JEE Chemistry Chemical Equilibrium MCQs Set 08

Answer: (a) \( S=\left(\dfrac{K_{sp}}{6912}\right)^{1/7} \)

Explanation:
1. \( \ce{Zr3(PO4)4 <=> 3Zr^{4+} + 4PO4^{3-}} \), so \( [\ce{Zr^{4+}}]=3S \) and \( [\ce{PO4^{3-}}]=4S \).
2. \( K_{sp}=(3S)^3(4S)^4=27\times256\,S^7=6912\,S^7 \).
3. \( S=\left(\dfrac{K_{sp}}{6912}\right)^{1/7} \).

Teacher's Note:
1. Write the dissociation, raise each ion concentration to its coefficient and multiply the numerical factors.

Answer: (b) \( 10^{25} \)

Explanation:
1. The target equals (reaction 2) minus 2 times (reaction 1): \( 2\ce{SO2} \) is obtained by reversing twice reaction 1.
2. \( K_3=\dfrac{K_2}{K_1^2}=\dfrac{10^{129}}{10^{104}} \).
3. \( K_3=10^{25} \).

Teacher's Note:
1. Reversing a reaction inverts K, doubling the coefficients squares K, and adding reactions multiplies the constants.

Answer: (d) I, II and III

Explanation:
1. I: \( 0.08 \) mol \( \ce{H+} \) minus \( 0.04 \) mol \( \ce{OH-} \) leaves 0.04 mol in 0.8 L, i.e. 0.05 M, pH = 1.3 (true).
2. II: \( K_w \) changes with temperature (true).
3. III: \( [\ce{H+}]=C\alpha=10^{-5} \) and \( K_a=\dfrac{C\alpha^2}{1-\alpha}=10^{-5} \) give \( \alpha=0.5 \) (true).
4. IV: the common-ion effect is an application of Le Chatelier's principle, so IV is false.

Teacher's Note:
1. Check each statement numerically.
2. The common-ion effect is a shift of equilibrium and follows Le Chatelier's principle.

Answer: (b) A large equilibrium constant means the reaction goes to completion, so no catalyst is required

Explanation:
1. The reaction is exothermic, so K decreases with rising temperature (correct).
2. An inert gas at constant volume does not change concentrations, so the equilibrium is undisturbed (correct).
3. \( \Delta n_g=2-3=-1 \), so more pressure favours the forward reaction (correct).
4. A large K says nothing about the rate; the reaction needs a catalyst (\( \ce{V2O5} \)) to be fast. So the last statement is incorrect.

Teacher's Note:
1. The equilibrium constant describes the position of equilibrium, not how fast it is reached.

Answer: (c) 5.35

Explanation:
1. For a salt of a strong acid and a weak base: \( \text{pH}=7-\tfrac12(\text{p}K_b+\log C) \).
2. \( =7-\tfrac12\left(5+\log0.02\right)=7-\tfrac12(5-1.699) \).
3. \( \text{pH}=7-1.65=5.35 \).

Teacher's Note:
1. Hydrolysis of the ammonium ion makes the solution acidic, so the pH is below 7.

Answer: (c) \( 3\times10^{-22}\ \text{M} \)

Explanation:
1. The common ion \( \ce{OH-} \) from NaOH is 0.2 M and dominates.
2. \( S=\dfrac{K_{sp}}{[\ce{OH-}]^3}=\dfrac{2.4\times10^{-24}}{(0.2)^3}=\dfrac{2.4\times10^{-24}}{8\times10^{-3}} \).
3. \( S=3\times10^{-22} \) M.

Teacher's Note:
1. Use the common-ion concentration in the Ksp expression and ignore the small contribution from the dissolving salt.

Answer: (b) \( 2.49\times10^{-10}\ \text{M} \)

Explanation:
1. \( K_{sp}=4S^3=4(1.84\times10^{-5})^3=2.49\times10^{-14} \).
2. At pH 12, \( [\ce{OH-}]=10^{-2} \) M.
3. \( S'=\dfrac{K_{sp}}{[\ce{OH-}]^2}=\dfrac{2.49\times10^{-14}}{10^{-4}}=2.49\times10^{-10} \) M.

