Presence of a chiral carbon is necessary but not sufficient for optical activity.

Why it is necessary: Optical activity requires a molecule to be chiral, i.e. non-superimposable on its mirror image. A chiral (asymmetric) carbon, bonded to four different groups, is the usual source of this asymmetry. Without any such centre, the molecule is generally achiral and optically inactive.

Why it is not sufficient: A compound may contain chiral carbons and still show no rotation:

1. Meso compounds (internal compensation): With two or more chiral carbons and an internal plane of symmetry, the rotation due to one half is cancelled by the other half. Example: meso-tartaric acid, HOOC–CH(OH)–CH(OH)–COOH, has two chiral carbons but is optically inactive.

2. Racemic mixtures (external compensation): An equimolar mixture of the (+) and (−) enantiomers rotates plane-polarised light equally in opposite directions, so the net rotation is zero. Example: (±)-lactic acid.

3. Rapid racemisation: If the enantiomers interconvert quickly (for example through a planar or rapidly inverting intermediate), an equilibrium racemic mixture forms and no net rotation is observed.

Conclusion: A chiral carbon provides the possibility of chirality, but the molecule must also lack a plane or centre of symmetry and must be present as a single enantiomer (or in excess of one) to show optical activity. Hence the presence of a chiral carbon is necessary but not sufficient.

$S_N2$ and $S_N1$ Reaction Mechanisms and Their Stereochemical Outcomes

1. $S_N2$ Reaction (Substitution Nucleophilic Bimolecular)

Mechanism: The reaction proceeds in a single step (concerted mechanism). The nucleophile attacks the carbon bearing the leaving group from the side opposite to the leaving group (backside attack, $180°$ to the C–X bond). As the nucleophile begins to bond, the leaving group departs simultaneously. A pentacoordinate transition state is formed in which the carbon is $sp^2$-hybridised with the three non-reacting groups in a plane and the nucleophile and leaving group partially bonded on opposite sides.

$\text{Nu}^- + \text{CH}_3\text{–X} \rightarrow [\text{Nu}\cdots\text{CH}_3\cdots\text{X}]^\ddagger \rightarrow \text{CH}_3\text{–Nu} + \text{X}^-$

Kinetics: Rate $= k[\text{Substrate}][\text{Nucleophile}]$ — second order. Substrate preference: $\text{CH}_3 >$ primary $>$ secondary $\gg$ tertiary (steric hindrance at the reaction centre).

Stereochemical outcome — Inversion of configuration (Walden inversion): Because the nucleophile enters from the face opposite the leaving group, the three remaining substituents are pushed through the plane and end up on the opposite side. Thus an optically active substrate gives the product with inverted configuration (R $\rightarrow$ S or S $\rightarrow$ R). If the carbon is not chiral (e.g. $\text{CH}_3\text{X}$), no stereochemical change is observed. The inversion is essentially 100% when the reaction is cleanly $S_N2$.

2. $S_N1$ Reaction (Substitution Nucleophilic Unimolecular)

Mechanism: The reaction proceeds in two steps via a carbocation intermediate.

• Step 1 (slow, rate-determining): The C–X bond breaks heterolytically, giving a planar carbocation and the leaving group. $\text{R–X} \rightarrow \text{R}^+ + \text{X}^-$
• Step 2 (fast): The nucleophile attacks the carbocation to give the product. $\text{R}^+ + \text{Nu}^- \rightarrow \text{R–Nu}$

Kinetics: Rate $= k[\text{Substrate}]$ — first order. Substrate preference: tertiary $>$ secondary $\gg$ primary (stability of carbocation: $3° > 2° > 1°$).

Stereochemical outcome — Racemisation (with partial inversion): The carbocation intermediate is planar ($sp^2$) with a vacant p-orbital perpendicular to the plane. The nucleophile can attack with equal probability from either face. Attack from the same side gives retention of configuration; attack from the opposite side gives inversion. The product is therefore obtained as a racemic mixture (50% R + 50% S) when the reaction goes fully through the free carbocation.

In practice, complete racemisation is rarely observed. The leaving group may still be loosely associated with one face of the carbocation (ion pair), partially shielding it. This causes the nucleophile to attack slightly more from the opposite face, so the product shows predominant inversion with some retention — i.e. partial racemisation.

3. Comparison Summary

4. Conclusion

In an $S_N2$ reaction the backside attack forces the three substituents to flip through the plane, giving inversion of configuration. In an $S_N1$ reaction the planar carbocation allows attack from both faces, giving racemisation, though ion-pair effects usually cause partial inversion rather than a perfectly racemic product.

Acid–base properties of methanol

Methanol, CH₃OH, is amphoteric: it is both a very weak acid and a very weak base, and it is neutral to litmus.

1. As an acid. The O–H hydrogen can be lost as a proton. Methanol reacts with active metals and very strong bases:

$2CH_3OH + 2Na \rightarrow 2CH_3ONa + H_2$

The methoxide ion is a strong base. Methanol has $pK_a \approx 15.5$, so it is about as acidic as water ($pK_a = 15.7$), and much weaker than carboxylic acids or phenol. It does not react appreciably with aqueous NaOH or turn litmus red. Alkyl groups with a +I effect make higher alcohols progressively weaker acids than methanol.

2. As a base. The lone pairs on oxygen accept a proton from strong acids to give the methyloxonium ion:

$CH_3OH + HCl \rightarrow CH_3OH_2^+ + Cl^-$

This property is the first step in many acid-catalysed reactions of alcohols.

Reaction with sulphuric acid

Methanol is protonated by concentrated H₂SO₄ and, on heating (about 140 °C) with the acid in excess of alcohol, undergoes intermolecular dehydration to dimethyl ether. The acid acts as a catalyst and dehydrating agent and is regenerated.

$2CH_3OH \xrightarrow[140\,^\circ C]{conc.\ H_2SO_4} CH_3OCH_3 + H_2O$

Mechanism:

$CH_3OH + H_2SO_4 \rightleftharpoons CH_3OH_2^+ + HSO_4^-$

$CH_3OH_2^+ + CH_3OH \rightarrow CH_3\text{–}O^+(H)\text{–}CH_3 + H_2O$

$CH_3\text{–}O^+(H)\text{–}CH_3 + HSO_4^- \rightarrow CH_3OCH_3 + H_2SO_4$

At lower temperature and with the acid in excess, methanol instead gives methyl hydrogen sulphate, $CH_3OH + H_2SO_4 \rightarrow CH_3OSO_3H + H_2O$.

Sodium tert-butoxide is used because the tert-butyl group must be supplied by the alkoxide, not by the alkyl halide.

The Williamson synthesis is an $S_N2$ reaction: an alkoxide attacks an alkyl halide and displaces the halide. It works well only with methyl or primary halides. With secondary and especially tertiary halides, the alkoxide acts as a strong base and the halide undergoes E2 elimination to an alkene.

To prepare tert-butyl ethyl ether, (CH₃)₃C–O–CH₂CH₃, there are two possible routes:

In the correct route, the electrophilic carbon is the unhindered primary carbon of ethyl bromide, so $S_N2$ substitution occurs readily. The bulky tert-butoxide, although a strong base, has no tertiary halide to eliminate and attacks the accessible CH₂ carbon, giving the ether in good yield.

Sodium ethoxide cannot be used with tert-butyl bromide because the tertiary carbon is too crowded for backside attack, and ethoxide abstracts a β-hydrogen to give isobutylene (2-methylpropene).

Rule: In an unsymmetrical ether, the bulky (tertiary) alkyl group should come from the alkoxide and the less hindered (primary) alkyl group from the halide.

Reaction: tert-Butyl bromide undergoes dehydrohalogenation (β-elimination) with alcoholic KOH on heating to give isobutylene (2-methylpropene).

$(CH_3)_3C\text{–}Br + KOH(alc) \xrightarrow{\Delta} (CH_3)_2C{=}CH_2 + KBr + H_2O$

Mechanism: E2 (bimolecular, concerted). Alcoholic KOH supplies a strong base ($OH^-$/$C_2H_5O^-$) in a medium of low polarity, which favours elimination over substitution. The rate is $k[(CH_3)_3CBr][OH^-]$, second order overall.

In a single step:

1. The base abstracts a β-hydrogen from one of the three methyl groups.
2. The C–H bond electrons shift to form the C=C π-bond.
3. The C–Br bond breaks and bromide leaves simultaneously, through an anti-periplanar transition state.

