$t _{\dfrac{3}{4}} = 2t _{\dfrac{1}{2}} = 32min.$
$t _{\dfrac{1}{2}} = 16min.$

Base hydrolysis of ethyl acetate is a second order reaction.
The other reactions are first order reactions.

$CH _{3}COOC _{2}H _{5}+H _{2}O\rightarrow CH _{3}COOH+C _{2}H _{5}OH$
$ rate=K[CH _{3}COO\ C _{2}H _{5}]^{1}[H _{2}O]^{0}$
thus, it is first order & bi molecular reaction.

It is an example of pseudo first order reaction because here $H _{2}O$ is in excess. Thus, rate of reaction depends on only concentration of acetic anhydride.
So, rate is given by
$=k[(CH _{3}CO) _{2}O]$

Rate constant $= 3 \times 10^{-5} sec^{-1}$

The hydrolysis of ethyl acetate in caustic soda aqueous solution is represented by the following reaction.
$CH _{3}COOC _{2}H _{5}+NaOH\rightarrow  CH _{3}COONa+C _{2}H _{5}OH$
The expression for the rate of the reaction is $r=k[CH _{3}COOC _{2}H _{5}][NaOH].$
The overall order of the reaction is 2 and it is a bimolecular reaction.

The decomposition of  $\mathrm{N} _{2}\mathrm{O} _{5}$ follows first order kinetics.

For $2H _2O _2(aq)\rightarrow 2H _2O(l)+O _2(g)$ rate of decomposition for $H _2O _2=k[H _2O _2]$. It is a first order reaction.  It proceeds through following mechanism.

Option (A),(B),(D) are correct.
(C) : The half-life of the reaction is independent of the initial concentration of the reactant. Half-life for first order reaction is :$t _{1/2} = 0.693/k$
A first-order reaction has a rate proportional to the concentration of one reactant.
First-order rate constants have units of $sec^{-1}$. In other words, a first-order reaction has a rate law in which the sum of the exponents is equal to 1. 

As we know,
$r=K[N _2O _5]=6.2\times 10^{-4}\times 1.25=7.75\times 10^{-4} mol L^{-1} s^{-1}$.

$[H _2SO _4]\, =\, 0.1\, M\, =\, 0.1\, \times\, 2\, =\, 0.2 N$

$[HCl]$ = $0.1 N$

In case of $[H _2SO _4]$ 

$r _1\, =\, k[H^{\oplus}][Ester]$ 

$\displaystyle k _{H _2SO _4}\, = \frac{r _1}{2\, N\, \times\, [Ester]}$ 

In case of HCl, $r _1\, =\, k[H^{\oplus}]\, [Ester]$ 

$\displaystyle k _{HCl}\, =\, \frac{r _2}{1\, N\, [Ester]}$ 

As we know,
for a reaction:
$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$
$\displaystyle -\, \frac{1}{2}\, \frac{d[N _2O _5]}{dt}\, =\, \frac{1}{4}\, \frac{d[NO _2]}{dt}\, =\, \frac{d[O _2]}{dt}$
So
$2k _1\, =\, k _2\, =\, 4k _3$

Given,
Since $t _{1/2}$ does not depends upon the sugar concentration means it is first order w.r.t [sugar]
$\therefore t _{1/2}\, \propto\, [sugar]^{1}$
$t _{1/2}\, \times\, a^{n\, -\, 1}\, =\, k$
$\displaystyle \frac{(t _1/2) _1}{(t _{1/2) _2}}\, =\, \frac{[H^{\oplus}] _1^{1-n}}{[H^{\oplus}] _2^{1-n}}$

$\displaystyle \frac{500}{50}\, =\, \left ( \frac{10^{-5}}{10^{-6}}\right )^{1-n}$

10 = $(10)^{1-n}\, \Rightarrow\, n\, =\, 0$

For a first order reaction, the graph of logarithm of the partial pressure of reactant to the time is a straight line with negative slope. 

Since $k _{H _2SO _4}\, >\, k _{HCl},$ hence $H _2SO _4$ is stronger acid than HCl.

