NCERT Solutions for Class 12 Chemistry Chapter 4 - Chemical Kinetics

Subtopics for Class 12 Chemistry Chapter 4 – Chemical Kinetics

Class 12 NCERT Solutions Chemistry Chapter 4 - Chemical Kinetics is fundamental to many of the concepts that you will study in the future. Given below are a few related subtopics that students should be aware of while studying Chapter 4 of Class 12 Chemistry.

1. Rate of a Chemical Reaction

2. Factors Influencing the Rate of a Reaction

1. Dependence of Rate on Concentration

2. Rate Expression and Rate Constant

3. Order of a Reaction

4. Molecularity of a Reaction

3. Integrated Rate Equations

1. Zero Order Reactions

2. First-Order Reactions

3. Half-Life of a Reaction

4. Pseudo First Order Reaction

5. Temperature Dependence of the Rate of a Reaction

1. Effect of Catalyst Ex 4.6 – Collision Theory of Chemical Reactions

Mastering Class 12 Chemistry Chapter 4: Chemical Kinetics - Question and Answers, and Tips for Success

Intext Exercise

1. For the reaction $\text{R}\to \text{P}$,  the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Ans: The average rate of the reaction can be calculated by dividing the change in the rate of decreasing the rate of reactant by the time taken. This is given below:

\[\text{Average rate = - }\frac{\text{ }\!\!\Delta\!\!\text{  }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}{\text{ }\!\!\Delta\!\!\text{ t}}\] 

This can be written as:

\[\text{Average rate = - }\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{2}}}\text{- }\!\![\!\!\text{ R}{{\text{ }\!\!]\!\!\text{ }}_{\text{1}}}}{{{\text{t}}_{\text{2}}}\text{-}{{\text{t}}_{\text{1}}}}\] 

\[{{\text{R}}_{\text{1}}}\text{ = 0}\text{.03 M}\] 

\[{{\text{R}}_{2}}\text{ = 0}\text{.02 M}\] 

\[{{\text{t}}_{\text{2}}}\text{ - }{{\text{t}}_{\text{1}}}\text{ = 25 min}\] 

Putting the values, we get:

\[\text{Average rate = - }\frac{0.02-0.03}{25}\] 

\[\text{Average rate = - }\frac{-0.01}{25}=4\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ M mi}{{\text{n}}^{-1}}\]

So, the average rate in minutes will be $4\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ M mi}{{\text{n}}^{-1}}$ 

Now, to find the average rate in seconds we have to divide the above answer by 60. So, the answer will be:

\[\text{Average rate = - }\frac{4\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ }}{60}=\text{ 6}\text{. 66  }\!\!\times\!\!\text{  1}{{\text{0}}^{-6}}\text{ M }{{\text{s}}^{-1}}\] 

Therefore, the average rate in seconds will be $6.66\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-6}}\text{ M }{{\text{s}}^{-1}}$ 

2. In a reaction, 2A $\to $ Products, the concentration of A decreases from $\text{0}\text{.5 mol }{{\text{L}}^{\text{-1}}}$ to $\text{0}\text{.4 mol }{{\text{L}}^{\text{-1}}}$ in 10 minutes. Calculate the rate during this interval.

Ans: The average rate of the reaction can be calculated by dividing the change in the rate of decreasing the rate of reactant by the time taken. This is given below: 

\[\text{Average rate = - }\frac{1}{2}\frac{\text{ }\!\!\Delta\!\!\text{  }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}{\text{ }\!\!\Delta\!\!\text{ t}}\] 

This is due the fact that the reaction given is:

2A $\to $ Products

So, the average rate will be written as:

\[\text{Average rate = - }\frac{1}{2}\frac{{{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}_{\text{2}}}\text{- }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}_{\text{1}}}}{{{\text{t}}_{\text{2}}}\text{-}{{\text{t}}_{\text{1}}}}\] 

\[{{\text{A}}_{\text{1}}}\text{ = 0}\text{.5 M}\] 

\[{{\text{A}}_{2}}\text{ = 0}\text{.4 M}\] 

\[{{\text{t}}_{\text{2}}}\text{ - }{{\text{t}}_{\text{1}}}\text{ = 10 min}\] 

Putting the values, we get:

\[\text{Average rate = - }\frac{1}{2}\text{  }\!\!\times\!\!\text{  }\frac{0.4-0.5}{10}\] 

\[\text{Average rate = - }\frac{1}{2}\text{  }\!\!\times\!\!\text{  }\frac{-0.1}{10}=5\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}\text{ M mi}{{\text{n}}^{-1}}\]

So, the average rate will be $\text{5  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}\text{ M mi}{{\text{n}}^{-1}}$

3. For a reaction, $\text{A + B }\to \text{ Product}$, the rate law is given by: $\text{r = k  }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{1/2}}}\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}$. What is the order of the reaction?

Ans: The order of the reaction can be calculated by adding the stoichiometry coefficients of the reactants in the given rate of the reaction.     

Given the rate is $\text{r = k  }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{1/2}}}\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}$ 

So, the order will be:

\[\text{Order = 2 + }\frac{\text{1}}{\text{2}}\text{ = 2}\text{.5}\] 

So, the order of the reaction is 2.5.

4. The conversion of the molecules X to Y follows second order kinetics. If concentration of x is increased to three times how will it affect the rate of formation of Y?

Ans: The reaction will be:

\[\text{X }\to \text{ Y}\] 

As the question says that this reaction follows the second order reaction, we can write the rate law equation as:

\[\text{Rate = k }\!\![\!\!\text{ X}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\text{ = k}{{\text{a}}^{\text{2}}}\]  

If [X] = a mol/ L

It is said that the concentration of X increases by three times, so we can write:

[X] = 3a mol/ L

Therefore, the rate of reaction will be:

\[\text{Rate = k (3a}{{\text{)}}^{\text{2}}}\text{ = 9 k}{{\text{a}}^{\text{2}}}\] 

Thus, the rate of the reaction will increase by 9 times or the rate formation will increase by 9 times.

5. A first order reaction has a rate constant $\text{1}\text{.15  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ }{{\text{s}}^{\text{-1}}}$. How long will 5 g reactant take to reduce to 3 g?  

Ans: The initial amount of the reactant is given as 5 g. We can write:

\[{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}\text{ = 5 g}\]

The final amount of the reactant is given as 3 g. We can write:

[R] = 3 g

We are also given the value of rate constant as:

\[\text{Rate constant = 1}\text{.15  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

We know that the reaction is a 1st order reaction, the time can be calculated by:

\[\text{t = }\frac{\text{2}\text{.303}}{\text{k}}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\] 

Putting the values in the above formula, we get:

\[\text{t = }\frac{\text{2}\text{.303}}{1.15\text{ x 1}{{\text{0}}^{-3}}}\text{ log}\frac{5}{3}\] 

\[\text{t = }\frac{\text{2}\text{.303}}{1.15\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}}\text{  }\!\!\times\!\!\text{  0}\text{.2219}\] 

t = 444 seconds

So, the time taken will be 444 seconds.

6. Time required to decompose $\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ to half its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Ans: We are given that the decomposition of $\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ is a first order reaction. So, we can write:

\[{{\text{t}}_{\text{1/2}}}\text{ = }\frac{\text{0}\text{.693}}{\text{k}}\] 

It is mentioned that the time required for the initial amount to become half is 60 minutes.

\[{{\text{t}}_{\text{1/2}}}\text{ = 60 min}\]

Therefore, $\text{k = }\frac{\text{0}\text{.693}}{{{\text{t}}_{\text{1/2}}}}$ 

Putting the value, we get:

\[\text{k = }\frac{\text{0}\text{.693}}{\text{60 }\!\!\times\!\!\text{ 60}}\text{ = 1}\text{.925 }\!\!~\!\!\text{  }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ }{{\text{s}}^{\text{-1}}}\]

Thus, the rate constant is $\text{1}\text{.925 }\!\!~\!\!\text{  }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ }{{\text{s}}^{\text{-1}}}$.

7. What will be the effect of temperature on rate constant?

Ans: An increase of 10 degrees in temperature causes a reaction's rate constant to almost double in size. In any case, the Arrhenius equation gives the exact temperature dependency of a chemical reaction rate.

The Arrhenius equation is given below:

\[\text{k = A }{{\text{e}}^{\text{-}{{\text{E}}_{\text{a}}}\text{/Rt}}}\] 

Where, A is the Arrhenius factor or the frequency factor,

T is the temperature,

R is the gas constant,

${{\text{E}}_{\text{a}}}$ is the activation energy.

8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate ${{\text{E}}_{\text{a}}}$.

