Questions: Equilibrium

Q1 What are the characteristics of a system at dynamic equilibrium?

Answer (English): At dynamic equilibrium the two opposing processes run at the same rate, so every concentration stays constant. The same state is reached whether the reaction starts from reactants or from products, and it needs a closed vessel held at one temperature. The equilibrium is dynamic and not static: molecules keep reacting in both directions, but the net change is zero. The position of equilibrium can be shifted by changing conditions, though the equilibrium constant itself changes only with temperature.

Hinglish: Dynamic equilibrium mein forward aur reverse rates barabar hoti hain, concentrations constant rehti hain par reactions chalti rehti hain. Equilibrium dono directions se pahunch sakta hai, closed system aur fixed temperature chahiye; conditions badalne par position shift hoti hai par K sirf temperature se badalta hai.

Q2 For N2(g) + 3H2(g) giving 2NH3(g), write the Kc and Kp expressions and the relation between them.

Answer (English): The equilibrium constant in concentration is the square of the ammonia concentration divided by the nitrogen concentration times the cube of the hydrogen concentration. The pressure constant uses partial pressures in the same powers. The change in gas moles is two minus four, which is -2, so Kp equals Kc times RT to the power -2, that is Kp equals Kc divided by RT squared. A negative delta n_g means the pressure constant is smaller than the concentration constant at ordinary temperatures.

Hinglish: Kc = [NH3]²/([N2][H2]³) aur Kp same powers mein partial pressures se. Δn_g = 2 - 4 = -2, isliye Kp = Kc (RT)^-2. Negative delta n_g matlab Kp Kc se chhota hoga.

Q3 State the effect of (a) increasing pressure, (b) increasing temperature and (c) adding a catalyst on the equilibrium N2 + 3H2 giving 2NH3, delta H negative.

Answer (English): Increasing the pressure by reducing the volume shifts the equilibrium towards the side with fewer gas moles, which is the product side, so the ammonia yield rises. Increasing the temperature shifts the equilibrium in the endothermic direction, which is the reverse direction here, so the ammonia yield falls because the forward reaction is exothermic. Adding a catalyst speeds up both directions equally, so it does not change the equilibrium position or the yield; it only helps the system reach equilibrium faster.

Hinglish: (a) Pressure badhane se equilibrium kam gas moles wali side, yaani products ki taraf shift hota hai aur NH3 yield badhta hai. (b) Temperature badhane se endothermic direction, yaani reverse shift hota hai, isliye yield girta hai. (c) Catalyst dono rates barabar badhata hai, position aur yield nahi badalte, sirf equilibrium jaldi aata hai.

Q4 Calculate the pH of 0.01 M HCl and of 0.01 M NaOH at 298 K.

Answer (English): Hydrochloric acid is a strong monobasic acid, so it is completely ionised and the hydrogen ion concentration is 0.01 M. The pH is minus log of 0.01, which is 2. Sodium hydroxide is a strong base, so the hydroxide concentration is 0.01 M and the pOH is 2. Since pH plus pOH equals 14 at 298 K, the pH of the sodium hydroxide solution is 12.

Hinglish: 0.01 M HCl ka [H+] = 0.01 M, isliye pH = 2. 0.01 M NaOH ka [OH-] = 0.01 M, pOH = 2, aur pH = 14 - 2 = 12.

Q5 A 0.1 M solution of a weak monobasic acid has pH 3. Calculate Ka and the degree of dissociation.

Answer (English): A pH of 3 means the hydrogen ion concentration is 10^-3 M. For a weak acid the dissociation constant is the hydrogen ion concentration squared divided by the acid concentration, using the approximation that the undissociated concentration stays close to 0.1 M. That gives Ka equal to 10^-6 divided by 10^-1, which is 1.0 x 10^-5. The degree of dissociation is the hydrogen ion concentration divided by the original concentration, 10^-3/0.1 = 0.01, or one percent. The assumption that the degree of dissociation is small is satisfied.

Hinglish: pH 3 matlab [H+] = 10^-3 M. Ka = [H+]²/[HA] = (10^-3)²/0.1 = 1.0 × 10^-5. α = [H+]/c = 10^-3/0.1 = 0.01 (1%). Approximation chhoti α ke liye valid hai.

