Understand Extent of Chemical Change - IB Chemistry

1. The decomposition of dinitrogen tetroxide is an endothermic process: $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2 \text{NO}_2\text{(g)}$, $\Delta H > 0$. The brown color of the equilibrium mixture is due to NO₂ gas. If a sealed tube containing an equilibrium mixture of these gases is placed in an ice bath, what observation would be made?
   Answer: The color becomes lighter (less brown) as the equilibrium shifts to the left. Placing the tube in an ice bath lowers the temperature. For an endothermic reaction, decreasing the temperature will cause the equilibrium to shift in the reverse, exothermic direction (to the left). A shift to the left consumes the brown NO₂ gas, so the color of the mixture will become lighter.
2. Two sealed flasks, X and Y, of equal volume are held at the same temperature. Flask X initially contains 1.0 mol of SO₂ and 0.5 mol of O₂. Flask Y initially contains 1.0 mol of SO₃. The reaction $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$ is allowed to reach equilibrium in both flasks. Which statement is correct once equilibrium is reached?
   Answer: The amounts of SO₂, O₂, and SO₃ will be identical in both flasks. The position of equilibrium is independent of the direction from which it is approached. Both flasks contain the same total number of S and O atoms, are at the same volume, and at the same temperature. Therefore, they will reach the exact same equilibrium state, with identical amounts of all three species. $K_c$ is a constant at a given temperature (B). Equilibrium is dynamic, so rates are non-zero (D).
3. What is the correct equilibrium constant expression, $K_c$, for the reaction: $2 \text{Fe(s)} + 3 \text{H}_2\text{O(g)} \rightleftharpoons \text{Fe}_2\text{O}_3\text{(s)} + 3 \text{H}_2\text{(g)}$?
   Answer: $K_c = \frac{[\text{H}_2]^3}{[\text{H}_2\text{O}]^3}$ The concentrations of pure solids (Fe(s) and Fe₂O₃(s)) are considered constant and are omitted from the equilibrium constant expression. The expression is products over reactants, with stoichiometric coefficients as powers. Thus, only the gaseous species are included.
4. The synthesis of methanol is represented by the equation: $\text{CO(g)} + 2 \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$; $\Delta H < 0$. Which change will increase the equilibrium yield of methanol (CH₃OH)?
   Answer: Decreasing the temperature and increasing the pressure. The forward reaction is exothermic ($\Delta H < 0$), so decreasing the temperature will shift the equilibrium to the right. The forward reaction has fewer moles of gas (3 moles of reactants vs. 1 mole of product), so increasing the pressure will shift the equilibrium to the right. Both changes favor a higher yield of methanol.
5. Consider the following equilibrium: $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2 \text{HI(g)}$. The volume of the sealed container is decreased at constant temperature. Which statement is correct?
   Answer: There is no shift in the equilibrium position, and the value of $K_c$ remains constant. A change in pressure (or volume) only affects the position of a gaseous equilibrium if the number of moles of gas on the reactant and product sides are different. In this reaction, there are 2 moles of gas on the left (1+1) and 2 moles of gas on the right. Since the moles are equal, there is no shift in the equilibrium position. $K_c$ is only affected by temperature.
6. Consider the gaseous equilibrium: $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$. If helium gas is added to the equilibrium mixture in a container of constant volume, what will be the effect on the equilibrium?
   Answer: The equilibrium position will not change because the partial pressures of the reacting gases remain constant. Adding an inert gas at constant volume increases the total pressure but does not change the volume available to the reacting gases. Therefore, the concentrations and partial pressures of PCl₅, PCl₃, and Cl₂ do not change. Since the reaction quotient Q remains equal to K, the equilibrium position is unaffected. The reason is the constancy of partial pressures, not just the inertness of helium (D).
7. The equilibrium constant expression for a gaseous reaction is given by $K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^3}$. Which chemical equation corresponds to this expression?
   Answer: $\text{A(g)} + 3\text{B(g)} \rightleftharpoons 2\text{C(g)}$ The equilibrium expression places product concentrations in the numerator and reactant concentrations in the denominator. The stoichiometric coefficients from the balanced equation become the powers in the expression. The numerator $[\text{C}]^2$ corresponds to 2C as a product. The denominator $[\text{A}][\text{B}]^3$ corresponds to A and 3B as reactants.
8. What can be deduced from the observation that the mixture turns more blue when heated?
   Answer: The forward reaction is endothermic, and heating increases the value of $K_c$. Heating the mixture causes a shift to the right, producing more of the blue product. According to Le Châtelier's principle, increasing the temperature favors the endothermic direction. Therefore, the forward reaction is endothermic. For an endothermic reaction, increasing the temperature causes the value of $K_c$ to increase, reflecting the greater proportion of products at the new equilibrium.
9. The decomposition of calcium carbonate, $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$, is an endothermic reversible reaction. Which statement explains why this reaction does not reach equilibrium in an open test tube?
   Answer: The carbon dioxide gas, a product, escapes from the system, preventing the reverse reaction. Dynamic equilibrium can only be established in a closed system. In an open test tube, the gaseous product CO₂ is free to escape. According to Le Châtelier's principle, the continual removal of a product forces the equilibrium to constantly shift to the right. The reverse reaction cannot occur at a significant rate, so a balance is never achieved.
10. The synthesis of hydrogen iodide has an equilibrium constant, $K_c$, of 54.3 at 430 °C. $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2 \text{HI(g)}$. Which conclusion can be drawn from this information about the system at equilibrium?
    Answer: The concentration of HI is significantly greater than the concentrations of H₂ and I₂. An equilibrium constant $K_c$ significantly greater than 1 indicates that at equilibrium, the products are favored. This means the concentration of the product (HI) will be much larger than the concentrations of the reactants (H₂ and I₂). A is incorrect because equilibrium implies that both reactants and products are present. C is incorrect because at equilibrium the rates are equal. D would be true if $K_c$ were much less than 1.
11. For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the equilibrium constant $K_c = 4.0 \times 10^6$ at 727°C. If the initial concentrations are $[SO_2] = 0.50 \text{ M}$, $[O_2] = 0.30 \text{ M}$, and $[SO_3] = 0.10 \text{ M}$, what can be concluded about the reaction's progress toward equilibrium?
    Answer: The reaction will proceed forward because $Q_c < K_c$ and the system has not reached equilibrium First calculate $Q_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(0.10)^2}{(0.50)^2(0.30)} = \frac{0.01}{0.075} = 0.133$. Since $Q_c = 0.133 < K_c = 4.0 \times 10^6$, the reaction will proceed forward to reach equilibrium. Choice B incorrectly states $Q_c > K_c$. Choice C assumes equilibrium without calculation. Choice D incorrectly introduces kinetics into a thermodynamic analysis.
12. The Haber process for ammonia synthesis is exothermic: $\text{N}_2\text{(g)} + 3 \text{H}_2\text{(g)} \rightleftharpoons 2 \text{NH}_3\text{(g)}$, $\Delta H^\ominus = -92 \, \text{kJ mol}^{-1}$. If the temperature of the system at equilibrium is increased, what will be the effect on the position of equilibrium and the value of the equilibrium constant, $K_c$?
    Answer: Equilibrium shifts left; $K_c$ decreases. For an exothermic reaction, heat is a product. Increasing the temperature adds heat, causing the equilibrium to shift to the left (the endothermic direction) to absorb it. This increases reactant concentrations and decreases product concentrations, which causes the value of $K_c$ to decrease.
13. The equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ has $K_c = 2.4 \times 10^{-3}$ at 1000 K. A mixture initially containing $N_2$ and $H_2$ in a 3:1 molar ratio at total concentration of 0.80 M reaches equilibrium. What fraction of the original $N_2$ remains unreacted?
    Answer: 0.91 because the small $K_c$ value indicates very limited conversion under these conditions Initial: $[H_2] = 0.60$ M, $[N_2] = 0.20$ M. Let x = mol/L of $N_2$ consumed. At equilibrium: $[N_2] = 0.20-x$, $[H_2] = 0.60-3x$, $[NH_3] = 2x$. $K_c = \frac{(2x)^2}{(0.20-x)(0.60-3x)^3} = 2.4 \times 10^{-3}$. Given the small $K_c$, assume x is small: $\frac{4x^2}{(0.20)(0.60)^3} ≈ 2.4 \times 10^{-3}$. Solving: $x ≈ 0.018$. Fraction unreacted = $\frac{0.20-0.018}{0.20} = 0.91$. Choices B and C overestimate conversion. Choice D underestimates the equilibrium extent.
14. A reversible reaction reaches a state of dynamic equilibrium in a closed container. Which statement correctly describes this state?
    Answer: The rate of the forward reaction is equal to the rate of the reverse reaction. The definition of dynamic equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the macroscopic properties, such as concentrations of all species, are constant, but not necessarily equal (A). The reactions are still occurring, so 'dynamic' means they have not stopped (B). The total moles of gas only remain constant if the stoichiometry of the reaction has an equal number of moles of gas on both sides (D).
15. The key step in the Contact process is $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$, $\Delta H < 0$. Which set of conditions would produce the greatest amount of SO₃ at equilibrium?
