MCQ
For the first order reaction, half life is $14\,sec$. The time required for the initial concentration to reduce to $\frac{1}{8}\,th$ of its value is .......... $\sec$
- A$28$
- ✓$42$
- C${\left( {14} \right)^3}$
- D${\left( {14} \right)^2}$
✓
Answer
Correct option: B.
$42$
b
$t _{1 / 2}=14 \,sec =\frac{\operatorname{ln} 2}{k} \ldots .[$ half-life for $1^{ st}$ order reaction] $k=\frac{\operatorname{In} 2}{14}-(1)$
$t _{1 / 2}=14 \,sec =\frac{\operatorname{ln} 2}{k} \ldots .[$ half-life for $1^{ st}$ order reaction] $k=\frac{\operatorname{In} 2}{14}-(1)$
For $1^{st}$ order reaction, $k=\frac{1}{t} \operatorname{ln} \frac{a_0}{a_t}$
given that $a_t=\frac{a_0}{8}$
$k=\frac{1}{t} \operatorname{ln} \frac{a_0}{a_0 / 8}=>k=\frac{1}{t} \operatorname{In} 8-(2)$
From $(1)$ and $(2)$,
$\frac{\operatorname{In} 2}{14}=\frac{1}{t} \operatorname{In}(2)^3=\frac{3}{t} \operatorname{In}(2)$
$\frac{1}{14}=\frac{3}{t}$
$t=42 \,sec$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In context of the lanthanoids, which of the following statements is not correct?
$9.8\,g$ of ${H_2}S{O_4}$ is present in $ 2 \,litres$ of a solution. The molarity of the solution is ....... $M$
→Number of complexes from the following with even number of unpaired "d" electrons is $......$
${\left[V\left(H_2 O\right)_6\right]^{3+}, \quad\left[Cr\left(H_2 O\right)_6\right]^{2+}, \quad\left[Fe\left(H_2 O\right)_6\right]^{3+}}$
${\left[Ni\left(H_2 O\right)_6\right]^{3+},\left[Cu\left(H_2 O\right)_6\right]^{2+}}$
$[$Given atomic numbers : $V =23, Cr =24, Fe =26$,$Ni=28, Cu=29 \text { ] }$
→${\left[V\left(H_2 O\right)_6\right]^{3+}, \quad\left[Cr\left(H_2 O\right)_6\right]^{2+}, \quad\left[Fe\left(H_2 O\right)_6\right]^{3+}}$
${\left[Ni\left(H_2 O\right)_6\right]^{3+},\left[Cu\left(H_2 O\right)_6\right]^{2+}}$
$[$Given atomic numbers : $V =23, Cr =24, Fe =26$,$Ni=28, Cu=29 \text { ] }$
For the dumb-bell shaped orbital, the value of $l$ is
→The most common minerals of phosphorus are
Indicate the nature of bonding in $CC{l_4}$ and $Ca{H_2}$
→Suitable reagent for above conversion
Pick the correct name of $[Co{(N{H_3})_5}Cl]C{l_2}$
→Match List$-I$ with List$-II$ :
→|
List$-I$ (Metal Ion) |
List$-II$ (Group in Qualitative analysis) |
| $(a)$ ${Mn}^{2+}$ | $(i)$ Group $- III$ |
| $(b)$ ${A} {s}^{3+}$ | $(ii)$ Group $- IIA$ |
| $(c)$ ${Cu}^{2+}$ | $(iii)$ Group $- IV$ |
| $(d)$ ${Al}^{3+}$ | $(iv)$ Group $- IIB$ |
Choose the most appropriate answer from the options given below :
$2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})$
→The above reaction is carried out in a vessel starting with partial pressure $\mathrm{P}_{\mathrm{SO}_{2}}=250\, \mathrm{~m}$ $bar,$ $\mathrm{P}_{0_{2}}=750 \,\mathrm{~m}$ $bar$ and $\mathrm{P}_{\mathrm{SO}_{3}}=0 \,\mathrm{bar}$. When the reaction is complete, the total pressure in the reaction vessel is $.....\mathrm{m}$ $bar.$ (Round off to the Nearest Integer).