(D) Given: $[A]_0 = 8[B]_0$. Half-lives: $(t_{1/2})_A = 10 \ min$,$(t_{1/2})_B = 40 \ min$. Rate constants: $k_A = \frac{\ln 2}{10}$,$k_B = \frac{\ln

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(D) Given: $[A]_0 = 8[B]_0$. Half-lives: $(t_{1/2})_A = 10 \ min$,$(t_{1/2})_B = 40 \ min$. Rate constants: $k_A = \frac{\ln 2}{10}$,$k_B = \frac{\ln 2}{40}$. For first-order kinetics,$[A]_t = [A]_0 e^{-k_A t}$ and $[B]_t = [B]_0 e^{-k_B t}$. Setting $[A]_t = [B]_t$: $[A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \implies \frac{[A]_0}{[B]_0} = e^{(k_A - k_B)t}$. Substituting values: $8 = e^{(\frac{\ln 2}{10} - \frac{\ln 2}{40})t}$. Taking natural log: $\ln 8 = (\frac{4\ln 2 - \ln 2}{40})t$. $3 \ln 2 = (\frac{3 \ln 2}{40})t$. $t = 40 \ min$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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In a reaction $A + B \rightarrow C$,initial concentrations of $A$ and $B$ are related as $[A]_0 = 8[B]_0$. The half-lives of $A$ and $B$ are $10 \ min$ and $40 \ min$,respectively. If they start to disappear at the same time,both following first-order kinetics,after how much time will the concentration of both the reactants be the same (in $min$)?

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A
$60$
B
$80$
C
$20$
D
$40$

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Chemical Kinetics

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First Order reaction

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The concentration of a substance undergoing a chemical reaction becomes one-half of its original value after time,regardless of the initial concentration. The reaction is an example of a

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In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

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The following results are obtained in one pseudo first order reaction:

Time $(s)$ $0$ $30$ $60$ $90$

Concentration $(mol \ L^{-1})$ $0.551$ $0.312$ $0.173$ $0.085$

$(a)$ Calculate the average rate of reaction between $30$ and $60$ seconds.
$(b)$ Calculate the rate constant $(k)$ of this first order reaction.

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$A$ substance $A$ decomposes by a first order reaction starting initially with $[A]_0 = 2.00 \, M$ and after $200 \, \min$,$[A]_t = 0.15 \, M$. For this reaction,what is the value of the rate constant $k$?

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$A$ first order reaction requires $30 \ min$ for $50\%$ completion. The time required to complete the reaction by $75\%$ will be .......... $\min.$

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Time $(s)$	$0$	$30$	$60$	$90$
Concentration $(mol \ L^{-1})$	$0.551$	$0.312$	$0.173$	$0.085$

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The concentration of a substance undergoing a chemical reaction becomes one-half of its original value after time,regardless of the initial concentration. The reaction is an example of a

In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

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