$HO^- + H\text{–}CH_2\text{–}C(CH_3)_2\text{–}Br \rightarrow [HO\cdots H\cdots CH_2\cdots C(CH_3)_2\cdots Br]^{\ddagger} \rightarrow H_2O + CH_2{=}C(CH_3)_2 + Br^-$

Why elimination and not $S_N2$: the tertiary carbon is shielded by three methyl groups, so backside attack is blocked. The base therefore removes a peripheral β-hydrogen instead. Nine equivalent β-hydrogens make this attack very probable, and the product alkene is stabilised.

Regiochemistry: all β-hydrogens are equivalent, so only one alkene, 2-methylpropene, can form. No Zaitsev/Hofmann question arises.

Note: without a strong base (for example in hot water or ethanol alone), the substrate ionises first and gives the $S_N1$/E1 products. The use of a strong base in alcoholic KOH makes E2 the pathway.

Acidity of terminal alkynes

A terminal alkyne has a hydrogen on an sp-hybridised carbon (R–C≡C–H). This hydrogen can be removed by a very strong base to give an alkynide ion:

$R\text{–}C{\equiv}C\text{–}H + B^- \rightleftharpoons R\text{–}C{\equiv}C^- + BH$

Factors responsible

1. High s-character of the sp orbital. The sp orbital has 50% s-character (sp² 33%, sp³ 25%). s-Electrons are held closer to the nucleus, so the carbon is effectively more electronegative and the C–H bond is more polarised.
2. Stability of the conjugate base. In the alkynide ion the lone pair sits in an sp orbital, close to the nucleus and well stabilised. A more stable anion means a stronger acid.
3. Order of acidity. $HC{\equiv}CH > CH_2{=}CH_2 > CH_3\text{–}CH_3$ ($pK_a$ about 25, 44, 50).
4. Substituents. Electron-releasing alkyl groups slightly decrease the acidity of R–C≡C–H compared with acetylene.

Acetylene and water

Water is the stronger acid by about nine $pK_a$ units, so acetylene is a weaker acid than water.

Justification

Acid strength depends on the stability of the conjugate base. In $OH^-$ the negative charge is on oxygen, which is much more electronegative than carbon. In $HC{\equiv}C^-$ the charge is on carbon, which, even with 50% s-character, is less electronegative than oxygen, so the anion is less stable.

Experimentally, sodium acetylide is made by treating acetylene with sodamide ($NaNH_2$), and it is decomposed by water:

$HC{\equiv}C^- Na^+ + H_2O \rightarrow HC{\equiv}CH + NaOH$

The reaction goes to the right because water is the stronger acid and hydroxide the weaker base. Hence acetylene does not react with aqueous NaOH or KOH.

Comprehensive Question

1. Molar heat capacity at constant pressure and constant volume

Molar heat capacity at constant pressure, (C_p):
It is the amount of heat required to raise the temperature of one mole of a gas through (1^\circ\text{C}) or (1\ \text{K}) at constant pressure.
Mathematically, [ C_p=\left(\frac{\partial H}{\partial T}\right)_p ]

Molar heat capacity at constant volume, (C_v):
It is the amount of heat required to raise the temperature of one mole of a gas through (1^\circ\text{C}) or (1\ \text{K}) at constant volume.
Mathematically, [ C_v=\left(\frac{\partial U}{\partial T}\right)_v ]

For an ideal gas, [ C_p-C_v=R ]

---

2. Kirchhoff’s equation

Kirchhoff’s equation gives the variation of heat of reaction at constant pressure with temperature.

For a general reaction, [ \Delta H = H_{\text{products}}-H_{\text{reactants}} ]

Differentiating with respect to temperature at constant pressure, [ \frac{d(\Delta H)}{dT}

\frac{dH_{\text{products}}}{dT}

\frac{dH_{\text{reactants}}}{dT} ]

But, [ \frac{dH}{dT}=C_p ]

Therefore, [ \boxed{\frac{d(\Delta H)}{dT}=\Delta C_p} ]

where, [ \Delta C_p

\sum C_p(\text{products})

\sum C_p(\text{reactants}) ]

This is Kirchhoff’s equation.

Integrating between temperatures (T_1) and (T_2), [ \Delta H_{T_2}

\Delta H_{T_1} + \int_{T_1}^{T_2}\Delta C_p,dT ]

If (\Delta C_p) is constant, [ \boxed{\Delta H_{T_2}

\Delta H_{T_1} + \Delta C_p(T_2-T_1)} ]

Explanation:
If (\Delta C_p>0), the heat of reaction increases with rise in temperature.
If (\Delta C_p<0), the heat of reaction decreases with rise in temperature.
For constant volume, the corresponding relation is [ \frac{d(\Delta U)}{dT}=\Delta C_v ]

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3. Calculation of enthalpy of formation of (NH_3) at (50^\circ\text{C})

Given reaction: [ \frac12 N_2(g)+\frac32 H_2(g)\rightarrow NH_3(g) ]

Given: [ \Delta H_f(NH_3)\text{ at }27^\circ\text{C}=-90.52\ \text{kJ/mol} ]

[ C_p(N_2)=27.19\ \text{J mol}^{-1}\text{K}^{-1} ]

[ C_p(H_2)=28.02\ \text{J mol}^{-1}\text{K}^{-1} ]

[ C_p(NH_3)=35.68\ \text{J mol}^{-1}\text{K}^{-1} ]

[ T_1=27+273=300\ \text{K} ]

[ T_2=50+273=323\ \text{K} ]

Now, [ \Delta C_p

C_p(NH_3)

\left[ \frac12 C_p(N_2)+\frac32 C_p(H_2) \right] ]

[

35.68- \left[ \frac12(27.19)+\frac32(28.02) \right] ]

[

35.68-[13.595+42.03] ]

[

35.68-55.625 ]

[

-19.945\ \text{J mol}^{-1}\text{K}^{-1} ]

[

-0.019945\ \text{kJ mol}^{-1}\text{K}^{-1} ]

Using Kirchhoff’s equation, [ \Delta H_{T_2}

\Delta H_{T_1} + \Delta C_p(T_2-T_1) ]

[

-90.52+(-0.019945)(323-300) ]

[

-90.52-0.458735 ]

[

-90.978735\ \text{kJ/mol} ]

[ \boxed{\Delta H_f(NH_3)\text{ at }50^\circ\text{C}\approx -90.98\ \text{kJ/mol}} ]

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OR

1. Relation between van der Waals constants and critical constants

For one mole of a gas, van der Waals equation is [ \left(P+\frac{a}{V^2}\right)(V-b)=RT ]

or, [ P=\frac{RT}{V-b}-\frac{a}{V^2} ]

At the critical point, the isotherm has a point of inflection. Hence, [ \left(\frac{\partial P}{\partial V}\right)_T=0 ]

and [ \left(\frac{\partial^2 P}{\partial V^2}\right)_T=0 ]

First derivative: [ \frac{dP}{dV}

-\frac{RT}{(V-b)^2} + \frac{2a}{V^3}

0 ]

[ \frac{RT}{(V-b)^2}

\frac{2a}{V^3} \quad (1) ]

Second derivative: [ \frac{d^2P}{dV^2}

\frac{2RT}{(V-b)^3}

\frac{6a}{V^4}

0 ]

[ \frac{RT}{(V-b)^3}

\frac{3a}{V^4} \quad (2) ]

Dividing equation (2) by equation (1), [ \frac{1}{V-b}

\frac{3}{2V} ]

[ 2V=3(V-b) ]

[ 2V=3V-3b ]

[ V=3b ]

Thus, [ \boxed{V_c=3b} ]

Substituting (V_c=3b) in equation (1), [ \frac{RT}{(3b-b)^2}

\frac{2a}{(3b)^3} ]

[ \frac{RT}{4b^2}

\frac{2a}{27b^3} ]

[ RT=\frac{8a}{27b} ]

[ \boxed{T_c=\frac{8a}{27Rb}} ]

Now, substituting (V_c=3b) and (T_c=\frac{8a}{27Rb}) in van der Waals equation, [ P_c

\frac{RT_c}{V_c-b}

\frac{a}{V_c^2} ]

[

\frac{R\left(\frac{8a}{27Rb}\right)}{3b-b}

\frac{a}{(3b)^2} ]

[

\frac{8a}{27b(2b)}

\frac{a}{9b^2} ]

[

\frac{4a}{27b^2}

\frac{a}{9b^2} ]

[

\frac{4a}{27b^2}

\frac{3a}{27b^2} ]

[

\frac{a}{27b^2} ]

Thus, [ \boxed{P_c=\frac{a}{27b^2}} ]

Hence, the required relations are: [ \boxed{V_c=3b,\qquad T_c=\frac{8a}{27Rb},\qquad P_c=\frac{a}{27b^2}} ]

