For a first-order reaction,the time taken for the concentration of the reactant to reach $3/4$ of its initial value is $t_{1/4}$. If the rate constant for the reaction is $K$,then $t_{1/4}$ can be expressed as: (in $/K$)
- A$0.29$
- B$0.10$
- C$0.75$
- D$0.69$
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The value of rate constant for a first order reaction is $2.303 \times 10^{-2} \text{ s}^{-1}$. What will be the time required to reduce the concentration to $\frac{1}{10}$th of its initial concentration (in $\text{ s}$)?
For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)
The decomposition of $N_2O_5$ occurs as $2N_2O_5 \to 4NO_2 + O_2$ and follows first-order kinetics. Hence,
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View SolutionFor a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).
$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$
$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$
For a first order reaction,the slope of the graph of $\log_{10}[A]_t$ versus time is equal to:
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