Equilibrium

Q1. (3)

• Dynamic equilibrium: the state of a reversible reaction in which the rate of the forward reaction equals the rate of the reverse reaction, so macroscopic concentrations remain constant. (1)
• Condition 1: rates of forward and reverse reactions are equal. (1)
• Condition 2: concentrations of reactants and products remain constant with time (reaction continues at molecular level). (1)

Q2. (4) (a) $K_c = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$ (2) (b) $K_p = K_c (RT)^{\Delta n}$ where $\Delta n =$ (moles gaseous product − moles gaseous reactant) (1) $\Delta n = 2 - (1+3) = -2$ (1)

Q3. (3)

• $Q_c = 3.0 > K_c = 1.8$. (1)
• Since $Q > K$, the ratio of products to reactants is too high; reaction proceeds in the reverse direction (right → left). (1)
• Products convert to reactants until $Q$ decreases to equal $K$. (1)

Q4. (4)

• Le Chatelier: if a system at equilibrium is subjected to a change (in concentration, pressure or temperature), the equilibrium shifts in the direction that tends to counteract the change. (1) (i) Increasing pressure shifts equilibrium toward the side with fewer gas moles (3 mol → 2 mol), i.e. forward — yield of $SO_3$ increases. (1.5) (ii) Reaction is exothermic ($\Delta H<0$); increasing temperature favours the endothermic (reverse) direction — yield of $SO_3$ decreases. (1.5)

Q5. (4) (a) $K_c = [CO_2]$ (or $K_p = p_{CO_2}$). (1) Pure solids have constant activity (=1), fixed density/concentration independent of amount, so they are omitted. (1) (b) Conjugate base of $HCO_3^-$ = $CO_3^{2-}$ (loss of $H^+$). (1) Conjugate acid of $HCO_3^-$ = $H_2CO_3$ (gain of $H^+$). (1)

Q6. (4) — 1 mark each

• $NH_3$: Brønsted base (accepts $H^+$).
• $BF_3$: Lewis acid (accepts electron pair).
• $H_2O$: amphoteric — acceptable as Brønsted acid or base.
• $HSO_4^-$: Brønsted acid (can donate $H^+$) [amphoteric acceptable].

Q7. (4) (a) HCl strong: $[H^+]=0.01$, $pH=-\log(10^{-2})=2$. (2) (b) NaOH strong: $[OH^-]=0.01$, $pOH=2$, $pH=14-2=12$. (2)

Q8. (5) (a) Ostwald: $\alpha=\sqrt{K_a/C}=\sqrt{1.8\times10^{-5}/0.10}=\sqrt{1.8\times10^{-4}}=1.34\times10^{-2}$. (2) (b) $[H^+]=C\alpha=0.10\times1.342\times10^{-2}=1.342\times10^{-3}$. (1.5) $pH=-\log(1.342\times10^{-3})=2.87$. (1.5)

Q9. (4) $pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} = 4.74 + \log\dfrac{0.30}{0.20}$ (2) $= 4.74 + \log(1.5) = 4.74 + 0.176 = 4.92$. (2)

Q10. (5) (a) Pure water: $s=\sqrt{K_{sp}}=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\ \text{M}$. (2) (b) In $0.10$ M NaCl: $[Cl^-]\approx0.10$, so $s=K_{sp}/[Cl^-]=1.8\times10^{-10}/0.10=1.8\times10^{-9}\ \text{M}$. (2) Comment: solubility drops (~10⁴-fold) due to the common-ion effect — added $Cl^-$ suppresses dissolution. (1)

[
  {"claim":"Q2 Delta n = -2","code":"result = (2-(1+3)) == -2"},
  {"claim":"Q7 pH HCl = 2 and NaOH = 12","code":"import sympy as sp\npH_a = -sp.log(sp.Rational(1,100),10)\npOH = -sp.log(sp.Rational(1,100),10)\npH_b = 14 - pOH\nresult = (pH_a==2) and (pH_b==12)"},
  {"claim":"Q8 alpha ~0.0134 and pH ~2.87","code":"import sympy as sp\nalpha = sp.sqrt(sp.Rational(18,1000000)/sp.Rational(1,10))\nalpha_v = float(alpha)\nH = 0.10*alpha_v\npH = -sp.log(H,10)\nresult = abs(alpha_v-0.01342)<1e-4 and abs(float(pH)-2.872)<0.02"},
  {"claim":"Q9 buffer pH ~4.92","code":"import sympy as sp\npH = 4.74 + sp.log(sp.Rational(30,20),10)\nresult = abs(float(pH)-4.916)<0.01"},
  {"claim":"Q10 solubilities","code":"import sympy as sp\ns_water = sp.sqrt(sp.Float(1.8e-10))\ns_ci = 1.8e-10/0.10\nresult = abs(float(s_water)-1.342e-5)<1e-7 and abs(s_ci-1.8e-9)<1e-12"}
]