(C) For a first order reaction,the concentration after time $t$ is given by $[A]_t = [A]_0 \times (1/2)^n$,where $n$ is the number of half-lives.Given

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(C) For a first order reaction,the concentration after time $t$ is given by $[A]_t = [A]_0 \times (1/2)^n$,where $n$ is the number of half-lives.Given that the concentration is reduced to $1/8$ of the initial concentration,we have $(1/2)^n = 1/8$.Since $1/8 = (1/2)^3$,we find that $n = 3$.This means $3$ half-lives have passed in $75 \ minutes$.Therefore,$3 \times t_{1/2} = 75 \ minutes$.$t_{1/2} = 75 / 3 = 25 \ minutes$.The closest option is $25.1 \ minutes$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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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 $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

AP EAMCET 2025 All AP EAMCET

A
$60.2$
B
$50.2$
C
$25.1$
D
$75.1$

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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 $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

Similar Questions

What is the slope of the graph plotted between half-life $(t_{1/2})$ and initial concentration $(a)$ for a $1^{st}$ order reaction?

The rate equation for a first-order reaction is given by $[R] = [R]_0 e^{-kt}$. $A$ straight line with a positive slope is obtained by plotting: ($[R]_0 =$ initial concentration of reactant,$[R] =$ concentration of reactant at time $t$)

For the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$,the rate constant is $3.0 \times 10^{-5} \ s^{-1}$. If the rate of reaction is $2.40 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,calculate the concentration of $N_2O_5$ in $mol \ L^{-1}$.

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The reaction $X \rightarrow$ products is a first order reaction. In $40 \ min$,the concentration of $X$ changes from $1.0 \ M$ to $0.25 \ M$. What is the rate of reaction when $[X] = 0.1 \ M$? $(\log 4 = 0.60)$

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