
For the reaction, \[3A+2B\to C+D\], the differential rate law can be written as:
A.\[3\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]
B.\[-\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]
C.\[+\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]
D.\[-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]
Answer
589.2k+ views
Hint: To solve this question, you need to apply the concepts of chemical kinetics. The differential rate law can be related to the rate of appearance (of product) and the rate of disappearance (of reactant).
Complete step by step answer:
The general formula for appearance or disappearance of a compound is calculated by the formula –
\[\pm \dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{reactant/product}}{\text{time}}\]
The rate of disappearance has a ‘-’ sign, because the rate decreases with time.
Rate of appearance of C \[=+\dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{product}}{\text{time}}\]
Rate of appearance of C \[=+\dfrac{\text{1}}{1}\text{x}\dfrac{\text{d }\!\![\!\!\text{ C }\!\!]\!\!\text{ }}{\text{dt}}\]
Rate of disappearance of A = \[-\dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{reactant}}{\text{time}}\]
Rate of disappearance of A \[=-\dfrac{\text{1}}{3}\text{x}\dfrac{\text{d }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}{\text{dt}}\]
The rate of appearance and disappearance is always the same.
Therefore, \[-\dfrac{\text{1}}{3}\dfrac{\text{d }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}{\text{dt}}=\dfrac{\text{d }\!\![\!\!\text{ C }\!\!]\!\!\text{ }}{\text{dt}}\]
Now, let us write the rate law –
Rate law = \[k={{[A]}^{n}}{{[B]}^{m}}\], where, n and m are the stoichiometric coefficients of the reacting elements.
For the given equation, the value of n and m are 3 and 2 respectively.
So, we can combine the two equations and write the rate of the reaction as –
\[-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]
Therefore, the answer is – option (d).
Additional Information: The word kinetic is derived from the Greek word ‘kinesis’ which means movement.
Note: As we can see, the rate of a reaction depends on concentration and time. Therefore, it is expressed in terms of concentration / time. If concentration is in mol/L, the unit of rate becomes \[mol{{L}^{-1}}{{s}^{-1}}\]. However, in case of gases, concentration is expressed in terms of its partial pressure. The rate becomes atm/s.
Complete step by step answer:
The general formula for appearance or disappearance of a compound is calculated by the formula –
\[\pm \dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{reactant/product}}{\text{time}}\]
The rate of disappearance has a ‘-’ sign, because the rate decreases with time.
Rate of appearance of C \[=+\dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{product}}{\text{time}}\]
Rate of appearance of C \[=+\dfrac{\text{1}}{1}\text{x}\dfrac{\text{d }\!\![\!\!\text{ C }\!\!]\!\!\text{ }}{\text{dt}}\]
Rate of disappearance of A = \[-\dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{reactant}}{\text{time}}\]
Rate of disappearance of A \[=-\dfrac{\text{1}}{3}\text{x}\dfrac{\text{d }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}{\text{dt}}\]
The rate of appearance and disappearance is always the same.
Therefore, \[-\dfrac{\text{1}}{3}\dfrac{\text{d }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}{\text{dt}}=\dfrac{\text{d }\!\![\!\!\text{ C }\!\!]\!\!\text{ }}{\text{dt}}\]
Now, let us write the rate law –
Rate law = \[k={{[A]}^{n}}{{[B]}^{m}}\], where, n and m are the stoichiometric coefficients of the reacting elements.
For the given equation, the value of n and m are 3 and 2 respectively.
So, we can combine the two equations and write the rate of the reaction as –
\[-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]
Therefore, the answer is – option (d).
Additional Information: The word kinetic is derived from the Greek word ‘kinesis’ which means movement.
Note: As we can see, the rate of a reaction depends on concentration and time. Therefore, it is expressed in terms of concentration / time. If concentration is in mol/L, the unit of rate becomes \[mol{{L}^{-1}}{{s}^{-1}}\]. However, in case of gases, concentration is expressed in terms of its partial pressure. The rate becomes atm/s.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Two Planoconcave lenses 1 and 2 of glass of refractive class 12 physics CBSE

The compound 2 methyl 2 butene on reaction with NaIO4 class 12 chemistry CBSE

Bacterial cell wall is made up of A Cellulose B Hemicellulose class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

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

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

