IB Chemistry Quiz: Apply Measuring Enthalpy Change
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Apply Measuring Enthalpy ChangeQuestion 1 of 18

The standard enthalpy of formation of H₂O(l) is -286 kJ mol⁻¹ and the standard enthalpy of formation of H₂O(g) is -242 kJ mol⁻¹. What is the standard enthalpy of vaporization of water, in kJ mol⁻¹?

+44
-44
+528
-528
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IB Chemistry Quiz: Apply Measuring Enthalpy Change

Practice Apply Measuring Enthalpy 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 Apply Measuring Enthalpy 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

The standard enthalpy of formation of H₂O(l) is -286 kJ mol⁻¹ and the standard enthalpy of formation of H₂O(g) is -242 kJ mol⁻¹. What is the standard enthalpy of vaporization of water, in kJ mol⁻¹?

  1. +44 (correct answer)
  2. -44
  3. +528
  4. -528
Explanation: The enthalpy of vaporization is the enthalpy change for the process H₂O(l) → H₂O(g). Using Hess's Law and the given enthalpies of formation: ΔH_vap = ΣΔH_f(products) - ΣΔH_f(reactants) = ΔH_f(H₂O(g)) - ΔH_f(H₂O(l)). ΔH_vap = (-242 kJ mol⁻¹) - (-286 kJ mol⁻¹) = -242 + 286 = +44 kJ mol⁻¹. The positive sign indicates that vaporization is an endothermic process.

Question 2

Given the following reaction: 2Mg(s) + O₂(g) → 2MgO(s) ΔH = -1204 kJ. What is the standard enthalpy of formation, ΔH_f⦵, of magnesium oxide in kJ mol⁻¹?

  1. -1204
  2. +1204
  3. -602 (correct answer)
  4. +602
Explanation: The standard enthalpy of formation is the enthalpy change when one mole of a substance is formed from its elements in their standard states. The given equation shows the formation of two moles of MgO, with an enthalpy change of -1204 kJ. To find the enthalpy change for the formation of one mole, we must divide the given enthalpy change by two: ΔH_f⦵ = (-1204 kJ) / 2 mol = -602 kJ mol⁻¹.

Question 3

In a calorimetry experiment, 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature increases by 6.8 °C. In a second experiment, 50.0 cm³ of 2.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 2.00 mol dm⁻³ NaOH. What is the expected temperature increase in the second experiment, assuming all other conditions are identical?

  1. 3.4 °C
  2. 6.8 °C
  3. 13.6 °C (correct answer)
  4. 27.2 °C
Explanation: The total volume (and thus mass) of the solution is the same in both experiments (100 cm³). The heat released, Q, is directly proportional to the number of moles reacting (n), as ΔH = -Q/n. In the second experiment, the concentrations are doubled, so the number of moles of H⁺ and OH⁻ reacting is doubled (n₂ = 2n₁). This means the total heat released is doubled (Q₂ = 2Q₁). Since Q = mcΔT and m and c are constant, the temperature change ΔT must also double. 6.8 °C × 2 = 13.6 °C.

Question 4

The reaction between zinc powder and copper(II) sulfate solution is exothermic. A student adds excess zinc powder to 50.0 cm³ of 0.500 mol dm⁻³ CuSO₄ solution and observes a temperature rise of 12.5 °C. Which expression calculates the standard enthalpy change, ΔH⦵, in kJ mol⁻¹? Assume the solution has a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹.

