JEE Chemistry Equilibrium MCQs Set 18

Answer: (c) All the physical processes stop at equilibrium

Explanation:
1. Equilibrium is dynamic: the forward and reverse processes continue at equal rates.
2. So the statement that all processes stop is wrong.

Teacher's Note:
1. Dynamic equilibrium means no net change but continuous activity.
2. Properties such as concentration and pressure stay constant.

Answer: (c) \(\Delta H > 0\)

Explanation:
1. On cooling, the equilibrium shifts towards the pink hexaaqua complex, which is the reverse reaction.
2. By Le Chatelier's principle, cooling favours the exothermic direction.
3. So the reverse is exothermic and the forward reaction is endothermic, \(\Delta H > 0\).

Teacher's Note:
1. Heat is a reactant for an endothermic forward reaction.
2. Cooling favours the exothermic direction.

Answer: (c) formic acid < acetic acid < hypochlorous acid

Explanation:
1. A larger Ka gives a higher [H+] and so a lower pH.
2. Ka order: formic > acetic > hypochlorous.
3. So the pH order is formic acid < acetic acid < hypochlorous acid.

Teacher's Note:
1. Stronger acid means lower pH at the same concentration.
2. [H+] is about \(\sqrt{K_a C}\) for weak acids.

Answer: (d) Lewis concept

Explanation:
1. BF3 has no H+ to donate but has an empty p orbital on boron.
2. It accepts an electron pair, so it is a Lewis acid.

Teacher's Note:
1. Lewis acids accept electron pairs.
2. BF3 forms an adduct with NH3.

Answer: (c) water < acetone < ether

Explanation:
1. A lower boiling point means weaker intermolecular forces and higher vapour pressure at a given temperature.
2. Boiling points: water > acetone > ether.
3. So the vapour pressure order is water < acetone < ether.

Teacher's Note:
1. Volatile liquids have a high vapour pressure.
2. Hydrogen bonding in water lowers its vapour pressure.

Answer: (d) The reaction is endothermic

Explanation:
1. K increases with temperature, which means the forward reaction is endothermic.
2. At 400 K, Q = 20^2/2 = 200, which is greater than K = 50, so the reaction goes backwards and more N2O4 forms.
3. The number of gas moles increases, so entropy increases.

Teacher's Note:
1. For an endothermic reaction K rises with T.
2. Compare Q with K to find the direction.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Down the group the H-X bond becomes weaker, so the acid loses H+ more easily.
2. This outweighs the decrease in bond polarity, so acidity increases.
3. The reason correctly explains the assertion.

Teacher's Note:
1. HI is the strongest of the halogen acids.
2. Bond length rises down the group, so the bond energy falls.

Answer: (b) Both A and R are true but R is not the correct explanation of A

Explanation:
1. The H+ from HCl is a common ion, which suppresses the ionisation of H2S, so A is true.
2. H2S is indeed a weak acid, so R is true.
3. However, the suppression is caused by the common ion, not by weakness alone, so R does not explain A.

Teacher's Note:
1. Common ion effect suppresses the ionisation of weak electrolytes.
2. This is used in group II qualitative analysis.

Answer: (c) A is true but R is false

Explanation:
1. At constant pressure, adding an inert gas increases the volume, which lowers the partial pressures.
2. The equilibrium shifts to the side with more moles of gas (the dissociation side), so A is true.
3. Helium does not remove Cl2, so R is false.

Teacher's Note:
1. Inert gas at constant pressure acts like a decrease in pressure.
2. At constant volume an inert gas has no effect.

Answer: (a) 9.0 L

Explanation:
1. pH 1 means [H+] = 0.1 M and pH 2 means 0.01 M, i.e. tenfold dilution.
2. Final volume must be 10 L, so water to be added = 10 - 1 = 9.0 L.

Teacher's Note:
1. Each pH unit needs a ten-fold dilution for a strong acid.
2. Added volume = final volume - initial volume.

Answer: (c) 1.8 atm

Explanation:
1. Let p(CO2) fall by x: p(CO2) = 0.5 - x and p(CO) = 2x.
2. Total = 0.5 + x = 0.8, so x = 0.3; p(CO2) = 0.2 atm and p(CO) = 0.6 atm.
3. \(K_p = (0.6)^2/0.2 = 1.8\) atm.

Teacher's Note:
1. Solid graphite is left out of the expression.
2. Total pressure rises by the net increase in moles of gas.

Answer: (a) II only

Explanation:
1. An acid donates a proton. In II, H2PO4- gives H+ to water to form HPO4 2-, so it acts as an acid.
2. In I it is formed as the conjugate base, and III would require it to accept a proton, so it acts as a base.

Teacher's Note:
1. Look at whether the species loses or gains a proton.
2. H2PO4- is amphiprotic.

Answer: (d) The concentrations of \(\ce{H+}\) and \(\ce{HCO3-}\) are approximately equal

Explanation:
1. K1 is much greater than K2, so nearly all H+ comes from the first ionisation: [H+] is about [HCO3-].
2. [CO3 2-] is approximately K2, which is very small (4.8 x 10^-11 M).

Teacher's Note:
1. For a diprotic weak acid, the second ionisation is negligible.
2. [A2-] is approximately K2 when [H+] = [HA-].

Answer: (d) 10

Explanation:
1. \([OH^-]^2 = K_{sp}/[Mg^{2+}] = 10^{-11}/10^{-3} = 10^{-8}\).
2. \([OH^-] = 10^{-4}\) M, so pOH = 4 and pH = 10.

