Q.iii. The rate constant for the reaction $\mathrm{2\ N_2O_5(g) \longrightarrow 2\ N_2O_4(g) + O_2(g)}$ is 4.98 $\times$ 10$^{-4}$ s$^{-1}$. The order of reaction is a. 2
b. 1
c. 0
d. 3

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 pressure: $k=\dfrac{2.303}{t}\log_{10}\dfrac{P_i}{2P_i-P}$, where $P_i$ is the initial pressure and $P$ the total pressure at time $t$. A useful derived result: the time for 99.9% completion of a first order reaction is exactly three times the time for 90% completion, since both are proportional to $\log_{10}$ of a fixed ratio.