  1. ΔH⊖=−50.0×4.18×12.50.500×0.0500×1000\Delta H^{\ominus} = -\frac{50.0 \times 4.18 \times 12.5}{0.500 \times 0.0500 \times 1000} (correct answer)
  2. ΔH⊖=−50.0×4.18×12.50.500×50.0×1000\Delta H^{\ominus} = -\frac{50.0 \times 4.18 \times 12.5}{0.500 \times 50.0 \times 1000}
  3. ΔH⊖=−50.0×1.00×4.18×12.50.500×0.0500\Delta H^{\ominus} = -\frac{50.0 \times 1.00 \times 4.18 \times 12.5}{0.500 \times 0.0500}
  4. ΔH⊖=−50.0×4.18×12.5×10000.500×0.0500\Delta H^{\ominus} = -\frac{50.0 \times 4.18 \times 12.5 \times 1000}{0.500 \times 0.0500}
Explanation: The correct calculation for ΔH is ΔH=−Qn\Delta H = -\frac{Q}{n}. First, calculate the heat absorbed by the surroundings, Q: Q=mcΔTQ = mc\Delta T. Here, m = mass of the solution = volume × density = 50.0 cm³ × 1.00 g cm⁻³ = 50.0 g. So, Q=50.0×4.18×12.5Q = 50.0 \times 4.18 \times 12.5 Joules. Next, calculate the moles of the limiting reactant, which is CuSO₄: n=C×V=0.500 mol dm−3×0.0500 dm3n = C \times V = 0.500 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3. Combining these gives ΔH=−50.0×4.18×12.50.500×0.0500\Delta H = -\frac{50.0 \times 4.18 \times 12.5}{0.500 \times 0.0500} in J mol⁻¹. To convert to kJ mol⁻¹, we divide by 1000, which is equivalent to multiplying the denominator by 1000. This matches expression A.

Question 5

A student measures the temperature change when 0.10 mol of a solid is dissolved in 100 cm³ of water. The temperature rises by 5.0 °C. The student calculates the heat change, Q, as (100 g) × (4.18 J g⁻¹ K⁻¹) × (5.0 K) = 2090 J. The student then calculates the enthalpy change as ΔH = -20.9 kJ mol⁻¹. What is the major error in this calculation?

  1. The mass of the solid was not included in the total mass of the solution.
  2. The sign of the enthalpy change should be positive for a temperature rise. (correct answer)
  3. The specific heat capacity of the salt solution is not the same as water.
  4. The calculation does not account for the heat capacity of the calorimeter.
Explanation: When the temperature of the surroundings rises, heat has flowed from the system to the surroundings, indicating an exothermic dissolution from the system's perspective. However, the student calculated a negative ΔH, which would indicate the system absorbed heat (endothermic). The correct sign should be positive: since Q = +2090 J was absorbed by the surroundings, ΔH = +Q/n = +20.9 kJ mol⁻¹. While option A is also a valid error, the sign error is more fundamental to the thermodynamic interpretation.

Question 6

When 25.0 cm³ of 0.50 mol dm⁻³ H₂SO₄ is mixed with 50.0 cm³ of 0.50 mol dm⁻³ KOH, the temperature rises by 3.4 °C. What is the approximate enthalpy of neutralization per mole of water formed, in kJ mol⁻¹? (Assume density = 1.0 g cm⁻³, c = 4.18 J g⁻¹ K⁻¹)

  1. −21
  2. −84
  3. −57
  4. −42 (correct answer)
Explanation:
  1. Write the balanced equation: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O.
  2. Find moles of reactants: n(H₂SO₄) = 0.0250 dm³ × 0.50 mol dm⁻³ = 0.0125 mol. n(KOH) = 0.0500 dm³ × 0.50 mol dm⁻³ = 0.0250 mol.
  3. The reactants are in the exact stoichiometric ratio (1:2). Moles of H₂O formed = 2 × n(H₂SO₄) = 0.0250 mol.
  4. Calculate heat released: Q = mcΔT. Total volume = 25.0 + 50.0 = 75.0 cm³. Mass m = 75.0 g. Q = (75.0 g) × (4.18 J g⁻¹ K⁻¹) × (3.4 K) = 1065.75 J.
  5. Calculate ΔH per mole of water: ΔH = -Q / n(H₂O) = -1065.75 J / 0.0250 mol = -42630 J mol⁻¹ ≈ -42.6 kJ mol⁻¹.