As we know, for a reaction: $2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$

$\displaystyle -\, \frac{1}{2}\, \frac{d[N _2O _5]}{dt}\, =\, \frac{1}{4}\, \frac{d[NO _2]}{dt}\, =\, \frac{d[O _2]}{dt}$

So, $2k _1\, =\, k _2\, =\, 4k _3.$

$\upsilon=k[ester][H _3O^+]$

$\displaystyle 0.2\, M \underset{t _{1/2}\, =\, 5hr}{\rightarrow} 0.1\, M \underset{t _{1/2}\, =\, 5hr}{\rightarrow} 0.05\, M$

$From\, 0.2\, M \underset{t\, =\, 10hr}{\rightarrow}\, 0.05\, M$
So $t _{1/2}$ is constant which is characteristic of first order reaction.
Hence, $t _{1/2}\, \propto\, (a)^0$.

As we know,
$\displaystyle t\, =\, \frac{2.303}{k}\, log\, \frac{V _0}{V _1}$

$\displaystyle =\, \frac{1}{k}\, ln\, \frac{V _0)}{V _1}$

$\displaystyle =\, \frac{1}{4.5\, \times\, 10^{-2}\, min^{-1}}\, In\, \frac{25mL}{5mL}$

$\displaystyle =\, \frac{log _{e}5}{4.5\, \times\, 10^{-2}}min$ 

For a first order reaction,

$\displaystyle k\, =\, \frac{2.303}{t}\, log\, \frac{V _{\infty}}{V _{\infty}\, -\, V _t}$

$\displaystyle =\, \frac{1}{t}\, log _e\, \frac{V _{\infty}}{V _{\infty}\, -\, V _t}$

$\displaystyle \frac{1}{15}\, log _e\, \frac{35mL}{(35\, -\, 9)\, mL}\, =\, \frac{1}{15}\, log _e\, \frac{35}{26}$

As the unit of rate constant is ${sec}^{-1}$, the reaction is first order reaction. 

${ C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 }+{ H } _{ 2 }O\xrightarrow [  ]{ \quad { H }^{ + }\quad  } { C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }+{ C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }$
Rate $=k\left[ { C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 } \right] \left[ { H } _{ 2 }O \right] $
When water is in excess, its concentration will be constant.
$\therefore $ Rate $={ k }^{ ' }\left[ { C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 } \right] $
The reaction is, therefore, pseudo first order or pseudo unimolecular reaction.

For the reaction,  $\displaystyle 2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2} $  the rate of reaction is  $\displaystyle \cfrac{1}{2}\cfrac{d}{dt}[{NO} _{2}]$

$Mol$ ${L}^{-1}$ of ${N} _{2}{O} _{5}$ reacted $=2\times 0.1=0.2$

$t=A.{ e }^{ -kt }\ So,\quad (A-{ A } _{ o })=A.({ e }^{ -kt }-1)\ \therefore 380=A.({ e }^{ -kt }-1)\ { A } _{ o }=\cfrac { 380 }{ { e }^{ -10k }-1 } $
 Otherewise,
$ { P } _{ o }=[{ A } _{ o }]RT\ { P } _{ o }={ [{ A }] } _{ 10 }RT\ \cfrac { { P } _{ 10 }-{ P } _{ o } }{ 7 } =\cfrac { ({ A } _{ 10 }-{ A } _{ o })RT }{ 7 } \ \cfrac { \cfrac { 760 }{ 380 }  }{ 10 } =\vartheta .RT\ \cfrac { 2 }{ 10RT } =\vartheta $
$ \vartheta \sim 0.01{ M. }{ min }^{ -1 }\longrightarrow$ Option (A)

Rate $=k{ \left[ { H } _{ 2 }{ O } _{ 2 } \right]  }^{ 1 }$
$2\times { 10 }^{ -4 }=\dfrac { 3\times { 10 }^{ -3 } }{ 60 } \times \left[ { H } _{ 2 }{ O } _{ 2 } \right] $
$\left[ { H } _{ 2 }{ O } _{ 2 } \right] =4 M$