Also, [ Z_c=\frac{P_cV_c}{RT_c}=\frac{3}{8} ]

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2. Calculation of critical constants for (HCl) gas

Given: [ a=3.716\ \text{L}^2\text{ atm mol}^{-2} ]

[ b=0.0408\ \text{L mol}^{-1} ]

[ R=0.0821\ \text{L atm mol}^{-1}\text{K}^{-1} ]

Critical volume: [ V_c=3b ]

[ V_c=3\times 0.0408 ]

[ \boxed{V_c=0.1224\ \text{L mol}^{-1}} ]

Critical temperature: [ T_c=\frac{8a}{27Rb} ]

[ T_c= \frac{8\times 3.716}{27\times 0.0821\times 0.0408} ]

[

\frac{29.728}{0.09044136} ]

[ \boxed{T_c\approx 328.7\ \text{K}} ]

In Celsius, [ T_c=328.7-273=55.7^\circ\text{C} ]

Critical pressure: [ P_c=\frac{a}{27b^2} ]

[ P_c= \frac{3.716}{27\times (0.0408)^2} ]

[

\frac{3.716}{27\times 0.00166464} ]

[

\frac{3.716}{0.04494528} ]

[ \boxed{P_c\approx 82.68\ \text{atm}} ]

Therefore, for (HCl) gas: [ \boxed{T_c\approx 328.7\ \text{K},\qquad P_c\approx 82.68\ \text{atm},\qquad V_c\approx 0.1224\ \text{L mol}^{-1}} ]

This item is the section heading Short Answer Questions, 6 × 4 = 24; the six individual questions are not recorded here, so none can be answered from this entry.

For a 4-mark short answer, write 8 to 10 lines: define the term in one line, give three or four points, and add a formula, labelled example or short table where relevant. Attempt any six questions in the paper.

Half-life of a reaction

The half-life of a reaction is the time required for the concentration of a reactant to fall to one-half of its initial value. It is denoted by $t_{1/2}$.

For a first-order reaction, $t_{1/2}$ is independent of the initial concentration.

Derivation

For a first-order reaction,

$k = \frac{2.303}{t}\log\frac{a}{a-x}$

where $a$ = initial concentration and $x$ = amount reacted at time $t$.

Half-life: at $t = t_{1/2}$, $x = a/2$, so $a - x = a/2$.

$k = \frac{2.303}{t_{1/2}}\log\frac{a}{a/2} = \frac{2.303}{t_{1/2}}\log 2$

$t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}$

Time for 75% completion: at $t = t_{75\%}$, $x = 0.75a$, so $a - x = 0.25a$.

$k = \frac{2.303}{t_{75\%}}\log\frac{a}{0.25a} = \frac{2.303}{t_{75\%}}\log 4$

$t_{75\%} = \frac{2.303 \times 0.6021}{k} = \frac{1.386}{k}$

Comparison

$\frac{t_{75\%}}{t_{1/2}} = \frac{1.386/k}{0.693/k} = 2$

$\therefore \; t_{75\%} = 2\,t_{1/2}$

Hence, the time taken to complete 75% of a first-order reaction is twice its half-life.

Mechanism of the reaction between iodine and propanone in acidic medium

The reaction is the acid-catalysed iodination of propanone:

$\mathrm{CH_{3}COCH_{3} + I_{2} \rightarrow [H+] CH_{3}COCH_{2}I + HI}$

The rate of this reaction is independent of the concentration of iodine — it is first order with respect to both propanone and the acid catalyst. This is because the slow step does not involve iodine.

Step 1 — Acid–base pre-equilibrium (fast, reversible):

The acid protonates the carbonyl oxygen of propanone, giving a resonance-stabilised protonated species:

$\mathrm{CH_{3}COCH_{3} + H+ \rightleftharpoons CH_{3}C(OH+)(CH_{3})}$

This protonation makes the carbonyl carbon more electrophilic and weakens the C–H bond of the methyl group, making the α-hydrogen acidic.

Step 2 — Rate-determining enolisation (slow):

A water molecule (acting as a base) removes an α-proton from the protonated propanone, and the C=C double bond forms as the C=O π-electrons shift onto oxygen. This gives the enol form of propanone:

$\mathrm{CH_{3}C(OH+)(CH_{3}) + H_{2}O \rightarrow CH_{3}C(OH)=CH_{2} + H_{3}O+}$

This step is the slowest step and therefore the rate-determining step. Since only propanone and H⁺ are involved here, the rate law is:

$\text{Rate} = k\,[\mathrm{CH_{3}COCH_{3}}][\mathrm{H+}]$

Step 3 — Attack of iodine on the enol (fast):

The electron-rich enol double bond attacks the iodine molecule, forming a cyclic iodonium-ion intermediate:

$\mathrm{CH_{3}C(OH)=CH_{2} + I_{2} \rightarrow [cyclic iodonium intermediate]^+ I-}$

Step 4 — Loss of proton to give the product (fast):

Loss of a proton from the intermediate gives iodo-propanone and HI:

$\mathrm{[intermediate]^+ \rightarrow CH_{3}COCH_{2}I + H+}$

The liberated H⁺ is regenerated, so the acid acts as a true catalyst. The iodo-propanone formed can undergo further rapid iodination to give di- and tri-iodo derivatives, but the kinetics are governed entirely by the slow enolisation step.

Key points to note:

• The slow step is enolisation, so the rate is independent of [I₂].
• The rate is first order in propanone and first order in H⁺ (second order overall).
• The reaction is general acid–base catalysed; the acid only increases the concentration of the protonated carbonyl, it is not consumed.
• Since the rate-determining step involves only propanone and H⁺, the rate law is Rate = k[CH₃COCH₃][H⁺].

Compressibility factor (Z) is defined as the ratio of the actual molar volume of a real gas to the molar volume of an ideal gas at the same temperature and pressure:

$Z = \frac{V_{\text{real}}}{V_{\text{ideal}}} = \frac{PV}{nRT}$

For an ideal gas, (Z = 1) under all conditions. For a real gas, (Z) deviates from unity, and the extent of deviation measures how far the gas is from ideal behaviour. The deviation arises because real gas molecules have finite volume and exert attractive/repulsive intermolecular forces, neither of which is accounted for in the ideal gas equation.

Effect of pressure on Z:

• At very low pressure, molecules are far apart, so both molecular volume and intermolecular forces are negligible. Hence (Z \to 1) and the gas behaves ideally.
• As pressure increases, molecules come closer. Attractive forces dominate initially, pulling molecules together and making the gas more compressible than ideal, so Z falls below 1 (e.g. for ( \text{CH}_4 ), ( \text{CO}_2 ), ( \text{NH}_3 )).
• At very high pressure, molecules are so close that their finite size (excluded volume) dominates. The gas becomes less compressible than ideal, so Z rises above 1 (e.g. for ( \text{H}_2 ), He, which show (Z > 1) even at moderate pressures because attractive forces are weak).
• Thus for most gases, Z first decreases, passes through a minimum, and then increases with rising pressure. For a few gases like hydrogen and helium, Z increases continuously from the start.

Effect of temperature on Z:

• At high temperature, molecules possess large kinetic energy, so attractive forces are easily overcome. The gas tends to behave ideally and Z approaches 1 over a wide pressure range.
• At low temperature, attractive forces become effective; Z deviates strongly from 1 and falls well below 1 at moderate pressures.
• At the Boyle temperature ((T_B)), the attractive and repulsive effects balance over a range of pressures, so (Z \approx 1) for a considerable pressure interval. Below (T_B), Z initially decreases; above (T_B), Z increases with pressure.

Thus Z is a function of both temperature and pressure, and its variation is conveniently shown by plotting Z against P at different constant temperatures (isotherms).

Crystalline vs Amorphous Solids

Note: Amorphous solids are regarded as super-cooled liquids because their particles are frozen in a disordered state, similar to a liquid, without attaining a regular lattice arrangement.

Surface Tension

Surface tension is the property of a liquid surface by virtue of which it behaves like a stretched elastic membrane and tends to minimise its surface area. Numerically, it is the force per unit length acting perpendicular to an imaginary line drawn on the liquid surface, in the plane of the surface:

$\gamma = \frac{F}{L}$

Its SI unit is N/m and its dimensional formula is ([MT^{-2}]).

Viscosity

Viscosity is the property of a liquid (or fluid) by which it opposes the relative motion between its adjacent layers. Numerically, the coefficient of viscosity (\eta) is the tangential viscous force per unit area required to maintain a unit velocity gradient between two parallel layers:

$F = -\eta A \frac{dv}{dx}$

Its SI unit is N s/m² (poiseuille) and its dimensional formula is ([ML^{-1}T^{-1}]).