Teacher's Note:
1. First get Ksp from the solubility in water, then divide by the fixed hydroxide concentration squared.

Answer: (b) \( \ce{2C(s) + O2(g) <=> 2CO(g)} \)

Explanation:
1. \( K_p=K_c(RT)^{\Delta n_g} \), so they differ only if \( \Delta n_g\ne0 \).
2. For the carbon reaction: \( \Delta n_g=2-1=1 \) (solid carbon is ignored).
3. The other three have equal moles of gas on both sides, so \( \Delta n_g=0 \).

Teacher's Note:
1. Count only the gaseous species when finding the change in moles.

Answer: (a) 9.0

Explanation:
1. Moles: \( \ce{H2SO4} \) = 2 mmol, \( \ce{NH4OH} \) = 6 mmol. Each mmol of \( \ce{H2SO4} \) neutralises 2 mmol of base: 4 mmol \( \ce{NH4OH} \) reacts.
2. Left: 2 mmol \( \ce{NH4OH} \) and 2 mmol \( \ce{(NH4)2SO4} \) = 4 mmol \( \ce{NH4+} \): a basic buffer.
3. \( \text{pOH}=4.7+\log\dfrac{4}{2}=4.7+0.3=5.0 \).
4. \( \text{pH}=14-5.0=9.0 \).

Teacher's Note:
1. Remember each formula unit of ammonium sulphate carries two ammonium ions, so the salt contributes twice its moles to the buffer ratio.

Answer: (a) \( 1,\ 24.62\ \text{dm}^3\text{atm mol}^{-1},\ 1.65\times10^{-3}\ \text{dm}^{-6}\text{atm}^{-2}\text{mol}^2 \)

Explanation:
1. \( K_p/K_c=(RT)^{\Delta n_g} \).
2. Reaction 1: \( \Delta n_g=0 \), so the ratio is 1. Reaction 2: \( \Delta n_g=1 \), so \( RT=24.62 \).
3. Reaction 3: \( \Delta n_g=-2 \), so \( (RT)^{-2}=\dfrac1{(24.62)^2}=1.65\times10^{-3} \).

Teacher's Note:
1. Use the relation between Kp and Kc for each reaction and keep track of the units of RT.

Answer: (b) 1.9 g, 0.28 mol/L

Explanation:
1. Moles of \( \ce{Na2SO4} \) \( =2/142=14.08 \) mmol, the limiting reagent.
2. \( \ce{Ca(OH)2 + Na2SO4 -> CaSO4 + 2Na+ + 2OH-} \): \( \ce{CaSO4} \) formed = 14.08 mmol, mass \( =14.08\times10^{-3}\times136=1.9 \) g.
3. \( \ce{OH-} \) formed = 28.16 mmol in 100 mL, i.e. 0.28 M.

Teacher's Note:
1. Find the limiting reagent, then use the 1 to 2 mole ratio of calcium sulphate to hydroxide.

Answer: (a) \( 0.242\ \text{atm}^2 \)

Explanation:
1. Moles of \( \ce{NH4SH} \) \( =5.1/51=0.1 \); 30% decomposed gives 0.03 mol each of \( \ce{NH3} \) and \( \ce{H2S} \).
2. Partial pressure of each gas: \( P=\dfrac{nRT}{V}=\dfrac{0.03\times0.082\times600}{3}=0.492 \) atm.
3. \( K_p=P_{\ce{NH3}}\times P_{\ce{H2S}}=(0.492)^2=0.242 \) atm\(^2\).

Teacher's Note:
1. Only gases appear in Kp.
2. Convert the temperature to kelvin and use the ideal gas law for each partial pressure.

Answer: (a) \( \dfrac{3^{3/2}K_p^{1/2}P^2}{16} \)

Explanation:
1. With almost all ammonia dissociated, \( p_{\ce{N2}}\approx P/4 \) and \( p_{\ce{H2}}\approx3P/4 \) (1 : 3 ratio).
2. \( K_p=\dfrac{p_{\ce{NH3}}^2}{p_{\ce{N2}}\,p_{\ce{H2}}^3}=\dfrac{p_{\ce{NH3}}^2}{(P/4)(27P^3/64)}=\dfrac{256\,p_{\ce{NH3}}^2}{27P^4} \).
3. \( p_{\ce{NH3}}=\dfrac{\sqrt{27K_p}\,P^2}{16}=\dfrac{3^{3/2}K_p^{1/2}P^2}{16} \).