$({ r } _{ t }-{ r } _{ 0 })=({ r } _{ 0 }-{ r } _{ \infty  }){ e }^{ -kt }\ \ln { \left( \cfrac { { r } _{ t }-{ r } _{ 0 } }{ { r } _{ 0 }-{ r } _{ \infty  } }  \right)  } =-kt\ k=\cfrac { 1 }{ t } \ln { \left( \cfrac { { r } _{ t }-{ r } _{ 0 } }{ { r } _{ 0 }-{ r } _{ \infty  } }  \right)  } $

Inversion of cane sugar follow Ist order reaction while its molecularity is 2 and reaction is given by
$ \implies C _{12}H _{22}O _{11} +H _2O \rightarrow C _6 H _{12}O _6 + C _6H _{12}O _6$

Here the rate of reaction is dependent on only $C _{12}H _{22}O _{11}$.

$k=0.06931{ s }^{ -1 }$

t = 0

A reaction which is not the first-order reaction naturally but made the first order by increasing or decreasing the concentration of one or the other reactant is known as Pseudo first-order reaction. 

The rate of formation of water is twice the rate of formation of oxygen.
$\frac {d[H _2O]} {dt}=2\frac {d[H _2O]} {dt}=2 \times 3.6  M  min^{-1} = 7.2 \, mol litre^{-1}min^{-1}$

$k _1 = \ \cfrac { 0.693 }{ 360 }= 1.92\times{ 10 }^{ -3 }{ min }^{ -1 }$

$\ log\cfrac { k _2 }{ k _1 } =\left( \cfrac { Ea }{ 2.303R }  \right) \left[ \left( \cfrac { { T } _{ 2 }-{ T } _{ 1 } }{ { T } _{ 1 }{ T } _{ 2 } }  \right)  \right]=  \cfrac { \left( 200\times { 10 }^{ 3 } \right)  }{ \left( 2.303\times 8.314 \right) \left[ \left( \cfrac { 723-653 }{ 653\times 723 }  \right)  \right]  } =  \cfrac { \left( 200\times { 10 }^{ 3 }\times 70 \right)  }{ \left( 2.303\times 8.314\times 653\times 723 \right)  } = 1.5487$

$\cfrac { k _2 }{ k _1 }  = Antilog (1.5487)= 35.38$, $k _2 = 35.38 \times 1.92 \times$ ${ 10 }^{ -3 } = 6.792 \times { 10 }^{ -2 }{ min }^{ -1 }$

Rate at $450^oC$, t = $\ \left( \cfrac { 2.303 }{ k _2 }  \right) log\left( \cfrac { 100 }{ 100-75 }  \right) $= $\ \left( \cfrac { 2.303 }{ 6.792\times { 10 }^{ -3 } }  \right) log\left( \cfrac { 100 }{ 25 }  \right) = \left( \cfrac { 2.303\times 0.6021 }{ 6.792\times { 10 }^{ 2 } }  \right)= 20.34$ minutes.

200 ml at 2 atm corresponds to 400 ml at 1 atm. This is the maximum value
In 15 minutes, the volume is 100 ml.
Thus one fourth of the reaction is complete in 15 minutes.
Hence, the volume of $KMnO _4$ will be $\displaystyle \frac{3}{4} \times 40 = 30 mL$

Inversion of cane sugar and Acid catalyzed by hydrolysis of ester are examples of pseudo unimolecular reactions.
Decomposition of ozone is a bimolecular reaction.
Decomposition of dintirogen pentoxide is also a bimolecular reaction.

(A) The rate of the reaction involving the conversion of ortho-hydrogen to parahydrogen is $\displaystyle -\, \frac{d[H _2]}{dt}\, =\, k[H _2]^{3/2}$
The order of the reaction is 1.5.
(B) The rate of the reaction involving the thermal decomposition of acetaldehyde is $k[CH _3CHO]^{3/2}$
The order of the reaction is1.5.
(C) In the formation of phosgene gas from CO and $Cl _2$, the rate of the reaction is $k[CO][Cl _2]^{1/2}$
The order of the reaction is 1.5.
(D) In the decomposition of $H _2O _2$, the rate of the reaction is $k[H _2O _2]$.
The order of the reaction is 1.

