Equilibrium practice · Hard

Option A (S is correct but E is wrong) is correct because the acidic nature of boron trihalides increases down the group due to increasing pπ-pπ back bonding, but the basic nature of nitrogen trihalides decreases down the group.

Option A (+3, +4, +5) is correct because these are the calculated oxidation states of P in the respective compounds based on standard oxidation rules.

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because $BaCO_3$ solubility increases in $HNO_3$ due to the reaction of carbonate with $H^+$, causing dissociation of the barium salt.

Option D ($2.7 \times {10^{ - 2}}\;M$) is correct because it accurately calculates the hydrogen ion concentration at the equivalence point considering the hydrolysis of the resulting salt.

Option C (40 ml) is correct because the volume of KMnO4 required is calculated based on the stoichiometry of the redox reaction.

Option A ($7.14 \times 10^{-3} \mathrm{M}$) is correct because it represents the calculated hydrogen ion concentration in the mixture of two weak acids.

Option B is correct. SO2 decolorizes iodine in the presence of starch, as it reduces iodine to iodide ions.

Option A (187.85) is correct because it's the calculated value of Kc based on the equilibrium concentrations of the reactants and products.

Option C ($25 \%$ ) is correct because the fraction of $A B_3({~g})$ dissociated is calculated based on the change in pressure at equilibrium.

Option A ($\left(K_1\right)^2=\frac{{1}}{{K_2}}$) is correct because the second equilibrium is the reverse of the first equilibrium multiplied by 2.

Option A ($2.08 \times 10^{-13}; 12.64 \times 10^{-8}$) is correct because it is calculated by using the dissociation constant and density to find the ionic product and degree of dissociation.

Option C is correct. The ratio [H₃A]/[A³⁻] at the second equivalent point is approximately 10⁻⁷, calculated using the given Ka values and the pH at the second equivalence point.

Option A (2.3) is correct because it accurately calculates the pH of the resulting solution after mixing two solutions with different pH values.

The correct answer is $\frac{{(n \alpha)^n \cdot P^{n-1}}}{{(1-\alpha) \cdot [1+(n-1) \alpha]^{n-1}}}$ because it correctly represents the equilibrium constant Kp in terms of degree of dissociation and total pressure.

Option D ($\sqrt{K \cdot V}$) is correct because the degree of ionization for a weak electrolyte is inversely proportional to the square root of the concentration, and concentration is inversely proportional to volume.

Option A (600) is correct because after adding chlorine and re-establishing equilibrium, the value of 90x is calculated to be 600.

Option B ($\mathrm{K}_3=\mathrm{K}_1 \mathrm{~K}_2$) is correct because when adding reactions, the equilibrium constants are multiplied.

Option C ($0.76\, torr$) is correct because the equilibrium vapor pressure is calculated from the given $K_p$ value.

Option C (1) is correct because after finding the roots and substituting them into the expression, the result simplifies to 1.

Option A ($5 \times 10^{-17}$) is correct because the ratio of [A^3-] to [H_3A] is calculated using the successive ionization constants.

Option C ($1.6 {~atm}$) is correct because it is calculated using the partial pressures of the gases at equilibrium and the expression for $K_p$.

Option C ($\mathrm{ZnS} < \mathrm{AgBr} < \mathrm{CuCO}_3 < \mathrm{PbSO}_4$) is correct because it represents the increasing order of solubility based on the calculated solubility from the given Ksp values.

Option C is correct because it correctly orders the pH values of the given solutions from most acidic to most basic.

Option C (0.029) is correct because the equilibrium constant Kc is calculated based on the stoichiometry of the reaction and the amount of ammonia produced, which is determined by the titration with sulfuric acid.

Option A ($-7.64 \times 10^4 \,J\,mol^{-1}$) is correct because it is calculated using the formula $\Delta G^{\circ} = -2.303RT\log K_c$.

Option A (Al3+) is correct because Al(OH)3 has the lowest Ksp, and therefore requires the lowest [OH-] concentration to precipitate.

The correct answer is B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) because both the assertion and the reason are true statements about the titration of sodium carbonate with hydrochloric acid, but the reason does not directly explain the assertion's observation regarding the volume difference with different indicators.

Option B ($10^{-2}$ M) is correct because after the reaction with HCl, the concentration of CH3COO- is determined by the equilibrium established with CH3COOH and H+.

Option C (Increase in temperature favours forward reaction) is correct because the reaction is endothermic ($\Delta H = +ve$), so increasing the temperature will shift the equilibrium towards the products according to Le Chatelier's principle.

Option C ($100 \mathrm{~mL}$) is correct because it is calculated using the Henderson-Hasselbalch equation and the given pH and pKa values to determine the volume of benzoic acid.