Answer

From the question, we can write down the following information:

Initial amount $=5 \mathrm{~g}$

Final concentration $=3 \mathrm{~g}$

Rate constant $=1.1510^{-3} \mathrm{~s}^{-1}$

We know that for a $1^{\text {storder }}$ reaction,

$ \begin{aligned} t & =\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ & =\frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3} \\ & =\frac{2.303}{1.15 \times 10^{-3}} \times 0.2219 \\ & =444.38 \mathrm{~s} \\ & =444 \mathrm{~s} \text { (approx) } \end{aligned} $

Answer

We know that for a $1^{\text {storder reaction, }}$

$t_{1 / 2}=\frac{0.693}{k}$

It is given that $t_{1 / 2}=60 \mathrm{~min}$

$ \begin{aligned} \therefore k & =\frac{0.693}{t_{1 / 2}} \\ & =\frac{0.693}{60} \\ & =0.01155 \mathrm{~min}^{-1} \\ & =1.155 \mathrm{~min}^{-1} \end{aligned} $

Or $k=1.925 \times 10^{-4} \mathrm{~s}^{-1}$