Tag: chemistry

Questions Related to chemistry

The rate constant for the hydrolysis reaction of an ester by dilute acid is $0.6931\times { 10 }^{ -3 }\ { s }^{ -1 }$. The time required to change the concentration of ester from $0.04$ $M$ to $0.01$ $M$ is:

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

  1. $6931$ sec

  2. $4000$ sec

  3. $2000$ sec

  4. $1000$ sec

Show answer
Correct Option: C
Explanation:

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

 So,
$ t=\cfrac { \ln { \left( \cfrac { 0.04 }{ 0.01 }  \right)  }  }{ 0.06931 } \ =2000{ s }^{ -1 }$

Benzene diazonium chloride (A) decomposes into chloro-benzene (B) and ${{\text{N}} _{\text{2}}}\left( {\text{g}} \right)$ in first order reaction volume of ${{\text{N}} _2}$ collected after 5 min and at the complete decomposition of A are 10 ml and 50 ml respectively. The rate constant for the reaction is:

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

  1. 0.446 ${\min ^{ - 1}}$

  2. 0.0446 ${\min ^{ - 1}}$

  3. 0.223 ${\min ^{ - 1}}$

  4. 0.112 ${\min ^{ - 1}}$

Show answer
Correct Option: B
Explanation:

t = 0

                              $\mathop {\text{A}}\limits _{50}  \to \mathop {\text{B}}\limits _0  + \mathop {{{\text{N}} _2}}\limits _0 $
At t = 5 min.    50 - 10          10 ml
t = complete                          50 ml
          $\ln  = \left( {\dfrac{{50}}{{40}}} \right) = {\kappa _1} \times 5$
  $\dfrac{{0.223}}{5} = {\kappa _1} \Rightarrow 0.0446\,{\min ^{ - 1}}$
Hence, option (B) is correct.

For the reaction of first order, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \rightleftharpoons 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, which of the following statements are correct?

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

  1. The concentration of the reactant decreases exponentially with time.

  2. The half-life of the reaction decreases with increasing temperature.

  3. The half-life of the reaction depends on the initial concentration of the reactant.

  4. The reaction proceeds to $99.6$% completion in eight half-life duration.

Show answer
Correct Option: A,B,D
Explanation:
For 1st order reaction,
$a _t=a _0e^{-kt}$
Also, $t _{1/2}=\frac{In2}{k}$ or $t _{1/2}\alpha \frac{1}{k}$
As the temperature increases, value of k also increases due to which $t _{1/2}$ decreases.
For 99.6% completion, $a _t=(\frac{100-99.6}{100})a _0=\frac{4a _0}{1000}$
$t=\frac{1}{k}In\frac{a _0}{4a-0/1000}=\frac{1}{k}In\frac{1000}{4}$
$=(\frac{t _{1/4}}{In2}).In250$
$t=8t _{1/2}$

For the reaction, ${\text{2}}{{\text{N}} _{\text{2}}}{{\text{O}} _{\text{5}}} \to {\text{4N}}{{\text{O}} _{\text{2}}} + {{\text{O}} _{\text{2}}}$, the value of rate and rate constant are $1.02\times 10^{-4} M/s$ and $3.4 \times {10^{ - 3}}{\sec ^{ - 1}}$ respectively. The concentration of ${{\text{N}} _{\text{2}}}{{\text{O}} _{\text{5}}}$ at that time will be: (in terms of molarity)

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

  1. $1.732$

  2. $3$

  3. ${\text{1}}{\text{.02}} \times {\text{1}}{{\text{0}}^{ - 4}}$

  4. $3.4 \times {10^4}$

Show answer
Correct Option: A

The hydrolysis of ethyl acetate,
$CH _{3}COOC _{2}H _{5} + H _{2}O\xrightarrow {H^{+}} CH _{3}COOH + C _{2}H _{5}OH$ is a reaction of:

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

  1. zero order

  2. pseudo first order

  3. second order

  4. third order

Show answer
Correct Option: B
Explanation:

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. 


Hydrolysis of ethyl acetate in presence of an excess of water:

$CH _3COOC _2H _5+H _2O(\text{excess}) \xrightarrow{H+} CH _3COOH+C _2H _5OH$

$r = k[CH _3COOC _2H _5]^2[H _2O]^0$

Excess $[H _2O]$ can cause the independency of reaction on $H _2O$.


Hence, it is a pseudo-first-order reaction.

Hence, the correct answer is option $\text{B}$.

In the following reaction $2H _{2}O _{2}\rightarrow 2H _{2}O+O _{2}$ rate of formation of $O _{2}$ is 3.6 M $ min^{-1}.$ The rate of formation of $H _{2}O$ is:

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

  1. $7.2 \, mol litre^{-1}min^{-1}$

  2. $7.8 \, mol litre^{-1}min^{-1}$

  3. $7.9 \, mol litre^{-1}min^{-1}$

  4. $7.5 \, mol litre^{-1}min^{-1}$

Show answer
Correct Option: A
Explanation:

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

At ${ 380 }^{ 0 }C$, the half life period for the first order decomposition of ${ H } _{ 2 }{ O } _{ 2 }$ is 360 minutes. The energy of activation of the reaction is 200 kJ ${ mol }^{ -1 }$. Calculate the time required for 75% decomposition at $450^{0}C$?

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

  1. 60 min

  2. 40 min

  3. 20.34 min

  4. 10 min

Show answer
Correct Option: C
Explanation:

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

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

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

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.

For the decomposition of $H _2O _2(aq.)$, it was found that $V _{O _2} (t=15 min.)$ was 100 mL (at 0$^oC$ and 1 atm) while $V _{O _2}$ (maximum) was 200 mL (at 0$^oC$ and 2 atm). If the same reaction had been followed by the titration method and if $V _{KMnO _4}^{cM} (t = 0)$ had been 40 mL, what would $V _{KMnO _4}^{cM}(t = 15 min)$ have been?

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

  1. 30 mL

  2. 25 mL

  3. 20 mL

  4. 15 mL

Show answer
Correct Option: A
Explanation:

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 $\displaystyle \frac{3}{4} \times 40 = 30 mL$

Which of the following are example of pseudo unimolecular reactions?
1. Inversion of cane sugar
2. Decomposition of ozone
3. Decomposition of $N _{2} O _{5}$
4. Acid catalysed by hydrolysis of ester

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

  1. 2 and 4

  2. 1 and 4

  3. 1, 2 and 4

  4. 1, 2, 3 and 4

Show answer
Correct Option: B
Explanation:

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.

Which of the following statement is/are correct ?

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

  1. The rate of the reaction involving the conversion of ortho-hydrogen to parahydrogen is $\displaystyle -\, \frac{d[H _2]}{dt}\, =\, k[H _2]^{3/2}$

  2. The rate of the reaction involving the thermal decomposition of acetaldehyde is $k[CH _3CHO]^{3/2}$

  3. In the formation of phosgene gas from CO and $Cl _2$, the rate of the reaction is $k[CO][Cl _2]^{1/2}$

  4. In the decomposition of $H _2O _2$, the rate of the reaction is $k[H _2O _2]$.

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

(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.