Ans: The formula that can be used to solve the question is:

\[\text{log}\frac{{{\text{k}}_{\text{1}}}}{{{\text{k}}_{\text{2}}}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303R}}\left[ \frac{{{\text{T}}_{\text{2}}}\text{-}{{\text{T}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}} \right]\] 

${{\text{T}}_{\text{1}}}$ temperature given is absolute temperature and it is equal to 298 K.

${{\text{T}}_{2}}$ = 298 + 10 = 308 K

It is given that the rate of the chemical reaction  double for an increase of 10 K, therefore, we can write the values of rate constant as:

\[{{\text{k}}_{\text{1}}}\text{ = x}\] 

\[{{\text{k}}_{\text{2}}}\text{ = 2x}\] 

Also, $\text{R = 8}\text{.314 J }{{\text{K}}^{\text{-1}}}\text{ mo}{{\text{l}}^{\text{-1}}}$ 

Now, putting all the values in the formula, we get:

\[\text{log}\frac{\text{2x}}{\text{x}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314}}\left[ \frac{\text{10}}{\text{298  }\!\!\times\!\!\text{  308}} \right]\] 

\[\text{log2 = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314}}\left[ \frac{\text{10}}{\text{298  }\!\!\times\!\!\text{  308}} \right]\] 

\[{{\text{E}}_{\text{a}}}\text{ = }\frac{2.303\text{  }\!\!\times\!\!\text{  8}\text{.314  }\!\!\times\!\!\text{  298  }\!\!\times\!\!\text{  308  }\!\!\times\!\!\text{  log 2}}{10}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 52897}\text{.78  J mo}{{\text{l}}^{-1}}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 52}\text{.89 kJ mo}{{\text{l}}^{-1}}\]

9. The activation energy for the reaction $\text{2H}{{\text{I}}_{\text{(g)}}}\to {{\text{H}}_{\text{2}}}_{_{\text{(g)}}}\text{+}{{\text{I}}_{\text{2}}}_{_{\text{(g)}}}$ is $\text{209}\text{.5 kJ mo}{{\text{l}}^{\text{-1}}}$  at 581 K. Calculate the fraction of molecules of molecules having energy equal to or greater than activation energy.

Ans: We are given the activation energy as $\text{209}\text{.5 kJ mo}{{\text{l}}^{\text{-1}}}$

T = 581 K

\[\text{R = 8}\text{.314 J }{{\text{K}}^{\text{-1}}}\text{ mo}{{\text{l}}^{\text{-1}}}\]

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

\[\text{x = }{{\text{e}}^{{{\text{E}}_{\text{a}}}\text{/RT}}}\] 

\[\text{ln x =}\frac{\text{-}{{\text{E}}_{\text{a}}}}{\text{RT}}\] 

\[\text{log x =}\frac{\text{-}{{\text{E}}_{\text{a}}}}{\text{2}\text{.303 RT}}\] 

\[\text{log x =}\frac{209500}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314  }\!\!\times\!\!\text{  581}}\text{ }=\text{ }18.8323\]

Now, taking the antilog:

x = Antilog (18.8323)

\[\text{x = 1}\text{.471  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-19}}}\] 

NCERT Exercise

1. From the rate expression for the following reactions, determine their order of reaction and the dimension of the rate constants.

(i). $\text{3 NO(g) }\to {{\text{N}}_{\text{2}}}\text{O(g)  Rate = k }\!\![\!\!\text{ NO}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}$ 

Ans: We are given:

\[\text{Rate = k }\!\![\!\!\text{ NO}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\] 

From this we can see that the order of the reaction = 2

\[\text{k = }\frac{\text{rate}}{{{\text{ }\!\![\!\!\text{ No }\!\!]\!\!\text{ }}^{\text{2}}}}\] 

Dimensions will be:

\[\text{k = }\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{{{\text{(mol }{{\text{L}}^{\text{-1}}}\text{)}}^{\text{2}}}}\] 

\[\text{=}\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{\text{mo}{{\text{l}}^{\text{2}}}\text{ }{{\text{L}}^{\text{2}}}}\] 

\[\text{= L mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}\]

(ii).${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{(aq)+3}{{\text{I}}^{\text{-}}}\text{(aq)+2}{{\text{H}}^{\text{+}}}\to \text{2}{{\text{H}}_{\text{2}}}\text{O(l)+I}_{\text{3}}^{\text{-}}\text{  Rate=k }\!\![\!\!\text{ }{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{ }\!\!]\!\!\text{  }\!\![\!\!\text{ }{{\text{I}}^{\text{-}}}\text{ }\!\!]\!\!\text{ }$  

Ans: Ans: \[Rate =k[{{H}_{2}}{{O}_{2}}][{{I}^{-}}]\] 

From this we can see that the order of the reaction = 2

\[k=\frac{rate}{[{{H}_{2}}{{O}_{2}}][{{I}^{-}}]}\]

Dimensions will be:

\[\text{k = }\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{{{\text{(mol }{{\text{L}}^{\text{-1}}}\text{)}}^{\text{2}}}}\] 

\[\text{=}\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{\text{mo}{{\text{l}}^{\text{2}}}\text{ }{{\text{L}}^{\text{2}}}}\] 

\[\text{= L mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}\]

(iii). $\text{C}{{\text{H}}_{\text{3}}}\text{CHO(g)}\to \text{C}{{\text{H}}_{\text{4}}}\text{(g)+ CO(g)  Rate = k }\!\![\!\!\text{ C}{{\text{H}}_{\text{3}}}\text{CHO}{{\text{ }\!\!]\!\!\text{ }}^{\text{3/2}}}$ 

Ans: \[\text{Rate = k }\!\![\!\!\text{ C}{{\text{H}}_{3}}CHO{{]}^{3/2}}\] 

From this we can see that the order of the reaction = $\frac{3}{2}$ 

\[\text{k = }\frac{\text{rate}}{{{\text{ }\!\![\!\!\text{ C}{{\text{H}}_{3}}CHO]}^{\frac{3}{2}}}}\] 

Dimensions will be:

\[\text{k = }\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{{{\text{(mol }{{\text{L}}^{\text{-1}}}\text{)}}^{3/2}}}\] 

\[\text{=}\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{\text{mo}{{\text{l}}^{\frac{3}{2}}}\text{ }{{\text{L}}^{\frac{3}{2}}}}\] 

\[\text{= }{{\text{L}}^{\frac{1}{2}}}\text{ mo}{{\text{l}}^{\text{-}\frac{1}{2}}}\text{ }{{\text{s}}^{\text{-1}}}\]

(iv). ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{Cl(g)}\to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{(g)+ HCl(g)  Rate = k }\!\![\!\!\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{Cl }\!\!]\!\!\text{ }$ 

Ans: \[\text{Rate = k }\!\![\!\!\text{ }{{\text{C}}_{2}}{{H}_{5}}Cl]\] 

From this we can see that the order of the reaction = 1

\[\text{k = }\frac{\text{rate}}{\text{ }\!\![\!\!\text{ }{{\text{C}}_{2}}{{H}_{5}}Cl]}\] 

Dimensions will be:

\[\text{k = }\frac{\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}}{\text{(mol }{{\text{L}}^{\text{-1}}})}\] 

\[\text{= }{{\text{s}}^{\text{-1}}}\]

2. For a reaction:

$\text{2A+B }\to \text{ }{{\text{A}}_{\text{2}}}\text{B}$ 

The $rate=k[A]{{[B]}^{2}}$ with $\text{k = 2}\text{.0 x 1}{{\text{0}}^{\text{-6}}}\text{ mo}{{\text{l}}^{\text{-2}}}\text{ }{{\text{L}}^{\text{2}}}\text{ }{{\text{s}}^{\text{-1}}}$.

Calculate the initial rate of the reaction when $\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{  = 0}\text{.1 mol }{{\text{L}}^{\text{-1}}}$, $\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{  = 0}\text{.2 mol }{{\text{L}}^{\text{-1}}}$. Calculate the rate of reaction after [A] is reduced to $\text{0}\text{.06 mol }{{\text{L}}^{\text{-1}}}$.

Ans: We are given the rate of the reaction as:

\[rate=k[A]{{[B]}^{2}}\]

Putting the values in this, we get the rate as:

\[\text{rate = 2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-6}}}\text{  }\!\!\times\!\!\text{  0}\text{.1  }\!\!\times\!\!\text{  (0}\text{.2}{{\text{)}}^{2}}\] 

\[\text{rate = 8}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-9}}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}\]  

When [A] is reduced from $0.10\text{ mol }{{\text{L}}^{-1}}$ to $0.06\text{ mol }{{\text{L}}^{-1}}$. So, the amount of [A] reacted will be:

\[=0.10-0.06=0.04\text{ mol }{{\text{L}}^{-1}}\]

Therefore, the concentration of B reacted will be:

\[=\frac{1}{2}\text{  }\!\!\times\!\!\text{  0}\text{.04 = 0}\text{.02 mol }{{\text{L}}^{-1}}\] 

Hence, new [B] = 0.2 – 0.02 = $0.18\text{ mol }{{\text{L}}^{-1}}$ 

Now, the new rate of the reaction will be:

\[\text{rate = 2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-6}}}\text{  }\!\!\times\!\!\text{  0}\text{.06  }\!\!\times\!\!\text{  (0}\text{.18}{{\text{)}}^{2}}\] 

\[\text{rate = 3}\text{.89  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-9}}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}\]  

Therefore, the rate of the reaction is $\text{3}\text{.89  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-9}}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}$.

