(ii) 

(i) The rate will become 4 times.
(ii) The overall order is equal to 2 if  'A' is in large excess.
(b) t_1/2 = 23.1 min
\(k=\frac{2.303}{t_{1 / 2}} \log \frac{[R]_{0}}{[R]_{0} / 2}\)
\(\Rightarrow \ t_{1 / 2}=\frac{2.303}{k} \log 2\)
\(\Rightarrow \quad k=\frac{2.303}{23.1} \times 0.3010 \mathrm{~min}^{-1}\)
Also \(t_{75 \%}=\frac{2.303}{k} \log \frac{[R]_{0}}{[R]_{0} / 4}\)
\(=\frac{2.303 \times 23.1}{2.303 \times 0.3010} \log 4\)
\(\Rightarrow \ t_{75 \%}=\frac{23.1}{0.3010} \times 0.6021=46.2 \mathrm{~min}\) 

(i) As the given compound with molecular formula C₉H₁₀O, forms a 2,4-DNP derivative and reduces Tollen's reagent, thus it must be an aldehyde.
(ii) As it undergoes Cannizzaro reaction, hence -CHO group is directly attached to the benzene ring.
(iii) On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Therefore, it must be an ortho-substituted benzaldehyde and the only o-substituted aromatic aldehyde which have C₉H₁₀O molecular formula is o-ethyl benzaldehyde.
Reactions involved

5.4\(\times\)10^-3mol L^-1s^-1

\(k={2.303\over t}log{[R]_0\over [R]}\)
\(={2.303\over 100}log{[R]_0\over {40\over 100}[R]_0}\)
[60% is complete, 40% is left]
\(k={2.303\over 100}(log5-log2)\)
\(k={2.303\over 100}(0.6990-0.3010)\)
\(k={2.303\over 100}\times0.3980\ min^{-1}\)
\(t_{90./.}={2.303\over k}log{[R]_0\over [R]}\)
[90% is complete, 10% is left]
\(t_{90./.}={2.303\over k}log{[R]_0\over {10\over 100}[R]_0}\)
\(={2.303\times100\over 2.303\times0.3980}log10\)
\(={100\over 0.3980}=251.26min\)

For first order reaction, \(t=\frac { 2.303 }{ k } \log { \frac { a }{ a-x } } \)
99% completion means that x = 99% of a = 0.99 a
\({ t }_{ 99 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 }^{ 2 } } =2\times \frac { 2.303 }{ k } \)
90% completion means that x = 90% of a = 0.90 a
\(\therefore \quad { t }_{ 90 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 } } =2\times \frac { 2.303 }{ k } \)
\(\therefore \quad \frac { { t }_{ 99 } }{ { t }_{ 90 } } ={ \left( \frac { 2\times 2.303 }{ k } \right) }/{ \left( \frac { 2.303 }{ k } \right) }=2\quad \)\({ t }_{ 99 }\%\)\(=2\times { t }_{ 90 }\%\)

(i) Order of reaction is zero
(ii) Slope = - k
(iii) mol L^-1 s^-1

PCC(Pyridinium chlorochromate)

Propanal and 2-methylpropanal both have a-hydrogens and hence can undergo cross aldol condensation in which each one of them can act either an electrophile or a nucleophile thereby giving two cross aldol products. Thus, 

Besides, cross aldol condensation each aldehyde can also undergo selfot simple aldol condensation giving two self or simple aldol products.

\(-\frac{d\left[N_2\right]}{d t}=\frac{1}{3} \frac{d\left[N_2\right]}{d t}=+\frac{1}{2} \frac{d\left[N H_3\right]}{d t}\)

HCOOH < CICH₂COOH < CCl₃COOH < CF₃COOH.

\((i) \ Rate \ =k{ \left[ A \right] }^{ 1 }{ \left[ B \right] }^{ 2 }\)
\(Rate \ =k{ \left[ A \right] }{ \left[ 3B \right] }^{ 2 }\)
\(Rate \ =9k{ \left[ A \right] }{ \left[ B \right] }^{ 2 }\)
If concentration of 'B' alone increases three times, the rate will increase nine times.
(ii) If concentration of 'A' as well as 'B' is doubled, rate \(=k{ \left[ 2A \right] }^{ 1 }{ \left[ 2B \right] }^{ 2 }\), the rate will increase eight times.
So rate \(=k{ \left[ 2A \right] }^{ 1 }{ \left[ 2B \right] }^{ 2 }\).

According to given information, the organic compound (A) with molecular formula C₉H₁₀O form 2-4 DNP derivative and also reduces Fehling solution. Hence, the compound (A) is an aldehyde.
Also, the compound (A) undergo Cannizzaro reaction hence, it does not contains an \(\alpha\)-H-atom. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. That means, -CHO group is directly attached to benzene ring and the compound is ortho-disubstituted benzene.
(a) Therefore according to above information the compound (A) is 2-ethylbenzaldehyde.

