NEET UG 2019 — Chemistry Physical Chemistry

For first order reaction, $$\mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$$t=k2.303​loga−xa​ Given : $\mathrm{k}=2.303 \times 10^{-3} \mathrm{~s}^{-1}, \mathrm{a}=40 \mathrm{~g}, \mathrm{a}-\mathrm{x}=10 \mathrm{~g}$k=2.303×10−3 s−1,a=40 g,a−x=10 g On substituting the given values in Eq. (i), we get $$\begin{aligned} t & =\frac{2.303}{2.303 \times 10^{-3}} \log \frac{40}{10} \\ & =10^3 \log 2^2=2 \times 10^3 \times \log 2 \\ & =2 \times 10^3 \times 0.3010=602 \mathrm{~s} \end{aligned}$$t​=2.303×10−32.303​log1040​=103log22=2×103×log2=2×103×0.3010=602 s​ Alternative method For first order reaction, $\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}$t1/2​=k0.693​ $$\mathrm{t}_{1 / 2}\left(\mathrm{t}_{50 \%}\right)=\frac{0.693}{2.303 \times 10^{-3}}=301 \mathrm{~s}$$t1/2​(t50%​)=2.303×10−30.693​=301 s Also, $\quad \mathrm{t}_{75 \%}=2 \mathrm{t}_{50 \%}$t75%​=2t50%​ $$\therefore \quad \mathrm{t}_{75 \%}=2 \times 301=602 \mathrm{~s}$$∴t75%​=2×301=602 s