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${t_{1/4}} = \frac{{2.303}}{K}\,\log \,\frac{3}{4}$
$\mathrm{t}_{1 / 4}=\frac{2.303}{\mathrm{k}} \log \frac{[\mathrm{A}]_{0}}{3[\mathrm{A}]_{0}} \times 4$
$t_{1 / 4}=\frac{2.303}{k} \log \frac{4}{3}$
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$2 C _{( s )}+ O _{2( g )} \rightarrow 2 CO ( g )$
When $12\,g$ carbon is burnt in $48\,g$ of oxygen, the volume of carbon monoxide produced is $......\times 10^{-1}\,L$ at STP [nearest integer]
[Given : Assume $CO$ as ideal gas, Mass of $C$ is $12\,g\,mol ^{-1}$, Mass of $O$ is $16\,g\,mol ^{-1}$ and molar volume of an ideal gas at STP is $22.7\,L\, mol ^{-1}$ ]