IB Chemistry Quiz: Understand Measuring Enthalpy Change
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Understand Measuring Enthalpy ChangeQuestion 1 of 20

The dissolution of 10.0 g of ammonium nitrate in 100.0 g of water causes the temperature of the solution to drop by 7.5 K. What is the enthalpy change of solution, in kJ mol⁻¹, for ammonium nitrate? (Mᵣ(NH₄NO₃) = 80.05; specific heat capacity of solution = 4.18 J g⁻¹ K⁻¹)

+27.6
-27.6
+25.1
-3.48
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Measuring Enthalpy Change

Practice Understand 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.

What this quiz covers

This quiz focuses on Understand 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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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The dissolution of 10.0 g of ammonium nitrate in 100.0 g of water causes the temperature of the solution to drop by 7.5 K. What is the enthalpy change of solution, in kJ mol⁻¹, for ammonium nitrate? (Mᵣ(NH₄NO₃) = 80.05; specific heat capacity of solution = 4.18 J g⁻¹ K⁻¹)

  1. +27.6 (correct answer)
  2. -27.6
  3. +25.1
  4. -3.48
Explanation: First, calculate the heat absorbed by the solution from the surroundings, Q. The total mass m = 100.0 g + 10.0 g = 110.0 g. Q = mcΔT = 110.0 g × 4.18 J g⁻¹ K⁻¹ × 7.5 K = 3448.5 J. Since the temperature dropped, the dissolution process is endothermic, so the enthalpy change for the system is positive. Next, calculate the moles of NH₄NO₃: n = mass / Mᵣ = 10.0 g / 80.05 g mol⁻¹ = 0.125 mol. Finally, calculate ΔH = +Q / n = +3448.5 J / 0.125 mol = +27588 J mol⁻¹ ≈ +27.6 kJ mol⁻¹.

Question 2

When 25.0 cm³ of 0.50 mol dm⁻³ H₂SO₄ reacts completely with 50.0 cm³ of 0.50 mol dm⁻³ KOH, the temperature rises by 3.4 K. What is the approximate heat energy (in J) transferred to the surroundings during this process? Assume the density and specific heat capacity of the solution are 1.0 g cm⁻³ and 4.18 J g⁻¹ K⁻¹, respectively.

  1. 1.1 kJ
  2. 440 J
  3. 1100 J (correct answer)
  4. 440 kJ
Explanation: The question asks for the heat energy transferred, Q. First, determine the total mass of the solution. The total volume is 25.0 cm³ + 50.0 cm³ = 75.0 cm³. Assuming a density of 1.0 g cm⁻³, the mass is 75.0 g. Now, use the formula Q = mcΔT. Q = 75.0 g × 4.18 J g⁻¹ K⁻¹ × 3.4 K = 1065.9 J. This value is approximately 1100 J.

Question 3

Which statement correctly describes the system and surroundings for the combustion of methane in a Bunsen burner?

  1. The methane and oxygen molecules are the system; the flame and the hot gases are the surroundings.
  2. The Bunsen burner is the system; the flame and chemicals are the surroundings.
  3. The reacting chemicals (methane and oxygen) are the system; the burner and the laboratory air are the surroundings. (correct answer)
  4. The flame is the system; the energy released as heat and light is the surroundings.
Explanation: In thermodynamics, the system is defined as the part of the universe being studied, which in a chemical reaction is the collection of reacting particles (reactants and products). The surroundings are everything else that can exchange energy with the system. For methane combusting, the system is the mixture of methane, oxygen, carbon dioxide, and water molecules. The burner, the air in the lab, and anything else in the vicinity constitute the surroundings.

Question 4

The equation for the combustion of ethanol is C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). Which of the following conditions are necessary for the enthalpy change of this reaction to be equal to the standard enthalpy change of combustion, ΔH⦵c?

  1. The reaction is carried out at 273 K and 1 atm pressure.
  2. Exactly one mole of ethanol is burned in three moles of oxygen.
  3. One mole of ethanol is completely burned in excess oxygen at 298 K and 100 kPa. (correct answer)
  4. The water produced is in the gaseous state, and the pressure is 100 kPa.
Explanation: The standard enthalpy change of combustion is defined as the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions. Standard conditions are defined as a pressure of 100 kPa and a specified temperature, usually 298 K. It is also crucial that all reactants and products are in their standard states; for water at 298 K, this is liquid.

Question 5

The combustion of 1.20 g of methane (CH₄, Mᵣ = 16.0) in a bomb calorimeter raises the temperature of the calorimeter system by 8.00 K. The heat capacity of the entire calorimeter system is C. If the accepted enthalpy of combustion for methane is -890 kJ mol⁻¹, what is the value of C in kJ K⁻¹?

  1. 8.34 (correct answer)
  2. 66.8
  3. 111
  4. 534
Explanation: First, calculate the moles of methane combusted: n = 1.20 g / 16.0 g mol⁻¹ = 0.0750 mol. Next, calculate the total heat released (Q) by this amount of methane using the accepted enthalpy value: Q = n × |ΔH| = 0.0750 mol × 890 kJ mol⁻¹ = 66.75 kJ. This is the heat that caused the temperature change. The relationship for a bomb calorimeter is Q = CΔT, where C is the total heat capacity. Rearranging for C: C = Q / ΔT = 66.75 kJ / 8.00 K ≈ 8.34 kJ K⁻¹.

Question 6

Which statement correctly describes energy changes during an endothermic reaction?

  1. The potential energy of the products is lower than that of the reactants, and heat is absorbed from the surroundings.
  2. The potential energy of the products is higher than that of the reactants, and heat is released to the surroundings.
  3. The potential energy of the products is higher than that of the reactants, and heat is absorbed from the surroundings. (correct answer)
  4. The potential energy of the products is lower than that of the reactants, and heat is released to the surroundings.
Explanation: In an endothermic reaction, the system absorbs energy from the surroundings. This absorbed energy increases the internal or potential energy of the chemical system. Therefore, the products of an endothermic reaction have a higher potential energy than the reactants. The enthalpy change (ΔH) is positive.

Question 7

The reaction 2H₂O₂(aq) → 2H₂O(l) + O₂(g) is exothermic. How does the addition of a catalyst, such as MnO₂, affect the enthalpy change (ΔH) and the activation energy (Eₐ) of the reaction?

  1. ΔH decreases and Eₐ decreases.
  2. ΔH remains unchanged and Eₐ decreases. (correct answer)
  3. ΔH decreases and Eₐ remains unchanged.
  4. ΔH remains unchanged and Eₐ remains unchanged.
Explanation: A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy (Eₐ). However, a catalyst does not affect the energy levels of the reactants or the products. The enthalpy change (ΔH) is the difference between the energy of the products and the energy of the reactants (ΔH = H_products - H_reactants). Since these energies are unaffected by the catalyst, ΔH remains unchanged.

Question 8

Consider the neutralization reaction: HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l). When 50.0 cm³ of 0.500 mol dm⁻³ HNO₃ is mixed with 25.0 cm³ of 0.800 mol dm⁻³ KOH, the temperature rises by 2.69 K. What is the enthalpy of neutralization per mole of water formed, in kJ mol⁻¹? (Assume solution density = 1.00 g cm⁻³ and specific heat capacity = 4.18 J g⁻¹ K⁻¹)

  1. -42.1 (correct answer)
  2. -52.7
  3. -33.7
  4. -21.1
Explanation: First, find the limiting reactant. n(HNO₃) = 0.0500 dm³ × 0.500 mol dm⁻³ = 0.0250 mol. n(KOH) = 0.0250 dm³ × 0.800 mol dm⁻³ = 0.0200 mol. Since the reaction is 1:1, KOH is the limiting reactant, and 0.0200 mol of water will be formed. Next, calculate Q. Total volume = 75.0 cm³, so mass = 75.0 g. Q = mcΔT = 75.0 g × 4.18 J g⁻¹ K⁻¹ × 2.69 K = 843.15 J. The reaction is exothermic. Finally, ΔH = -Q / n = -843.15 J / 0.0200 mol = -42158 J mol⁻¹ ≈ -42.1 kJ mol⁻¹.

Question 9

The enthalpy change for the complete combustion of 1.00 mol of propan-1-ol, C₃H₇OH(l), is -2021 kJ. A spirit burner containing propan-1-ol is used to heat 200.0 g of water, causing the temperature to rise by 40.0 °C. What mass of propan-1-ol was burned to cause this temperature change? (Mᵣ(C₃H₇OH) = 60.1; c(water) = 4.18 J g⁻¹ K⁻¹)

  1. 0.99 g (correct answer)
  2. 33.5 g
  3. 1.66 g
  4. 0.016 g
Explanation: First, calculate the heat absorbed by the water: Q = mcΔT = 200.0 g × 4.18 J g⁻¹ K⁻¹ × 40.0 K = 33440 J or 33.44 kJ. This heat was supplied by the combustion of propan-1-ol. Now, find the moles of propan-1-ol needed to produce this much heat: n = Q / |ΔH| = 33.44 kJ / 2021 kJ mol⁻¹ = 0.01655 mol. Finally, convert moles to mass: mass = n × Mᵣ = 0.01655 mol × 60.1 g mol⁻¹ ≈ 0.99 g.

