Answer
Verified
405k+ views
Hint: To solve this question, you have to simply apply the concept of rate law or rate equation which generally gives an idea about the relationship between the rate of a chemical reaction and the concentration of the reactants in the reaction.
Complete step by step solution:
Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the reactants concentration. The rate law is given as:
$rate=\text{ k }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}^{y}}$ , where
k is the rate constant,
[A] and [B] are the molar concentrations of the reactants and
x and y represents the reaction order.
Given that, Ethanal, dimerises in an alkaline solution according to the following equation.
$2C{{H}_{3}}CHO\to C{{H}_{3}}CH(OH)C{{H}_{2}}CHO$
So, here the rate law can be written as:
$rate=\text{ k }\!\![\!\!\text{ C}{{\text{H}}_{3}}\text{CHO}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]}^{y}}$, where
$[C{{H}_{3}}CHO]$ represents the concentration of $C{{H}_{3}}CHO$ and
$[O{{H}^{-}}]$ represents the concentration of $O{{H}^{-}}$.
As per the information provided,
For first case, the rate law will be:
$1=\text{ k }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{\text{ }\!\![\!\!\text{ 0}\text{.015}]}^{y}}$ (i)
For second case, the rate law will be written as:
$2=\text{k }\!\![\!\!\text{ }\text{.2}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{[0.015]}^{y}}$ (ii)
For third case, the rate law will be:
$8=\text{k }\!\![\!\!\text{ }\text{.4}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{[0.030]}^{y}}$ (iii)
(i) Dividing equation (ii) from (i), we get:
$2={{\left[ \dfrac{0.2}{0.1} \right]}^{x}}$
Then, $2={{2}^{x}}$
So, $x=1$.
Thus, the order of the reaction with respect to $C{{H}_{3}}CHO$ is first order.
(ii) Dividing equation (iii) from (i), we get:
$8={{\left[ \dfrac{0.4}{0.1} \right]}^{x}}{{[2]}^{y}}$
Putting the value of x from above solution, we get:
$8={{\left[ 4 \right]}^{1}}{{[2]}^{y}}$
Then, $y=1$.
Thus, the order of the reaction with respect to $O{{H}^{-}}$ is first order.
(iii) The overall rate equation or rate law for this reaction is $rate=\text{ }\!\![\!\!\text{ C}{{\text{H}}_{3}}\text{CHO }\!\!]\!\!\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]$.
(iv) The unit for the rate constant is $mo{{l}^{-1}}d{{m}^{3}}{{s}^{-1}}$ as per any suitable time unit.
Note: It is important to note that; the rate of a reaction is a vital property because it says whether a reaction can occur during a lifetime. The rate law says about the concentration of the reactants and the order of the reaction.
Complete step by step solution:
Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the reactants concentration. The rate law is given as:
$rate=\text{ k }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}^{y}}$ , where
k is the rate constant,
[A] and [B] are the molar concentrations of the reactants and
x and y represents the reaction order.
Given that, Ethanal, dimerises in an alkaline solution according to the following equation.
$2C{{H}_{3}}CHO\to C{{H}_{3}}CH(OH)C{{H}_{2}}CHO$
So, here the rate law can be written as:
$rate=\text{ k }\!\![\!\!\text{ C}{{\text{H}}_{3}}\text{CHO}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]}^{y}}$, where
$[C{{H}_{3}}CHO]$ represents the concentration of $C{{H}_{3}}CHO$ and
$[O{{H}^{-}}]$ represents the concentration of $O{{H}^{-}}$.
As per the information provided,
For first case, the rate law will be:
$1=\text{ k }\!\![\!\!\text{ 0}\text{.1}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{\text{ }\!\![\!\!\text{ 0}\text{.015}]}^{y}}$ (i)
For second case, the rate law will be written as:
$2=\text{k }\!\![\!\!\text{ }\text{.2}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{[0.015]}^{y}}$ (ii)
For third case, the rate law will be:
$8=\text{k }\!\![\!\!\text{ }\text{.4}{{\text{ }\!\!]\!\!\text{ }}^{x}}{{[0.030]}^{y}}$ (iii)
(i) Dividing equation (ii) from (i), we get:
$2={{\left[ \dfrac{0.2}{0.1} \right]}^{x}}$
Then, $2={{2}^{x}}$
So, $x=1$.
Thus, the order of the reaction with respect to $C{{H}_{3}}CHO$ is first order.
(ii) Dividing equation (iii) from (i), we get:
$8={{\left[ \dfrac{0.4}{0.1} \right]}^{x}}{{[2]}^{y}}$
Putting the value of x from above solution, we get:
$8={{\left[ 4 \right]}^{1}}{{[2]}^{y}}$
Then, $y=1$.
Thus, the order of the reaction with respect to $O{{H}^{-}}$ is first order.
(iii) The overall rate equation or rate law for this reaction is $rate=\text{ }\!\![\!\!\text{ C}{{\text{H}}_{3}}\text{CHO }\!\!]\!\!\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]$.
(iv) The unit for the rate constant is $mo{{l}^{-1}}d{{m}^{3}}{{s}^{-1}}$ as per any suitable time unit.
Note: It is important to note that; the rate of a reaction is a vital property because it says whether a reaction can occur during a lifetime. The rate law says about the concentration of the reactants and the order of the reaction.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE