(B) For a reaction of order $n$,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.Giv

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proportional to the second power of concentration

(B) For a reaction of order $n$,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.Given that the concentration decreases from $0.1 \, M$ to $0.01 \, M$,the concentration becomes $\frac{1}{10}$ of the initial value.The half-life increases by $10$ times,so $10 = (\frac{1}{10})^{-(n-1)}$.$10^1 = 10^{n-1}$.Comparing the exponents,$1 = n - 1$,which gives $n = 2$.Thus,the rate of reaction is proportional to the second power of the concentration.

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

Medium

In the reaction $A + B \to \text{Products}$,the initial concentration of both $A$ and $B$ is $0.1 \, M$. When the concentration decreases to $1.0 \times 10^{-2} \, M$,the half-life period increases by ten times. The rate of the reaction is:

A
proportional to the first power of concentration
B
proportional to the second power of concentration
C
independent of concentration
D
proportional to the third power of concentration

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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 mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$
Then the correct statement is:

Difficult

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For the reaction $A + B \xrightarrow{K} C$,identify the incorrect order of reaction indicated against the rate expression.

Easy

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Consider a reaction $aG + bH \rightarrow$ Products. When the concentration of both reactants $G$ and $H$ is doubled,the rate increases by $8$ times. However,when the concentration of $G$ is doubled keeping the concentration of $H$ fixed,the rate is doubled. The overall order of the reaction is:

MediumIIT 2007

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$A$ chemical reaction proceeds through the following steps:
Step-$I$: $2A \rightleftharpoons X$ (fast)
Step-$II$: $X + B \rightarrow Y$ (slow)
Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)
The rate law for the overall reaction is:

Medium

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The reaction $3ClO^{-} \rightarrow ClO_{3}^{-} + 2Cl^{-}$ occurs in the following two steps:
$(i)$ $ClO^{-} + ClO^{-} \xrightarrow{K_{1}} ClO_{2}^{-} + Cl^{-}$ (Slow step)
$(ii)$ $ClO_{2}^{-} + ClO^{-} \xrightarrow{K_{2}} ClO_{3}^{-} + Cl^{-}$ (Fast step)
Then the rate of the given reaction is equal to . . . . . . .

EasyGUJCET 2015

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In the reaction $A + B \to \text{Products}$,the initial concentration of both $A$ and $B$ is $0.1 \, M$. When the concentration decreases to $1.0 \times 10^{-2} \, M$,the half-life period increases by ten times. The rate of the reaction is:

Similar Questions

The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$Then the correct statement is:

For the reaction $A + B \xrightarrow{K} C$,identify the incorrect order of reaction indicated against the rate expression.

Consider a reaction $aG + bH \rightarrow$ Products. When the concentration of both reactants $G$ and $H$ is doubled,the rate increases by $8$ times. However,when the concentration of $G$ is doubled keeping the concentration of $H$ fixed,the rate is doubled. The overall order of the reaction is:

$A$ chemical reaction proceeds through the following steps:Step-$I$: $2A \rightleftharpoons X$ (fast)Step-$II$: $X + B \rightarrow Y$ (slow)Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)The rate law for the overall reaction is:

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The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$
Then the correct statement is:

$A$ chemical reaction proceeds through the following steps:
Step-$I$: $2A \rightleftharpoons X$ (fast)
Step-$II$: $X + B \rightarrow Y$ (slow)
Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)
The rate law for the overall reaction is: