Q.vi. Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.

While the differential rate law describes how fast concentration is CHANGING, the integrated rate law gives concentration itself as a function of elapsed time -- what is actually measured in the lab. For a first order reaction $A\rightarrow\text{product}$ ($\text{rate}=-d[A]/dt=k[A]$), integrating gives $k=\dfrac{2.303}{t}\log_{10}\dfrac{[A]_0}{[A]_t}$, equivalently $[A]_t=[A]_0e^{-kt}$; its rate constant always carries units of (time)$^{-1}$. Rearranged forms give straight-line graphs: rate vs $[A]_t$ (through origin, slope $k$), $\log_{10}[A]_t$ vs $t$ (slope $-k/2.303$), or $\log_{10}([A]_0/[A]_t)$ vs $t$ (through origin, slope $k/2.303$). For a gas-phase reaction $A(g)\rightarrow B(g)+C(g)$, the same law can be rewritten entirely in terms of total p …