IB Chemistry Quiz: Understand Extent Of Chemical Change
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Understand Extent Of Chemical ChangeQuestion 1 of 20

For the reaction SO2(g)+12O2(g)⇌SO3(g)\text{SO}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightleftharpoons \text{SO}_3\text{(g)}, a V₂O₅ catalyst is used. Which statement best describes the effect of the catalyst?

It increases the equilibrium yield of SO₃ by shifting the equilibrium to the right.
It increases the value of the equilibrium constant, KcK_c, resulting in more product.
It increases the rate of the forward reaction more than the rate of the reverse reaction.
It increases the rates of both the forward and reverse reactions equally, leading to no change in the equilibrium position.
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Extent Of Chemical Change

Practice Understand Extent Of Chemical Change in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Extent Of Chemical Change, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Question 1

For the reaction SO2(g)+12O2(g)⇌SO3(g)\text{SO}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightleftharpoons \text{SO}_3\text{(g)}, a V₂O₅ catalyst is used. Which statement best describes the effect of the catalyst?

  1. It increases the equilibrium yield of SO₃ by shifting the equilibrium to the right.
  2. It increases the value of the equilibrium constant, KcK_c, resulting in more product.
  3. It increases the rate of the forward reaction more than the rate of the reverse reaction.
  4. It increases the rates of both the forward and reverse reactions equally, leading to no change in the equilibrium position. (correct answer)
Explanation: A catalyst provides an alternative reaction pathway with a lower activation energy for both the forward and reverse reactions. This increases the rates of both reactions equally. As a result, equilibrium is reached faster, but the position of the equilibrium and the value of KcK_c are not affected.

Question 2

What is the correct equilibrium constant expression, KcK_c, for the reaction: 2Fe(s)+3H2O(g)⇌Fe2O3(s)+3H2(g)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)}?

  1. Kc=[Fe2O3][H2]3[Fe]2[H2O]3K_c = \frac{[\text{Fe}_2\text{O}_3][\text{H}_2]^3}{[\text{Fe}]^2[\text{H}_2\text{O}]^3}
  2. Kc=[H2]3[H2O]3K_c = \frac{[\text{H}_2]^3}{[\text{H}_2\text{O}]^3} (correct answer)
  3. Kc=[Fe]2[H2O]3[Fe2O3][H2]3K_c = \frac{[\text{Fe}]^2[\text{H}_2\text{O}]^3}{[\text{Fe}_2\text{O}_3][\text{H}_2]^3}
  4. Kc=[H2][H2O]K_c = \frac{[\text{H}_2]}{[\text{H}_2\text{O}]}
Explanation: 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.

Question 3

Consider the equilibrium: N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2 \text{NO}_2\text{(g)}. If a small amount of NO₂ is added to the sealed vessel at constant temperature, what happens as the system moves to re-establish equilibrium?

  1. The concentration of N₂O₄ increases and the value of KcK_c increases.
  2. The concentration of N₂O₄ decreases and the value of KcK_c remains constant.
  3. The concentration of N₂O₄ increases and the value of KcK_c remains constant. (correct answer)
  4. The concentration of N₂O₄ decreases and the value of KcK_c decreases.
Explanation: According to Le Châtelier's principle, adding a product (NO₂) will cause the equilibrium to shift to the left to consume the added substance. This shift increases the concentration of the reactant (N₂O₄). The equilibrium constant, KcK_c, is only affected by changes in temperature, so it remains constant.

Question 4

The Haber process for ammonia synthesis is exothermic: N2(g)+3H2(g)⇌2NH3(g)\text{N}_2\text{(g)} + 3 \text{H}_2\text{(g)} \rightleftharpoons 2 \text{NH}_3\text{(g)}, ΔH⊖=−92 kJ mol−1\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, KcK_c?

  1. Equilibrium shifts left; KcK_c decreases. (correct answer)
  2. Equilibrium shifts right; KcK_c increases.
  3. Equilibrium shifts left; KcK_c remains constant.
  4. Equilibrium shifts right; KcK_c decreases.
Explanation: 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 KcK_c to decrease.