Question 10

In an experiment, 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH in a polystyrene cup. A temperature increase of ΔT is recorded. The experiment is repeated using 50.0 cm³ of 2.00 mol dm⁻³ HCl and 50.0 cm³ of 2.00 mol dm⁻³ NaOH. What is the expected temperature increase in the second experiment, assuming the heat capacity of the cup and heat loss are negligible?

  1. 0.5 ΔT
  2. ΔT
  3. 2 ΔT (correct answer)
  4. 4 ΔT
Explanation: The heat evolved, Q, is proportional to the number of moles of reactants (Q = nΔH). In the second experiment, the concentrations are doubled while volumes remain the same, so the number of moles of H⁺ and OH⁻ reacting is doubled. This means the heat evolved (Q) will be doubled. The total volume of the solution is the same in both experiments (100.0 cm³), so the mass (m) being heated is approximately the same. Since ΔT = Q / (mc), if Q is doubled and m and c are constant, ΔT will also be doubled.

Question 11

A calorimetry experiment is performed in a simple polystyrene cup. Which of the following assumptions, if incorrect, would lead to the magnitude of the calculated exothermic enthalpy change being artificially high?

  1. Assuming no heat is lost to the surroundings.
  2. Assuming the specific heat capacity of the solution is the same as that of pure water. (correct answer)
  3. Assuming no heat is absorbed by the polystyrene cup.
  4. Assuming the volume measurement of the solutions is perfectly accurate.
Explanation: If the actual specific heat capacity of the salt solution is lower than that of water (4.18 J g⁻¹ K⁻¹), using the higher value for water in the calculation Q = mcΔT would result in a calculated Q that is larger than the actual heat released. This would lead to a calculated magnitude for ΔH (|ΔH| = Q/n) that is artificially high. Errors in A and C both lead to calculated |ΔH| values that are too low. An error in D could lead to a value that is too high or too low.

Question 12

A student determines the enthalpy of solution of a salt using a coffee-cup calorimeter. The calculated enthalpy change is -25.3 kJ mol⁻¹. If a significant amount of heat was absorbed by the calorimeter itself, but this was not accounted for in the calculation, how does the calculated value compare to the true enthalpy change?

  1. The calculated value is more exothermic (more negative) than the true value.
  2. The calculated value is less exothermic (less negative) than the true value. (correct answer)
  3. The calculated value is equal to the true value because the calorimeter's heat capacity is negligible.
  4. The effect cannot be determined without knowing the heat capacity of the calorimeter.
Explanation: The total heat released by the reaction (Q_total) is split between heating the solution (Q_solution) and heating the calorimeter (Q_calorimeter). The student's calculation only uses Q_solution = mcΔT. Therefore, the calculated Q is less than the actual Q_total. Since ΔH = -Q/n, a smaller value for Q will result in a calculated |ΔH| that is smaller than the true value. For an exothermic reaction, this means the calculated value is less negative (less exothermic) than the true value.

Question 13

When 2.43 g of magnesium (Aᵣ = 24.3) is added to 100.0 cm³ of 1.00 mol dm⁻³ CuSO₄(aq), the temperature increases by 45.0 °C. Assuming the solution has a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹, what is the enthalpy change, in kJ mol⁻¹, for the reaction Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)?

  1. +188
  2. -18.8
  3. +18.8
  4. -188 (correct answer)
Explanation: First, find the limiting reactant. n(Mg) = 2.43 g / 24.3 g mol⁻¹ = 0.100 mol. n(CuSO₄) = 0.100 dm³ × 1.00 mol dm⁻³ = 0.100 mol. The stoichiometry is 1:1, so neither is limiting. We use n = 0.100 mol. Second, calculate heat evolved, Q. The mass of the solution is 100.0 cm³ × 1.00 g cm⁻³ = 100.0 g. Q = mcΔT = 100.0 g × 4.18 J g⁻¹ K⁻¹ × 45.0 K = 18810 J. Third, calculate ΔH. Since temperature increased, the reaction is exothermic (ΔH is negative). ΔH = -Q / n = -18810 J / 0.100 mol = -188100 J mol⁻¹ = -188.1 kJ mol⁻¹.

Question 14

In an experiment, 25.0 cm³ of 1.0 mol dm⁻³ HCl is mixed with 25.0 cm³ of 1.0 mol dm⁻³ NaOH. A temperature rise of 6.5 °C is observed. When the experiment is repeated with 50.0 cm³ of 1.0 mol dm⁻³ HCl and 50.0 cm³ of 1.0 mol dm⁻³ NaOH, what will be the temperature rise?