Effect of Temperature

Surface tension: Surface tension decreases with rise in temperature. As temperature increases, the average kinetic energy of the molecules increases, the intermolecular cohesive force weakens, and hence surface tension falls. The variation is nearly linear and is given by

$\gamma_t = \gamma_0 (1 - \alpha t)$

where (\gamma_0) is the surface tension at 0 °C and (\alpha) is the temperature coefficient. At the critical temperature, surface tension becomes zero. (Impurities also affect it — soluble impurities generally lower it.)

Viscosity: Viscosity of a liquid decreases with rise in temperature. With increase in temperature, the cohesive force between liquid molecules decreases and molecular momentum transfer becomes easier, so the layers slide more freely. The variation follows

$\eta_t = \frac{\eta_0}{1 + \alpha t + \beta t^2}$

where (\eta_0) is the viscosity at 0 °C. (For gases, however, viscosity increases with temperature, since it depends mainly on molecular momentum transfer.)

Raoult's law of relative lowering of vapour pressure

Statement. In a solution of a non-volatile solute in a volatile solvent, the vapour pressure of the solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution:

$p = p^\circ\, x_A$

where $p^\circ$ is the vapour pressure of the pure solvent and $x_A$ is the mole fraction of the solvent.

Since the solute is non-volatile, the vapour pressure of the solution is always less than that of the pure solvent. The lowering of vapour pressure is $p^\circ - p$, and the relative lowering of vapour pressure is

$\frac{p^\circ - p}{p^\circ} = \frac{p^\circ - p^\circ x_A}{p^\circ} = 1 - x_A = x_B$

i.e. the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute.

For a dilute solution,

$\frac{p^\circ - p}{p^\circ} = x_B = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A} = \frac{w_B/M_B}{w_A/M_A}$

$\boxed{M_B = \frac{w_B \times M_A \times p^\circ}{w_A \times (p^\circ - p)}}$

where $w_A, M_A$ are the mass and molar mass of the solvent and $w_B, M_B$ those of the solute. Thus the molecular mass of a non-volatile solute can be found from the relative lowering of vapour pressure (Ostwald–Walker method).

---

Determination of molecular mass from depression of boiling point

Principle. The elevation of boiling point of a solvent produced by a non-volatile solute is directly proportional to the molality of the solution:

$\Delta T_b = T_b - T_b^\circ = K_b\, m = K_b \cdot \frac{w_B \times 1000}{M_B \times w_A}$

where $K_b$ is the molal elevation constant (ebullioscopic constant) of the solvent, $w_B$ the mass of solute, $w_A$ the mass of solvent in grams. Hence

$\boxed{M_B = \frac{K_b \times w_B \times 1000}{\Delta T_b \times w_A}}$

$K_b$ is a constant for a given solvent and is calculated from

$K_b = \frac{R\,T_b^{\circ 2}}{1000\,L_v} = \frac{R\,T_b^{\circ 2} M_A}{1000\,\Delta H_{vap}}$

where $L_v$ is the latent heat of vaporisation per gram and $\Delta H_{vap}$ the molar enthalpy of vaporisation of the solvent. For water $K_b = 0.52\ \text{K kg mol}^{-1}$, for benzene $2.53$, for chloroform $3.63\ \text{K kg mol}^{-1}$.

Method (Cottrell's ebullioscopic method).

1. A known mass of the pure solvent ($w_A$) is taken in the inner boiling tube and heated so that it boils steadily; the thermometer is placed so that its bulb is just above the liquid surface.
2. The boiling point of the pure solvent, $T_b^\circ$, is noted (the thermometer is read when the liquid condenses on the bulb and the reading becomes steady).
3. A weighed amount of the non-volatile solute ($w_B$) is added through the side tube and dissolved.
4. The boiling point of the solution, $T_b$, is noted in the same way. The difference gives $\Delta T_b = T_b - T_b^\circ$.
5. The experiment is repeated with different masses of the solute; $\Delta T_b$ is found to be proportional to the molality, confirming the relation.
6. Substituting $K_b$, $w_B$, $w_A$ and $\Delta T_b$ in the formula gives the molecular mass $M_B$ of the solute.

Precautions. The thermometer must be calibrated and read with a magnifying lens (Beckmann thermometer is used for accuracy); the solute must be non-volatile and must not dissociate or associate in the solvent; the same thermometer and apparatus must be used for both readings; the boiling must be steady and the solution well stirred (Cottrell pump ensures this).

Note. If the solute associates or dissociates, the value obtained is the abnormal molecular mass; the true value is obtained by applying the van't Hoff factor $i$: $M_B = i\,K_b w_B \times 1000/(\Delta T_b\, w_A)$.

Buffer solution: A buffer solution is a solution that resists a marked change in its pH when a small amount of a strong acid or a strong base is added to it, or when it is diluted with water. It generally consists of a mixture of a weak acid and its salt with a strong base (acidic buffer, e.g. CH₃COOH + CH₃COONa), or a weak base and its salt with a strong acid (basic buffer, e.g. NH₄OH + NH₄Cl).

Acetic acid–sodium acetate buffer: In this mixture, acetic acid (CH₃COOH) is the weak acid and sodium acetate (CH₃COONa) is its salt. The salt is completely ionised while the acid is only slightly ionised:

$\mathrm{CH_{3}COONa \rightarrow CH_{3}COO- + Na+} \quad \text{(complete)}$ $\mathrm{CH_{3}COOH \rightleftharpoons CH_{3}COO- + H+} \quad \text{(partial)}$

Because of the common ion (CH₃COO⁻) supplied by the salt, the ionisation of the acid is suppressed, so the solution contains a large reserve of undissociated CH₃COOH and a large reserve of CH₃COO⁻ ions. Its pH is given by the Henderson equation:

$\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}$

1. When a small amount of acid (H⁺) is added: The added H⁺ ions are neutralised by the acetate ions already present in the solution:

$\mathrm{CH_{3}COO- + H+ \rightarrow CH_{3}COOH}$

The strong acid is thus converted into the weak acid, which is only slightly ionised. Hence the H⁺ concentration of the solution hardly increases and the pH remains almost unchanged.

2. When a small amount of base (OH⁻) is added: The added OH⁻ ions are neutralised by the undissociated acetic acid:

$\mathrm{CH_{3}COOH + OH- \rightarrow CH_{3}COO- + H_{2}O}$

The strong base is thus converted into the acetate ion (the conjugate base), which is a very weak base. Hence the OH⁻ concentration hardly increases and the pH again remains almost unchanged.

Thus, the buffer works by converting any added strong acid or strong base into a weak acid or weak base respectively, so that the pH is kept practically constant.

Hydrolysis of salt

Hydrolysis of a salt is the reaction of the salt (or more precisely, of its ions) with water, whereby the salt solution becomes acidic or basic. The ion of the weak acid or weak base present in the salt reacts with water to produce either H₃O⁺ or OH⁻ ions, so the solution is no longer neutral.

Salt of weak acid and strong base (e.g. CH₃COONa)

Let the weak acid be HA with dissociation constant Kₐ, and the salt be BA (B⁺ from the strong base, A⁻ from the weak acid). In water:

$\mathrm{BA \rightarrow B+ + A-} \qquad \text{(complete)}$

$\mathrm{A- + H_{2}O \rightleftharpoons HA + OH-} \qquad \text{(hydrolysis)}$

The cation B⁺ does not react with water (it is the conjugate of a strong base).

Derivation of hydrolysis constant, K_h

$K_h = \frac{[\mathrm{HA}][\mathrm{OH-}]}{[\mathrm{A-}]}$

Multiply numerator and denominator by [H⁺]:

$K_h = \frac{[\mathrm{HA}]}{[\mathrm{H+}][\mathrm{A-}]} \times [\mathrm{H+}][\mathrm{OH-}] = \frac{K_w}{K_a}$

So, $K_h = \frac{K_w}{K_a}$

Degree of hydrolysis, h

Let C be the concentration of the salt and h the degree of hydrolysis. At equilibrium:

$K_h = \frac{(Ch)(Ch)}{C(1-h)} = \frac{Ch^2}{1-h}$

For a weak acid–strong base salt, h is very small, so 1 − h ≈ 1:

$K_h = Ch^2 \quad\Rightarrow\quad h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a C}}$

Result

$K_h = \frac{K_w}{K_a}, \qquad h = \sqrt{\frac{K_w}{K_a C}}$

The solution is basic, and the pH is given by

$\mathrm{pOH} = \frac{1}{2}\left(\mathrm{p}K_w - \mathrm{p}K_a - \log C\right)$

Derivation of maximum work done during reversible expansion of an ideal gas

Consider $n$ moles of an ideal gas in a cylinder fitted with a frictionless, weightless piston. Let the gas expand from initial volume $V_1$ to final volume $V_2$ at constant temperature $T$.