3. The decomposition of $\text{N}{{\text{H}}_{\text{3}}}$ on the platinum surface is zero order reaction. What are the rates of production of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ if $\text{k = 2}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}\text{ mo}{{\text{l}}^{\text{-1}}}\text{ L }{{\text{s}}^{\text{-1}}}$ ?

Ans: The following equation represents the breakdown of ammonia on the platinum surface.

\[\text{2N}{{\text{H}}_{{{\text{3}}_{\text{(g)}}}}}\xrightarrow{\text{Pt}}{{\text{N}}_{\text{2}}}_{_{\text{(g)}}}\text{+3}{{\text{H}}_{\text{2}}}_{_{\text{(g)}}}\] 

Therefore, we can write the rate of the reaction as:

\[\text{Rate = -}\frac{\text{1}}{\text{2}}\frac{\text{d }\!\![\!\!\text{ N}{{\text{H}}_{\text{3}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = }\frac{\text{d }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = }\frac{\text{1}}{\text{3}}\frac{\text{d }\!\![\!\!\text{ }{{\text{H}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\] 

But we are given that the reaction is a zero order reaction.

So, \[\text{Rate = -}\frac{\text{1}}{\text{2}}\frac{\text{d }\!\![\!\!\text{ N}{{\text{H}}_{\text{3}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = }\frac{\text{d }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = }\frac{\text{1}}{\text{3}}\frac{\text{d }\!\![\!\!\text{ }{{\text{H}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = 2}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}\] 

Thus, the rate of production of ${{\text{N}}_{\text{2}}}$ will be:

\[\text{ }\frac{\text{d }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = 2}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}\]

And the rate of production of ${{\text{H}}_{\text{2}}}$ will be:

\[\frac{\text{d }\!\![\!\!\text{ }{{\text{H}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }}{\text{dt}}\text{ = 3  }\!\!\times\!\!\text{  2}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}=7.\text{5  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}\] 

4. The decomposition of dimethyl ether leads to the formation of $\text{C}{{\text{H}}_{\text{4}}}$, ${{\text{H}}_{\text{2}}}$ and CO and the reaction rate is given by $\text{Rate = k  }\!\![\!\!\text{ C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{3/2}}}$ 

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

$\text{Rate = k (}{{\text{P}}_{\text{C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}}}{{\text{)}}^{\text{3/2}}}$ 

If the pressure is measured in bar and time in minutes, then what are the units of rate and rate   constants?

Ans: If the pressure is measured in bar and time in minutes, then unit of rate will be:

\[\text{= bar mi}{{\text{n}}^{\text{-1}}}\] 

We are given the rate of the reaction as: $\text{Rate = k  }\!\![\!\!\text{ C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{3/2}}}$ 

Therefore, $\text{k = }\frac{\text{Rate}}{\text{  }\!\![\!\!\text{ C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{3/2}}}}$ 

So, we can write the units of rate constant as:

\[\text{k = }\frac{\text{bar mi}{{\text{n}}^{\text{-1}}}}{\text{ba}{{\text{r}}^{\text{3/2}}}}\text{ = ba}{{\text{r}}^{\text{-1/2}}}\text{ mi}{{\text{n}}^{\text{-1}}}\]

So, the units are \[\text{ba}{{\text{r}}^{\text{-1/2}}}\text{ mi}{{\text{n}}^{\text{-1}}}\]

6. Mention the factors that affect the rate of a chemical reaction.

Ans: Factors that influence a reaction's speed include.

• Reactant nature: The rate of reaction is affected by the kind of reactant. For example, ionic compound reactions are quicker than covalent compound reactions.

• The state of the reactants: Solid reactions are sluggish, liquid reactions are rapid, and gas reactions are very quick.

• Temperature: In addition, temperature has a significant impact on response rate. Every 10 degrees Celsius increase in temperature doubles the pace of reaction.

• Presence of catalyst: The rate of reaction is also affected by the presence of a catalyst in the reaction. It enhances the pace of reaction by increasing reaction surface area, by generating unstable intermediates with the substrate, and by offering a lower activation energy alternative path.

7. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is:

(i). Doubled

Ans: Let us assume that the concentration of the reactant be [A] = a

Rate of the reaction will be:

\[\text{R = k }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\text{ = k}{{\text{a}}^{\text{2}}}\] 

It is said that the concentration of A increases by two times, so we can write:

[A] = 2a mol/ L

Therefore, the rate of reaction will be:

\[\text{Rate = k (2a}{{\text{)}}^{\text{2}}}\text{ = 4 k}{{\text{a}}^{\text{2}}}\] 

Thus, the rate of the reaction will increase by 4 times.

(ii). Reduced to half

It is said that the concentration of A reduced to half, so we can write:

[X] = $\frac{1}{2}$ a mol/ L

Therefore, the rate of reaction will be:

\[\text{Rate = k }{{\left[ \frac{1}{2}a \right]}^{2}}\text{ = }\frac{1}{4}\text{ k}{{\text{a}}^{\text{2}}}\] 

Thus, the rate of the reaction will reduce by ¼ times.

8. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

Ans: An increase of 10 degrees in temperature causes a reaction's rate constant to almost double in size. In any case, the Arrhenius equation gives the exact temperature dependency of a chemical reaction rate.

The Arrhenius equation is given below:

\[\text{k = A }{{\text{e}}^{\text{-}{{\text{E}}_{\text{a}}}\text{/Rt}}}\] 

Where, A is the Arrhenius factor or the frequency factor,

T is the temperature,

R is the gas constant,

${{\text{E}}_{\text{a}}}$ is the activation energy.

The formula can also be written as:

\[\text{log}\frac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303R}}\left[ \frac{{{\text{T}}_{\text{2}}}\text{-}{{\text{T}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}} \right]\] 

Where ${{\text{k}}_{\text{2}}}$is the rate constant at temperature \[{{\text{T}}_{\text{2}}}\] 

${{\text{k}}_{1}}$ is the rate constant at temperature ${{\text{T}}_{\text{1}}}$ 

9. In a pseudo first order hydrolysis of ester in water, the following results were obtained

(i). Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Ans: Between the time interval of 30 to 60 sec the average rate of reaction will be calculated as:

\[\text{= }\frac{\text{d }\!\![\!\!\text{ Ester }\!\!]\!\!\text{ }}{\text{dt}}\] 

Putting the values from the data given in the question, we can write:

\[\text{= }\frac{0.31-0.17}{60-30}=\frac{0.14}{30}\] 

\[\text{= 4}\text{.67  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

(ii). Calculate  the pseudo first order rate constant for the hydrolysis of ester.

Ans: The rate law formula for pseudo first order reaction will be:

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\]

Now, when t = 30 s, then we rate constant will be:

\[{{\text{k}}_{1}}\text{ = }\frac{\text{2}\text{.303}}{30}\text{ log}\frac{0.55}{0.31}=1.91\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ }{{\text{s}}^{-1}}\] 

Now, when t = 60 s, then we rate constant will be:

\[{{\text{k}}_{2}}\text{ = }\frac{\text{2}\text{.303}}{60}\text{ log}\frac{0.55}{0.17}=1.96\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ }{{\text{s}}^{-1}}\]

Now, when t = 90 s, then we rate constant will be:

\[{{\text{k}}_{3}}\text{ = }\frac{\text{2}\text{.303}}{90}\text{ log}\frac{0.55}{0.085}=2.075\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ }{{\text{s}}^{-1}}\]

So, we can calculate the average rate constant as:

\[\text{k = }\frac{{{\text{k}}_{\text{1}}}\text{ + }{{\text{k}}_{\text{2}}}\text{ + }{{\text{k}}_{\text{3}}}}{\text{3}}\] 

\[\text{=1}\text{.98  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-1}}}\]

10. A reaction is first order in A and second order in B.

(i). Write the differential rate equation

Ans: The differential rate equation can be written as:

\[\text{-}\frac{\text{d }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}{\text{dt}}\text{ = k }\!\![\!\!\text{ A }\!\!]\!\!\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\] 

(ii). How is the rate affected by increasing the concentration of B three times?

