Chemistry - Chemical Equilibrium Question with Solution | TestHub

Option A is incorrect. Diamond (d=3.5 g/mL) is denser than graphite (d=2.3 g/mL). For $\Delta H=-1.9$ KJ/mol (exothermic), high pressure favors the denser phase (diamond), and low temperature favors the exothermic reaction. Thus, high pressure and low temperature favor diamond formation, not graphite.

Option B is correct. For ${\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})\rightleftharpoons 2N{\mathrm{O}}_2(\mathrm{g})$, if $\alpha$ is the degree of dissociation, then $P_{{\mathrm{N}}_2{\mathrm{O}}_4}=\frac{1-\alpha }{1+\alpha }P$ and $P_{\mathrm{N}{\mathrm{O}}_2}=\frac{2\alpha }{1+\alpha }P$.

$K_p=\frac{P_{\mathrm{N}{\mathrm{O}}_2}^2}{P_{{\mathrm{N}}_2{\mathrm{O}}_4}}=\frac{(\frac{2\alpha }{1+\alpha }P{)}^2}{\frac{1-\alpha }{1+\alpha }P}=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$.

For small $\alpha$, $1-{\alpha }^2\approx 1$, so $K_p\approx 4{\alpha }^{2}P$, which gives $\alpha =\sqrt{\frac{K_p}{4P}}$.

The given formula $\alpha =\sqrt{\frac{K_p}{4P+K_p}}$ is derived from

$$\begin{gathered} \text{displaylines}{K}_p(1-{\alpha }^2)=4{\alpha }^{2}P \\ ⟹K_p-K_p{\alpha }^2=4{\alpha }^{2}P \\ ⟹K_p={\alpha }^2(4P+K_p) \\ ⟹{\alpha }^2=\frac{K_p}{4P+K_p} \\ ⟹\alpha =\sqrt{\frac{K_p}{4P+K_p}} \end{gathered}$$.

This formula is correct.

Option C is correct. For $\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+3{\mathrm{F}}_2(\mathrm{g})\rightleftharpoons 2Cl{\mathrm{F}}_3(\mathrm{g})$; $\Delta H=-329$ kJ.

Dissociation of $\mathrm{Cl}{\mathrm{F}}_3(\mathrm{g})$ means the reverse reaction: $2Cl{\mathrm{F}}_3(\mathrm{g})\rightleftharpoons \mathrm{C}{\mathrm{l}}_2(\mathrm{g})+3{\mathrm{F}}_2(\mathrm{g})$.

For the reverse reaction, $\Delta H=+329$ kJ (endothermic).

Addition of an inert gas at constant pressure increases the total volume, which decreases the partial pressures of reactants and products. This shifts the equilibrium towards the side with more moles of gas to counteract the decrease in pressure. In the reverse reaction, there are 2 moles of gas on the left and 4 moles of gas on the right. Thus, the equilibrium shifts to the right, favoring dissociation.

Option D is incorrect. A reversible reaction at equilibrium is dynamic, meaning the forward and reverse reactions continue at equal rates. It does not stop microscopically.