(B) The rate law is given by $Rate = k[A]^1[B]^0$. Since the sum of the exponents of the concentration terms in the rate law is $1 + 0 = 1$,the order

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Molecularity of the reaction is $2$ but order is $1$.

(B) The rate law is given by $Rate = k[A]^1[B]^0$. Since the sum of the exponents of the concentration terms in the rate law is $1 + 0 = 1$,the order of the reaction is $1$. Molecularity is defined as the number of reacting species taking part in an elementary reaction. Since $A$ and $B$ are involved in the reaction $A + B \to \text{products}$,the molecularity is $2$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

For the reaction $A + B \to \text{products}$,it is found that the rate of the reaction is proportional to the concentration of $A$,but it is independent of the concentration of $B$. Then:

A
The order of the reaction is $2$ and molecularity is $1$.
B
Molecularity of the reaction is $2$ but order is $1$.
C
Order is $2$ and molecularity is $2$.
D
Order of the reaction is $2$ but molecularity is $0$.

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Class 12

Chemistry Questions

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

For the first order decomposition reaction of $N_2O_5$,it is found that -
$(a)$ $2N_2O_5 \rightarrow 4NO_{2(g)} + O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$(b)$ $N_2O_5 \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k'[N_2O_5]$
Which of the following is true?

Medium

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For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:

MediumTS EAMCET 2024

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Write about elementary and complex reactions.

Medium

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In a reaction,the concentration of reactant is increased two times and three times,then the increases in the rate of reaction were four times and nine times respectively. The order of reaction is:

Medium

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Higher order $(> 3)$ reactions are rare due to:

DifficultJEE MAIN 2015

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For the reaction $A + B \to \text{products}$,it is found that the rate of the reaction is proportional to the concentration of $A$,but it is independent of the concentration of $B$. Then:

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For the first order decomposition reaction of $N_2O_5$,it is found that -$(a)$ $2N_2O_5 \rightarrow 4NO_{2(g)} + O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$(b)$ $N_2O_5 \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k'[N_2O_5]$Which of the following is true?

For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$The correct relation between $K_1, K_2$ and $K_3$ is:

Write about elementary and complex reactions.

In a reaction,the concentration of reactant is increased two times and three times,then the increases in the rate of reaction were four times and nine times respectively. The order of reaction is:

Higher order $(> 3)$ reactions are rare due to:

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For the first order decomposition reaction of $N_2O_5$,it is found that -
$(a)$ $2N_2O_5 \rightarrow 4NO_{2(g)} + O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$(b)$ $N_2O_5 \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k'[N_2O_5]$
Which of the following is true?

For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is: