Let $\mathrm{m}$ be the mass of substance at time $\mathrm{t}$. Then,
$\begin{aligned} & \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km} \text {, where } \mathrm{k}>0 \\ & \Rightarrow \frac{\mathrm{dm}}{\mathrm{m}}=-\mathrm{kdt}\end{aligned}$
Integrating on both sides, we get
$\log m=-k t+c$
When $\mathrm{t}=0, \mathrm{~m}=\mathrm{m}_0$
$\therefore \quad \log \mathrm{m}_0=0+\mathrm{c}$
$\Rightarrow \mathrm{c}=\log \mathrm{m}_0$
$\therefore \quad \log m=-k t+\log m_0$
$\Rightarrow \log \frac{\mathrm{m}}{\mathrm{m}_0}=-\mathrm{kt}$
When $\mathrm{t}=\mathrm{h}, \mathrm{m}=\frac{1}{2} \mathrm{~m}_0$
$\therefore \quad \log \left(\frac{\frac{1}{2} m_0}{m_0}\right)=-k h$
$\begin{aligned} & \Rightarrow \log \frac{1}{2}=-\mathrm{kh} \\ & \Rightarrow \log 2=\mathrm{kh}\end{aligned}$
$\Rightarrow \mathrm{k}=\frac{\log 2}{\mathrm{~h}}$ ...(i)
Initial decay rate,
$\frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km}_0$
$=\frac{-\mathrm{m}_0}{\mathrm{~h}} \log 2$ ...[From (i)]