Ans: The concentration of B is increased by 3 times, then B =3B

Therefore, the rate will be:

 \[\text{-}\frac{\text{d }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}{\text{dt}}\text{ = k }\!\![\!\!\text{ A }\!\!]\!\!\text{  }\!\![\!\!\text{ 3B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}=9\text{ }\times \text{  }\!\![\!\!\text{ A }\!\!]\!\!\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\]  

Therefore, the rate will increase by 9 times.

(iii). How is the rate affected when the concentration of both A and B are doubled?

Ans: The concentration of A is doubled, then A = 2A

The concentration of B is doubled, then B = 2B

Therefore, the rate will be:

\[\text{-}\frac{\text{d }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}{\text{dt}}\text{ = k }\!\![\!\!\text{ 2A }\!\!]\!\!\text{  }\!\![\!\!\text{ 2B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}=8\text{ }\times \text{  }\!\![\!\!\text{ A }\!\!]\!\!\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\] 

Therefore, the rate will increase by 8 times.

11. In a reaction between A and B the initial rate of reaction $\text{(}{{\text{r}}_{\text{0}}}\text{)}$ was measured for different concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?

Ans: Let us assume that the order of the reaction with respect to A be x and with respect to B be y.

Therefore, we can write:

\[{{\text{r}}_{\text{0}}}\text{ = k  }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\] 

\[\text{5}\text{.07  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ = k  }\!\![\!\!\text{ 0}\text{.20}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.30}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{    }.........(i)\] 

\[\text{5}\text{.07  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ = k  }\!\![\!\!\text{ 0}\text{.20}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.10}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{    }.........(ii)\] 

\[\text{1}\text{.43  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ = k  }\!\![\!\!\text{ 0}\text{.40}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.05}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{    }.........(iii)\]

Let us divide (i) by (ii), we get:

\[\frac{\text{5}\text{.07  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}}{\text{5}\text{.07  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ }}\text{ = }\frac{\text{k  }\!\![\!\!\text{ 0}\text{.20}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.30}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}{\text{ k  }\!\![\!\!\text{ 0}\text{.20}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.10}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}\text{ }\] 

\[\text{=1 = }\frac{{{\text{ }\!\![\!\!\text{ 0}\text{.30 }\!\!]\!\!\text{ }}^{\text{y}}}}{{{\text{ }\!\![\!\!\text{ 0}\text{.10 }\!\!]\!\!\text{ }}^{\text{y}}}}\] 

\[\text{= }{{\left( \frac{\text{0}\text{.30}}{\text{0}\text{.10}} \right)}^{\text{x}}}\text{ = }{{\left( \frac{\text{0}\text{.30}}{\text{0}\text{.10}} \right)}^{\text{y}}}\] 

Therefore, y = 0

Now, dividing (iii) by (i), we get:

\[\frac{\text{1}\text{.43  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}}{\text{5}\text{.07  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ }}\text{ = }\frac{\text{k  }\!\![\!\!\text{ 0}\text{.40}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.05}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}{\text{ k  }\!\![\!\!\text{ 0}\text{.20}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.30}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}\text{ }\] 

\[\frac{\text{1}\text{.43  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}}{\text{5}\text{.07  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ }}\text{ = }\frac{\text{k  }\!\![\!\!\text{ 0}\text{.40}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{ }}{\text{ k  }\!\![\!\!\text{ 0}\text{.20}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{ }}\] 

Because y = 0

\[\text{= 2}\text{.821 = }{{\text{2}}^{\text{x}}}\] 

Now, taking log on both sides.

log 2.821 = x log 2

\[\text{x = }\frac{\text{log 2}\text{.821}}{\text{log 2}}\] 

x = 1.496

x = 1.5

Thus, the order of the reaction according to A is 1.5 and according to B is 0.

12. The following results have been obtained during the kinetic studies of the reaction.

\[\text{2A + B}\to \text{ C + D}\]

Determine the rate law and the rate constant for the reaction.

Ans: Let us assume that the order of the reaction with respect to A be x and with respect to B be y.

Therefore, we can write:

\[\text{Rate = k  }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\] 

\[\text{6}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ = k  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{    }.........(i)\] 

\[\text{7}\text{.2  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ = k  }\!\![\!\!\text{ 0}\text{.3}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.2}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{    }.........(ii)\] 

\[\text{2}\text{.88 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{ = k }\!\!~\!\!\text{   }\!\![\!\!\text{ 0}\text{.3}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{ }\!\!~\!\!\text{   }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{ }\!\!~\!\!\text{   }\!\!~\!\!\text{  }.........\text{(iii)}\]

\[\text{2}\text{.40  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ = k  }\!\![\!\!\text{ 0}\text{.4}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}\text{    }.........(iv) \] 

Let us divide (iv) by (i), we get:

\[\frac{\text{2}\text{.40  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}}{\text{6}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ }}\text{ = }\frac{\text{k  }\!\![\!\!\text{ 0}\text{.4}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}{\text{ k  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}\text{ }\] 

\[\text{=4 = }\frac{{{\text{ }\!\![\!\!\text{ 0}\text{.4 }\!\!]\!\!\text{ }}^{x}}}{{{\text{ }\!\![\!\!\text{ 0}\text{.1 }\!\!]\!\!\text{ }}^{x}}}\] 

\[\text{= 4 = }{{\left( \frac{\text{0}\text{.4}}{\text{0}\text{.1}} \right)}^{x}}\] 

Therefore, x = 1

Now, dividing (iii) by (ii), we get:

\[\frac{\text{2}\text{.88  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-1}}}}{\text{7}\text{.2  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }}\text{ = }\frac{\text{k  }\!\![\!\!\text{ 0}\text{.3}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.4}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}{\text{ k  }\!\![\!\!\text{ 0}\text{.3}{{\text{ }\!\!]\!\!\text{ }}^{\text{x}}}\text{  }\!\![\!\!\text{ 0}\text{.2}{{\text{ }\!\!]\!\!\text{ }}^{\text{y}}}}\text{ }\] 

\[\text{= 4 = }{{\left( \frac{\text{0}\text{.4}}{\text{0}\text{.2}} \right)}^{y}}\] 

\[\text{= 4 = }{{\text{2}}^{\text{y}}}\] 

\[\text{= }{{\text{2}}^{2}}\text{ = }{{\text{2}}^{\text{y}}}\] 

y = 2

Thus, the order of the reaction according to A is 1 and according to B is 2.

So, the rate law is:

\[\text{Rate = k }\!\![\!\!\text{ A }\!\!]\!\!\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\] 

\[\text{k = }\frac{\text{Rate}}{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{  }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}}\] 

Now, putting the values for each experiment, we get:

From experiment I:

\[\text{k = }\frac{6.0\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}}{\text{ }\!\![\!\!\text{ 0}\text{.1 }\!\!]\!\!\text{  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}}\] 

\[\text{= 6}\text{.0 }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{\text{-2}}}\text{ mi}{{\text{n}}^{\text{-1}}}\]

From experiment II:

\[\text{k = }\frac{\text{7}\text{.3  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}}{\text{ }\!\![\!\!\text{ 0}\text{.3 }\!\!]\!\!\text{  }\!\![\!\!\text{ 0}\text{.2}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}}\] 

\[\text{= 6}\text{.0 }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{\text{-2}}}\text{ mi}{{\text{n}}^{\text{-1}}}\]

From experiment III:

\[\text{k = }\frac{\text{2}\text{.88  }\!\!\times\!\!\text{  1}{{\text{0}}^{-1}}}{\text{ }\!\![\!\!\text{ 0}\text{.3 }\!\!]\!\!\text{  }\!\![\!\!\text{ 0}\text{.4}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}}\]

\[\text{= 6}\text{.0 }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{\text{-2}}}\text{ mi}{{\text{n}}^{\text{-1}}}\] 

From experiment IV:

\[\text{k = }\frac{\text{2}\text{.40  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}}{\text{ }\!\![\!\!\text{ 0}\text{.4 }\!\!]\!\!\text{  }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}}\] 

\[\text{= 6}\text{.0 }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{\text{-2}}}\text{ mi}{{\text{n}}^{\text{-1}}}\]

Therefore, the rate constant will be k $\text{= 6}\text{.0 }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{\text{-2}}}\text{ mi}{{\text{n}}^{\text{-1}}}$

13. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Ans: We are given that, the reaction between A and B is first order with respect to A and zero order with respect to B.

