Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 5

The standard entropy of reaction isUsing ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$,The formation enthalpy of elemental fluorine gas is zero in its standard state. ThusThe large negative entropy change is consistent with converting two gas molecules to one; it is already included in the Gibbs-energy relation.

For ideal gases and standard pressure $p^{\circ }=1\mathrm{bar}$,Since $p_{\mathrm{ClF}}/p_{\mathrm{Cl}{\mathrm{F}}_3}=9$, this gives $p_{{\mathrm{F}}_2}/p^{\circ }=1/(9K_p)$. ThereforeMore exactly, subtract the tiny fluorine pressure from $1\mathrm{bar}$ and divide the remainder in the ratio $9:1$. The very small fluorine activity reconciles the reactant-rich ClF ratio with the very large equilibrium constant.

For each ideal gas, the chemical potential relative to the pure gas at standard pressure is $\Delta {\mu }_i=RT\ln (p_i/p^{\circ })$. HenceAll three shifts are negative because their partial pressures are below $1\mathrm{bar}$. As a check, ${\Delta }_r{G}^{\circ }+\Delta {\mu }_{\mathrm{Cl}{\mathrm{F}}_3}-\Delta {\mu }_{\mathrm{ClF}}-\Delta {\mu }_{{\mathrm{F}}_2}=0$ at equilibrium, within rounding.

Subtract the second supplied reaction from the first. Chlorine trifluoride and aqueous HCl cancel, while one water and two HF remain on their respective sides, giving the required reaction. By Hess's law for Gibbs energy,The subtraction reverses the second reaction before adding it; adding both supplied Gibbs energies without changing its sign would correspond to a different chemical equation.

With the stipulated $-5.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ value and ideal-solution approximations, $\boxed{[\mathrm{HF}]\approx 1.94\mathrm{M}}$ and the initial oxygen-difluoride pressure is $\boxed{p_0\approx 9.01\mathrm{bar}}$. These are estimates for the hypothetical reaction energetics specified in this part.

Take liquid-water activity as one, ideal gases, and aqueous HF activity as $[\mathrm{HF}]/c^{\circ }$, with $c^{\circ }=1\mathrm{M}$. The equilibrium constant isThus $[\mathrm{HF}]=c^{\circ }\sqrt{K/2}=1.9395\mathrm{M}$.

Let the fixed liquid volume be $V_l$ and headspace volume $V_g=4V_l$. If $\xi$ moles react, then $n_{{\mathrm{O}}_2}=\xi$, $n_{\mathrm{O}{\mathrm{F}}_2}=n_0-\xi$, and the pressure ratio gives $\xi /(n_0-\xi )=2$, so $n_0=3\xi /2$. The HF amount is $2\xi =[\mathrm{HF}]V_l$. Thereforeusing $R=0.083145\mathrm{L}\mathrm{bar}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$ with molarity in $\mathrm{mol}{\mathrm{L}}^{-1}$. The equilibrium gas pressures are approximately $6.01$ bar oxygen and $3.00$ bar oxygen difluoride; gas stoichiometry is one molecule to one, so their sum equals the initial gas pressure under these approximations.

Assume negligible gas dissolution, HF vapour, HF dissociation and nonideality, and negligible change of liquid/headspace volumes on reaction. Water consumption is about $0.97$ mol per litre of original liquid, small compared with its roughly $55.5$ mol of water. At nearly $2\mathrm{M}$ HF, the ideal-solution approximation is nevertheless a substantive limitation: without activity coefficients and an acid-speciation model, a precise real-solution pressure cannot be deduced from the supplied data.