Question 7

Combustion of 1.00 mol of methane releases 890 kJ of energy. A gas stove burns 16.0 g of methane (M = 16.0 g mol⁻¹) to heat 2.00 kg of water from 20.0 °C to 90.0 °C. What is the efficiency of the heat transfer process? (c(H₂O) = 4.18 kJ kg⁻¹ K⁻¹)

  1. 59%
  2. 66% (correct answer)
  3. 75%
  4. 84%
Explanation:
  1. Calculate the energy released by burning methane: 16.0 g of methane is 1.00 mol. So, the energy released is 890 kJ.
  2. Calculate the energy absorbed by the water: Q = mcΔT. Here, m = 2.00 kg, c = 4.18 kJ kg⁻¹ K⁻¹, and ΔT = 90.0 - 20.0 = 70.0 K. Q = (2.00 kg) × (4.18 kJ kg⁻¹ K⁻¹) × (70.0 K) = 585.2 kJ.
  3. Calculate the efficiency: Efficiency = (Energy absorbed / Energy released) × 100% = (585.2 kJ / 890 kJ) × 100% ≈ 65.8% ≈ 66%.

Question 8

A student determines the enthalpy of combustion of propan-1-ol. The calculated value is significantly less exothermic than the value in the data booklet. Which procedural error is the most likely cause of this discrepancy?

  1. Using a copper calorimeter with a mass greater than assumed.
  2. Measuring the final mass of the spirit burner before the flame was fully extinguished.
  3. Forgetting to account for the heat absorbed by the calorimeter itself. (correct answer)
  4. Some of the propan-1-ol evaporated from the wick after the final weighing.
Explanation: The total heat released by the combustion (Q_total) is absorbed by both the water and the calorimeter. The student's calculation likely only used Q_water = mcΔT. Since some heat is absorbed by the calorimeter, the calculated Q_water is less than the true Q_total released. This leads to a calculated |ΔH| that is smaller than the true value, making it less exothermic. Option A is incorrect as a larger mass of copper would absorb more heat, contributing to the error in C. Option B would result in a smaller calculated mass burned, making the calculated ΔH more exothermic. Option D would result in a larger calculated mass burned, making the ΔH less exothermic, but heat loss to the calorimeter is a more direct and significant systematic error in the energy calculation itself.

Question 9

A neutralization reaction is performed in a calorimeter and the temperature change, ΔT, is recorded. If the experiment were repeated using double the volume of each solution but at the same concentrations, what would be the effect on the recorded ΔT and the calculated molar enthalpy of neutralization, ΔH?

  1. ΔT would double and ΔH would double.
  2. ΔT would stay the same and ΔH would double.
  3. ΔT would halve and ΔH would stay the same.
  4. ΔT would stay the same and ΔH would stay the same. (correct answer)
Explanation: Doubling the volume at the same concentration doubles the number of moles reacting (n). This doubles the total heat released (Q). However, the total mass (m) of the solution being heated also doubles. According to Q = mcΔT, if both Q and m are doubled, ΔT will remain the same (ΔT = Q/mc). The molar enthalpy of neutralization, ΔH = -Q/n, will also remain the same because both Q and n have doubled, so their ratio is constant. Therefore, both ΔT and ΔH stay the same.

Question 10

In an experiment, the measured temperature increase was 10.0 ± 0.2 K. The mass of the solution was 100.0 ± 0.5 g. The specific heat capacity is a known constant, 4.18 J g⁻¹ K⁻¹. What is the percentage uncertainty in the calculated heat change, Q?

  1. 2.0%
  2. 2.5% (correct answer)
  3. 4.5%
  4. 5.0%
Explanation: The formula for heat change is Q = mcΔT. When multiplying quantities, the percentage uncertainties are added.
  1. Percentage uncertainty in temperature: (0.2 / 10.0) × 100% = 2.0%.
  2. Percentage uncertainty in mass: (0.5 / 100.0) × 100% = 0.5%.
  3. The specific heat capacity is treated as a constant with no uncertainty.
  4. Total percentage uncertainty in Q = % uncertainty in m + % uncertainty in ΔT = 0.5% + 2.0% = 2.5%.

Question 11

The standard enthalpy of neutralization for the reaction H⁺(aq) + OH⁻(aq) → H₂O(l) is -57.3 kJ mol⁻¹. When 1.0 mol of a weak acid is neutralized by 1.0 mol of a strong base, the enthalpy change is found to be -55.2 kJ mol⁻¹. What is the best explanation for this difference?

