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Questions Related to chemistry

The inversion of a sugar follows first-order rate equation which can be followed by noting the change in the rotation of the plane of polarization of light in the polarimeter. If $r _{\propto},\, r _{\zeta}$ and $r _0$ are the rotations at $t\, =\, \propto$, t = t, and t = 0, then the first order reaction can be written as:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $\displaystyle k\, =\, \frac{1}{t}\, log\, \frac{r _{1}\, -\, r _{\propto}}{r _{0}\, -\, r _{\propto}}$

  2. $\displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{0}\, -\, r _{\propto}}{r _{1}\, -\, r _{\propto}}$

  3. $\displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{\propto}\, -\, r _{0}}{r _{\propto}\, -\, r _{1}}$

  4. $\displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{\propto}\, -\, r _{1}}{r _{\propto}\, -\, r _{0}}$

Show answer
Correct Option: B
Explanation:

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 $\displaystyle a-x = r _{t}\, -\, r _{\infty}$

The reaction, $Sucrose\xrightarrow [  ]{ { H }^{ + } } Glucose+Fructose$, takes  place at certain temperature while the volume of solution is maintained at $1$ litre. At time zero the initial rotation of the mixture is ${ 34 }^{ o }C$.After $30$ minutes the total rotation of solution is ${ 19 }^{ o }C$ and after a very long time, the total rotation is ${ -11 }^{ o }C$. Find the time when solution was optically inactive?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $135$ min

  2. $103.7$ min

  3. $38.7$ min

  4. $45$ min

Show answer
Correct Option: B
Explanation:

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. 

So, time taken is $ t =\dfrac{ 2.303}{k}log(45/11) = 103.7$ min.

Inversion of a sugar folllows first order rate equation which can be followed by nothing the change in rotation of the plane of polarization of light in the polarimeter. If $r _{\infty,}:r _t:and:r _0$ are the rotations at
 $t\,=\,\infty,t\,=\,t:and:t\,=\,0,$ then, first order reaction can be written as:

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $\;k\,=\,\displaystyle\frac{1}{t}log\displaystyle\frac{r _t-r _{\infty}}{r _0-r _{\infty}}$

  2. $\;k\,=\,\displaystyle\frac{1}{t}\,ln\displaystyle\frac{r _0-r _{\infty}}{r _t-r _0}$

  3. $\;k\,=\,\displaystyle\frac{1}{t}\,ln\displaystyle\frac{r _{\infty}-r _0}{r _{\infty}-r _t}$

  4. None of these

Show answer
Correct Option: C
Explanation:

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


Initially               a                Excess                  0                0 
After time t        a-x            Constant                x                x
At infinity           0               Constant               a                 a
If $r _0,r _t$ and $r _{\infty}$ be the observed angle of rotations of the sample at zero times, times $t$ and infinity respectively, and $k _1,k _2$ and $k _3$ proportionate in terms of sucrose,glucose and fructose, respectively.
Then,
$r _0=k _1a$
$r _t=k _1(a-x)+k _2x+k _3x$
$r _{\infty}=k _2a+k _3a$
From these equations it can be shown that
$\dfrac{a}{a-x}=\dfrac{r _0-r _{\infty}}{r _t-r _{\infty}}$
So, the expression for the rate constant rate of this reaction in terms of the optical rotational data may be 
put as $k=\dfrac{2.303}{t}\log \dfrac{r _0-r _{\infty}}{r _t-r _{\infty}}$

In the following reaction $2H _2O _2\rightarrow2H _2O+O _2$ rate of formation of $O _2$ is 3.6 M min$^{-1}$.


(a) What is rate of formation of $H _2O\ ?$        
(b) What is rate of disappearance of $H _2O _2$?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. (i) $7.2$ mol litre$^{-1}$ min$^{-1},$ (ii) $7.2$ mol litre$^{-1}$ min$^{-1}$

  2. (i) $3.6$ mol litre$^{-1}$ min$^{-1},$ (ii) $3.6$ mol litre$^{-1}$ min$^{-1}$

  3. (i) $14.4$ mol litre$^{-1}$ min$^{-1},$ (ii) $14.4$ mol litre$^{-1}$ min$^{-1}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$ (a)2H _{2}O _{2}\rightarrow 2H _{2}O+O _{2} $

We know, $ \dfrac{-1}{2}\dfrac{d[H _{2}O _{2}]}{dt} = \dfrac{-1}{2}\dfrac{d[H _{2}O]}{dt} = \dfrac{d[O _{2}]}{dt} $

$ \dfrac{d[H _{2}O]}{dt} = 2\dfrac{d[O _{2}]}{dt} = 2\times 3.6\,M\,min^{-1} $

$ \dfrac{d[H _{2}O]}{dt} = 7.2\,M\,min^{-1} $

$(b) \dfrac{-d[H _{2}O _{2}]}{dt} = \dfrac{2}{2}\dfrac{d[H _{2}O]}{dt} = \dfrac{-d[H _{2}O _{2}]}{dt} = 7.2\,M\,min^{-1} $ 

Option A is correct.

