MCQ
$t_{1/4}$ for first order reaction is given as
- A${t_{1/4}} = \frac{{2.303}}{K}\,\log \,4$
- B${t_{1/4}} = \frac{{2.303}}{K}\,\log \,2$
- ✓${t_{1/4}} = \frac{{2.303}}{K}\,\log \,\frac{4}{3}$
- D${t_{1/4}} = \frac{{2.303}}{K}\,\log \,\frac{3}{4}$
✓
Answer
Correct option: C.
${t_{1/4}} = \frac{{2.303}}{K}\,\log \,\frac{4}{3}$
c
$\mathrm{t}_{1 / 4}=\frac{2.303}{\mathrm{k}} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{A}] \mathrm{t}}$
$\mathrm{t}_{1 / 4}=\frac{2.303}{\mathrm{k}} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{A}] \mathrm{t}}$
$\mathrm{t}_{1 / 4}=\frac{2.303}{\mathrm{k}} \log \frac{[\mathrm{A}]_{0}}{3[\mathrm{A}]_{0}} \times 4$
$t_{1 / 4}=\frac{2.303}{k} \log \frac{4}{3}$
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