⇒Chemical Kinetics

Correct Option is (D)

$[A{]}_0=100\mathrm{\%},[A{]}_t=100-20=80\mathrm{\%}$

Substitution of these in above

$\begin{array}{rl}\mathrm{k}& =\frac{2.303}{\mathrm{t}}{\log }_{10}\frac{[\mathrm{A}{]}_0}{[\mathrm{A}{]}_{\mathrm{t}}}\\ \mathrm{k}& =\frac{2.303}{\mathrm{t}}{\log }_{10}\frac{100}{80}\\ & =\frac{2.303}{23.03\min }{\log }_{10}(1.25)\\ & =\frac{2.303}{23.03\min }\times 0.0969\\ & =9.69\times {10}^{-3}{\text{ minute }}^{-1}\end{array}$

Correct Option is (B)

$99.9\mathrm{\%}$ of the reaction is complete.

So, if $[A{]}_0=100$, then $[A{]}_t=100-99.9=0.1$

$\begin{array}{rl}\mathrm{t}& =\frac{2.303}{\mathrm{k}}{\log }_{10}\frac{[\mathrm{A}{]}_0}{[\mathrm{A}{]}_{\mathrm{t}}}\\ & =\frac{2.303}{0.576}{\log }_{10}\frac{100}{0.1}=\frac{2.303}{0.576}{\log }_{10}(1000)\\ & =\frac{2.303}{0.576}\times 3\\ & =11.99\approx 12\text{ minutes }\end{array}$

Correct Option is (A)

For a first order reaction, $k=\frac{0.693}{t_{1/2}}$

$\mathrm{k}=\frac{0.693}{40}=1.733\times {10}^{-2}{\text{ minute }}^{-1}$

Correct Option is (A)

$80\mathrm{\%}$ of the reactant has reacted.

So, if $[A{]}_0=100$, then $[A{]}_{\mathrm{t}}=100-80=20$

$\begin{array}{rl}\mathrm{k}& =\frac{2.303}{\mathrm{t}}{\log }_{10}\frac{[\mathrm{A}{]}_0}{[\mathrm{A}{]}_{\mathrm{t}}}\\ & =\frac{2.303}{15}{\log }_{10}\frac{100}{20}\\ & =\frac{2.303}{15}{\log }_{10}(5)\\ & =\frac{2.303}{15}\times 0.699\\ & =0.1073\\ & \approx 0.11{\text{ minute }}^{-1}\end{array}$