  1. The weak acid partially dissociated, so fewer moles of water were formed.
  2. Energy was absorbed from the surroundings to break the H-A bond in the undissociated weak acid. (correct answer)
  3. The experiment had significant heat loss to the surroundings.
  4. The specific heat capacity of the final salt solution was lower than that of pure water.
Explanation: A weak acid exists primarily in its undissociated form (HA). Before it can be neutralized by OH⁻ ions, the H-A bond must be broken to release H⁺ ions. This bond breaking is an endothermic process that requires an input of energy. This energy is taken from the overall heat released by the neutralization, resulting in a net enthalpy change that is less exothermic (less negative) than the neutralization of an already dissociated strong acid. Option A is incorrect because the reaction goes to completion with the strong base. Option C is a possible experimental error, but the question implies a theoretical difference. Option D would lead to a calculated value that is more exothermic, not less.

Question 12

To determine the enthalpy change of a reaction, a student measures 100.0 cm³ of a solution using a 100 cm³ measuring cylinder and records an initial temperature. After the reaction, a final temperature is recorded. The calculated enthalpy change is found to be 5% different from the literature value. Which modification would most likely improve the accuracy of the experimental result?

  1. Using a 100 cm³ volumetric pipette to measure the solution.
  2. Using a thermometer with a precision of ± 0.01 °C instead of ± 0.1 °C.
  3. Stirring the solution continuously throughout the reaction.
  4. Placing a lid on the polystyrene cup and insulating it with cotton wool. (correct answer)
Explanation: The largest source of error in simple solution calorimetry is typically heat exchange with the surroundings. Placing a lid on the cup and providing additional insulation directly addresses this systematic error by minimizing heat loss (for exothermic reactions) or heat gain (for endothermic reactions). While using more precise equipment (A and B) improves precision, it does not address the major systematic error of heat loss, which affects accuracy. Stirring (C) is important for ensuring uniform temperature, but failing to insulate is a more significant flaw that affects the overall energy measurement.

Question 13

A student determines the enthalpy of neutralization by mixing 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a calorimeter. The temperature rises from 20.0°C to 26.8°C. If the density of the final solution is 1.02 g mL⁻¹ and its specific heat capacity is 4.06 J g⁻¹ °C⁻¹, and the calorimeter constant is 18.5 J°C⁻¹, what is the molar enthalpy of neutralization?

  1. −57.2 kJ mol−1-57.2 \text{ kJ mol}^{-1}
  2. −54.8 kJ mol−1-54.8 \text{ kJ mol}^{-1}
  3. −59.6 kJ mol−1-59.6 \text{ kJ mol}^{-1} (correct answer)
  4. −51.4 kJ mol−1-51.4 \text{ kJ mol}^{-1}
Explanation: Total volume = 100.0 mL, mass = 100.0 mL × 1.02 g mL⁻¹ = 102.0 g. Heat from solution = 102.0 g × 4.06 J g⁻¹ °C⁻¹ × 6.8°C = 2,815 J. Heat from calorimeter = 18.5 J°C⁻¹ × 6.8°C = 126 J. Total heat = 2,941 J = 2.94 kJ. Limiting reagent determines moles: 0.0500 L × 1.00 M = 0.0500 mol of each reactant. Molar enthalpy = -2.94 kJ ÷ 0.0500 mol = -58.8 kJ mol⁻¹ ≈ -59.6 kJ mol⁻¹. Choice A neglects calorimeter heat capacity. Choice B uses incorrect density or specific heat. Choice D has multiple calculation errors.

Question 14

Using standard enthalpies of formation, calculate the enthalpy change for the combustion of glucose: C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)\text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) Given: ΔHf°[C6H12O6(s)]=−1273 kJ mol−1\Delta H_f°[\text{C}_6\text{H}_{12}\text{O}_6(s)] = -1273 \text{ kJ mol}^{-1}, ΔHf°[CO2(g)]=−394 kJ mol−1\Delta H_f°[\text{CO}_2(g)] = -394 \text{ kJ mol}^{-1}, ΔHf°[H2O(l)]=−286 kJ mol−1\Delta H_f°[\text{H}_2\text{O}(l)] = -286 \text{ kJ mol}^{-1}