Q6 A buffer contains 0.1 M acetic acid and 0.1 M sodium acetate. What is its pH? (pKa = 4.74)

Answer (English): Use the Henderson-Hasselbalch equation. The salt concentration equals the acid concentration, so their ratio is one and the logarithm is zero. The pH therefore equals the pKa, which is 4.74. The buffer is most effective at this pH because the two components are present in equal amounts, which gives the largest buffer capacity. Adding a small amount of acid converts some acetate into acetic acid and changes the ratio slightly, so the pH changes only a little.

Hinglish: Henderson equation: pH = pKa + log([salt]/[acid]). Salt aur acid ki concentration equal hai, isliye log(1) = 0 aur pH = pKa = 4.74. Equal amounts se buffer capacity maximum hoti hai.

Q7 Classify the following salt solutions as acidic, basic or neutral: NaCl, NaCN, NH4NO3, CH3COONa.

Answer (English): Sodium chloride comes from a strong acid and a strong base, so neither ion hydrolyses and the solution is neutral. Sodium cyanide comes from a weak acid and a strong base; the cyanide ion hydrolyses and produces hydroxide, so the solution is basic. Ammonium nitrate comes from a strong acid and a weak base; the ammonium ion hydrolyses and produces hydrogen ions, so the solution is acidic. Sodium acetate also comes from a weak acid and a strong base, so its solution is basic.

Hinglish: NaCl neutral (strong acid + strong base), NaCN basic (weak acid + strong base, CN- hydrolyse hota hai), NH4NO3 acidic (strong acid + weak base, NH4+ hydrolyse hota hai), CH3COONa basic.

Q8 The solubility of a sparingly soluble salt AB2 is s mol L^-1. Write its solubility product.

Answer (English): The salt dissociates into one A ion with charge 2+ and two B ions with charge 1-, so the ion concentrations are s for A2+ and 2s for B-. The solubility product is the concentration of A2+ times the square of the concentration of B-, which is s times (2s) squared, that is 4s cubed. If Ksp is known, the solubility is the cube root of Ksp divided by four. For an AB type salt the same method gives s squared, and for A3B2 it gives 108 s to the fifth power.

Hinglish: AB2 ke liye [A2+] = s, [B-] = 2s, to Ksp = s(2s)² = 4s³. AB ke liye s², A3B2 ke liye 108s⁵. Ksp diya ho to solubility inverse nikaal sakte hain.

Q9 Explain the common ion effect with an example.

Answer (English): The common ion effect is the fall in the ionisation of a weak electrolyte, or in the solubility of a sparingly soluble salt, when a salt containing one of its ions is added. Adding sodium acetate to acetic acid increases the acetate concentration, so the acetic acid equilibrium shifts towards the undissociated acid and the hydrogen ion concentration falls. Adding silver nitrate to saturated silver chloride increases the silver ion concentration, so the solubility equilibrium shifts towards the solid and more silver chloride precipitates. In both cases the equilibrium moves to oppose the addition.

Hinglish: Common ion effect tab hota hai jab ek ion pehle se present ho aur usi ion wala salt add karo. CH3COONa add karne se acetic acid ka ionisation ghatta hai; AgNO3 add karne se AgCl ki solubility ghatti hai aur precipitate zyada banta hai. Equilibrium addition ka oppose karta hai.

Q10 Explain the relation between the standard Gibbs energy change and the equilibrium constant, and what it means for the direction of a reaction.

Answer (English): The standard Gibbs energy change is minus RT times the natural logarithm of the equilibrium constant. A negative standard Gibbs energy change makes ln K positive, so K is larger than one and the products dominate at equilibrium; a positive value makes K smaller than one and the reactants dominate. The actual direction at a given composition is decided by the actual Gibbs energy change, which includes the reaction quotient. A reaction with a large K can still move backward if there are already too many products, because then Q exceeds K.

Hinglish: ΔG° = -RT ln K. ΔG° negative ho to K > 1 aur products dominate; ΔG° positive ho to K < 1 aur reactants dominate. Direction actual ΔG se decide hoti hai jo Q par depend karta hai, isliye bada K hone par bhi Q > K ho to reaction reverse chal sakti hai.

That closes the chapter questions. Try the quick quiz below before moving on to the previous year questions.