(C) For a first order reaction,the integrated rate equation is $\ln(P/P_0) = -kt$.Comparing this with the equation of a straight line $y = mx + c$,whe

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(C) For a first order reaction,the integrated rate equation is $\ln(P/P_0) = -kt$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln(P/P_0)$ and $x = t$,the slope $m = -k$.From the given graph,the slope is $-3.465 \times 10^4 \ s^{-1}$.Therefore,$-k = -3.465 \times 10^4 \ s^{-1}$,which gives $k = 3.465 \times 10^4 \ s^{-1}$.Given $k = x \times 10^4 \ s^{-1}$,we have $x = 3.465 \approx 3$ (rounding to the nearest integer as per options).

Class 12 Chemistry Chemical Kinetics First Order reaction

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For the decomposition of azomethane,$CH_3N_2CH_{3(g)} \rightarrow CH_3CH_{3(g)} + N_{2(g)}$,a first order reaction,the variation in partial pressure with time at $600 \ K$ is given as shown in the graph. The value of the rate constant $k$ in $s^{-1}$ is $x \times 10^4$. Find the value of $x$.

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$1$
B
$2$
C
$3$
D
$9$

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For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

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For a first order gas phase reaction:
$A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$
Let $P_0$ be the initial pressure of $A$ and $P_t$ be the total pressure at time $t$. The integrated rate equation is:

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In a first order reaction,the concentration of the reactant is reduced to $25\%$ in one hour. The half-life period of the reaction is: (in $hr$)

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For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is:

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The integrated rate law equation for a first-order gas-phase reaction $A(g) \rightarrow B(g) + C(g)$ is given by (where $P_i$ is the initial pressure and $P_t$ is the total pressure at time $t$):

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For the decomposition of azomethane,$CH_3N_2CH_{3(g)} \rightarrow CH_3CH_{3(g)} + N_{2(g)}$,a first order reaction,the variation in partial pressure with time at $600 \ K$ is given as shown in the graph. The value of the rate constant $k$ in $s^{-1}$ is $x \times 10^4$. Find the value of $x$.

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For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

For a first order gas phase reaction: $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$ Let $P_0$ be the initial pressure of $A$ and $P_t$ be the total pressure at time $t$. The integrated rate equation is:

In a first order reaction,the concentration of the reactant is reduced to $25\%$ in one hour. The half-life period of the reaction is: (in $hr$)

For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is:

The integrated rate law equation for a first-order gas-phase reaction $A(g) \rightarrow B(g) + C(g)$ is given by (where $P_i$ is the initial pressure and $P_t$ is the total pressure at time $t$):

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For a first order gas phase reaction:
$A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$
Let $P_0$ be the initial pressure of $A$ and $P_t$ be the total pressure at time $t$. The integrated rate equation is: