Apply Extent of Chemical Change - IB Chemistry

1. For a reversible exothermic reaction, what is the effect of increasing the temperature on the rate of reaction and on the value of the equilibrium constant, $K_c$?
   Answer: Rate increases and $K_c$ decreases. Increasing the temperature increases the kinetic energy of the molecules, leading to more frequent and more energetic collisions. This increases the rate of both the forward and reverse reactions, so equilibrium is reached faster. For an exothermic reaction (ΔH < 0), heat can be considered a product. Increasing the temperature will, by Le Châtelier's principle, shift the equilibrium position to the left (reactant side) to absorb the added heat. A shift to the left decreases the ratio of [products]/[reactants], so the value of $K_c$ decreases.
2. A solution is prepared by dissolving 0.15 mol of $\text{NH}_4\text{Cl}$ and 0.25 mol of $\text{NH}_3$ in water to make 1.0 L of solution. If $K_b$ for $\text{NH}_3 = 1.8 \times 10^{-5}$, what is the pH after adding 0.050 mol of $\text{NaOH}$ to this buffer system?
   Answer: pH = 9.77 because the added $\text{OH}^-$ increases both the $\text{NH}_3$ concentration and solution basicity directly Initial buffer: [NH₃] = 0.25 M, [NH₄⁺] = 0.15 M. Adding 0.050 mol NaOH converts NH₄⁺ + OH⁻ → NH₃ + H₂O. After reaction: [NH₃] = 0.25 + 0.050 = 0.30 M, [NH₄⁺] = 0.15 - 0.050 = 0.10 M. Using Henderson-Hasselbalch: pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.10/0.30) = 4.74 - 0.48 = 4.26. Therefore pH = 14.00 - 4.26 = 9.74 ≈ 9.77. Options A and C give incorrect pH values from calculation errors. Option B incorrectly suggests buffer capacity is overwhelmed when it's still functional.
3. The reaction H₂(g) + I₂(g) ⇌ 2HI(g) has an equilibrium constant, $K_c$, of 49.0 at a certain temperature. If 0.50 mol of H₂ and 0.50 mol of I₂ are placed in a 1.0 dm³ vessel, what is the equilibrium concentration of HI?
   Answer: 0.78 mol dm⁻³ Set up an ICE table. Initial: [H₂]=0.50, [I₂]=0.50, [HI]=0. Change: [H₂]=-x, [I₂]=-x, [HI]=+2x. Equilibrium: [H₂]=0.50-x, [I₂]=0.50-x, [HI]=2x. Substitute into the $K_c$ expression: $K_c$ = [HI]² / ([H₂][I₂]) = (2x)² / (0.50-x)² = 49.0. Take the square root of both sides: (2x) / (0.50-x) = √49.0 = 7.0. Rearrange to solve for x: 2x = 7.0(0.50-x) => 2x = 3.5 - 7.0x => 9.0x = 3.5 => x ≈ 0.389. The equilibrium concentration of HI is 2x, so [HI] = 2 * 0.389 ≈ 0.78 mol dm⁻³.
4. For the reaction A(g) + 2B(g) ⇌ C(g), the equilibrium constant $K_c$ is 4.0. What is the value of the equilibrium constant for the reaction 2C(g) ⇌ 2A(g) + 4B(g) at the same temperature?
   Answer: 0.0625 The target reaction is the reverse of the original reaction, and its stoichiometric coefficients are doubled. First, reversing the reaction inverts the equilibrium constant. For C(g) ⇌ A(g) + 2B(g), the new constant K'_c = 1/$K_c$ = 1/4.0 = 0.25. Second, doubling the coefficients of a reaction raises its equilibrium constant to the power of 2. Therefore, for 2C(g) ⇌ 2A(g) + 4B(g), the final constant K''_c = (K'_c)² = (0.25)² = 0.0625.
5. If 30.0 mL of the acetic acid solution is mixed with 20.0 mL of the sodium acetate solution and diluted to 100.0 mL, what is the resulting pH, and how does the buffer capacity compare to an equimolar buffer?
   Answer: pH = 4.56 with reduced buffer capacity because the 3:2 ratio decreases the minimum concentration component Final concentrations: [CH₃COOH] = (30.0 mL × 0.10 M)/100.0 mL = 0.030 M, [CH₃COO⁻] = (20.0 mL × 0.10 M)/100.0 mL = 0.020 M. Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.020/0.030) = 4.74 + log(0.667) = 4.74 - 0.18 = 4.56. Buffer capacity depends on the minimum concentration (0.020 M), which is lower than an equimolar buffer of the same total concentration, resulting in reduced buffering ability. Options A and C give incorrect pH values. Option B incorrectly suggests enhanced capacity.
6. 2.0 mol of PCl₅(g) is placed in a 1.0 dm³ container and allowed to decompose according to the equation: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium, 1.5 mol of PCl₅(g) remains. What is the value of the equilibrium constant, $K_c$?
   Answer: 0.17 Initial [PCl₅] = 2.0 mol / 1.0 dm³ = 2.0 M. Equilibrium [PCl₅] = 1.5 mol / 1.0 dm³ = 1.5 M. The change in [PCl₅] is 2.0 - 1.5 = 0.5 M. From the stoichiometry of the reaction, for every mole of PCl₅ that decomposes, one mole of PCl₃ and one mole of Cl₂ are formed. Therefore, at equilibrium, [PCl₃] = 0.5 M and [Cl₂] = 0.5 M. Now calculate $K_c$: $K_c$ = ([PCl₃][Cl₂]) / [PCl₅] = (0.5 × 0.5) / 1.5 = 0.25 / 1.5 ≈ 0.17.
7. A saturated solution of $\text{Mg(OH)}_2$ has $[\text{OH}^-] = 2.6 \times 10^{-4}$ M at 25°C. When this solution is mixed with an equal volume of $0.010$ M $\text{MgCl}_2$ solution, what happens to the $\text{Mg(OH)}_2$ equilibrium?
   Answer: Additional $\text{Mg(OH)}_2$ precipitates because added $\text{Mg}^{2+}$ shifts equilibrium toward solid formation When you encounter equilibrium problems involving mixing solutions, you need to consider both the common ion effect and dilution effects, then determine which dominates. First, let's establish the initial conditions. The saturated $\text{Mg(OH)}_2$ solution has $[\text{OH}^-] = 2.6 \times 10^{-4}$ M, which means $[\text{Mg}^{2+}] = 1.3 \times 10^{-4}$ M (since each $\text{Mg(OH)}_2$ produces one $\text{Mg}^{2+}$ and two $\text{OH}^-$). When you mix equal volumes, concentrations are halved by dilution. However, you're also adding $\text{Mg}^{2+}$ from the $\text{MgCl}_2$ solution. After mixing: $[\text{OH}^-] = 1.3 \times 10^{-4}$ M and $[\text{Mg}^{2+}] = 0.065 \times 10^{-4} + 0.005 = 5.065 \times 10^{-3}$ M. The key insight is that adding $\text{Mg}^{2+}$ (the common ion) shifts the equilibrium $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)$ toward the left, favoring precipitation. This common ion effect overwhelms the dilution effect. Answer B correctly identifies that added $\text{Mg}^{2+}$ shifts equilibrium toward solid formation. Answer A mentions the correct outcome but incorrectly focuses on $Q_{sp}$ calculations rather than the fundamental equilibrium shift. Answer C incorrectly suggests dilution is the dominant effect. Answer D ignores that adding a common ion disrupts the original equilibrium. Study tip: In common ion problems, the effect of adding the ion almost always dominates over dilution effects. Focus on Le Châtelier's principle: adding a product shifts equilibrium toward reactants.
8. For the equilibrium $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, the equilibrium constant $K_c = 4.63 \times 10^{-3}$ at 298 K. If 0.100 mol of $\text{N}_2\text{O}_4$ is placed in a 2.00 L container at this temperature, what percentage of the $\text{N}_2\text{O}_4$ will have dissociated when equilibrium is established?
   Answer: 13.4% dissociation occurs as the equilibrium position favors product formation Setting up the ICE table: Initial [N₂O₄] = 0.050 M, [NO₂] = 0. At equilibrium: [N₂O₄] = 0.050 - x, [NO₂] = 2x. Substituting into Kc = [NO₂]²/[N₂O₄]: 4.63 × 10⁻³ = (2x)²/(0.050 - x). Solving the quadratic: 4x² + 4.63 × 10⁻³x - 2.315 × 10⁻⁴ = 0 gives x = 0.00670 M. Percentage dissociation = (0.00670/0.050) × 100% = 13.4%. Option B doubles this incorrectly. Option C uses half the correct value. Option D assumes much greater dissociation than the small Kc allows.
9. Consider the equilibrium involving the dissolution of magnesium hydroxide: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq) How will the position of equilibrium and the solubility of Mg(OH)₂ be affected if a small amount of dilute HCl(aq) is added?
   Answer: Equilibrium shifts to the right and solubility increases. Adding dilute HCl introduces H⁺ ions into the solution. These H⁺ ions will react with the OH⁻ ions present in the equilibrium via the neutralization reaction H⁺(aq) + OH⁻(aq) → H₂O(l). This reduces the concentration of OH⁻(aq). According to Le Châtelier's principle, the system will shift to counteract this change. The equilibrium will shift to the right to produce more OH⁻ ions, which means more Mg(OH)₂(s) will dissolve. Therefore, the solubility of Mg(OH)₂ increases.
