Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / c

Convert the supplied concentrations to $[\mathrm{NO}]=1.00\times 10^{-7}\mathrm{M}$ and $[{\mathrm{O}}_2]=3.00\times 10^{-5}\mathrm{M}$. ThenThe combined loss coefficient is $6600{\mathrm{s}}^{-1}$. Neglecting the smaller NO-dependent loss would give $8.72\times 10^{-14}\mathrm{M}$, a close low-NO approximation, but the full expression is just as easy to evaluate.

The estimate is $\boxed{t_{10\%}=1.60\times 10^{-5}\mathrm{s}\approx 16\mu \mathrm{s}}$. During the initial buildup, assume instantaneous mixing, initially zero ONOO, constant temperature, and negligible depletion of NO and oxygen. Their concentrations can then be held fixed while solving for $X=[\mathrm{ONOO}]$:The initial-value solution is $X(t)=X_{\mathrm{ss}}(1-e^{-\lambda t})$. Setting $X=0.1X_{\mathrm{ss}}$ givesA constant-initial-rate estimate gives $0.1X_{\mathrm{ss}}/f=1.52\times 10^{-5}\mathrm{s}$, consistent to about 5%. The intermediate concentration and total reaction progress over this interval are tiny relative to $10^{-7}\mathrm{M}$ NO, supporting the negligible-depletion approximation. The actual onset could be limited by mixing or instrumental response; a large vessel does not itself guarantee homogeneous mixing in microseconds.