO^- etc.
Lesser the electronegativity of the elements, greater is the +I effect. The relative order of +I effect of some alkyl groups is given below
\(-\mathrm{C}\left(\mathrm{CH}_3\right)_3>-\mathrm{CH}\left(\mathrm{CH}_3\right)_2>-\mathrm{CH}_2 \mathrm{CH}_3>-\mathrm{CH}_3\)
Let us understand the influence of inductive effect on some properties of organic compounds.
Reactivity :
When a highly electronegative atom such as halogen is attached to a carbon then it makes the C-X bond polar. In such cases the -I effect of halogen facilitates the attack of an incoming nucleophile at the polarised carbon, and hence increases the reactivity.

If a - I group is attached nearer to a carbonyl carbon, it decreases the availability of electron density-on the carbonyl carbon, and hence increases the rate of the nucelophilic addition reaction.
Aciditv of carborvlic acids :
When a halogen atom is attached to the carbon which is nearer to the carboxylic acid group, its -I effect withdraws the bonded electrons towards itself and makes the ionisation of H⁺ easy. The acidity of various chloro acetic acid is in the following order. The strength of the acid increases with increase in the -I effect of the group attached to the carboxyl group.
Tiichloro acetic acid > Dichloro acetic acid > Chloro acetic acid > acetic acid

(i) BOD and COD

(ii) Viable and non-viable particulate pollutants

\(\overset { +3 }{ C } _{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow \overset { +4 }{ C } { O }_{ 2 }\)
\(\overset { +6 }{ Cr } _{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { Cr }^{ 3+ }\)
(1)  \(\Rightarrow \) \({ C }_{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow { 2CO }_{ 2 }+{ 2e }^{ - }\)
\({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 7H }^{ 2 }O\)

The balanced equation for the reaction is, 2AI + Fe₂O₃\(\rightarrow\) Al₂O₃ + 2Fe
                                                                          (2 x 27) (160)               (102)
Number of mol of aluminum taken =\({mass \ of \ A1 \ taken \over gram \ atomic \ weight}={124\over27}=4.6mol \)
Number of mol of Fe₂O₃ taken = \({mass \ of \ Fe_2O_3 \ taken \over gram \ molecular \ weight}={601\over160}=3.75mol \)
According to the equation,
2 mol of Al reacts with = 1 mol of Fe₂O₃
4.6 mol of Al will react with = \(1\over2\) x 4.6 = 2.3 mol of Fe₂O₃
number of mol of Fe₂O₃ taken = 3.75 mol
number of mol Fe₂O₃ reacted = 2.30 mol
Hence, Al is the limiting reagent and 1.45 mol of Fe₂O₃ will be left behind, unreacted.
2 mol of Aluminium = 54g Al = 102 g Al₂O₃
124.g of Aluminium = \(102\over54\)x 124
= 234 g of Al₂O₃
Mass of Fe₂O₃ left behind = No. of mol x gram molecular unit
= 1.45 x 160
= 232 g