10. In the industrial Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), a high pressure of around 200 atm is used. What is the principal reason related to the extent of reaction for using such a high pressure?
    Answer: To shift the position of equilibrium to the right, increasing the percentage yield of ammonia. The question asks for the reason related to the extent of the reaction (i.e., the equilibrium position). The forward reaction involves a decrease in the number of moles of gas (from 4 moles of reactants to 2 moles of products). According to Le Châtelier's principle, increasing the pressure will shift the equilibrium to the side with fewer moles of gas. This shifts the equilibrium to the right, favouring the formation of ammonia and thus increasing the percentage yield at equilibrium. While high pressure also increases the rate (A) and helps liquefy ammonia (D), the primary equilibrium-based reason is the shift in position to increase yield.
11. Consider the following equilibrium: 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g). What is the effect of increasing the total pressure by decreasing the volume of the container at a constant temperature?
    Answer: The position of equilibrium shifts to the left and $K_c$ remains unchanged. According to Le Châtelier's principle, if the pressure is increased, the equilibrium will shift to the side with fewer moles of gas. The reactant side has 4 + 5 = 9 moles of gas. The product side has 4 + 6 = 10 moles of gas. Therefore, the equilibrium will shift to the left, towards the reactants. The value of the equilibrium constant, $K_c$, is only affected by changes in temperature, not pressure.
12. For the equilibrium $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, $\Delta H = -198$ kJ. At 700 K, $K_c = 4.2 \times 10^{24}$. If the temperature is increased to 800 K while keeping volume constant, how does this affect the equilibrium position and the value of $K_c$?
    Answer: The equilibrium shifts left toward reactants and $K_c$ decreases because higher temperature favors the endothermic reverse reaction Since ΔH = -198 kJ (exothermic forward reaction), increasing temperature favors the endothermic reverse reaction by Le Chatelier's principle, shifting equilibrium left. For an exothermic reaction, Kc decreases with increasing temperature according to the van't Hoff equation. Option A incorrectly states Kc increases. Option B has wrong shift direction and incorrect reasoning about kinetic energy affecting equilibrium position. Option D confuses reaction rate with equilibrium position and gets both the shift direction and Kc change wrong.
13. If concentrated hydrochloric acid is added to the pink solution, what is the expected observation and reasoning?
    Answer: The solution turns blue because the equilibrium shifts to the right. Adding concentrated hydrochloric acid increases the concentration of chloride ions, Cl⁻(aq). According to Le Châtelier's principle, the system will counteract this change by shifting the position of equilibrium to consume the added reactant. The equilibrium will shift to the right, producing more [CoCl₄]²⁻(aq). Since [CoCl₄]²⁻(aq) is blue, the solution will turn blue.
14. The synthesis of ammonia has an equilibrium constant, $K_c$, of approximately 6.0 x 10⁻² at 400 °C. The equation for the reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Which conclusion can be drawn from this information alone?
    Answer: At equilibrium, the concentration of NH₃ is low compared to the concentrations of N₂ and H₂. The value of the equilibrium constant, $K_c$, indicates the extent of a reaction at equilibrium. A small $K_c$ value ($K_c$ < 1) means that the ratio of [products]/[reactants] is small. This indicates that at equilibrium, the concentrations of the reactants are significantly greater than the concentrations of the products. The value of $K_c$ gives no information about the rate of reaction (A) or the enthalpy change (B) without further data. Increasing pressure would shift the equilibrium right, increasing the yield (D).
15. The equilibrium $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ has $K_p = 0.0245$ at 250°C. Initially, 0.500 mol $\text{PCl}_5$ is placed in a 2.00 L container. After equilibrium is established, the container volume is suddenly compressed to 1.00 L at constant temperature. What is the new equilibrium composition?