    Answer: Low temperature, high pressure, presence of a catalyst. The question asks for the conditions that maximize the equilibrium yield. The forward reaction is exothermic, so a low temperature favors the products. The forward reaction involves a decrease in moles of gas (3 to 2), so high pressure favors the products. A catalyst does not affect the position of the equilibrium, only the rate at which it is achieved. Therefore, low temperature and high pressure give the highest equilibrium yield.
16. The equilibrium constant for the reaction $2\text{O}_3\text{(g)} \rightleftharpoons 3\text{O}_2\text{(g)}$ is $K_c = 2.0 \times 10^{57}$ at 298 K. If a system is at equilibrium, which statement is the most accurate?
    Answer: The system contains almost entirely O₂, and adding more O₂ would cause a significant shift to the left. A very large $K_c$ value indicates that the equilibrium position lies far to the right, meaning the concentration of products (O₂) is vastly greater than that of reactants (O₃). The system consists almost entirely of O₂. According to Le Châtelier's principle, adding more product (O₂) will cause the equilibrium to shift to the left to consume it. Given the huge concentration of O₂, even a small addition is a significant disturbance, causing a noticeable shift.
17. For a hypothetical reaction $2X(g) + Y(g) \rightleftharpoons Z(g)$, the equilibrium constant $K_p = 1.44 \times 10^{-3}$ at 400°C. A reaction vessel initially contains $X$, $Y$, and $Z$ at partial pressures of 0.50 atm, 0.30 atm, and 0.040 atm respectively. Which statement correctly describes the system's approach to equilibrium?
    Answer: The reaction quotient $Q_p = 1.07 \times 10^{-3}$, so the forward reaction will be slightly favored until equilibrium Calculate $Q_p = \frac{P_Z}{P_X^2 \cdot P_Y} = \frac{0.040}{(0.50)^2(0.30)} = \frac{0.040}{0.075} = 1.07 \times 10^{-3}$. Since $Q_p = 1.07 \times 10^{-3} < K_p = 1.44 \times 10^{-3}$, the forward reaction is favored. Choice A miscalculates $Q_p$ by inverting the expression. Choice C incorrectly assumes near-equilibrium when $Q_p/K_p ≈ 0.74$. Choice D unnecessarily complicates with unit conversion when comparing reaction quotient to equilibrium constant.
18. The equilibrium constant, $K_c$, for the reaction $\text{A} + 2\text{B} \rightleftharpoons \text{C}$ is 4.0. What is the value of $K_c$ for the reaction $\text{C} \rightleftharpoons \text{A} + 2\text{B}$ at the same temperature?
    Answer: 0.25 The second reaction is the reverse of the first reaction. The equilibrium constant for a reverse reaction ($K_{rev}$) is the reciprocal of the equilibrium constant for the forward reaction ($K_{fwd}$). Therefore, $K_{rev} = 1 / K_{fwd} = 1 / 4.0 = 0.25$.
19. In the industrial Haber process, $\text{N}_2\text{(g)} + 3 \text{H}_2\text{(g)} \rightleftharpoons 2 \text{NH}_3\text{(g)}$, $\Delta H < 0$, typical conditions are 400–450 °C and 150–250 atm. Why is a relatively high temperature used when a low temperature favours the equilibrium yield?
    Answer: The high temperature provides a sufficient rate of reaction, which is a necessary economic compromise. While a low temperature would thermodynamically favor a higher yield of ammonia (since the reaction is exothermic), the rate of reaction at low temperatures is impractically slow. A compromise temperature is used to achieve an economically viable rate of reaction, even though it reduces the maximum possible equilibrium yield. This is a classic example of balancing kinetics and thermodynamics.
20. The degree of dissociation ($\alpha$) of $PCl_5$ according to $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ is 0.25 at 2.0 atm total pressure and 523 K. If the pressure is increased to 8.0 atm at the same temperature, what will be the new degree of dissociation?
    Answer: 0.145 because $K_p$ remains constant and the relationship $\alpha \propto P^{-1/2}$ applies to this dissociation For the dissociation $PCl_5 \rightleftharpoons PCl_3 + Cl_2$, $\Delta n = 1$. At equilibrium: $K_p = \frac{4\alpha^2 P}{1-\alpha^2}$. Initially: $K_p = \frac{4(0.25)^2(2.0)}{1-(0.25)^2} = \frac{0.50}{0.9375} = 0.533$. At 8.0 atm: $0.533 = \frac{4\alpha^2(8.0)}{1-\alpha^2}$. Solving: $\alpha^2 = 0.0208$, so $\alpha = 0.144 ≈ 0.145$. This follows $\alpha \propto P^{-1/2}$ for simple dissociation. Choice A uses wrong proportionality. Choice C overestimates pressure effect. Choice D incorrectly applies kinetic reasoning.

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