Question 5

The decomposition of dinitrogen tetroxide is an endothermic process: N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2 \text{NO}_2\text{(g)}, ΔH>0\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?

  1. The color becomes darker brown as the equilibrium shifts to the right.
  2. The color becomes lighter (less brown) as the equilibrium shifts to the left. (correct answer)
  3. The color becomes lighter (less brown) as the equilibrium shifts to the right.
  4. The color does not change, but the pressure inside the tube increases.
Explanation: 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.

Question 6

What is the correct expression for the equilibrium constant, KcK_c, for the reaction of ethanoic acid with water? CH3COOH(aq)+H2O(l)⇌CH3COO−(aq)+H3O+(aq)\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}

  1. Kc=[CH3COO−][H3O+][CH3COOH][H2O]K_c = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}][\text{H}_2\text{O}]}
  2. Kc=[CH3COO−][H3O+][CH3COOH]K_c = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} (correct answer)
  3. Kc=[CH3COOH][H2O][CH3COO−][H3O+]K_c = \frac{[\text{CH}_3\text{COOH}][\text{H}_2\text{O}]}{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}
  4. Kc=[CH3COOH][CH3COO−][H3O+]K_c = \frac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}
Explanation: In aqueous solutions, water (H₂O) is the solvent, and its concentration is large and essentially constant. Therefore, it is omitted from the equilibrium constant expression. This particular constant is known as the acid dissociation constant, KaK_a.

Question 7

The equilibrium involving cobalt(II) complexes is shown below. The hexaaquacobalt(II) ion is pink, and the tetrachloridocobaltate(II) ion is blue.

[Co(H2O)6]2+(aq)+4Cl−(aq)⇌[CoCl4]2−(aq)+6H2O(l)[\text{Co(H}_2\text{O)}_6]^{2+}\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightleftharpoons [\text{CoCl}_4]^{2-}\text{(aq)} + 6\text{H}_2\text{O(l)}

When the equilibrium mixture is gently heated, it turns a more intense blue.

What can be deduced from the observation that the mixture turns more blue when heated?

  1. The forward reaction is exothermic, and heating decreases the value of KcK_c.
  2. The forward reaction is endothermic, and heating increases the value of KcK_c. (correct answer)
  3. The forward reaction is exothermic, and heating increases the value of KcK_c.
  4. The forward reaction is endothermic, and heating decreases the value of KcK_c.
Explanation: 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 KcK_c to increase, reflecting the greater proportion of products at the new equilibrium.

Question 8

The equilibrium constant for the reaction 2O3(g)⇌3O2(g)2\text{O}_3\text{(g)} \rightleftharpoons 3\text{O}_2\text{(g)} is Kc=2.0×1057K_c = 2.0 \times 10^{57} at 298 K. If a system is at equilibrium, which statement is the most accurate?

  1. The rates of the forward and reverse reactions are both extremely large.
  2. The concentration of O₃ is nearly zero, so removing O₂ would cause a negligible shift.
  3. The system contains almost entirely O₂, and adding more O₂ would cause a significant shift to the left. (correct answer)
  4. The concentrations of O₃ and O₂ are approximately equal since the system is at equilibrium.
Explanation: A very large KcK_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.

Question 9

The equilibrium constant expression for a gaseous reaction is given by Kc=[C]2[A][B]3K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^3}. Which chemical equation corresponds to this expression?

  1. A(g)+3B(g)⇌2C(g)\text{A(g)} + 3\text{B(g)} \rightleftharpoons 2\text{C(g)} (correct answer)
  2. 2C(g)⇌A(g)+3B(g)2\text{C(g)} \rightleftharpoons \text{A(g)} + 3\text{B(g)}
  3. A(g)+B(g)3⇌C(g)2\text{A(g)} + \text{B(g)}^3 \rightleftharpoons \text{C(g)}^2
  4. A(g)+B(g)⇌C(g)\text{A(g)} + \text{B(g)} \rightleftharpoons \text{C(g)}
Explanation: 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 [C]2[\text{C}]^2 corresponds to 2C as a product. The denominator [A][B]3[\text{A}][\text{B}]^3 corresponds to A and 3B as reactants.

Question 10

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 2SO2(g)+O2(g)⇌2SO3(g)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?

  1. The amount of SO₃ in flask X will be greater than in flask Y.
  2. The value of the equilibrium constant, KcK_c, will be greater in flask X than in flask Y.
  3. The rate of the forward reaction will be zero in flask Y.
  4. The amounts of SO₂, O₂, and SO₃ will be identical in both flasks. (correct answer)
Explanation: 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. KcK_c is a constant at a given temperature (B). Equilibrium is dynamic, so rates are non-zero (D).

Question 11

The equilibrium constant, KcK_c, for the reaction A+2B⇌C\text{A} + 2\text{B} \rightleftharpoons \text{C} is 4.0. What is the value of KcK_c for the reaction C⇌A+2B\text{C} \rightleftharpoons \text{A} + 2\text{B} at the same temperature?

  1. 4.0
  2. -4.0
  3. 2.0
  4. 0.25 (correct answer)
Explanation: The second reaction is the reverse of the first reaction. The equilibrium constant for a reverse reaction (KrevK_{rev}) is the reciprocal of the equilibrium constant for the forward reaction (KfwdK_{fwd}). Therefore, Krev=1/Kfwd=1/4.0=0.25K_{rev} = 1 / K_{fwd} = 1 / 4.0 = 0.25.

Question 12

The decomposition of calcium carbonate, CaCO3(s)⇌CaO(s)+CO2(g)\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?

  1. The reverse reaction rate is too slow at atmospheric pressure for equilibrium to establish.
  2. The carbon dioxide gas, a product, escapes from the system, preventing the reverse reaction. (correct answer)
  3. The reaction is only spontaneous at very high temperatures not achievable in a test tube.
  4. The solid reactants and products have low surface area, preventing efficient collisions.
Explanation: 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.

Question 13

The key step in the Contact process is 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}, ΔH<0\Delta H < 0. Which set of conditions would produce the greatest amount of SO₃ at equilibrium?

  1. High temperature, high pressure, presence of a catalyst.
  2. Low temperature, high pressure, absence of a catalyst.
  3. Low temperature, high pressure, presence of a catalyst. (correct answer)
  4. High temperature, low pressure, presence of a catalyst.
Explanation: 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.

Question 14

The synthesis of hydrogen iodide has an equilibrium constant, KcK_c, of 54.3 at 430 °C. H2(g)+I2(g)⇌2HI(g)\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?

  1. The reaction proceeds to completion, and only HI is present.
  2. The concentration of HI is significantly greater than the concentrations of H₂ and I₂. (correct answer)
  3. The rate of the forward reaction is much faster than the rate of the reverse reaction.
  4. The concentrations of H₂ and I₂ are significantly greater than the concentration of HI.
Explanation: An equilibrium constant KcK_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 KcK_c were much less than 1.

Question 15

Consider the gaseous equilibrium: PCl5(g)⇌PCl3(g)+Cl2(g)\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?

  1. The equilibrium will shift to the right because the total pressure increases.
  2. The equilibrium will shift to the left because the total pressure increases.
  3. The equilibrium position will not change because the partial pressures of the reacting gases remain constant. (correct answer)
  4. The equilibrium position will not change because helium is an unreactive noble gas.
Explanation: 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).

Question 16

Consider the following equilibrium: H2(g)+I2(g)⇌2HI(g)\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?

  1. The equilibrium shifts to the right, and the value of KcK_c increases.
  2. The equilibrium shifts to the left, and the value of KcK_c decreases.
  3. There is no shift in the equilibrium position, but the value of KcK_c increases.
  4. There is no shift in the equilibrium position, and the value of KcK_c remains constant. (correct answer)
Explanation: 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. KcK_c is only affected by temperature.

Question 17

The synthesis of methanol is represented by the equation: CO(g)+2H2(g)⇌CH3OH(g)\text{CO(g)} + 2 \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}; ΔH<0\Delta H < 0. Which change will increase the equilibrium yield of methanol (CH₃OH)?

