What is the difference between rate law expression and the rate obtained from law of mass action?
A. Rate law expression is experimentally obtained, while law of mass action is theoretically obtained
B. Rate law expression is theoretically obtained, while law of mass action is experimentally obtained
C. Law of mass action depends on the stoichiometric coefficient of the reactants, while rate law expression may or may not
D. Rate law expression depends on the stoichiometric coefficient of the reactants, while law of mass action doesn’t
Answer
636.9k+ views
Hint: Law of mass action is a concept of masses and concentrations but rate law is a concept of chemical kinetics. Both the terms are related, the rate of a chemical reaction is directly proportional to the concentration of reactants. Like this, $\text{r}\propto {{\left[ \text{A} \right]}^{\text{n}}}$. Write the differences between the two terms and the answer will be obtained.
Complete step by step answer:
Let us discuss the differences between the rate law expression and rate from law of mass action:
The correct answer of this question is Rate law expression is experimentally obtained, while law of mass action is theoretically obtained and Law of mass action depends on the stoichiometric coefficient of the reactants, while rate law expression may or may not.
The correct options are ‘a’ and ‘c’.
Note: Some of examples where rate law is not equal to law of mass action are-
(1) $\text{CHC}{{\text{l}}_{3}}+\text{C}{{\text{l}}_{2}}\to \text{CC}{{\text{l}}_{4}}+\text{HCl}$. Rate expression will be rate = k$\left[ \text{CHC}{{\text{l}}_{3}} \right]{{\left[ \text{C}{{\text{l}}_{2}} \right]}^{{}^{1}/{}_{2}}}$.
(2) $\text{C}{{\text{H}}_{3}}\text{COO}{{\text{C}}_{2}}{{\text{H}}_{5}}+{{\text{H}}_{2}}\text{O}\to \text{C}{{\text{H}}_{3}}\text{COOH}+{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}$. Rate expression will be rate = k${{\left[ \text{C}{{\text{H}}_{3}}\text{COO}{{\text{C}}_{2}}{{\text{H}}_{5}} \right]}^{1}}{{\left[ {{\text{H}}_{2}}\text{O} \right]}^{0}}$.
Complete step by step answer:
Let us discuss the differences between the rate law expression and rate from law of mass action:
| RATE LAW | LAW OF MASS ACTION |
| It is an expression that tells the rate of a reaction | It is an equation which represents the ratio between the masses of products and masses of reactants whose power is raised according to their stoichiometric coefficients. |
| Its expression contains the rate constant, concentration of reactants and the order of the reaction. Has no specific expression. | Its expression is written as Reaction is $\text{2A}+\text{3B}\to $ productsExpression is $\text{r}\propto {{\left[ \text{A} \right]}^{2}}{{\left[ \text{B} \right]}^{3}}$ which includes the concentration of products and reactants and their stoichiometric powers. |
| Concentrations of reactants are raised to powers which may or may not be equal to the stoichiometric coefficient of the reaction. It is experimental data. | Concentrations of reactants are raised to powers of the stoichiometric coefficient of the reaction. It is a theoretical concept. |
| Gives the rate of a particular reaction. | Gives the favouring direction in which reaction will proceed. |
| Products of a reaction are not included in the expression | Products of a reaction are included in the expression |
| A proportionality constant $\left( \text{k} \right)$ is used | A proportionality constant $\left( \text{k} \right)$ is not used |
The correct answer of this question is Rate law expression is experimentally obtained, while law of mass action is theoretically obtained and Law of mass action depends on the stoichiometric coefficient of the reactants, while rate law expression may or may not.
The correct options are ‘a’ and ‘c’.
Note: Some of examples where rate law is not equal to law of mass action are-
(1) $\text{CHC}{{\text{l}}_{3}}+\text{C}{{\text{l}}_{2}}\to \text{CC}{{\text{l}}_{4}}+\text{HCl}$. Rate expression will be rate = k$\left[ \text{CHC}{{\text{l}}_{3}} \right]{{\left[ \text{C}{{\text{l}}_{2}} \right]}^{{}^{1}/{}_{2}}}$.
(2) $\text{C}{{\text{H}}_{3}}\text{COO}{{\text{C}}_{2}}{{\text{H}}_{5}}+{{\text{H}}_{2}}\text{O}\to \text{C}{{\text{H}}_{3}}\text{COOH}+{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}$. Rate expression will be rate = k${{\left[ \text{C}{{\text{H}}_{3}}\text{COO}{{\text{C}}_{2}}{{\text{H}}_{5}} \right]}^{1}}{{\left[ {{\text{H}}_{2}}\text{O} \right]}^{0}}$.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

