Question #44186

Consider the complex gas-phase oxidation of nitric oxide at 298 K: $2\ce{NO}(g) + \ce{O2}(g) \rightarrow 2\ce{NO2}(g)$ The proposed mechanism is: Step 1: $\ce{NO} + \ce{NO} \xrightarrow{k_1} \ce{N2O2}$ Step -1: $\ce{N2O2} \xrightarrow{k_{-1}} \ce{NO} + \ce{NO}$ Step 2: $\ce{N2O2} + \ce{O2} \xrightarrow{k_2} 2\ce{NO2}$ The rate constants are $k_1 = 4 \times 10^5 \, \text{M}^{-1}\text{s}^{-1}$, $k_{-1} = 2 \times 10^6 \, \text{s}^{-1}$, and $k_2 = 10^3 \, \text{M}^{-1}\text{s}^{-1}$. The reaction is carried out with a constant excess concentration of oxygen, $[\ce{O2}] = 0.5 \, \text{M}$. Applying the Steady State Approximation (SSA) for the intermediate $\ce{N2O2}$ (assume $k_{-1} \gg k_2[\ce{O2}]$), a kinetic plot of $\frac{1}{[\ce{NO}]}$ versus time $t$ (in seconds) yields a straight line with slope $S$. The numerical value of $S$ is exactly equal to the uncertainty in velocity ($\Delta v$, in m/s) of a microscopic particle of mass $m_p = \frac{h}{160\pi}$ kg (where $h$ is Planck's constant). Calculate the minimum uncertainty in the position ($\Delta x$) of this particle.