  1. 3.25 °C
  2. 6.5 °C (correct answer)
  3. 13.0 °C
  4. 26.0 °C
Explanation: In the second experiment, the volumes of both solutions are doubled, while the concentrations remain the same. This means the number of moles of reactants (n) is doubled. Consequently, the heat evolved (Q) is also doubled (since Q ∝ n). However, the total volume of the final solution is also doubled, which means its mass (m) is doubled. According to the formula ΔT = Q / (mc), if both Q and m are doubled, the ratio Q/m remains the same. Therefore, the temperature change, ΔT, will be the same.

Question 15

In a calorimetry experiment to determine the enthalpy of combustion of magnesium, a student obtains a value that is significantly less exothermic than the accepted literature value. Which of the following is the most likely systematic error responsible for this discrepancy?

  1. Some of the magnesium reacted with nitrogen in the air to form magnesium nitride. (correct answer)
  2. The initial mass of magnesium was measured to be lower than its actual mass.
  3. The final temperature was read after the water had started to cool down significantly.
  4. The specific heat capacity of water was assumed to be 4.2 J g⁻¹ K⁻¹ instead of 4.18 J g⁻¹ K⁻¹.
Explanation: The accepted value for the enthalpy of combustion assumes complete reaction with oxygen (MgO formation). The formation of magnesium nitride (Mg₃N₂) from magnesium and nitrogen is also exothermic, but less so than the formation of MgO. If some magnesium forms Mg₃N₂ instead of MgO, the total heat released will be lower, leading to a calculated enthalpy of combustion that is less exothermic (less negative) than the true value.

Question 16

For a certain exothermic reaction, the activation energy of the forward reaction is +75 kJ mol⁻¹ and the activation energy of the reverse reaction is +120 kJ mol⁻¹. What is the standard enthalpy change, ΔH⦵, for the forward reaction?

  1. +195 kJ mol⁻¹
  2. +45 kJ mol⁻¹
  3. -195 kJ mol⁻¹
  4. -45 kJ mol⁻¹ (correct answer)
Explanation: The relationship between the activation energies and the enthalpy change is given by ΔH = Eₐ(forward) - Eₐ(reverse). Substituting the given values: ΔH = (+75 kJ mol⁻¹) - (+120 kJ mol⁻¹) = -45 kJ mol⁻¹. The negative sign is consistent with the information that the reaction is exothermic.

Question 17

A 0.50 g sample of a solid is dissolved in 50.0 g of water in an insulated calorimeter. The temperature is recorded over time. The initial temperature of the water is 21.0 °C. The solid is added at t = 60 s. The temperature reaches a maximum of 28.5 °C at t = 180 s and then begins to cool. What is the best method to determine the value of ΔT for calculating the enthalpy change?

  1. Use the maximum temperature reading, so ΔT = 28.5 °C − 21.0 °C.
  2. Extrapolate the cooling curve back to the time of mixing (t = 60 s) to find a corrected maximum temperature. (correct answer)
  3. Take the average temperature between t = 60 s and t = 180 s to account for heat loss.
  4. Use the temperature reading at a very long time after mixing, once thermal equilibrium is reached.
Explanation: Since the calorimeter is not a perfect insulator, heat is lost to the surroundings as soon as the temperature rises above ambient temperature. This heat loss occurs simultaneously with the heat generation from the reaction. To account for this, the cooling portion of the temperature-time graph should be extrapolated back to the time of mixing. This gives a theoretical maximum temperature that would have been reached if the reaction were instantaneous and no heat was lost, providing a more accurate ΔT.

Question 18

The standard enthalpy change of neutralization is defined for the reaction H⁺(aq) + OH⁻(aq) → H₂O(l). The enthalpy change for the reaction between aqueous ethanoic acid and aqueous potassium hydroxide is found to be less exothermic than for hydrochloric acid and potassium hydroxide. What is the correct explanation?

  1. Potassium hydroxide is a weaker base than sodium hydroxide, releasing less energy.
  2. Energy is absorbed to ionize the weak ethanoic acid before neutralization can occur. (correct answer)
  3. The salt potassium ethanoate is unstable and absorbs energy upon its formation.
  4. The reaction involving ethanoic acid is slower, so less heat is produced per second.
Explanation: Hydrochloric acid is a strong acid and is fully ionized in solution. Ethanoic acid is a weak acid and exists mostly as undissociated molecules. Before it can neutralize the hydroxide ions, the ethanoic acid molecules must first dissociate (CH₃COOH ⇌ CH₃COO⁻ + H⁺). This dissociation is an endothermic process that requires energy input. This energy input partially offsets the exothermic neutralization reaction, making the overall enthalpy change less exothermic than that for a strong acid.