For the reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow \mathrm{4NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)
overall rate of reaction is
\( -\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \)
\(=+\frac{d\left[\mathrm{O}_{2}\right]}{d t} \)
Given,
\(\frac{d\left[\mathrm{NO}_{2}\right]}{d t}=2.8 \times 10^{-3} \mathrm{Ms}^{-1} \)
\(-\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=? \)
\(-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \)
Putting the given values in the above equation. we get
\(\frac{-d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{1}{4} \times 2 \times 2.8 \times 10^{-3} \mathrm{Ms}^{-1}\)
\(= 1.4 \times 10^{-3} \mathrm{Ms}^{-1} \)
(b)
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right)\)
\(\log { 4 } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 1 }{ 300 } -\frac { 1 }{ 320 } \right) \)
\({ E }_{ a }=\frac { 19.147\times 0.6021\times 300\times 320 }{ 20 }\)
\(\)= 55.336 KJ mol^-1

(i) Propanal and propanone :
Propanal gives silver mirror test with Tollen's reagent. Propanone does not gives this test.
Propanal gives red ppt with Fehlings solution but propanone does not give this test.
(ii) Acetophenone and Benzophenone :
Acetophenone gives yellow ppt of idoform with alkaline solution of iodine (iodoform test) and Benzophenone doe not give this test.
(iii) Phenol and Benzoic acid :
Phenol does not react with sodium hydrogen carbonate and benzoic acid gives effervescence of carbon dioxide gas upon reaction with sodium hydrogen carbonate.
Phenol gives violet colour with ferric chloride solution but benzoic acid does not give such colour.
(iv) Benzoic acid and Ethyl benzoate :
 Ethyl benzoate does not react with sodium hydrogen carbonate and benzoic acid gives effervescence of carbon dioxide gas upon reaction with sodium hydrogen carbonate.
When ethyl benzoate is boiled with excess NaOH, it gives ethanol. Ethanol on heating with iodine gives yellow ppt of iodoform. This test is not given by benzoic acid.
(v) Pentan-2-one and Pentan-3-one :
Pentan-2-one gives iodoform test and Pentan-3-one  does not gives iodoform test
Pentan-2-one gives white ppt with sodium bisulphite whereas pentan-3-one  does not give ppt.
(vi) Benzaldehyde and Acetophenone :
Benzaldehyde forms silver mirror with ammoniacal silver nitrate solution (Tollen\s reagent) and Acetophenone does not give this test.
Acetophenone gives iodoform test. Benzaldehyde does not react.
(vii) Ethanal and Propanal :
Ethanal gives iodoform test but propanal does not give such test.

(i)

(ii) \((CH_{ 3 })_{ 2 }C=CHCOCH_{ 3 }\)
(b) (i) On adding sodium bicarbonate to benzoic acid, brisk effervescence of CO₂ is evolved. Whereas ethylbenzoate does not. 1
(ii) Acetophenone having at least one group on heating with alkaline solution of iodine forms yellow colored precipitate of iodoform.

Whereas benzaldehyde does not 1
(iii) Benzoic acid reacts with NaHC03 giving CO2 gas with effervescence whereas phenol does not.

 \({ t }_{ { 1 }/{ 2 } }=5730 \ years\)
\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
\(k=\frac { 0.693 }{ 5730 } { years }^{ -1 }\)
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ N \right] }_{ 0 } }{ \left[ N \right] } }\)
\(t=\frac { 2.303 }{ k } \log { \frac { 100 }{ 80 } }\)
\(=\frac { 2.303 }{ 0.693 } \times 5730\left[ \log { 5 } -\log { 4 } \right] \)
\(=\frac { 5730 }{ 0.3010 } \times \left[ 0.6990-0.6021 \right] \)
\(=\frac { 5730 }{ 0.3010 } \times 0.0969=\frac { 55.237 }{ 0.3010 } \)
\(=1844.64 \ years\approx 1845 \ years\)

(a) (i) Order of a reaction. The sum of the exponents (powers) of the concentration of reactants in the rate law is termed as order of the reaction. It can be in fraction. It can be zero also.
(ii) Molecularity: Total number of atoms, ions or molecules of the reactants involved in the reaction is termed as its molecularity. It is always in whole number. It is never more than three. It cannot be zero.
(b)
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right)\)
\(\log { 4 } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 1 }{ 300 } -\frac { 1 }{ 320 } \right) \)
\({ E }_{ a }=\frac { 19.147\times 0.6021\times 300\times 320 }{ 20 }\)
\(=55.336\quad kJ\ { mol }^{ -1 }\)

Methyl ketones are characterized by iodoform test

acetaldehyde contains \(\alpha \)- hydrogens and hence does not undergo aldol condensation

(a) Rate = \(\frac {dx}{dt} = k[A_o]^{o} = k \) or \(k = \frac {dx}{dt} = \frac {conc}{Time} =\frac {mol L^{-1}}{s}\)mol L^-1 s^-1.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(c) Assertion is correct but reason is incorrect. For effective collision orientation during collision is irrespective.

(b): Aldehydes have higher molecular weight than parent alkanes as well as polarity in aldehydes shows higher boiling point than parent alkanes. Aldehydes do not have any hydrogen atom attached directly to the oxygen so they cannot form hydrogen bond with each other.

(d) Only reactions with zero order do not have the order in fractions. Thus, (A) is incorrect but (R) is correct.

(a) 

(b) 4-hydroxy- 3-methoxy benzaldehyde
(c) 

(d)

(e)
​​​​​​
Acetic anhydride is formed.

(a) Zero order reaction.
(b) \(\frac{k}{2.303}\) where 'k' is rate constant.
(c) t _1/2 is directly proportional to initial concentration.
(d)  t _{99. 9 %}= 10 t _1/2 = 10 x 40 = 400 minutes
(e) \(t_{1 / 2}=\frac{[\mathrm{R}]_{0}}{2 k}\) for zero order reaction.