Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 5

For the stated single elementary step and constant reaction volume,The equal magnitudes follow from the unit stoichiometric coefficients. In the buffered case this hydroxide consumption is balanced by buffer chemistry; the measured total hydroxide concentration can then remain constant.

because a second-order concentration product must give a rate in $\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$.

At $25^{\circ }\mathrm{C}$, $[\mathrm{O}{\mathrm{H}}^{-}]=10^{\mathrm{pH}-14}\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ in the usual dilute-solution thermodynamic activity approximation. Therefore A: $10^{-2}\mathrm{M}$; B: $10^{-4}\mathrm{M}$; C: $10^{-4}\mathrm{M}$. These are initial concentrations; only C is actively buffered at its stated value.

Let $x=[\mathrm{MeX}]$ and $x_0=10^{-4}\mathrm{M}$.

A and C: integrate $dx/x=-k{h}_{\mathrm{const}}dt$:using $h_{\mathrm{const}}=10^{-2}\mathrm{M}$ for A and $10^{-4}\mathrm{M}$ for C.

B: integrate $dx/x^2=-kdt$:For completeness, unbuffered unequal initial concentrations obey $h=x+\delta$, $\delta =h_0-x_0$, and the exact expression isIt approaches the equimolar second-order expression as $\delta \to 0$ and the excess-hydroxide exponential when depletion is negligible.

The unequal-concentration expression is the unequal-concentration second-order integration formula.

Set $x=0.1x_0$:The corresponding numerical concentration factors are approximately $230.3$, $90000$ and $23026{\mathrm{M}}^{-1}$, each divided by $k$. Thus A is about 100 times faster than C at this conversion, while B takes about $9/\ln 10=3.91$ times as long as C because its hydroxide is depleted.

Draw one maximum for the concerted SN2 reaction step, with no intermediate minimum. Its transition structure iswith O-C forming and C-X breaking, both shown as partial bonds. O-C-X is approximately linear, and the three C-H bonds lie in the equatorial plane. The overall charge is $-1$, distributed over the entering and leaving groups.

The sketch illustrates an exergonic case; the elementary-step description alone does not supply a numerical reaction energy or determine the product-reactant energy difference for every possible X.

A transition structure is a saddle point on the potential-energy surface: motion along the reaction coordinate lowers energy in either direction. It has one unstable vibrational direction and no intermediate residence basin.

An intermediate is a local minimum, separated from neighboring states by barriers, with a finite lifetime. It may be short-lived, but that does not make it a transition structure. The elementary SN2 reaction step has a transition structure and no distinct bonded intermediate.