(B) For a first-order reaction,the rate constant $K$ is given by $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right)$.For $t_{1/8}$,the concentration

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(B) For a first-order reaction,the rate constant $K$ is given by $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right)$.For $t_{1/8}$,the concentration $C_t = C_0 / 8$,so $K t_{1/8} = \ln(8) = 3 \ln(2)$.For $t_{1/10}$,the concentration $C_t = C_0 / 10$,so $K t_{1/10} = \ln(10)$.Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{3 \ln(2)}{\ln(10)} = 3 \log_{10} 2$.Given $\log_{10} 2 = 0.3$,we have $\frac{t_{1/8}}{t_{1/10}} = 3 \times 0.3 = 0.9$.Therefore,$\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$.

Class 12 Chemistry Chemical Kinetics First Order reaction

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

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$8$
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$9$
C
$7$
D
$6$

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Chemical Kinetics

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First Order reaction

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Decomposition of $H_2O_2$ follows a first order reaction. In $50 \ min$,the concentration of $H_2O_2$ decreases from $0.5 \ M$ to $0.125 \ M$. For such decomposition,when the concentration of $H_2O_2$ reaches $0.05 \ M$,the rate of formation of $O_2$ will be:

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$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

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State whether the following sentences are true $(T)$ or false $(F)$.
$(a)$ The reaction of an ester with water produces an alcohol and an acid.
$(b)$ The concentration of water remains constant during the hydrolysis of an ester.
$(c)$ In the hydrolysis of an ester,$[H_2O]$ remains constant.

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The half-life period for a first-order reaction is . . . . . . .

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The rate constant for the decomposition of $H_2O_2$ is $3.66 \times 10^{-3} \ s^{-1}$. If the initial concentration of $H_2O_2$ is $0.882 \ M$,then in how many seconds will its concentration become $0.600 \ M$?

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An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

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Decomposition of $H_2O_2$ follows a first order reaction. In $50 \ min$,the concentration of $H_2O_2$ decreases from $0.5 \ M$ to $0.125 \ M$. For such decomposition,when the concentration of $H_2O_2$ reaches $0.05 \ M$,the rate of formation of $O_2$ will be:

$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

State whether the following sentences are true $(T)$ or false $(F)$.$(a)$ The reaction of an ester with water produces an alcohol and an acid.$(b)$ The concentration of water remains constant during the hydrolysis of an ester.$(c)$ In the hydrolysis of an ester,$[H_2O]$ remains constant.

The half-life period for a first-order reaction is . . . . . . .

The rate constant for the decomposition of $H_2O_2$ is $3.66 \times 10^{-3} \ s^{-1}$. If the initial concentration of $H_2O_2$ is $0.882 \ M$,then in how many seconds will its concentration become $0.600 \ M$?

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State whether the following sentences are true $(T)$ or false $(F)$.
$(a)$ The reaction of an ester with water produces an alcohol and an acid.
$(b)$ The concentration of water remains constant during the hydrolysis of an ester.
$(c)$ In the hydrolysis of an ester,$[H_2O]$ remains constant.