Dinitropentaoxide decomposes as follows :
    $N _2O _5:(g)\rightarrow2:NO _2(g)+\frac{1}{2}O _2:(g)$
Given that         

$ _d:[N _2O _5]:/:dt=k _1[N _2O _5]$
$d:[NO _2]:/:dt=k _2[N _2O _5]$
$d:[O _2]:/:dt=k _3[N _2O _5]$
What is the relation between $k _1,:k _2:and:k _3$?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $2k _1=k _2=4k _3$

  2. $2k _2=k _1=4k _3$

  3. $2k _3=k _2=4k _1$

  4. $2k _1=k _2=4k _2$

Show answer
Correct Option: A
Explanation:

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

The following data were obtained in experiment on inversion of cane sugar.
Time (minutes)         0        60        120      180      360     $\infty $
Angle of rotation  +13.1   +11.6   +10.2   +9.0   +5.87  -3.8
   (degree)
Determine total time ?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. 966 min

  2. 483 min

  3. 1932 min

  4. None of these

Show answer
Correct Option: A
Explanation:

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 $

Derive an expression for the Rate (k) of reaction :
$2N _{2}O _{5}(g)\rightarrow 4NO _{2}(g)+O _{2}(g)$


With the help of following mechanism:

$N _{2}O _{5}\overset{K _a}{\rightarrow}NO _{2}+NO _{3}$
$NO _{3}+NO _{2}\overset{K _{-a}}{\rightarrow}N _{2}O _{5}$
$NO _{2}+NO _{3}\overset{K _b}{\rightarrow}NO _{2}+O _{2}+NO$
$NO+NO _{3}\overset{K _c}{\rightarrow}2NO _{2}$

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+2k _{b}}[N _{2}O _{5}]$

  2. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}-2k _{b}}[N _{2}O _{5}]$

  3. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+k _{b}}[N _{2}O _{5}]$

  4. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{2k _{-a}-2k _{b}}[N _{2}O _{5}]$

Show answer
Correct Option: A
Explanation:

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

But $\displaystyle \dfrac {[NO _2][NO _3]}{[N _2O _5]}=  \dfrac {K _a}{K _{-a} + 2k _b}$

Hence $\displaystyle [NO _2][NO _3]  =\dfrac {K _a}{K _{-a}+2k _b} [N _2O _5]$......(2)

Substitute equation (2) in equation (1):

$\displaystyle \displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+2k _{b}}[N _{2}O _{5}]$

The rate constant for the reaction, ${ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 2N{ O } _{ 2 }\left( g \right) +\dfrac { 1 }{ 2 } { O } _{ 2 }\left( g \right) $, is $2.3\times { 10 }^{ -2 }\ { sec }^{ -1 }$. Which equation given below describes the change of $\left[ { N } _{ 2 }{ O } _{ 5 } \right] $ with time, ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }$ and ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }$ corresponds to concentration of ${ N } _{ 2 }{ O } _{ 5 }$ initially and time $t$ respectively?

chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

  1. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }{ e }^{ kt }$

  2. $\log _{ e }{ \dfrac { { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 } }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t } } } =kt$

  3. $\log _{ 10 }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t } } =\log _{ 10 }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 } } -kt$

  4. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }+kt$

Show answer
Correct Option: A,B,C
Explanation:

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

Among the following pairs of compounds, the one that illustrates the law of multiple proportions is:

chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

  1. $NH _{3}$ and $NCl _{3}$

  2. $H _{2}S$ and $SO _{2}$

  3. $CuO$ and $Cu _{2}O$

  4. $CS _{2}$ and $FeSO _{4}$

Show answer
Correct Option: C
Explanation:

According to the concept of law of multiple proportions, if two elements chemically combine to give two or more compounds, then the weight of one element which combines with the fixed weight of the other element in those compound, bear simple multiple ratios to one another.


Hence, $CuO$ and $CU _2O$ form two different compounds in a ratio of 1:2 with the fixed weight of oxygen.

The % composition of four hydrocarbons are as follows:


(i) (ii) (iii) (iv)
%C 75 80 85.7 91.3
%H 25 20 14.3 8.7


The data illustrates the law of:

chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

  1. constant proportion

  2. conservation of mass

  3. multiple proportion

  4. reciprocal proportion

Show answer
Correct Option: C