  1. Increasing the temperature and increasing the pressure.
  2. Decreasing the temperature and increasing the pressure. (correct answer)
  3. Increasing the temperature and decreasing the pressure.
  4. Decreasing the temperature and decreasing the pressure.
Explanation: The forward reaction is exothermic (ΔH<0\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.

Question 18

In the industrial Haber process, N2(g)+3H2(g)⇌2NH3(g)\text{N}_2\text{(g)} + 3 \text{H}_2\text{(g)} \rightleftharpoons 2 \text{NH}_3\text{(g)}, ΔH<0\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?

  1. The high temperature is required to increase the value of the equilibrium constant, KcK_c.
  2. The high pressure is only effective at shifting the equilibrium at high temperatures.
  3. The high temperature provides a sufficient rate of reaction, which is a necessary economic compromise. (correct answer)
  4. The catalyst used in the process is only active at temperatures above 400 °C, which is the primary reason.
Explanation: 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.

Question 19

For the equilibrium 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), the equilibrium constant Kc=4.0×106K_c = 4.0 \times 10^6 at 727°C. If the initial concentrations are [SO2]=0.50 M[SO_2] = 0.50 \text{ M}, [O2]=0.30 M[O_2] = 0.30 \text{ M}, and [SO3]=0.10 M[SO_3] = 0.10 \text{ M}, what can be concluded about the reaction's progress toward equilibrium?

  1. The reaction will proceed forward because Qc<KcQ_c < K_c and the system has not reached equilibrium (correct answer)
  2. The reaction will proceed in reverse because Qc>KcQ_c > K_c and there is excess product present
  3. The system is at equilibrium because the forward and reverse rates are equal at this temperature
  4. The reaction direction cannot be determined without knowing the activation energy for both directions
Explanation: First calculate Qc=[SO3]2[SO2]2[O2]=(0.10)2(0.50)2(0.30)=0.010.075=0.133Q_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 Qc=0.133<Kc=4.0×106Q_c = 0.133 < K_c = 4.0 \times 10^6, the reaction will proceed forward to reach equilibrium. Choice B incorrectly states Qc>KcQ_c > K_c. Choice C assumes equilibrium without calculation. Choice D incorrectly introduces kinetics into a thermodynamic analysis.

Question 20

The equilibrium N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) has Kc=2.4×10−3K_c = 2.4 \times 10^{-3} at 1000 K. A mixture initially containing N2N_2 and H2H_2 in a 3:1 molar ratio at total concentration of 0.80 M reaches equilibrium. What fraction of the original N2N_2 remains unreacted?

  1. 0.91 because the small KcK_c value indicates very limited conversion under these conditions (correct answer)
  2. 0.85 because equilibrium favors reactants but significant NH3NH_3 formation still occurs at high temperature
  3. 0.78 because the 3:1 stoichiometric ratio promotes more complete utilization of nitrogen gas
  4. 0.94 because high temperature and small KcK_c combine to minimize product formation significantly
Explanation: Initial: [H2]=0.60[H_2] = 0.60 M, [N2]=0.20[N_2] = 0.20 M. Let x = mol/L of N2N_2 consumed. At equilibrium: [N2]=0.20−x[N_2] = 0.20-x, [H2]=0.60−3x[H_2] = 0.60-3x, [NH3]=2x[NH_3] = 2x. Kc=(2x)2(0.20−x)(0.60−3x)3=2.4×10−3K_c = \frac{(2x)^2}{(0.20-x)(0.60-3x)^3} = 2.4 \times 10^{-3}. Given the small KcK_c, assume x is small: 4x2(0.20)(0.60)3≈2.4×10−3\frac{4x^2}{(0.20)(0.60)^3} ≈ 2.4 \times 10^{-3}. Solving: x≈0.018x ≈ 0.018. Fraction unreacted = 0.20−0.0180.20=0.91\frac{0.20-0.018}{0.20} = 0.91. Choices B and C overestimate conversion. Choice D underestimates the equilibrium extent.