Expression for work done

For an infinitesimal reversible expansion, let the external pressure be $P_{\text{ext}}$ and the piston move through a small distance $dl$. If $A$ is the area of the piston, the force is

$F = P_{\text{ext}} \cdot A$

The small work done is

$dW = F \cdot dl = P_{\text{ext}} \cdot A \cdot dl = P_{\text{ext}} \, dV$

where $dV = A\,dl$ is the small increase in volume.

For a reversible process, the external pressure differs from the gas pressure by an infinitesimal amount, so $P_{\text{ext}} \approx P_{\text{gas}} = P$. Hence

$dW = P \, dV$

For an ideal gas, $P = \dfrac{nRT}{V}$. Therefore

$dW = \frac{nRT}{V} \, dV$

Integrating from $V_1$ to $V_2$ at constant temperature $T$:

$W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \int_{V_1}^{V_2} \frac{dV}{V}$

$W = nRT \ln \frac{V_2}{V_1}$

Since $P_1 V_1 = P_2 V_2$ (Boyle's law at constant $T$), this can also be written as

$W = nRT \ln \frac{P_1}{P_2}$

Why this is the maximum work

In a reversible process the opposing pressure at every instant is only infinitesimally less than the gas pressure, so the gas works against the greatest possible opposing pressure throughout the expansion. Any irreversible (sudden) expansion works against a smaller constant external pressure, giving less work. Hence the work obtained in a reversible isothermal expansion is the maximum work obtainable from the given change of state.

Sign convention: Since the gas does work on the surroundings during expansion, $W$ is negative in the thermodynamic convention ($W = -nRT \ln(V_2/V_1)$); the magnitude of the maximum work is $nRT \ln(V_2/V_1)$.

Result

$\boxed{W_{\max} = nRT \ln \frac{V_2}{V_1} = nRT \ln \frac{P_1}{P_2}}$

Racemisation in Sₙ1 with Net Inversion — Justification

The basic picture

In an Sₙ1 reaction the leaving group departs first, giving a planar carbocation intermediate. A planar (sp²) carbon is attacked by the nucleophile equally well from either face, so a single enantiomer should give a 50 : 50 mixture of the two enantiomers — i.e. complete racemisation.

In practice the product is only partly racemised, and the inverted product usually predominates slightly. This is because the intermediate is not truly free and symmetric at the moment of attack. The reasons are:

1. Ion pair (intimate ion pair) mechanism

The leaving group does not move far away at once. It stays close to the carbocation as a tight ion pair:

$\text{R–X} \;\longrightarrow\; [\text{R}^+ \cdots \text{X}^-] \;\longrightarrow\; \text{R}^+ + \text{X}^-$

• In the intimate (tight) ion pair, the leaving group X⁻ shields the face of the carbocation from which it departed.
• The nucleophile therefore attacks mainly from the opposite (front) face, giving inversion.
• Only when the ion pair separates into free ions can the nucleophile attack both faces equally, giving racemisation.

So the product is a mixture of inverted and retained (racemised) forms, with inversion in excess.

2. Shielding by the leaving group and solvent

Even in the solvent-separated ion pair, the departing group and solvent molecules are not symmetrically arranged around the carbocation. The side vacated by the leaving group is partly blocked, so backside attack is favoured → net inversion.

3. Internal return

Sometimes the leaving group re-attacks the carbocation from the same face before diffusing away (internal return), regenerating the retained configuration. This retained fraction reduces the amount of racemisation and leaves an excess of inverted product.

4. Stability and lifetime of the carbocation

• A short-lived, unstable carbocation (e.g. primary, or in a poorly ionising solvent) is attacked before it can become symmetrically solvated → more inversion.
• A long-lived, well-stabilised carbocation (e.g. tertiary, benzylic, in a polar protic solvent) becomes fully solvated and symmetric → more racemisation.

Conclusion

Complete racemisation would require a free, planar, symmetrically solvated carbocation. Because the leaving group remains associated as an ion pair and shields one face, and because internal return can occur, the nucleophile attacks the two faces unequally. Hence Sₙ1 gives predominantly inversion with partial racemisation, not complete racemisation.

Chemoselective oxidation of primary alcohols to aldehydes

A primary alcohol can give either an aldehyde (partial oxidation) or a carboxylic acid (further oxidation). The problem is that the aldehyde formed is more easily oxidised than the alcohol itself, so with strong oxidants in aqueous medium it is oxidised straight through to the acid. To stop at the aldehyde, two conditions must be met:

1. Use a mild, selective oxidant that cannot oxidise the aldehyde (no water, no strong acid).
2. Remove the aldehyde as it is formed (by distillation) so it never stays in contact with the oxidant.

Reagents used

• PCC (Pyridinium chlorochromate) — ( \text{C}_5\text{H}_5\text{NH}^+\text{CrO}_3\text{Cl}^- ) in anhydrous ( \text{CH}_2\text{Cl}_2 ). Anhydrous conditions prevent formation of the hydrate ( \text{RCH(OH)}_2 ), which is the species actually oxidised to the acid. PCC therefore gives the aldehyde.
• PDC (Pyridinium dichromate) in anhydrous ( \text{CH}_2\text{Cl}_2 ) — same principle.
• Collins reagent (( \text{CrO}_3 \cdot 2\text{C}_5\text{H}_5\text{N} )) in ( \text{CH}_2\text{Cl}_2 ).
• Dess–Martin periodinane — a very mild, modern reagent; highly chemoselective for primary alcohols.
• Swern oxidation (( \text{DMSO}/(\text{COCl})_2 ), then ( \text{Et}_3\text{N} )) — anhydrous, mild, no over-oxidation.
• MnO₂ — selective for allylic and benzylic alcohols.

Example

$$\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}OH \rightarrow [PCC][CH_{2}Cl_{2}] CH_{3}CH_{2}CH_{2}CHO}$$

Butan-1-ol gives butanal. If ( \text{KMnO}_4/\text{H}^+ ) or ( \text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4 ) were used instead, butanoic acid ( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} ) would be obtained.

A second example with a double bond shows the chemoselectivity clearly:

$$\mathrm{HOCH_{2}CH=CHCH_{3} \rightarrow [PCC][CH_{2}Cl_{2}] OHCCH=CHCH_{3}}$$

The ( \text{C=C} ) bond is untouched — PCC does not attack alkenes, whereas hot alkaline ( \text{KMnO}_4 ) would destroy it.

Why PCC works

PCC is a mild ( \text{Cr(VI)} ) reagent in a non-aqueous solvent. It oxidises the alcohol to the aldehyde, but the aldehyde has no ( \text{–OH} ) group to form the chromate ester needed for further oxidation, and no water is present to form the gem-diol. Hence oxidation stops at the aldehyde stage.

Why cleavage of ether linkage by acids requires protonation first

An ether has the structure R–O–R′. The oxygen atom carries two lone pairs but no positive charge, and the C–O bond is not particularly weak. For a nucleophile such as the iodide ion (I⁻) to attack the carbon and displace the alkoxide, the carbon must first be made electrophilic enough. This is achieved only by protonation of the ether oxygen:

$\mathrm{R-O-R' + H+ \rightarrow [fast] R-\overset{+}{O}H-R'}$

Protonation converts the poor leaving group $\mathrm{RO-}$ (or $\mathrm{R'O-}$) into a neutral, much better leaving group — an alcohol molecule, $\mathrm{ROH}$. The positively charged oxonium ion strongly withdraws electron density from the adjacent carbons, so the C–O bond becomes weak and polarised. Only now can the halide ion $\mathrm{I-}$ carry out an $\mathrm{S_{N}2}$ attack on the less hindered carbon, expelling neutral alcohol. Without protonation, the strongly basic alkoxide ion cannot leave and the reaction does not proceed under mild acid conditions. Hence acid cleavage of an ether is possible only after protonation.