Question 19

In a constant-pressure calorimetry experiment, 2.50 g of zinc metal is added to 150.0 mL of 1.00 M CuSO₄ solution at 22.5°C. The reaction proceeds according to: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). The final temperature reaches 35.8°C. If the experiment is repeated using zinc powder instead of zinc pellets, while keeping all other conditions identical, what is the most likely outcome?

  1. The final temperature will be higher because increased surface area accelerates the reaction rate
  2. The final temperature will be lower because powder form leads to incomplete reaction due to rapid heat loss
  3. The final temperature will be the same because the total enthalpy change depends only on initial and final states (correct answer)
  4. The final temperature will be higher because zinc powder has a different standard enthalpy of formation than zinc pellets
Explanation: This question tests your understanding of enthalpy as a state function in thermochemistry. When analyzing calorimetry experiments, you need to distinguish between factors that affect reaction kinetics (how fast) versus thermodynamics (how much energy is released). The key insight is that enthalpy change depends only on the initial and final states of the system, not the pathway taken between them. Whether you use zinc pellets or zinc powder, you're still converting the same amount of Zn(s) and CuSO₄(aq) into the same products: ZnSO₄(aq) and Cu(s). Since the chemical equation and quantities remain identical, the total heat released (ΔH\Delta H) must be the same, leading to the same final temperature. Option A incorrectly assumes that faster reaction rate means more heat released. While zinc powder does increase surface area and reaction rate, this only affects how quickly the temperature rises, not the maximum temperature reached. Option B suggests incomplete reaction due to heat loss, but this contradicts the premise that "all other conditions" remain identical. The same amount of limiting reagent will be consumed regardless of zinc's physical form. Option D demonstrates a fundamental misunderstanding of state functions. The standard enthalpy of formation for zinc metal is the same whether it's powder or pellets - they're both pure zinc in the solid state. Remember: In thermochemistry problems, always distinguish between kinetics (rate) and thermodynamics (energy). The physical form of reactants affects reaction speed but not the total enthalpy change when comparing identical chemical transformations.

Question 20

A student performs a calorimetry experiment to determine the specific heat of an unknown metal. A 45.2 g sample of the metal is heated to 95.0°C and then quickly transferred to 75.0 g of water at 18.0°C in an insulated container. The final equilibrium temperature is 22.4°C. However, during the transfer process, the metal cools to 89.0°C before being placed in water. How does this error affect the calculated specific heat of the metal?

  1. The calculated specific heat will be too low because less heat was actually transferred to the water
  2. The calculated specific heat will be too low because the heat lost during transfer is not accounted for in the calculation
  3. The calculated specific heat will be unaffected because the final equilibrium temperature accounts for all heat transfers
  4. The calculated specific heat will be too high because the student assumes a larger temperature change for the metal (correct answer)
Explanation: When analyzing calorimetry errors, you need to trace how each mistake propagates through the heat transfer equation: qmetal=−qwaterq_{metal} = -q_{water} and q=mcΔTq = mc\Delta T. In this experiment, the student assumes the metal started at 95.0°C when calculating its temperature change, but it actually started at only 89.0°C when placed in water. This means the actual temperature change of the metal was smaller than assumed in the calculation. Here's why answer D is correct: When using cmetal=−mwater×cwater×ΔTwatermmetal×ΔTmetalc_{metal} = \frac{-m_{water} \times c_{water} \times \Delta T_{water}}{m_{metal} \times \Delta T_{metal}}, the student uses a larger ΔTmetal\Delta T_{metal} (95.0°C - 22.4°C = 72.6°C) instead of the actual smaller value (89.0°C - 22.4°C = 66.6°C). Since ΔTmetal\Delta T_{metal} appears in the denominator, using a larger value makes the calculated specific heat artificially high. A is wrong because the same amount of heat was transferred to water—the final temperature tells us exactly how much heat the water gained. B incorrectly suggests the lost heat isn't accounted for, but the water temperature change already reflects all heat it actually received. C is wrong because while the equilibrium temperature is correct, the calculation incorrectly assumes the metal's starting temperature, leading to error. Study tip: In calorimetry error analysis, always track which variable in your equation is affected and whether it's in the numerator or denominator. Errors that increase the denominator will decrease your calculated value, and vice versa.