Therefore, the rate of the reaction is given by:

\[\text{Rate = k }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{1}}}{{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}^{\text{0}}}\] 

So, we can write:

\[\text{Rate = k }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\]

According to the experiment I we can write:

\[\text{2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ = k(0}\text{.1)}\] 

\[\text{k = 0}\text{.2 mi}{{\text{n}}^{\text{-1}}}\] 

According to the experiment II we can write:

\[4.0\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}=(0.2)\text{  }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\] 

\[\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{  = 0}\text{.2 mol }{{\text{L}}^{\text{-1}}}\] 

According to the experiment III we can write:

\[\text{Rate = 0}\text{.2  }\!\!\times\!\!\text{  0}\text{.4}\]

\[\text{= 0}\text{.08 mol }{{\text{L}}^{\text{-1}}}\text{ mi}{{\text{n}}^{\text{-1}}}\] 

According to the experiment III we can write:

\[2.0\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}=(0.2)\text{  }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\] 

\[\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{  = 0}\text{.1 mol }{{\text{L}}^{\text{-1}}}\]

14. Calculate the half-life of a first order reaction from their rate constants given below:

(i). $\text{200 }{{\text{s}}^{\text{-1}}}$ 

Ans: Half-life of the reaction can be related with the rate constant of the reaction as:

\[{{\text{t}}_{\text{1/2}}}\text{ = }\frac{\text{0}\text{.693}}{\text{k}}\] 

Putting the value of time, we get:

\[\text{k = }\frac{\text{0}\text{.693}}{\text{200}}\text{ = 3}\text{.46  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ s}\] 

So, the rate of the reaction is $\text{3}\text{.46  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ s}$

(ii). $\text{2 mi}{{\text{n}}^{\text{-1}}}$ 

Ans: Half-life of the reaction can be related with the rate constant of the reaction as:

\[{{\text{t}}_{\text{1/2}}}\text{ = }\frac{\text{0}\text{.693}}{\text{k}}\] 

Putting the value of time, we get:

\[\text{k = }\frac{\text{0}\text{.693}}{\text{2}}\text{ = 0}\text{.346 min}\] 

So, the rate of the reaction is 0.346 min.

(iii). $\text{4 year}{{\text{s}}^{\text{-1}}}$ 

Ans: Half-life of the reaction can be related with the rate constant of the reaction as:

\[{{\text{t}}_{\text{1/2}}}\text{ = }\frac{\text{0}\text{.693}}{\text{k}}\] 

Putting the value of time, we get:

\[\text{k = }\frac{\text{0}\text{.693}}{4}\text{ = 0}\text{.173 years}\] 

So, the rate of the reaction is 0.173 years.

15. The half-life for radioactive decay of $^{\text{14}}\text{C}$ is 5730 years. An archeological artifact containing wood had only 80% of the $^{\text{14}}\text{C}$ found in a living tree. Estimate the age of the sample.

Ans: The given reaction in the above question is radioactive decay and the radioactive decay follows the first order kinetics. Therefore, the decay constant:

\[\text{ }\!\!\lambda\!\!\text{  = }\frac{\text{0}\text{.693}}{{{\text{t}}_{\text{1/2}}}}\] 

We are given a half-life of 5730 years. 

\[\text{ }\!\!\lambda\!\!\text{  = }\frac{\text{0}\text{.693}}{5730}\text{ year}{{\text{s}}^{-1}}\]  

To find the age of the sample, we can write:

\[\text{t = }\frac{\text{2}\text{.303}}{\text{ }\!\!\lambda\!\!\text{ }}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\] 

80% of the wood is found so, the initial amount can be taken as 100 and the final amount as 80. Putting the values, we get:

\[\text{t = }\frac{\text{2}\text{.303}}{\frac{0.693}{5730}}\text{ log}\frac{100}{80}\] 

t = 1845 years

Therefore, the age of the sample is 1845 years.

16. The experiment data for decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$

\[\text{ }\!\![\!\!\text{ 2}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\to \text{4N}{{\text{O}}_{\text{2}}}\text{+}{{\text{O}}_{\text{2}}}\text{ }\!\!]\!\!\text{ }\] 

In gas phase at 318 K are given below:

(i). Plot [${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$] against t

Ans: The graph is given below:

(ii). Find the half-life period for the reaction

Ans: Time corresponding to the concentration, 

$\frac{\text{1}\text{.630  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}}{\text{2}}\text{ mol }{{\text{L}}^{\text{-1}}}\text{=0}\text{.815  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ mol }{{\text{L}}^{\text{-1}}}$  is the half-life.

From the graph, the half life obtained is 1440 s.

(iii). Draw the graph between $\text{log }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\!\!]\!\!\text{ }$ and t

Ans: 

(iv). What is the rate law?

Ans: The rate law of the reaction will be:

\[\text{Rate = k }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\!\!]\!\!\text{ }\] 

(v). Calculate the rate constant.

Ans: From the plot $\text{ }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\!\!]\!\!\text{ }$ v/s t, is given by:

\[\text{= -}\frac{\text{k}}{\text{2}\text{.303}}\] 

Therefore, we obtain:

\[\text{= -}\frac{\text{k}}{\text{2}\text{.303}}=\frac{0.67}{3200}\] 

\[\text{= 4}\text{.82  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}\text{ mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

(vi). Calculate the half-life period from k and compare it will (ii)

Ans: Half-life is given by:

\[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{4.82\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-4}}}s\]

= 1438 seconds.

The value of half-life calculated from the k is very close to that obtained from the graph.

16. The rate constant for a first order reaction is $\text{60 }{{\text{s}}^{\text{-1}}}$ . How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Ans: As we know that:

\[\text{t = }\frac{\text{2}\text{.303}}{\text{k}}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\] 

The initial value of the reactant has become 1/16th. Now, putting the values, we get:

\[\text{t = }\frac{\text{2}\text{.303}}{60}\text{ log}\frac{16}{1}\] 

\[\text{t = }\frac{\text{2}\text{.303}}{\text{60}}\text{ log16 = 4}\text{.62  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ s}\] 

Therefore, the time required will be $4.62\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ s}$.

17. During a nuclear explosion, one of the products is $^{\text{90}}\text{Sr}$ with a half-life of 28.1 years. If 1 $\text{ }\!\!\mu\!\!\text{ g}$ of $^{\text{90}}\text{Sr}$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically. 

Ans: As radioactive disintegration follows first order kinetics.

Decay constant of $^{\text{90}}\text{Sr}$ (k) = $\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{28.1}=2.466\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ }{{\text{y}}^{-1}}$ 

To calculate the amount left after 10 years.

\[\text{a = 1  }\!\!\mu\!\!\text{ g}\] 

$t = 10 years$

\[\text{k=2}\text{.466  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{y}}^{\text{-1}}}\] 

$(a-x) = ?$

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{\text{a}}{\text{a-x}}\] 

\[\text{2}\text{.466  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ = }\frac{\text{2}\text{.303}}{10}\text{ log}\frac{1}{\text{a-x}}\] 

\[\text{log(a-x) = -0}\text{.1071}\] (a-x) = Antilog -0.1071 = 0.7814 \[\text{ }\!\!\mu\!\!\text{ g}\]

To calculate the amount left after 60 years.

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{\text{a}}{\text{a-x}}\] 

\[\text{2}\text{.466 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-2}}\text{ = }\frac{\text{2}\text{.303}}{60}\text{ log}\frac{1}{\text{a-x}}\] \[\text{log(a-x) = -0}\text{.6425}\]

(a-x) = Antilog -0.6425 = 0.2278 $\text{ }\!\!\mu\!\!\text{ g}$ 

18. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. 

Ans: For first order reaction, we can write:

\[{{t}_{99\text{ % } \!\!\!\!\text{ }}}=\frac{2.303}{k}\log \frac{a}{a-x}\] 

99% completion means that x = 99% of a = 0.99 a

So, we can write:

\[{{t}_{99\text{ % } \!\!\!\!\text{ }}}=\frac{2.303}{k}\log \frac{a}{a-0.99a}\]

\[{{{t}}_{\text{99 % }\!\!\!\!\text{ }}}\text{ = }\frac{\text{2}\text{.303}}{\text{k}}\text{ log 1}{{\text{0}}^{2}}\] 

\[{{t}_{\text{99%}\text{ }}}\text{ = 2}\times \frac{\text{2}\text{.303}}{\text{k}}\]

90% completion means that x = 90% of a = 0.90 a

\[{{t}_{99\text{ % } \!\!\!\!\text{ }}}=\frac{2.303}{k}\log \frac{a}{a-0.90a}\]

\[{{{t}}_{\text{99 % }\!\!\!\!\text{ }}}\text{ = }\frac{\text{2}\text{.303}}{\text{k}}\text{ log 1}{{\text{0}}^{}}\] 

\[{{t}_{\text{99%}\text{ }}}\text{ =}\frac{\text{2}\text{.303}}{\text{k}}\]

Now, we can take the ratio as given below:

\[\frac{{{t}_{99\text{%}\!\!\!\!\text{ }}}}{{{t}_{90\text{%}\!\!\!\!\text{ }}}}=\frac{\left( \frac{2\text{ x 2}\text{.303}}{k} \right)}{\left( \frac{2.303}{k} \right)}\] 

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

19. A first order reaction takes 40 min for 30% decomposition. Calculate ${{\text{t}}_{\text{1/2}}}$.

Ans: 30% decomposition means that x = 30% of a = 0.30 a

Since, the reaction is of 1st order, we can write:

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{\text{a}}{\text{a-x}}\] 

Time is given as 40 min. So, putting the values, we get:

\[\text{k = }\frac{\text{2}\text{.303}}{40}\text{ log}\frac{\text{a}}{\text{a-0}\text{.30a}}\] 

\[\text{k = }\frac{\text{2}\text{.303}}{40}\text{ log}\frac{10}{7}{{\min }^{-1}}\] 

\[\text{k = }\frac{\text{2}\text{.303}}{40}\text{  }\!\!\times\!\!\text{  0}\text{.1549 mi}{{\text{n}}^{-1}}=8.918\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}{{\min }^{-1}}\]  

Now, we can calculate the half-life period as we have the rate constant value.

