Question
For the first-order reaction half-life is $14 \mathrm{~s}$. The time required for the initial concentration to reduce to $1 / 8$ th of its value is:(a) $21 \mathrm{~s}$(b) $32 \mathrm{~s}$(c) $42 \mathrm{~s}$(d) $142 \mathrm{~s}$
Step 1
693}{k}$, where $k$ is the rate constant. Given that the half-life is 14 seconds, we can solve for $k$: \[k = \frac{0.693}{14} = 0.0495 \, \text{s}^{-1}\] Show more…
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For the first-order reaction half-life is $14 \mathrm{~s}$. The time required for the initial concentration to reduce to $1 / 8$ th of its value is (a) $21 \mathrm{~s}$ (b) $32 \mathrm{~s}$ (c) $42 \mathrm{~s}$ (d) $14^{2} \mathrm{~s}$
In a certain first-order reaction it takes 240 s for the reactant concentration to decrease to 1/16 of its initial value. The half-life of this reaction is a. 120 s b. 30 s c. 60 s d. 70 s e.15 s f. 480 s
The reaction $\mathrm{A}(g) \longrightarrow \mathrm{B}(g)+\mathrm{C}(g)$ is known to be first order in $\mathrm{A}(g)$. It takes $28 \mathrm{~s}$ for the concentration of $\mathrm{A}(g)$ to decrease by one-half of its initial value. How long does it take for the concentration of $\mathrm{A}(g)$ to decrease to one-fourth of its initial value? to one-eighth of its initial value?
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