What is the rate of reaction for $2A \to B$ ?
A. $ - \dfrac{1}{2}[\dfrac{{d[A]}}{{dt}}]$
B. $ - \dfrac{{d[A]}}{{dt}}$
C. $ - \dfrac{{d[B]}}{{dt}}$
D. $ + \dfrac{{d[A]}}{{dt}}$
Answer
562.5k+ views
Hint: We know that rate of a reaction of a chemical reaction is the speed at which the chemical reaction is taking place. The concentration of reactants always decreases and the concentration of the products always increase in a given chemical reaction.
Complete step-by-step answer:As we already know that rate of reaction is an important concept of chemical kinetics which is used to determine how fast a chemical reaction is taking place. For a chemical reaction $aA + bB \to cC + dD$, the rate of reaction can be written as $R.O.R = - \dfrac{{\dfrac{{d[A]}}{{dt}}}}{a} = - \dfrac{{\dfrac{{d[B]}}{{dt}}}}{b} = + \dfrac{{\dfrac{{d[C]}}{{dt}}}}{c} = + \dfrac{{\dfrac{{d[D]}}{{dt}}}}{d}$. Here, $a,b,c,d$ are the coefficients of $A,B,C,D$ respectively. Here $A,B$ are the reactants and $C,D$ are the products of the reaction $aA + bB \to cC + dD$. For reactants we will write a negative sign because the concentration of the reactants are decreasing. For products we will write a positive sign because the concentration of the products is increasing. $[A],[B],[C],[D]$ are the concentrations of $A,B,C,D$ respectively .
So in the question the reaction given is $2A \to B$, so the rate of reaction of the this chemical equation can be written as
$
Rate of reaction = - \dfrac{{\dfrac{{d[A]}}{{dt}}}}{2} = + \dfrac{{d[B]}}{{dt}} \\
Rate of reaction = - \dfrac{1}{2}\dfrac{{d[A]}}{{dt}} = + \dfrac{{d[B]}}{{dt}} \\
$
We have derived the rate of reaction of the reaction $2A \to B$. So from the above explanation and calculation it is clear to us that the correct answer of the given question is option: A. $ - \dfrac{1}{2}[\dfrac{{d[A]}}{{dt}}]$.
Note: Always remember that for a chemical reaction $aA + bB \to cC + dD$, the rate of reaction can be written as $R.O.R = - \dfrac{{\dfrac{{d[A]}}{{dt}}}}{a} = - \dfrac{{\dfrac{{d[B]}}{{dt}}}}{b} = + \dfrac{{\dfrac{{d[C]}}{{dt}}}}{c} = + \dfrac{{\dfrac{{d[D]}}{{dt}}}}{d}$.We can increase or decrease the rate of reaction by using suitable catalyst and inhibitors. Rate of reaction depends upon the concentration of the reactants and the physical state of the reactant. Always try to solve the question carefully and avoid silly mistakes while solving the numerical.
Complete step-by-step answer:As we already know that rate of reaction is an important concept of chemical kinetics which is used to determine how fast a chemical reaction is taking place. For a chemical reaction $aA + bB \to cC + dD$, the rate of reaction can be written as $R.O.R = - \dfrac{{\dfrac{{d[A]}}{{dt}}}}{a} = - \dfrac{{\dfrac{{d[B]}}{{dt}}}}{b} = + \dfrac{{\dfrac{{d[C]}}{{dt}}}}{c} = + \dfrac{{\dfrac{{d[D]}}{{dt}}}}{d}$. Here, $a,b,c,d$ are the coefficients of $A,B,C,D$ respectively. Here $A,B$ are the reactants and $C,D$ are the products of the reaction $aA + bB \to cC + dD$. For reactants we will write a negative sign because the concentration of the reactants are decreasing. For products we will write a positive sign because the concentration of the products is increasing. $[A],[B],[C],[D]$ are the concentrations of $A,B,C,D$ respectively .
So in the question the reaction given is $2A \to B$, so the rate of reaction of the this chemical equation can be written as
$
Rate of reaction = - \dfrac{{\dfrac{{d[A]}}{{dt}}}}{2} = + \dfrac{{d[B]}}{{dt}} \\
Rate of reaction = - \dfrac{1}{2}\dfrac{{d[A]}}{{dt}} = + \dfrac{{d[B]}}{{dt}} \\
$
We have derived the rate of reaction of the reaction $2A \to B$. So from the above explanation and calculation it is clear to us that the correct answer of the given question is option: A. $ - \dfrac{1}{2}[\dfrac{{d[A]}}{{dt}}]$.
Note: Always remember that for a chemical reaction $aA + bB \to cC + dD$, the rate of reaction can be written as $R.O.R = - \dfrac{{\dfrac{{d[A]}}{{dt}}}}{a} = - \dfrac{{\dfrac{{d[B]}}{{dt}}}}{b} = + \dfrac{{\dfrac{{d[C]}}{{dt}}}}{c} = + \dfrac{{\dfrac{{d[D]}}{{dt}}}}{d}$.We can increase or decrease the rate of reaction by using suitable catalyst and inhibitors. Rate of reaction depends upon the concentration of the reactants and the physical state of the reactant. Always try to solve the question carefully and avoid silly mistakes while solving the numerical.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

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

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