We can write:

\[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{8.918\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}}=7.77\text{ }\min \]

So, the half-life is 7.77 min.

20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained:

Calculate the rate constant.

Ans:  The decomposition of azoisopropane to hexane and nitrogen at 54.3 k is represented by the following equation.

Total pressure after time t, we will be:

\[{{\text{P}}_{\text{t}}}\text{ = (}{{\text{P}}_{\text{o}}}\text{-p) + p + p}\] 

\[{{\text{P}}_{\text{t}}}\text{=}{{\text{P}}_{\text{o}}}\text{+p}\] 

\[\text{p=}{{\text{P}}_{\text{t}}}-{{\text{P}}_{\text{o}}}\] 

Now, we can substitute the value of p for the pressure of reactant at time t

\[\text{=}{{\text{P}}_{\text{o}}}\text{-p}\] 

\[\text{=}{{\text{P}}_{\text{o}}}\text{-(}{{\text{P}}_{\text{t}}}\text{-}{{\text{P}}_{\text{o}}}\text{)}\]

\[\text{= 2}{{\text{P}}_{\text{o}}}\text{ - }{{\text{P}}_{\text{t}}}\] 

Now, we can apply the rate constant formula of 1st order reaction.

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{\text{P}}{\text{2}{{\text{P}}_{\text{o}}}\text{-}{{\text{P}}_{\text{t}}}}\] 

When t = 360 s,

Putting the values, we get:

\[\text{k = }\frac{\text{2}\text{.303}}{360}\text{ log}\frac{35.0}{\text{2  }\!\!\times\!\!\text{  35 - 54}}=2.175\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}\text{ }{{\text{s}}^{-1}}\]  

When t = 720 s,

Putting the values, we get:

\[\text{k = }\frac{\text{2}\text{.303}}{720}\text{ log}\frac{35.0}{\text{2  }\!\!\times\!\!\text{  35 - 63}}=2.235\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}\text{ }{{\text{s}}^{-1}}\]

Now, we can find the average value:

\[\text{k = }\frac{\text{(2}\text{.175  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{)+(2}\text{.235  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{)}}{\text{2}}\text{ }{{\text{s}}^{\text{-1}}}\] 

\[\text{k = 2}\text{.20  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ }{{\text{s}}^{\text{-1}}}\]

21. The following data were obtained during the first order thermal decomposition of $\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ at a constant volume.

\[\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{{{\text{2}}_{\text{(g)}}}}}\to \text{S}{{\text{O}}_{{{\text{2}}_{\text{(g)}}}}}\text{+C}{{\text{l}}_{{{\text{2}}_{\text{(g)}}}}}\]    

Calculate the rate of the reaction when total pressure is 0.65 atm.

Ans: The given reaction shows the thermal decomposition of $\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ at constant volume.

Total pressure after time t, we will be:

\[{{\text{P}}_{\text{t}}}\text{ = (}{{\text{P}}_{\text{o}}}\text{-p) + p + p}\] 

\[{{\text{P}}_{\text{t}}}\text{=}{{\text{P}}_{\text{o}}}\text{+p}\] 

\[\text{p=}{{\text{P}}_{\text{t}}}-{{\text{P}}_{\text{o}}}\] 

Now, we can substitute the value of p for the pressure of reactant at time t

\[\text{=}{{\text{P}}_{\text{o}}}\text{-p}\] 

\[\text{=}{{\text{P}}_{\text{o}}}\text{-(}{{\text{P}}_{\text{t}}}\text{-}{{\text{P}}_{\text{o}}}\text{)}\]

\[\text{= 2}{{\text{P}}_{\text{o}}}\text{ - }{{\text{P}}_{\text{t}}}\] 

Now, we can apply the rate constant formula of 1st order reaction.

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{\text{P}}{\text{2}{{\text{P}}_{\text{o}}}\text{-}{{\text{P}}_{\text{t}}}}\] 

When the t = 100 s

\[\text{k = }\frac{\text{2}\text{.303}}{100}\text{ log}\frac{0.5}{\text{2  }\!\!\times\!\!\text{  0}\text{.5 - 0}\text{.6}}\] 

\[\text{k = 2}\text{.231  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

When ${{\text{P}}_{\text{t}}}$ = 0.65 atm 

Therefore, pressure of $\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ at time t total pressure is 0.65 atm, is

${{\text{P}}_{\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}\text{=2}{{\text{P}}_{\text{o}}}\text{-}{{\text{P}}_{\text{t}}}$ 

= 2 $\text{ }\!\!\times\!\!\text{ }$ 0.50 – 0.65

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by:

\[\text{Rate = k(}{{\text{P}}_{\text{S}{{\text{O}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}\text{)}\] 

\[\text{Rate = (2}\text{.33 x 1}{{\text{0}}^{\text{-3}}}\text{)(0}\text{.354) = 7}\text{.8  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}\text{atm }{{\text{s}}^{\text{-1}}}\] 

22. The rate constant for the decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ at various temperature is given below:

Draw a graph between ln K and 1/T and calculate the values of A and ${{\text{E}}_{\text{a}}}$. Predict the rate constant at $\text{3}{{\text{0}}^{\text{o}}}$ and $\text{5}{{\text{0}}^{\text{o}}}$C.

Ans: As the data is given we can write:

The graph is draw below:

Slope of the line, will be given as:

\[\frac{{{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{1}}}}{{{\text{x}}_{\text{2}}}\text{-}{{\text{x}}_{\text{1}}}}\text{ = 12}\text{.301 K}\] 

According to the Arrhenius equation,

\[\text{Slope = -}\frac{{{\text{E}}_{\text{a}}}}{\text{R}}\] 

\[\text{= }{{\text{E}}_{\text{a}}}\text{= -slope  }\!\!\times\!\!\text{  R}\]

= (-12.301) (8.314)

\[\text{=102}\text{.27 KJ mo}{{\text{l}}^{\text{-1}}}\] 

Again,

\[\text{ln k = ln A - }\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}\] 

\[\text{ln A = ln k + }\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}\] 

As T = 273 K and ln k = -7.147

Applying this in the formula, we get:

\[\text{ln A = -7}\text{.147 - }\frac{102.27\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{3}}}{8.314\text{ }\!\!\times\!\!\text{ 273}}=37.911\] So, \[\text{A=2}\text{.91 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\]

When T = 30 + 273 K = 303 K

\[\frac{\text{1}}{\text{T}}\text{ = 0}\text{.0033K = 3}\text{.3  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{K}\] 

Now, at $\frac{\text{1}}{\text{T}}\text{ = 0}\text{.0033K = 3}\text{.3  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{K}$

ln k = - 2.8

Therefore, $\text{k = 6}\text{.08  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-1}}}$  

When T = 50 + 273 K = 323 K

\[\frac{\text{1}}{\text{T}}\text{ = 0}\text{.0031K = 3}\text{.1  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{K}\] 

Now, at $\frac{\text{1}}{\text{T}}\text{ = 0}\text{.0031K = 3}\text{.1  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{K}$

ln k = - 0.5

Therefore, $\text{k = 0}\text{.607 }{{\text{s}}^{\text{-1}}}$

23. The rate constant for the decomposition of hydrocarbons is $\text{2}\text{.418  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ }{{\text{s}}^{\text{-1}}}$ at 546 K. If the energy of activation is 179.9 kJ /mol, what will be the value of pre-exponential factor.