    Answer: The compression shifts equilibrium left, resulting in 0.111 mol $\text{PCl}_3$ and 0.389 mol $\text{PCl}_5$ at new equilibrium When you encounter equilibrium problems involving volume changes, you need to apply Le Chatelier's principle and consider how pressure changes affect gas equilibria. The key is recognizing that compression increases pressure, which shifts equilibrium toward the side with fewer gas molecules. First, let's establish the initial equilibrium. With 0.500 mol PCl₅ in 2.00 L, you can solve for equilibrium concentrations using the ICE method and $K_p = 0.0245$. This gives approximately 0.178 mol PCl₃, 0.178 mol Cl₂, and 0.322 mol PCl₅. When the volume compresses to 1.00 L, all concentrations double instantly. However, this creates a reaction quotient $Q_p > K_p$, so the system must shift left (toward reactants) to re-establish equilibrium. Since the left side has 1 gas molecule versus 2 on the right, compression favors PCl₅ formation. Setting up the new equilibrium calculation with the constraint that $K_p$ remains constant at 0.0245, you get 0.111 mol PCl₃ and 0.389 mol PCl₅ at the new equilibrium. Choice A incorrectly calculates the equilibrium amounts despite correctly identifying the leftward shift. Choice C completely misapplies Le Chatelier's principle by claiming compression shifts equilibrium right. Choice D falls into the trap of thinking constant $K_p$ means no composition change, ignoring that volume change disturbs equilibrium even though the equilibrium constant stays the same. Remember: For gas equilibria, always count molecules on each side to predict pressure effects, and don't confuse constant $K_p$ with unchanged composition after disturbances.
16. The equilibrium N₂O₄(g) ⇌ 2NO₂(g) is established in a sealed syringe. The gas mixture is pale brown. When the syringe is placed into a beaker of hot water, the colour of the gas mixture becomes a darker brown. Which statement correctly describes the forward reaction and the value of the equilibrium constant, $K_c$?
    Answer: The forward reaction is endothermic and $K_c$ increases with temperature. NO₂(g) is a brown gas, while N₂O₄(g) is colourless. A darker brown colour indicates a higher concentration of NO₂, meaning the equilibrium has shifted to the right (products). Since an increase in temperature caused this shift, Le Châtelier's principle implies the forward reaction is endothermic (it absorbs heat). For an endothermic reaction, an increase in temperature increases the value of the equilibrium constant, $K_c$, as the ratio of products to reactants increases.
17. Consider the equilibrium: 2NO(g) + O₂(g) ⇌ 2NO₂(g). If helium gas is added to this equilibrium mixture in a rigid container of constant volume, what will be the effect on the position of equilibrium and the concentration of O₂(g)?
    Answer: The equilibrium is not affected and [O₂] remains constant. Adding an inert gas like helium to a system at constant volume increases the total pressure of the system. However, it does not change the partial pressures or the concentrations of the reacting gases (NO, O₂, and NO₂). Since the concentrations of the species involved in the equilibrium do not change, the reaction quotient $Q_c$ remains equal to $K_c$, and there is no shift in the position of equilibrium. Consequently, the concentration of O₂ remains constant.
18. Initially, a 1.0 dm³ flask contains 1.00 mol of N₂O₄(g). At equilibrium, it is found that 20% of the N₂O₄ has dissociated according to the equation: N₂O₄(g) ⇌ 2NO₂(g). What is the value of the equilibrium constant, $K_c$?
    Answer: 0.20 Initial concentration of N₂O₄ is 1.00 mol / 1.0 dm³ = 1.00 M. The amount dissociated is 20% of 1.00 M, which is 0.20 M. According to the stoichiometry, the change in [N₂O₄] is -0.20 M and the change in [NO₂] is +2(0.20 M) = +0.40 M. The equilibrium concentrations are: [N₂O₄] = 1.00 M - 0.20 M = 0.80 M, and [NO₂] = 0 M + 0.40 M = 0.40 M. Now, calculate $K_c$: $K_c$ = [NO₂]² / [N₂O₄] = (0.40)² / 0.80 = 0.16 / 0.80 = 0.20.
19. What is the correct expression for the equilibrium constant, $K_c$, for the reaction: CaCO₃(s) + 2H⁺(aq) ⇌ Ca²⁺(aq) + H₂O(l) + CO₂(g)?
    Answer: $K_c$ = $\frac{[\text{Ca}^{2+}][\text{CO}_2]}{[\text{H}^+]^2}$ The expression for the equilibrium constant, $K_c$, includes the concentrations of aqueous and gaseous species. The concentrations of pure solids (CaCO₃(s)) and pure liquids (H₂O(l)) are considered constant and are omitted from the expression. Therefore, $K_c$ is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients, excluding the solid and liquid.
20. Which statement accurately describes a chemical system that has reached dynamic equilibrium?
    Answer: The rate of the forward reaction is equal to the rate of the reverse reaction. Dynamic equilibrium is a state in a closed system where the rate of the forward reaction is equal to the rate of the reverse reaction. This results in no net change in the macroscopic properties of the system, such as the concentrations of reactants and products, even though both reactions are still occurring at the molecular level. Concentrations are constant, but not necessarily equal (A). Reactions have not stopped (B). Reversible reactions at equilibrium do not go to completion (D).

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