For a first order reaction $\displaystyle A \rightarrow P$, the expression for the rate constant is
$\displaystyle \displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{ a}{a-x}$
Here, A is reactant, P is product, a is the initial concentration of A and $a-x$ is the concentration of A at time t.

The inversion of a sugar follows first order rate equation which is given below.
$\displaystyle \displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{0}\, -\, r _{\infty}}{r _{t}\, -\, r _{\infty}}$
Here, $\displaystyle a = r _{0}\, -\, r _{\infty}$ and $\displaystyle a-x = r _{t}\, -\, r _{\infty}$

rate constant $k = \dfrac{2.303}{t}log\dfrac{(r _0 - r _\infty) }{(r _t-r _\infty)} = 0.0135$
At the point of optical inactiveness, rotation is zero. 

$\underset{d-Sucrose}{C _{12}H _{22}O _{11}}+H _2O\xrightarrow{H+}\underset{d-Glucose}{C _6H _{12}O _6}+\underset{l-Fructose}{C _6H _{12}O _6}$

$\displaystyle N _{2}O _{5}:(g)\rightarrow2:NO _{2}:(g)+\frac{1}{2}O _{2}:(g)$
$\displaystyle -d:[N _{2}O _{5}]/\mathrm{d} t=k _{1}[N _{2}O _{5}]$
$\displaystyle d:[NO _{2}]/\mathrm{d} t=k _{2}[N _{2}O _{5}]$
$\displaystyle d[O _{2}]/\mathrm{d} t=k _3[N _{2}O _{5}]$
$-\displaystyle \frac{\mathrm{d} N _{2}O _{5}}{\mathrm{d} t}=\frac{1}{2}\frac{\mathrm{d} NO _{2}}{\mathrm{d} t}
=2\frac{\mathrm{d} O _{2}}{\mathrm{d} t}$
$\displaystyle k _{1}=\frac{k _{2}}{2}=2k _{3}$
$\displaystyle 2k _{1}=k _{2}=4k _{3}$

The integrated rate law expression for the inversion of can sugar (assuming first order kinetics) is as shown.
$\displaystyle k = \frac {2.303}{t} log \frac {r _0 - r _{\infty}}{r _t - r _{\infty}} $
For 60 minutes
$\displaystyle k = \frac {2.303}{60} log \frac {13.1 - (-3.8)}{11.6 - (3.8)} = 0.001549 $
For 120 minutes
$\displaystyle k = \frac {2.303}{120} log \frac {13.1 - (-3.8)}{10.2 - (3.8)} = 0.001569 $
For 180 minutes
$\displaystyle k = \frac {2.303}{180} log \frac {13.1 - (-3.8)}{9.0 - (3.8)} = 0.001544 $
For 360 minutes
$\displaystyle k = \frac {2.303}{360} log \frac {13.1 - (-3.8)}{5.87 - (3.8)} = 0.001551 $
Since, the value of k is constant, the reaction follows first order reaction.
The average value of k is $\displaystyle  \frac {0.001549+0.001569+0.001544+0.001551}{4} = \frac {0.0062135}{4} = 0.001553 : min^{-1}$
To determine the total time, substitute $\displaystyle r _t = 0 $ in the above expression.
$\displaystyle t = \frac {2.303}{k} log \frac {r _0 - r _{\infty}}{r _t - r _{\infty}} $
$\displaystyle t = \frac {2.303}{0.001553} log \frac {13.1 - (-3.8)}{0 - (-3.8)} = 966 : min $

Rate $\displaystyle = k _b[NO _2][NO _3] $ .....(1)

${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }{ e }^{ -kt }\ { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }{ e }^{ kt }\ \ln { \left( \cfrac { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 } }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t } }  \right)  } ={ e }^{ kt }\ \ln { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t } } =\ln { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 } } -kt$