Ans: We are given some values as:

\[\text{K = 2}\text{.418  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ }{{\text{s}}^{\text{-1}}}\]   

\[T = 546 K\]

\[{{\text{E}}_{\text{a}}}\text{ = 179}\text{.9 kJ mo}{{\text{l}}^{\text{-1}}}=179.9\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{3}}\text{ J mo}{{\text{l}}^{-1}}\] 

We the Arrhenius equation is:

\[\text{k = A }{{\text{e}}^{\text{-}{{\text{E}}_{\text{a}}}\text{/RT}}}\] 

In the log form, this can be written as:

\[\text{ln k = ln A - }\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}\] 

\[\text{log k = log A - }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303 RT}}\] 

\[\text{log k = log (2}\text{.418  }\!\!\times\!\!\text{  1}{{\text{0}}^{-5}}\text{) - }\frac{179.9\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{3}}}{\text{2}\text{.303 }\!\!\times\!\!\text{  8}\text{.314  }\!\!\times\!\!\text{  546}}\] 

= (0.3835 – 5) + 17.2082 = 12.5917

Therefore, A = antilog (12.5917)

\[\text{A = 3}\text{.912  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{12}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

24. Consider a certain reaction $\text{A }\to \text{ Products}$ with $\text{k = 2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-1}}}$. Calculate the concentration of A remaining after 100s if the initial concentration of A is $\text{1}\text{.0 mol }{{\text{L}}^{\text{-1}}}$.

Ans: We are given some values, that are given below:

\[\text{k = 2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

\[t = 100 s\]

\[{{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}_{\text{o}}}\text{=1}\text{.0 mol }{{\text{L}}^{\text{-1}}}\] 

As we can see that the units of k are given in ${{\text{s}}^{\text{-1}}}$, this means that the reaction is a first order reaction.

Therefore, we can write:

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}\] 

Putting the values, we get:

\[\text{2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ = }\frac{\text{2}\text{.303}}{100}\text{ log}\frac{1.0}{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}\] 

\[\text{2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{ = }\frac{\text{2}\text{.303}}{100}\text{ (-log }\!\![\!\!\text{ A }\!\!]\!\!\text{ )}\] 

\[\text{(-log }\!\![\!\!\text{ A }\!\!]\!\!\text{ ) = }\frac{\text{2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{  }\!\!\times\!\!\text{  100}}{2.303}\] 

\[\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{  =antilog }\left( \frac{\text{2}\text{.0  }\!\!\times\!\!\text{  1}{{\text{0}}^{-2}}\text{  }\!\!\times\!\!\text{  100}}{2.303} \right)\] 

\[=0.135\text{ mol }{{\text{L}}^{-1}}\] 

Therefore, the remaining amount of A is $0.135\text{ mol }{{\text{L}}^{-1}}$.

25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with${{\text{t}}_{\text{1/2}}}\text{ = 3}\text{.00 hours}$. What fraction of a sample of sucrose remains after 8 hours?

Ans: The given reaction is a first order reaction. So, we can write:

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\] 

It is given that the half-life is 3 hours. Therefore, we can write:

\[\text{k = }\frac{\text{0}\text{.693}}{{{\text{t}}_{\text{1/2}}}}\] 

So, putting the values in this, we get:

\[\text{k = }\frac{\text{0}\text{.693}}{\text{3}}\text{ = 0}\text{.231 }{{\text{h}}^{\text{-1}}}\] 

Now, we can put this value of rate constant in the first order reaction formula.

\[\text{0}\text{.231 = }\frac{\text{2}\text{.303}}{8}\text{ log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\] 

\[\text{log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\text{ = }\frac{\text{0}\text{.231  }\!\!\times\!\!\text{  8}}{2.303}\text{ }\] 

\[\text{log}\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\text{ = 0}\text{.8024}\] 

\[\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\text{ = antilog (0}\text{.8024)}\] 

\[\frac{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{\text{0}}}}{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}\text{ = 6}\text{.3445}\] 

Or we can write:

\[\frac{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}{{{\text{ }\!\![\!\!\text{ R }\!\!]\!\!\text{ }}_{0}}}\text{ = 0}\text{.158}\] 

Therefore, the fraction of sample of sucrose that remains after 8 hours is 0.158.

26. The decomposition of hydrocarbon follows the equation $\text{k = (4}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{11}}}\text{ }{{\text{s}}^{\text{-1}}}\text{)}{{\text{e}}^{\text{-280000K/T}}}$. Calculate ${{\text{E}}_{\text{a}}}$.

Ans: According to the Arrhenius equation,

\[\text{k = A}{{\text{e}}^{\text{-}{{\text{E}}_{\text{a}}}\text{/RT}}}\]   

We are given the equation as:

\[\text{k = (4}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{11}}}\text{ }{{\text{s}}^{\text{-1}}}\text{)}{{\text{e}}^{\text{-280000K/T}}}\] 

Therefore, the formula can be written as:

\[\text{-}\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}\text{ = -}\frac{\text{28000 K}}{\text{T}}\] 

This can be written as:

\[{{\text{E}}_{\text{a}}}\text{ = 28000  }\!\!\times\!\!\text{  R}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 28000  }\!\!\times\!\!\text{  8}\text{.314 = 232}\text{.79 kJ mo}{{\text{l}}^{\text{-1}}}\] 

Therefore, the value of ${{\text{E}}_{\text{a}}}$ is $\text{232}\text{.79 kJ mo}{{\text{l}}^{\text{-1}}}$

27. The rate constant for the first order decomposition of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is given by the following equation: 

$\text{log k = 14}\text{.34 - 1}\text{.25  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ K/T}$ 

Calculate ${{\text{E}}_{\text{a}}}$ for this reaction and at what temperature will its half-period be 256 minutes?    

Ans: According to the Arrhenius equation,

\[\text{k = A}{{\text{e}}^{\text{-}{{\text{E}}_{\text{a}}}\text{/RT}}}\]

This can be written as:

\[\text{ln k = ln A - }\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}\] 

In the log form it can be written as:

\[\text{log k = log A - }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303 RT}}\] 

We are given:

\[\text{log k = 14}\text{.34 - 1}\text{.25  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ K/T}\] 

Comparing these two, we get:

\[\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303 RT}}\text{ = }\frac{\text{1}\text{.25  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ K}}{\text{T}}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 2}\text{.303 R  }\!\!\times\!\!\text{  1}\text{.25  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ K}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314  }\!\!\times\!\!\text{  1}\text{.25  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ K}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 239}\text{.34 kJ mo}{{\text{l}}^{-1}}\] 

We are given a half-life of 256 minutes.

\[\text{k = }\frac{\text{0}\text{.693}}{{{\text{t}}_{\text{1/2}}}}\] 

\[\text{k = }\frac{\text{0}\text{.693}}{\text{256  }\!\!\times\!\!\text{  60}}\text{ = 4}\text{.51  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{ }{{\text{s}}^{\text{-1}}}\] 

Now, we have the value of rate constant, we can put in the equation:

\[\text{log(4}\text{.51  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-5}}}\text{) = 14}\text{.34 - }\frac{\text{1}\text{.25  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}}{\text{T}}\] 

T = 669 K

28. The decomposition of A into product has value of k as $\text{4}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{3}}}\text{ }{{\text{s}}^{\text{-1}}}$  at $\text{1}{{\text{0}}^{\text{o}}}\text{C}$ and energy of activation $\text{60 kJ mo}{{\text{l}}^{\text{-1}}}$. At what temperature would k be $\text{1}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ }{{\text{s}}^{\text{-1}}}$?

Ans: We are some information:

\[{{\text{k}}_{\text{1}}}\text{ = 4}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{3}}}\] 

\[{{\text{T}}_{\text{1}}}\text{ = 10 + 273 = 283 K}\] 

\[{{\text{k}}_{2}}\text{ = 1}\text{.5  }\!\!\times\!\!\text{  1}{{\text{0}}^{4}}\] 

\[{{\text{T}}_{\text{2}}}=?\] 

\[{{\text{E}}_{\text{a}}}\text{ = 60 kJ mo}{{\text{l}}^{\text{-1}}}\] 

Applying Arrhenius equation:

\[\text{log}\frac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303R}}\left( \frac{{{\text{T}}_{\text{2}}}\text{-}{{\text{T}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}} \right)\] 

Putting the values, we can write:

\[\text{log}\frac{1.5\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{4}}}{4.5\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{3}}}\text{ = }\frac{60}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314}}\left( \frac{{{\text{T}}_{\text{2}}}\text{-283}}{\text{283}{{\text{T}}_{\text{2}}}} \right)\] 

\[\text{log 3}\text{.333 = 3133}\text{.63}\left( \frac{{{\text{T}}_{\text{2}}}\text{-283}}{\text{283}{{\text{T}}_{\text{2}}}} \right)\] 

\[\frac{0.5228}{3133.63}=\left( \frac{{{\text{T}}_{\text{2}}}\text{-283}}{\text{283}{{\text{T}}_{\text{2}}}} \right)\] 

\[\text{0}\text{.0472}{{\text{T}}_{\text{2}}}\text{ = }{{\text{T}}_{\text{2}}}\text{ - 283}\] 

\[{{\text{T}}_{\text{2}}}\text{ = 297 K }\] 

Or we can write:

\[{{\text{T}}_{\text{2}}}\text{ = 2}{{\text{4}}^{\text{o}}}\text{C}\] 

29. The time required for 10% completion of a first order reaction at 298 k is equal to that required for its 25% completion at 308 K. If the value of A is $\text{4  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{10}}}\text{ }{{\text{s}}^{\text{-1}}}$ , Calculate k at 318 K and ${{\text{E}}_{\text{a}}}$.