Teacher's Note:
1. Use Ksp = [Mg2+][OH-]^2.
2. pH = 14 - pOH.

Answer: (a) 1/RT

Explanation:
1. \(K_p/K_c = (RT)^{\Delta n_g}\), with \(\Delta n_g = 1 - 2 = -1\).
2. So \(K_p/K_c = 1/RT\).

Teacher's Note:
1. Count gas moles on each side.
2. Units of R must match the units of Kc.

Answer: (b) \(\ce{HPO4^{2-}}\)

Explanation:
1. The conjugate base is obtained by removing one proton.
2. H2PO4- minus H+ gives HPO4 2-.

Teacher's Note:
1. H3PO4 is the conjugate acid of H2PO4-.
2. H2PO4- is amphiprotic.

Answer: (d) lowering the temperature and increasing the pressure

Explanation:
1. The forward reaction is exothermic, so low temperature favours it.
2. It reduces the moles of gas (3 to 2), so high pressure favours it.

Teacher's Note:
1. Exothermic reactions are favoured by cooling.
2. Pressure favours the side with fewer gas moles.

Answer: (d) \(\ce{HClO4}\) and \(\ce{NaClO4}\)

Explanation:
1. An acidic buffer contains a weak acid and its salt with a strong base.
2. HClO4 is a strong acid, so it cannot form a buffer with its conjugate base.

Teacher's Note:
1. Buffers need a weak acid or weak base.
2. Strong acids and strong bases do not form buffers.

Answer: (c) \(\ce{Na2CO3}\)

Explanation:
1. KCl and NaCl are salts of strong acid and strong base, giving pH 7.
2. CuSO4 is acidic by hydrolysis of Cu2+.
3. Na2CO3 is a salt of a weak acid and a strong base, giving a basic solution (CO3 2- hydrolyses).

Teacher's Note:
1. The anion of a weak acid hydrolyses to give OH-.
2. Na2CO3 has pH about 11 at 0.1 M.

Answer: (b) \(\ce{BF3}\)

Explanation:
1. BF3 has an incomplete octet and an empty orbital, so it is a Lewis acid.
2. PF3, CO and F- all have lone pairs available for donation.

Teacher's Note:
1. Lewis bases donate electron pairs.
2. Boron halides are Lewis acids.

Answer: (c) 10 : 1

Explanation:
1. pKa = 4 and the target pH = 5.
2. pH = pKa + log([salt]/[acid]) gives log ratio = 1.
3. [Salt]/[Acid] = 10 : 1.

Teacher's Note:
1. Use the Henderson-Hasselbalch equation.
2. A 10 : 1 ratio changes pH by one unit.

Answer: (c) \(x = (2K_p/P)^{1/3}\)

Explanation:
1. Start with 2 mol: AB2 = 2(1-x), AB = 2x, B2 = x; total = 2 + x, about 2.
2. Partial pressures: \(p_{AB_2} = 2(1-x)P/(2+x)\), \(p_{AB} = 2xP/(2+x)\), \(p_{B_2} = xP/(2+x)\). For small x, \(K_p = (2x)^2 x P/[2(2)^2] = x^3P/2\).
3. So \(x = (2K_p/P)^{1/3}\).

Teacher's Note:
1. Use mole fractions times total pressure.
2. Small x lets you drop x from (1 - x) and 2 + x.

Answer: (c) \(\ce{CH3COO-}\)

Explanation:
1. The weaker the acid, the stronger its conjugate base.
2. Acetic acid is a weak acid, while HCl, HNO3 and HSO4- are strong, so acetate is the strongest base.

Teacher's Note:
1. Conjugate base strength is inversely related to acid strength.
2. Cl- and NO3- are extremely weak bases.

Answer: (a) constant temperature

Explanation:
1. The vapour pressure of a liquid is a function of temperature only.
2. At a fixed temperature it has a fixed value at equilibrium.

Teacher's Note:
1. Vapour pressure rises with temperature.
2. It is independent of the amount of liquid.

Answer: (a) 0.0064 g

Explanation:
1. s = \(\sqrt{K_{sp}} = \sqrt{2.5\times10^{-9}} = 5\times10^{-5}\) mol/L.
2. Mass = \(5\times10^{-5} \times 128 = 6.4\times10^{-3}\) g.

Teacher's Note:
1. For a 1:1 salt, s = √Ksp.
2. Multiply by molar mass for g/L.

Answer: (a) 1

Explanation:
1. \(pH = pK_a + \log([salt]/[acid])\).
2. A tenfold increase adds log 10 = 1 to the pH.

Teacher's Note:
1. Buffer pH changes with the logarithm of the ratio.
2. The pKa remains constant.

Answer: (a) \(4\times10^{-10}\)

Explanation:
1. OH- from NaOH is 0.1 M and OH- from Mg(OH)2 is negligible.
2. \([Mg^{2+}] = K_{sp}/[OH^-]^2 = 4\times10^{-12}/(0.1)^2 = 4\times10^{-10}\) M.

Teacher's Note:
1. The common ion suppresses the solubility.
2. Use the NaOH concentration as [OH-].

Answer: (a) 1.5

Explanation:
1. ZnS precipitates when \([S^{2-}] = 10^{-21}/10^{-2} = 10^{-19}\) M.
2. \([H^+]^2 = 10^{-22}/10^{-19} = 10^{-3}\), so \([H^+] = 10^{-1.5}\) M.
3. The minimum pH is 1.5.

Teacher's Note:
1. Higher pH raises [S2-] through the H2S equilibrium.
2. Find [S2-] from the Ksp first.