Chapter 3: Chemical Kinetics

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 $6.66\text{  }\!\!\times\!\!\text{  1}{{\text{0}}^{-6}}\text{ M }{{\text{s}}^{-1}}$

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 $\text{5  }\!\!\times\!\!\text{  1}{{\text{0}}^{-3}}\text{ M mi}{{\text{n}}^{-1}}$

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.

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.

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.

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 $\text{1}\text{.925 }\!\!~\!\!\text{  }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ }{{\text{s}}^{\text{-1}}}$.

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.

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:

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

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

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

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

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

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)

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