Ans: There are two cases in this question. As the reaction given is first order reaction, we can use:

\[\text{k = }\frac{\text{2}\text{.303}}{\text{t}}\text{ log}\frac{\text{a}}{\text{a-x}}\] 

For case 1:

\[{{\text{k}}_{\text{298K}}}\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{\text{1}}}}\text{ log}\frac{\text{a}}{\text{a - 0}\text{.10a}}\] 

\[{{\text{k}}_{\text{298K}}}\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{\text{1}}}}\text{ log}\frac{10}{9}\] 

\[{{\text{k}}_{\text{298K}}}\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{\text{1}}}}\text{  }\!\!\times\!\!\text{  0}\text{.0458}\] 

\[{{\text{t}}_{\text{1}}}\text{ = }\frac{\text{0}\text{.1055}}{{{\text{k}}_{\text{298K}}}}\] 

For case 2:

\[{{\text{k}}_{\text{308K}}}\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{2}}}\text{ log}\frac{\text{a}}{\text{a - 0}\text{.25a}}\] 

\[{{\text{k}}_{\text{308K}}}\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{2}}}\text{ log}\frac{4}{3}\] 

\[{{\text{k}}_{\text{308K}}}\text{ = }\frac{\text{2}\text{.303}}{{{\text{t}}_{2}}}\text{  }\!\!\times\!\!\text{  0}\text{.125}\] 

\[{{\text{t}}_{2}}\text{ = }\frac{\text{0}\text{.2879}}{{{\text{k}}_{\text{308K}}}}\]

But ${{\text{t}}_{\text{1}}}\text{ = }{{\text{t}}_{\text{2}}}$ 

Hence,

\[\frac{\text{0}\text{.1055}}{{{\text{k}}_{\text{298K}}}}\text{ = }\frac{\text{0}\text{.2879}}{{{\text{k}}_{\text{308K}}}}\] 

\[\frac{{{\text{k}}_{\text{308K}}}}{{{\text{k}}_{\text{298K}}}}\text{ = 2}\text{.7289}\]  

Now, applying the Arrhenius equation,

\[\text{log}\frac{{{\text{k}}_{\text{308K}}}}{{{\text{k}}_{\text{298K}}}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303R}}\left( \frac{{{\text{T}}_{\text{2}}}\text{-}{{\text{T}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}} \right)\] 

\[\text{log(2}\text{.7289) = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314}}\left( \frac{\text{308-298}}{298\text{  }\!\!\times\!\!\text{  308}} \right)\] 

\[{{\text{E}}_{\text{a}}}\text{ = 76}\text{.623 kJ mo}{{\text{l}}^{\text{-1}}}\] 

Now, the calculation of k at 318 K

\[\text{log k = log A - }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303RT}}\]

\[\text{log k = log (4  }\!\!\times\!\!\text{  1}{{\text{0}}^{10}}\text{) - }\frac{7623}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314  }\!\!\times\!\!\text{  318}}\] 

\[\text{log k = 10}\text{.6021 - 12}\text{.5843 = -1}\text{.9822}\] 

k = Antilog (-1.9822) = $\text{antilog (\bar{2}}\text{.0178) = 1}\text{.042  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-1}}}$ 

Therefore, k is $\text{1}\text{.042  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-1}}}$.

30. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Ans: We are given that:

\[{{\text{k}}_{\text{2}}}\text{ = 4 }{{\text{k}}_{\text{1}}}\] 

\[{{\text{T}}_{\text{1}}}\text{ = 293K}\]

\[{{\text{T}}_{2}}\text{ = 313K}\]  

According the Arrhenius equation, we get:

\[\text{log}\frac{{{\text{k}}_{2}}}{{{\text{k}}_{1}}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303R}}\left( \frac{{{\text{T}}_{\text{2}}}\text{-}{{\text{T}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}} \right)\] 

Putting the values, we get:

\[\text{log}\frac{\text{4}{{\text{k}}_{1}}}{{{\text{k}}_{1}}}\text{ = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314}}\left( \frac{\text{313-293}}{293\text{  }\!\!\times\!\!\text{  313}} \right)\] 

\[\text{0}\text{.6021 = }\frac{{{\text{E}}_{\text{a}}}}{\text{2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314}}\left( \frac{\text{313-293}}{293\text{  }\!\!\times\!\!\text{  313}} \right)\] 

\[{{\text{E}}_{\text{a}}}\text{=}\frac{\text{0}\text{.6021  }\!\!\times\!\!\text{  2}\text{.303  }\!\!\times\!\!\text{  8}\text{.314  }\!\!\times\!\!\text{  293  }\!\!\times\!\!\text{  313}}{\text{20}}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 52863}\text{.00 J mo}{{\text{l}}^{\text{-1}}}\] 

\[{{\text{E}}_{\text{a}}}\text{ = 52}\text{.863 kJ mo}{{\text{l}}^{\text{-1}}}\] 

Therefore, the required activation energy is $\text{52}\text{.863 kJ mo}{{\text{l}}^{\text{-1}}}$.

Chemical Kinetics Class 12 NCERT Solutions PDF Download

Physical Chemistry which is a crucial part of the subject requires thorough reading and understanding of the concepts. Therefore, focussing on qualitative ways of preparation rather than quantitative measures would turn out to be fruitful for an aspiring board examinee.

To solve this issue of preparing from qualitative sources, educational sites like Vedantu provide free solutions in PDF format. The Chemical Kinetics Class 12 NCERT Solutions PDF is free to download and provides an elaborate explanation of the topic. Prepared under expert supervision, these solutions are the perfect partner of high scoring students.

CBSE Class 12 Chemistry Chemical Kinetics NCERT Solutions

The students will learn about the practical methods of determining equations by initial rate data method, logarithm data method, half-line method, by integrated rate equation and isolation method in Chapter 4 Chemistry Class 12. They will also learn the integrated rate equations which include zero-order reaction, first line, and half-line methods. Moreover, there are eight sub-topics which are prepared to help in the all-round development of students.

For a better understanding of topics, the solution has been divided into in-text questions and additional NCERT exercises. There are nine in-text questions in ch 4 Chemistry Class 12 book, with 30 additional exercises. To make the best out of their preparation, students can refer to the Class 12th Chemistry Chapter 4 NCERT solution which gives answers apt for scoring well.

In-text Questions: NCERT Solution of Chemical Kinetics Class 12th

The nine intext question makes students access their reasoning and logical skill by asking questions like average rate of reaction, the effect of rate of formation of y in a reaction, rate constant, a fraction of molecules and rate of reaction when the concentration of reactants is doubled or reduced. Furthermore, students can follow Class 12 Chemistry ch 4 NCERT solutions to understand the pattern of writing answers that can help fetch a better score.

NCERT Exercises: Chemical Kinetics Class 12th

Apart from the nine intext questions, Chapter 4 Chemistry Class 12 solutions contain an additional 30 exercises designed for comprehensive knowledge of students. The Chemical Kinetics Class 12 solutions contain questions like rate expressions for chemical reaction, an initial rate of reaction for an equation, units of rate, and rate constant when pressure is measured in bar, time and energy activation of the reaction assuming that it doesn't change with temperature, etc.

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Marks Distribution of NCERT Solutions Class 12 Chemistry Chemical Kinetics

NCERT Solutions Class 12 Chemistry - Related Links

Check out the links provided below to access CBSE Class 12 Chemistry syllabus, important questions, revision notes,  previous years’ questions papers and sample papers.

• Class 12 Chemistry Revision Notes

• CBSE Class 12 Chemistry Previous Year Question Papers 

• CBSE Sample Papers for Class 12 Chemistry with Solutions

• CBSE Syllabus for Class 12 Chemistry Syllabus

• Important Questions for CBSE Class 12 Chemistry

• NCERT Exemplar for Class 12 Chemistry

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• Revision Notes for Class 12

• Important Questions for Class 12

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