Reaction of methoxyethane with cold HI

Methoxyethane is an unsymmetrical ether, $\mathrm{CH_{3}-O-CH_{2}CH_{3}}$. In cold HI, protonation occurs on oxygen, and the iodide ion attacks the less hindered carbon (the methyl carbon), so the product is methyl iodide and ethanol:

$\mathrm{CH_{3}-O-CH_{2}CH_{3} + HI (cold) \rightarrow CH_{3}I + CH_{3}CH_{2}OH}$

Mechanism:

1. $\mathrm{CH_{3}-O-CH_{2}CH_{3} + H+ \rightarrow CH_{3}-\overset{+}{O}H-CH_{2}CH_{3}}$ (fast protonation)
2. $\mathrm{I- + CH_{3}-\overset{+}{O}H-CH_{2}CH_{3} \rightarrow CH_{3}I + HOCH_{2}CH_{3}}$ ($\mathrm{S_{N}2}$ attack at $\mathrm{CH_{3}}$)

The ethanol formed can react with a further molecule of HI to give ethyl iodide and water, but with cold HI the reaction stops mainly at the alcohol stage. This confirms that the ether linkage is broken only because the oxygen was protonated first, making the C–O bond cleavable.

Product formed: Propan-1-ol ($CH_3CH_2CH_2OH$)

This is the hydroboration–oxidation reaction (Brown hydroboration). Propene reacts with diborane ($B_2H_6$, which acts as $BH_3$) to give tripropylborane, which on oxidation with alkaline $H_2O_2$ gives propan-1-ol.

$CH_3-CH=CH_2 \xrightarrow{B_2H_6} (CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2/OH^-} CH_3CH_2CH_2OH$

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Orientation (anti-Markovnikov addition)

The net addition of H and OH follows anti-Markovnikov orientation: OH ends up on the less substituted carbon (C-1) and H on the more substituted carbon (C-2). This is opposite to acid-catalysed hydration, which gives propan-2-ol.

Reasons:

• Steric effect (main reason): the bulky $BH_2$ group attaches to the less hindered terminal carbon C-1.
• Electronic effect: boron is electron-deficient (empty p-orbital) and acts as the electrophile. In the transition state, the partial positive charge develops on the more substituted carbon C-2, which is better able to stabilise it, while the hydrogen (hydride character) moves to C-2. Boron therefore goes to C-1.

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Mechanism

Step 1: Addition of $BH_3$ (concerted, syn). The empty p-orbital of boron accepts the $\pi$ electrons of propene and, through a four-centre transition state, B bonds to C-1 and H to C-2 on the same face of the double bond.

$CH_3-CH=CH_2 + BH_3 \rightarrow CH_3CH_2CH_2-BH_2$

The process repeats with two more propene molecules to give the trialkylborane:

$3\,CH_3CH=CH_2 + BH_3 \rightarrow (CH_3CH_2CH_2)_3B$

Step 2: Oxidation. In alkaline medium $H_2O_2$ gives the hydroperoxide ion $HOO^-$, which attacks boron. An alkyl group then migrates from boron to oxygen with loss of $OH^-$, with retention of configuration. Repetition for the other two alkyl groups and hydrolysis give the alcohol.

$(CH_3CH_2CH_2)_3B + 3H_2O_2 + 3OH^- \rightarrow 3\,CH_3CH_2CH_2OH + B(OH)_3 + 3H_2O$

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Conclusion: Propene on hydroboration–oxidation gives propan-1-ol by anti-Markovnikov, syn addition.

Stereochemistry of reduction of alkynes to alkenes

An alkyne can be reduced to an alkene by two methods, and each gives a different stereoisomer, because the two hydrogen atoms add either to the same face (syn) or to opposite faces (anti) of the triple bond.

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1. Catalytic hydrogenation: syn addition giving the cis-alkene

Reagents: $H_2$ with a poisoned catalyst, such as Lindlar's catalyst (Pd on $CaCO_3$ poisoned with quinoline) or P-2 nickel ($Ni_2B$).

Mechanism: The alkyne is adsorbed flat on the catalyst surface and both hydrogen atoms are delivered from the same side. The poisoned catalyst is too weak to reduce the alkene further, so the reaction stops at the alkene.

Result: the cis (Z)-alkene.

$R-C\equiv C-R' \xrightarrow{H_2,\ \text{Lindlar}} \underset{cis}{R-CH=CH-R'}$

Example: but-2-yne gives cis-but-2-ene.

$CH_3-C\equiv C-CH_3 \xrightarrow{H_2/\text{Lindlar}} cis\text{-}CH_3CH=CHCH_3$

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2. Dissolving-metal reduction: anti addition giving the trans-alkene

Reagents: sodium or lithium in liquid ammonia ($Na/NH_3(l)$).

Mechanism:

1. An electron from sodium is transferred to the alkyne, forming a radical anion.
2. The radical anion takes a proton from $NH_3$, giving a vinyl radical.
3. A second electron is added, giving a vinyl anion. The vinyl radical and anion prefer the trans arrangement, in which the bulky groups R and R' are far apart and repulsion is least.
4. The vinyl anion takes a second proton from $NH_3$, giving the alkene.

Result: the trans (E)-alkene.

$R-C\equiv C-R' \xrightarrow{Na/NH_3(l)} \underset{trans}{R-CH=CH-R'}$

Example:

$CH_3-C\equiv C-CH_3 \xrightarrow{Na/NH_3(l)} trans\text{-}CH_3CH=CHCH_3$

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Summary

By choosing the reagent, either the cis or the trans alkene can be prepared from the same alkyne.

Part A — Physics

1. Pressure of an ideal gas from kinetic theory

Consider a cubical vessel of side (L), volume (V=L^3), containing (N) ideal gas molecules, each of mass (m). Let one molecule have velocity components (u,v,w) along the (x,y,z) axes.

For the face perpendicular to the (x)-axis:

• Momentum before collision (=mu)
• Momentum after collision (=-mu)
• Change in momentum of molecule (=-2mu)
• Momentum transferred to the wall (=2mu)

The molecule travels between two successive collisions with the same face through distance (2L), so time taken is [ t=\frac{2L}{u} ] Force exerted by one molecule on the face: [ F_1=\frac{2mu}{2L/u}=\frac{mu^2}{L} ] Total force due to all (N) molecules: [ F=\frac{m}{L}\sum_{i=1}^{N}u_i^2 ] Pressure is force per unit area: [ P=\frac{F}{L^2}=\frac{m}{L^3}\sum u_i^2=\frac{m}{V}\sum u_i^2 ] For any molecule, [ c_i^2=u_i^2+v_i^2+w_i^2 ] Since molecular motion is random, [ \overline{u^2}=\overline{v^2}=\overline{w^2}=\frac{\overline{c^2}}{3} ] Hence, [ \sum u_i^2=\frac{N\overline{c^2}}{3} ] Therefore, [ P=\frac{1}{3}\frac{Nm}{V}\overline{c^2} ] or [ PV=\frac{1}{3}Nm\overline{c^2} ]

2. Kinetic energy of one mole of an ideal gas

Total translational kinetic energy of all molecules is [ E=N\left(\frac{1}{2}m\overline{c^2}\right) ] From the pressure expression, [ PV=\frac{1}{3}Nm\overline{c^2} ] But [ Nm\overline{c^2}=2E ] So, [ PV=\frac{2}{3}E ] For one mole of an ideal gas, [ PV=RT ] Thus, [ RT=\frac{2}{3}E ] [ E=\frac{3}{2}RT ] Hence, the kinetic energy of the molecules in one mole of an ideal gas is [ \boxed{E=\frac{3}{2}RT} ]

3. Root mean square velocity of oxygen at (27^\circ\text{C})

Given: [ T=27+273=300\ \text{K} ] For oxygen, (O_2): [ M=32\ \text{g mol}^{-1}=32\times 10^{-3}\ \text{kg mol}^{-1} ] [ R=8.314\ \text{J mol}^{-1}\text{K}^{-1} ]

Root mean square velocity: [ v_{\text{rms}}=\sqrt{\frac{3RT}{M}} ] Substitute: [ v_{\text{rms}}=\sqrt{\frac{3\times 8.314\times 300}{32\times 10^{-3}}} ] [ v_{\text{rms}}=\sqrt{\frac{7482.6}{0.032}} ] [ v_{\text{rms}}=\sqrt{233831.25} ] [ \boxed{v_{\text{rms}}\approx 483.6\ \text{m s}^{-1}} ]

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OR — Chemistry

1. Integrated rate law for (2A\rightarrow B)

Given: [ -\frac{d[A]}{dt}=K[A]^2 ] Separate the variables: [ \frac{d[A]}{[A]^2}=-K,dt ] Integrate between (t=0), ([A]=[A]0) and time (t), ([A]=[A]): [ \int{[A]0}^{[A]}\frac{d[A]}{[A]^2}=-K\int_0^t dt ] [ \left[-\frac{1}{[A]}\right]{[A]_0}^{[A]}=-Kt ] [ -\frac{1}{[A]}+\frac{1}{[A]_0}=-Kt ] [ \frac{1}{[A]}-\frac{1}{[A]_0}=Kt ] Therefore, [ \boxed{K=\frac{1}{t}\left(\frac{1}{[A]}-\frac{1}{[A]_0}\right)} ] or [ \boxed{\frac{1}{[A]}=Kt+\frac{1}{[A]_0}} ]

2. Graphical determination of (K)

Plot (\frac{1}{[A]}) against time (t). The equation [ \frac{1}{[A]}=Kt+\frac{1}{[A]_0} ] is of the form (y=mx+c). Hence the graph is a straight line.