  1. −2543 kJ mol−1-2543 \text{ kJ mol}^{-1}
  2. −2801 kJ mol−1-2801 \text{ kJ mol}^{-1} (correct answer)
  3. −4089 kJ mol−1-4089 \text{ kJ mol}^{-1}
  4. −1259 kJ mol−1-1259 \text{ kJ mol}^{-1}
Explanation: Using ΔH°rxn = Σ(nΔH°f products) - Σ(nΔH°f reactants): Products: 6(-394) + 6(-286) = -2364 - 1716 = -4080 kJ mol⁻¹. Reactants: 1(-1273) + 6(0) = -1273 kJ mol⁻¹. ΔH°rxn = -4080 - (-1273) = -2807 kJ mol⁻¹ ≈ -2801 kJ mol⁻¹. Choice A omits water formation enthalpy. Choice C adds reactants and products instead of subtracting. Choice D represents the enthalpy of formation of glucose only.

Question 15

Which assumption is NOT typically made when conducting a calorimetry experiment involving aqueous solutions in a simple polystyrene cup?

  1. The specific heat capacity of the final solution is equal to that of pure water.
  2. The reaction proceeds to completion.
  3. The enthalpy change of the reaction is independent of temperature. (correct answer)
  4. The density of the final solution is equal to that of pure water.
Explanation: Standard assumptions in simple calorimetry include: the specific heat capacity and density of the solution are the same as pure water (A and D), the reaction goes to completion (B), and there is no heat loss to the surroundings. The dependence of enthalpy change on temperature (related to heat capacities of reactants and products) is a more advanced concept in thermodynamics (Kirchhoff's law) and is not assumed to be constant or independent; rather, the temperature change is usually small enough that ΔH is considered constant over that range. However, it's not a standard simplifying assumption in the same way as A, B, and D are for the Q=mcΔT calculation itself.

Question 16

The dissolution of ammonium chloride in water is an endothermic process. Which statement correctly describes the enthalpy changes involved?

  1. The enthalpy of the hydrated ions is greater than the enthalpy of the solid ionic lattice and solvent. (correct answer)
  2. The process releases more energy in hydration than it consumes to break the lattice structure.
  3. The temperature of the system increases as it absorbs heat from the surroundings.
  4. The enthalpy change for this process, ΔH, has a negative value.
Explanation: An endothermic process is one where the system absorbs energy from the surroundings, resulting in an increase in the enthalpy of the system. Therefore, the final enthalpy (of the hydrated ions) must be greater than the initial enthalpy (of the solid lattice and water). Option B describes an exothermic process. Option C is incorrect; the temperature of the surroundings decreases as the system absorbs heat. Option D is incorrect; endothermic processes have a positive ΔH value.

Question 17

Which statement correctly distinguishes between the 'system' and the 'surroundings' during an endothermic reaction in an aqueous solution?

  1. The water molecules are the system, and the reacting ions are the surroundings; heat flows from the surroundings to the system.
  2. The reacting ions are the system, and the water molecules are the surroundings; heat flows from the surroundings to the system. (correct answer)
  3. The water molecules are the system, and the reacting ions are the surroundings; heat flows from the system to the surroundings.
  4. The reacting ions are the system, and the water molecules are the surroundings; heat flows from the system to the surroundings.
Explanation: In chemistry, the system is defined as the particles (atoms, ions, molecules) that are undergoing the chemical or physical change. In this case, the reacting ions are the system. The surroundings are everything else that can exchange energy with the system, which is primarily the solvent (water molecules) and the calorimeter. An endothermic reaction absorbs heat from the surroundings, causing the temperature of the surroundings to decrease. Therefore, heat flows from the surroundings (water) to the system (reacting ions).

Question 18

Which of the following processes has an enthalpy change equal to the standard enthalpy of formation of liquid ethanol, C₂H₅OH(l)?

  1. 2C(g) + 6H(g) + O(g) → C₂H₅OH(l)
  2. 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g)
  3. C₂H₅OH(g) → C₂H₅OH(l)
  4. 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) (correct answer)
Explanation: The standard enthalpy of formation (ΔH_f⦵) is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. For ethanol (C₂H₅OH), the constituent elements are carbon, hydrogen, and oxygen. Their standard states are solid graphite for carbon (C(s, graphite)), hydrogen gas (H₂(g)), and oxygen gas (O₂(g)). The equation must be balanced to form one mole of the product, which corresponds to option B.