(A) The rate law for the reaction is given by $Rate = k[C_6H_5N_2^+Cl^-]^n$,where $n$ is the order of the reaction.Given that when the initial concent

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(A) The rate law for the reaction is given by $Rate = k[C_6H_5N_2^+Cl^-]^n$,where $n$ is the order of the reaction.Given that when the initial concentration is doubled,the rate of evolution of $N_2$ (which is the rate of reaction) becomes two times faster.Let the initial rate be $R_1 = k[C]^n$ and the new rate be $R_2 = k[2C]^n$.According to the problem,$R_2 = 2R_1$.Therefore,$k[2C]^n = 2 \times k[C]^n$.$2^n = 2^1$.Thus,$n = 1$.Hence,it is a first order reaction.

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

Easy

The diazonium salt decomposes as $C_6H_5N_2^+Cl^- \to C_6H_5Cl + N_2$. At $0 \ ^\circ C$,the evolution of $N_2$ becomes two times faster when the initial concentration of the salt is doubled. Therefore,it is:

A
$A$. $A$ first order reaction
B
$B$. $A$ second order reaction
C
$C$. Independent of the initial concentration of the salt
D
$D$. $A$ zero order reaction

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Class 12

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

The following are the rate constants of two different reactions. What is the overall order of reaction for each?
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$

Medium

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The rate law for the reaction between substances $A$ and $B$ is given by $\text{Rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what is the ratio of the new rate to the initial rate?

Easy

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For a chemical reaction,$2A + 2B \to C + D$,the order of reaction is $1$ with respect to $A$ and $1$ with respect to $B$. The initial rate of the reaction is $4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$. When $50\%$ of the reactants are converted into products,the rate of the reaction would become:

Medium

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Which of the following options correctly represents the relationship between $t_{7/8}$ and $t_{1/2}$ where $t_{7/8}$ represents the time required for the concentration to become $1/8$ of the original for a reaction of order $n$?

Difficult

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From the rate expression for the following reaction,determine its order of reaction and the dimensions of the rate constant: $(iv) \ C_{2}H_{5}Cl_{(g)} \rightarrow C_{2}H_{4(g)} + HCl_{(g)} \quad \text{Rate} = k[C_{2}H_{5}Cl]$

Medium

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The diazonium salt decomposes as $C_6H_5N_2^+Cl^- \to C_6H_5Cl + N_2$. At $0 \ ^\circ C$,the evolution of $N_2$ becomes two times faster when the initial concentration of the salt is doubled. Therefore,it is:

Similar Questions

The following are the rate constants of two different reactions. What is the overall order of reaction for each?$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$

The rate law for the reaction between substances $A$ and $B$ is given by $\text{Rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what is the ratio of the new rate to the initial rate?

For a chemical reaction,$2A + 2B \to C + D$,the order of reaction is $1$ with respect to $A$ and $1$ with respect to $B$. The initial rate of the reaction is $4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$. When $50\%$ of the reactants are converted into products,the rate of the reaction would become:

Which of the following options correctly represents the relationship between $t_{7/8}$ and $t_{1/2}$ where $t_{7/8}$ represents the time required for the concentration to become $1/8$ of the original for a reaction of order $n$?

From the rate expression for the following reaction,determine its order of reaction and the dimensions of the rate constant: $(iv) \ C_{2}H_{5}Cl_{(g)} \rightarrow C_{2}H_{4(g)} + HCl_{(g)} \quad \text{Rate} = k[C_{2}H_{5}Cl]$

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The following are the rate constants of two different reactions. What is the overall order of reaction for each?
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$