Chemical Equilibrium

$ \mathrm{PCI}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) $

Case 1 : At constant pressure, the addition of inert gas (helium) will shift the reaction where number of moles of gaseous substances are more hence, the reaction will proceed in forward direction. Then answer will be $A$.

Case 2 : At constant volume, addition of He will not affect the equilibrium so reaction will not shift in any direction then answer will be $B$.

The thermal decomposition of limestone (CaCO₃) is represented as:

$ \mathrm{CaCO}_3(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}), \Delta_r \mathrm{H} = \text {positive} $

The reaction is endothermic, meaning it requires heat to proceed.

(i) The forward reaction is endothermic (ΔH > 0), indicating heat absorption, while the reverse reaction is exothermic (ΔH < 0), indicating heat release.

Therefore, increasing the temperature will favor the forward reaction, leading to more CaCO₃ dissociation. Conversely, decreasing the temperature will favor the reverse reaction. Option (A) is correct.

The equilibrium constant (K) varies with temperature but does not depend on the concentration of any species in an aqueous solution or the pressure of gases such as CO₂. Option (B) is correct.

(ii) The equilibrium constant (K) for the reaction is expressed as:

$ \mathrm{K}_{\mathrm{C}} = [\mathrm{CO}_2] $

Because CaCO₃ and CaO are solids, their concentrations do not affect the equilibrium constant. Option (B) is correct.

(iii) At a given temperature, the equilibrium constant in terms of pressure (K_P) depends on the vapour pressure of CO₂.

$ \mathrm{K}_{\mathrm{P}} = \mathrm{P}_{\mathrm{CO}_2} $

The equilibrium constant (K_C) also depends on the pressure of CO₂:

$ \mathrm{K}_{\mathrm{C}} = \mathrm{P}_{\mathrm{CO}_2} (\mathrm{RT})^{-\Delta n} $

Option (C) is correct.

The temperature dependence of the Arrhenius constant is given by:

$ \ln \frac{\mathrm{K}_2}{\mathrm{K}_1} = \frac{\Delta_r \mathrm{H}}{\mathrm{R}} \left[ \frac{1}{\mathrm{T}_2} - \frac{1}{\mathrm{T}_1} \right] $

(iv) A catalyst alters the activation energy of the reaction, thus affecting its speed. However, it does not change the enthalpy or heat of the reactants and products. As a result, the enthalpy of the reaction ($\Delta_r H$) remains unchanged. Option (D) is correct.

Δ_rH is the enthalpy of the reaction.

For the equilibrium reaction in aqueous medium :

$ 2 \mathrm{Cu}^{+} \rightleftharpoons \mathrm{Cu}^0+\mathrm{Cu}^{2+} $

(i) If concentration of copper (I) is reduced, then according to Le-chatelier's principle the equilibrium reaction will shift backward.

(ii) The copper (II) ion reacts with chloride $\left(\mathrm{Cl}^{-}\right)$ion forming copper (III) chloride, which reacts with copper to produce the precipitate of CuCl .

$ \begin{aligned} & \mathrm{Cu}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CuCl}_2 \\\\ & \mathrm{CuCl}_2+\mathrm{Cu} \rightarrow 2 \mathrm{CuCl} \end{aligned} $

Since, the end product involves consumption of copper (I) ion to form CuCl , hence, reactive shifts backward.

(iii) The copper (II) ion also reacts with cyanide $\left(\mathrm{CN}^{-}\right)$ion forming copper (II) cyanide which on further reactions form $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$ with copper in (I) oxidation state shifting reaction backward.

$ \begin{aligned} \mathrm{Cu}^{2+}+2 \mathrm{CN}^{-} & \rightarrow \mathrm{Cu}(\mathrm{CO})_2 \\\\ 2 \mathrm{Cu}(\mathrm{CN})_2 & \rightarrow 2 \mathrm{CuCN}+(\mathrm{CN})_2 \\\\ \mathrm{CuCN}+3 \mathrm{CN}^{-} & \rightarrow\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-} \end{aligned} $

(iv) Reaction of copper (II) ion with thiocyanide $\left(\mathrm{SCN}^{-}\right)$gives a complex $\left(\mathrm{Cu}(\mathrm{SCN})_4\right]^{3-}$ with copper in +1 oxidation state. This shifts reaction backward.

$ \mathrm{Cu}^{2+}+4 \mathrm{SCN}^{-} \rightarrow\left[\mathrm{Cu}(\mathrm{SCN})_4\right]^{3-} $

Hence, cyanide, thiocyanide and chloride forms complexes with copper in I oxidation state. Hence, reaction shifts backward to generate more copper (I) ions.