Teacher's Note:
1. Use the 1 : 3 stoichiometric split of the total pressure for the products and solve for the small ammonia pressure.

Answer: (d) 0.4736 V

Explanation:
1. At equilibrium \( E^\circ_{cell}=\dfrac{0.0592}{n}\log K_c \) with \( n=2 \).
2. \( \log(10\times10^{15})=16 \).
3. \( E^\circ_{cell}=\dfrac{0.0592}{2}\times16=0.4736 \) V.

Teacher's Note:
1. At equilibrium the cell potential is zero, which links the standard potential to the logarithm of K through the Nernst equation.

Answer: (a) 80 kJ/mol

Explanation:
1. \( K_w=10^{-14} \) at 298 K.
2. \( \Delta G^\circ=-2.303RT\log K_w=2.303\times8.314\times298\times14 \) J/mol.
3. \( \Delta G^\circ\approx79.9\) kJ/mol \(\approx80 \) kJ/mol.

Teacher's Note:
1. A tiny equilibrium constant means a large positive standard Gibbs energy change.

Answer: (a) \( 2\sqrt{x+y}\ \text{atm} \)

Explanation:
1. Let \( p_B=a \), \( p_E=b \); then \( p_C=a+b \).
2. \( x=a(a+b) \) and \( y=b(a+b) \), so \( x+y=(a+b)^2 \) and \( a+b=\sqrt{x+y} \).
3. Total pressure \( =a+b+(a+b)=2\sqrt{x+y} \).

Teacher's Note:
1. Common gas C appears in both equilibria with pressure equal to the sum of the other two partial pressures.

Answer: (b) \( 3\times10^{-20}\ \text{M} \)

Explanation:
1. Overall: \( \ce{H2S <=> 2H+ + S^{2-}} \), \( K=K_1K_2=1.2\times10^{-20} \).
2. \( [\ce{S^{2-}}]=\dfrac{K[\ce{H2S}]}{[\ce{H+}]^2}=\dfrac{1.2\times10^{-20}\times0.10}{(0.20)^2} \).
3. \( [\ce{S^{2-}}]=3\times10^{-20} \) M.

Teacher's Note:
1. The strong acid fixes the hydrogen-ion concentration; multiply the two dissociation constants for the overall equilibrium.

Answer: (a) \( 1.1\times10^{-9}\ \text{M} \)

Explanation:
1. After mixing, \( [\ce{SO4^{2-}}]=\dfrac{1\times50}{500}=0.1 \) M.
2. \( [\ce{Ba^{2+}}]_{final}=\dfrac{K_{sp}}{0.1}=1\times10^{-9} \) M in 500 mL.
3. The original solution was 450 mL: \( M_1\times450=10^{-9}\times500 \), so \( M_1=1.1\times10^{-9} \) M.

Teacher's Note:
1. Account for the dilution: the barium solution was originally 450 mL, since 50 mL of sodium sulphate was added to reach 500 mL.

Answer: (b) \( \ce{BCl3}\ \text{and}\ \ce{AlCl3} \)

Explanation:
1. A Lewis acid accepts an electron pair.
2. \( \ce{BCl3} \) and \( \ce{AlCl3} \) have incomplete octets (vacant p orbital), so they are the standard Lewis acids.
3. \( \ce{PH3} \) has a lone pair and is a Lewis base.

Teacher's Note:
1. Look for electron-deficient atoms with a vacant orbital for the best-defined Lewis acids.

Answer: (c) \( \ce{CH3COOK} \)

Explanation:
1. \( \ce{CH3COOK} \) is the salt of a strong base (KOH) and a weak acid, so its anion hydrolyses to give \( \ce{OH-} \).
2. \( \ce{FeCl3} \) is acidic, and the other two salts involve weak-base cations, so they are less basic.

Teacher's Note:
1. A salt of a strong base and a weak acid gives a basic solution.