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

The decadic Beer-Lambert law can be writtenHere $l$ is the path length in metres, $c$ is concentration in $\mathrm{mol}{\mathrm{L}}^{-1}$ or M, and $\epsilon$ is the decadic molar attenuation coefficient in ${\mathrm{M}}^{-1}{\mathrm{m}}^{-1}$. The intensity ratio and the exponent are dimensionless. Equivalently $I=I_0{e}^{-\kappa lc}$ with $\kappa =(\ln 10)\epsilon$ in the same units. Defining the logarithm convention avoids a missing factor of $\ln 10$.

The absorbance is $\boxed{A={\log }_{10}(I_0/I)=-{\log }_{10}(I/I_0)}$. It is dimensionless. It differs from the absorbed fraction $1-I/I_0$.

For a first-order reaction with no appreciable reverse contribution in this decay experiment,With fixed path length and attenuation coefficient, and absorbance attributable to ONOO after blank/background correction, the Beer-Lambert law gives $A(t)=\epsilon l{c}_0{e}^{-kt}$. ThereforeThe gradient is $-k$, in ${\mathrm{s}}^{-1}$ when time is measured in seconds. Constant illumination wavelength, homogeneous sampling and negligible product absorbance at that wavelength are required for this direct interpretation.

The intermediate balance isSetting this derivative approximately to zero under the quasi-steady-state approximation givesBoth dissociation and reaction with NO remove the intermediate. Its steady concentration is not zero and does not require the first reversible pair to be at equilibrium.

The last elementary reaction creates two nitrogen dioxide molecules per event, soThe factor of two is required for the production rate requested. The conventional reaction rate for $2NO+{\mathrm{O}}_2\to 2N{\mathrm{O}}_2$ would be one half of this production rate.

“Low NO” means $k_2[\mathrm{NO}]\ll k_{-1}$, or $[\mathrm{NO}]\ll 6.5\times 10^{-6}\mathrm{M}$. ThenThe effective orders are two in NO and one in oxygen, giving overall order three. For the production-rate convention,If the conventional reaction rate $v=\frac{1}{2}d[\mathrm{N}{\mathrm{O}}_2]/dt$ is used, its coefficient is $2.91\times 10^7{\mathrm{M}}^{-2}{\mathrm{s}}^{-1}$. These are the same kinetics under explicitly different normalizations. At high NO the denominator changes, so the effective NO order approaches one rather than remaining two.

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.