• Slope (=K)
• Intercept (=\frac{1}{[A]_0})

So the rate constant (K) is obtained from the slope of the straight-line graph.

3. Rate constant at (323\ \text{K})

Given: [ K_1=1.2\times 10^{-3}\ \text{s}^{-1} ] [ T_1=303\ \text{K},\qquad T_2=323\ \text{K} ] [ E_a=54.2\ \text{kJ mol}^{-1}=54200\ \text{J mol}^{-1} ] [ R=8.314\ \text{J mol}^{-1}\text{K}^{-1} ]

Using the Arrhenius equation: [ \ln\frac{K_2}{K_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) ] [ \ln\frac{K_2}{K_1}=\frac{54200}{8.314}\left(\frac{1}{303}-\frac{1}{323}\right) ] [ \ln\frac{K_2}{K_1}=6519.1\left(\frac{323-303}{303\times 323}\right) ] [ \ln\frac{K_2}{K_1}=6519.1\left(\frac{20}{97869}\right) ] [ \ln\frac{K_2}{K_1}=1.3322 ] [ \frac{K_2}{K_1}=e^{1.3322}=3.789 ] [ K_2=1.2\times 10^{-3}\times 3.789 ] [ \boxed{K_2\approx 4.55\times 10^{-3}\ \text{s}^{-1}} ]

So the rate constant at (323\ \text{K}) is approximately [ \boxed{4.55\times 10^{-3}\ \text{s}^{-1}} ]

Miller indices are the three smallest integers (h, k, l) which represent the orientation of a crystal plane. They are proportional to the reciprocals of the intercepts made by the plane on the three crystallographic axes (a, b, c), and are written as ((hkl)).

Given:
Intercepts of the plane are (2a, 3b, 2c), i.e. along the axes the intercepts are (2, 3, 2).

Steps:

1. Write intercepts in terms of axial units: [ p=2,\quad q=3,\quad r=2 ]

2. Take reciprocals: [ \frac{1}{p}:\frac{1}{q}:\frac{1}{r} =\frac{1}{2}:\frac{1}{3}:\frac{1}{2} ]

3. Multiply by the LCM of denominators, i.e. (6): [ \frac{1}{2}\times 6 : \frac{1}{3}\times 6 : \frac{1}{2}\times 6 =3:2:3 ]

4. Reduce to smallest integers: [ h:k:l=3:2:3 ]

Hence, the Miller indices of the given plane are:

[ \boxed{(323)} ]

Raoult's Law

Statement: At a given temperature, the partial vapour pressure of each volatile component of a solution is directly proportional to its mole fraction in the solution.

For a binary solution of two volatile liquids A and B,

$p_A \propto x_A \quad\Rightarrow\quad p_A = p_A^\circ x_A$ $p_B \propto x_B \quad\Rightarrow\quad p_B = p_B^\circ x_B$

where $p_A^\circ, p_B^\circ$ are the vapour pressures of the pure components and $x_A, x_B$ their mole fractions. Total pressure $p = p_A + p_B$.

For a solution of a non-volatile solute in a volatile solvent, the vapour pressure is due only to the solvent, and Raoult's law gives

$p = p^\circ x_1 = p^\circ\frac{n_1}{n_1+n_2}$

so the relative lowering of vapour pressure equals the mole fraction of the solute:

$\frac{p^\circ - p}{p^\circ} = x_2 = \frac{n_2}{n_1+n_2}$

Determination of Molecular Mass of Solute

Since the solution is dilute, $n_2 \ll n_1$, so $n_1 + n_2 \approx n_1$:

$\frac{p^\circ - p}{p^\circ} = \frac{n_2}{n_1} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_2 M_1}{M_2 w_1}$

Rearranging,

$\boxed{M_2 = \frac{w_2 \, M_1 \, p^\circ}{w_1 \,(p^\circ - p)}}$

where $w_2$ = mass of solute, $w_1$ = mass of solvent, $M_1$ = molecular mass of solvent, $M_2$ = molecular mass of solute, $p^\circ$ = vapour pressure of pure solvent and $p$ = vapour pressure of the solution.

Procedure: The lowering of vapour pressure $(p^\circ - p)$ is measured experimentally — by the Ostwald–Walker dynamic method (air is bubbled through the solvent and then through the solution; the loss in mass of the two vessels gives the ratio $p^\circ/p$) or by the static method using a manometer. Knowing $w_1, w_2, M_1$ and the ratio $p^\circ/(p^\circ - p)$, $M_2$ is calculated from the above relation.

Note: This is a colligative property, so the value obtained is the apparent molecular mass; it agrees with the true value only for dilute solutions of non-dissociating, non-associating solutes.

Calorific value of a fuel is the amount of heat energy liberated when a unit mass (or unit volume, for gases) of the fuel is completely burnt in excess oxygen. It is usually expressed in $\text{kJ g}^{-1}$ or $\text{kJ kg}^{-1}$. A fuel with a higher calorific value gives more heat per unit mass and is therefore more efficient.

Given:

• Enthalpy of combustion of methane, $\Delta H_c = -890.3\ \text{kJ mol}^{-1}$
• Enthalpy of combustion of ethane, $\Delta H_c = -1560\ \text{kJ mol}^{-1}$

Calorific value is calculated as:

$$\text{Calorific value} = \frac{|\Delta H_c|}{\text{Molar mass}}$$

1. Methane, $\text{CH}_4$

Molar mass of $\text{CH}_4 = 12 + 4(1) = 16\ \text{g mol}^{-1}$

$$\text{Calorific value} = \frac{890.3}{16} = 55.64\ \text{kJ g}^{-1} = 55{,}640\ \text{kJ kg}^{-1}$$

2. Ethane, $\text{C}_2\text{H}_6$

Molar mass of $\text{C}_2\text{H}_6 = 2(12) + 6(1) = 30\ \text{g mol}^{-1}$

$$\text{Calorific value} = \frac{1560}{30} = 52.00\ \text{kJ g}^{-1} = 52{,}000\ \text{kJ kg}^{-1}$$

Conclusion:
Methane has a calorific value of $55.64\ \text{kJ g}^{-1}$, while ethane has $52.00\ \text{kJ g}^{-1}$. Therefore, methane has better fuel efficiency because it produces more heat per gram on complete combustion.

Given:

• Bond enthalpy of H–H = 436 kJ mol⁻¹
• Bond enthalpy of N–N (actually N≡N) = 941 kJ mol⁻¹
• Bond enthalpy of N–H = 391 kJ mol⁻¹

Formation reaction of ammonia:

$$\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightarrow NH_3$$

Bonds broken:

$$\frac{1}{2}(N \equiv N) + \frac{3}{2}(H-H) = \frac{1}{2}(941) + \frac{3}{2}(436) = 470.5 + 654 = 1124.5 \text{ kJ}$$

Bonds formed:

$$3(N-H) = 3 \times 391 = 1173 \text{ kJ}$$

Enthalpy of formation:

$$\Delta H_f = \text{Energy absorbed} - \text{Energy released}$$ $$\Delta H_f = 1124.5 - 1173 = -48.5 \text{ kJ mol}^{-1}$$

Therefore, the enthalpy of formation of ammonia is:

$$\boxed{\Delta H_f(NH_3) = -48.5 \text{ kJ mol}^{-1}}$$

Given: [ M_{H_2}=2\ \text{g/mol},\quad M_{I_2}=254\ \text{g/mol},\quad M_{HI}=128\ \text{g/mol} ]

Initial moles: [ n_{H_2}=\frac{10}{2}=5\ \text{mol} ] [ n_{I_2}=\frac{760}{254}\approx 3\ \text{mol} ]

Let (x) mol of (I_2) react. Since (H_2) is in excess, (I_2) is the limiting reactant.

At equilibrium: [ H_2=5-x,\quad I_2=3-x,\quad HI=2x ]

For the reaction, [ H_2+I_2\rightleftharpoons 2HI ] [ K_c=\frac{[HI]^2}{[H_2][I_2]}=32 ]

Since volume cancels, [ 32=\frac{(2x)^2}{(5-x)(3-x)} ] [ 4x^2=32(5-x)(3-x) ] [ 4x^2=32(15-8x+x^2) ] [ 4x^2=480-256x+32x^2 ] [ 28x^2-256x+480=0 ] [ 7x^2-64x+120=0 ]

Solving, [ x=\frac{64-\sqrt{64^2-4(7)(120)}}{2(7)} ] [ x=\frac{64-\sqrt{736}}{14} ] [ x=\frac{64-27.13}{14}=2.633\ \text{mol} ]

Moles of (HI) formed: [ n_{HI}=2x=2(2.633)=5.266\ \text{mol} ]

Mass of (HI) formed: [ m_{HI}=5.266\times 128=674.0\ \text{g} ]

[ \boxed{\text{Mass of }HI\text{ formed}\approx 674\ \text{g}} ]

Using exact (n_{I_2}=760/254=2.992) mol gives about (673\ \text{g}); the usual textbook approximation (I_2\approx 3) mol gives (674\ \text{g}).

Buffer capacity: It is the measure of the ability of a buffer solution to resist a change in pH when a small amount of strong acid or base is added. Quantitatively, it is the number of moles of strong acid or base required to change the pH of 1 litre of buffer by one unit. It is maximum when ([Salt]/[Acid]=1), i.e. when (pH=pK_a).

Buffer range: It is the pH range over which a buffer solution can effectively resist pH change. Practically, a buffer is effective within (pK_a \pm 1), i.e. from (pK_a-1) to (pK_a+1).

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Calculation of pH

Given:

[ CH_3COOH:\quad N/2,\ V=200\ \text{mL} ]

[ NaOH:\quad N/4,\ V=200\ \text{mL} ]

For (CH_3COOH), since it is monobasic,

[ \text{Moles of } CH_3COOH = \frac{1}{2}\times \frac{200}{1000}=0.1\ \text{mol} ]

For (NaOH),

[ \text{Moles of } NaOH = \frac{1}{4}\times \frac{200}{1000}=0.05\ \text{mol} ]

Reaction:

[ CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O ]

NaOH is limiting, so it neutralizes (0.05) mol of acetic acid.

Remaining (CH_3COOH = 0.1-0.05=0.05) mol

Acetate formed, (CH_3COONa = 0.05) mol

Total volume (=200+200=400) mL (=0.4) L

[ [CH_3COOH]=\frac{0.05}{0.4}=0.125\ \text{M} ]

[ [CH_3COONa]=\frac{0.05}{0.4}=0.125\ \text{M} ]

Using Henderson’s equation:

[ pH=pK_a+\log\frac{[Salt]}{[Acid]} ]

[ pK_a=-\log K_a=-\log(1.74\times10^{-5})=4.76 ]

Since ([Salt]=[Acid]),

[ pH=4.76+\log 1=4.76 ]

[ \boxed{pH=4.76} ]

Effect of temperature, pressure and concentration with reference to the law of mass action

The law of mass action states that the rate of a chemical reaction is proportional to the product of the molar concentrations of the reactants, each raised to a power equal to its stoichiometric coefficient. For a reversible reaction

$aA + bB \rightleftharpoons cC + dD$

the equilibrium constant is

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

1. Effect of concentration At constant temperature, $K_c$ is constant. If the concentration of a reactant is increased, the value of the reaction quotient $Q$ becomes less than $K_c$, so the forward reaction is favoured until $Q = K_c$ again. If a product is added, $Q > K_c$ and the backward reaction is favoured. Thus a change in concentration shifts the equilibrium position but does not change the value of $K_c$.

2. Effect of pressure Pressure affects only gaseous equilibria, and only when the number of gaseous moles differs on the two sides. On increasing pressure, the system shifts to the side having fewer moles of gas (Le Chatelier's principle), because that reduces the volume and relieves the stress. For example, in

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

increasing pressure favours the forward reaction ($\Delta n = 2 - 4 = -2$). If $\Delta n = 0$, pressure has no effect on the equilibrium position. The value of $K_c$ itself remains unchanged; only $K_p$ and $K_x$ relations change.

3. Effect of temperature Temperature is the only factor that changes the value of the equilibrium constant. For an exothermic forward reaction, increasing temperature decreases $K_c$ (backward reaction favoured); for an endothermic forward reaction, increasing temperature increases $K_c$ (forward reaction favoured). This follows from the van't Hoff equation:

$\frac{d \ln K}{dT} = \frac{\Delta H}{RT^2}$

Conclusion: Concentration and pressure change only the equilibrium position, while temperature changes the equilibrium constant itself.

Collision frequency (Z): The average number of collisions suffered by a single molecule per unit time. For a gas of $N$ molecules in volume $V$ with average speed $\bar{c}$ and collision cross-section $\sigma = \pi d^2$,

$Z = \sqrt{2}\,\sigma\,\bar{c}\,\frac{N}{V} = \sqrt{2}\,\pi d^2 \bar{c}\,\frac{N}{V}$

The factor $\sqrt{2}$ appears because, in the frame of a given molecule, the other molecules are also moving, so the relative speed is $\sqrt{2}\,\bar{c}$.

Collision diameter (d): The distance between the centres of two molecules at the instant of their closest approach in a collision. It is treated as the effective diameter of the rigid sphere representing the molecule. The collision cross-section is $\sigma = \pi d^2$. For a typical gas, $d$ is of the order of $10^{-10}\ \text{m}$.

Mean free path ($\lambda$): The average distance travelled by a molecule between two successive collisions. Since a molecule travels $\bar{c}$ per second and suffers $Z$ collisions in that time,

$\lambda = \frac{\bar{c}}{Z} = \frac{1}{\sqrt{2}\,\pi d^2}\cdot\frac{V}{N} = \frac{k_B T}{\sqrt{2}\,\pi d^2 p}$

using $\dfrac{N}{V} = \dfrac{p}{k_B T}$. Thus $\lambda$ is inversely proportional to pressure and to the square of the collision diameter, and directly proportional to temperature. At STP it is of the order of $10^{-7}\ \text{m}$, about a thousand times the molecular diameter.

Surface Tension

Definition: Surface tension is the property of a liquid surface by virtue of which it behaves like a stretched elastic membrane and tends to contract to the minimum possible area. Numerically, it is the force per unit length acting perpendicular to an imaginary line drawn on the free surface of the liquid, in the plane of the surface.

$\gamma = \frac{F}{l} \quad \text{(unit: N m}^{-1}\text{, dimension: } MT^{-2})$

Determination by Drop Weight Method

Principle: When a liquid is allowed to escape slowly through a capillary tube of radius $r$, a drop is formed at the tip. The drop grows until its weight just balances the upward force due to surface tension acting along the circumference of the tip. At the instant of detachment,

$W = mg = 2\pi r\,\gamma$

so that

$\gamma = \frac{mg}{2\pi r}$

Since the radius $r$ of the tip is difficult to measure accurately, the method is used as a comparison method with a reference liquid (usually water) using the same tube.

Apparatus: A stalagmometer (a capillary tube with a bulb and a flattened tip), a specific gravity bottle, a weighing balance, and the two liquids.

Procedure:

1. The stalagmity is cleaned and filled with the experimental liquid up to a fixed mark, and the liquid is allowed to fall drop by drop.
2. The drops collected in a weighed specific gravity bottle are counted as $n_1$ for a fixed volume $V$ of the liquid.
3. The experiment is repeated with water (reference liquid), giving $n_2$ drops for the same volume $V$.
4. The densities $\rho_1$ (liquid) and $\rho_2$ (water) are measured with the specific gravity bottle.

Calculation: For the same volume $V$,

$W_1 = \frac{V\rho_1 g}{n_1}, \qquad W_2 = \frac{V\rho_2 g}{n_2}$

Since the same tube is used, $\gamma \propto W$, hence

$\frac{\gamma_1}{\gamma_2} = \frac{W_1}{W_2} = \frac{\rho_1 n_2}{\rho_2 n_1}$

$\boxed{\gamma_1 = \gamma_2 \cdot \frac{\rho_1}{\rho_2} \cdot \frac{n_2}{n_1}}$

Taking $\gamma_2$ for water as known ($0.072$ N m$^{-1}$ at $20^\circ$C), $\gamma_1$ is obtained.

Note: In practice only a part of the drop (about $f$ times) actually falls, so a correction factor $f$ is introduced (Tate's law): $mg = 2\pi r\gamma f$. In the comparison method this factor cancels out, which is why the drop weight method is used as a relative method.