- 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?
Answer: Energy is absorbed to ionize the weak ethanoic acid before neutralization can occur.
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.
- 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?
Answer: 6.5 °C
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.
- 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?
Answer: The final temperature will be the same because the total enthalpy change depends only on initial and final states
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) 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. - 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?
Answer: Some of the magnesium reacted with nitrogen in the air to form magnesium nitride.
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.
- If the temperature of a chemical reaction in solution is observed to decrease, what can be concluded about the reaction?
Answer: It is endothermic, and the system absorbs energy from the surroundings.
The solution (part of the surroundings) gets colder because the chemical reaction (the system) is absorbing heat energy from it. A process that absorbs heat from the surroundings is defined as endothermic. Therefore, the system gains potential energy at the expense of the kinetic energy of the surroundings, causing the observed temperature drop.
- In a bomb calorimeter experiment, 1.25 g of glucose undergoes complete combustion, causing the temperature to rise from 21.50°C to 26.74°C. The heat capacity of the entire calorimeter system is 9.85 kJ°C⁻¹. Which statement correctly explains why this measured value differs from the standard enthalpy of combustion found in data tables?
Answer: The measured value represents ΔU while tabulated values represent ΔH, and ΔH = ΔU + ΔnRT for this reaction
Bomb calorimeters operate at constant volume, measuring internal energy change (ΔU). Standard enthalpies of combustion in tables are at constant pressure (ΔH). For glucose combustion: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, Δn = 0, but the conversion still matters because ΔH = ΔU + ΔnRT. The other options incorrectly identify which quantity is measured or focus on secondary effects.
- 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 coffee cup calorimeter. The temperature increases from 20.0°C to 26.7°C. If the student repeats the experiment using 50.0 mL of 1.00 M HCl with 75.0 mL of 1.00 M NaOH, which outcome is most likely?
Answer: The temperature change will be less than 26.7°C due to dilution effect and excess base
The limiting reagent is still HCl (0.050 mol), so the same amount of heat is released. However, the total mass of solution increases from 100.0 g to 125.0 g, meaning more mass absorbs the same heat, resulting in a smaller temperature change. Additionally, the excess NaOH contributes to heat absorption without generating additional heat. ΔT = q/(m × c), so larger m gives smaller ΔT.
- A student uses a bomb calorimeter with a heat capacity of 8.25 kJ°C⁻¹ to determine the enthalpy of combustion of benzoic acid. When 1.150 g of benzoic acid burns completely, the temperature increases by 3.26°C. The student then uses this calorimeter to burn 0.895 g of an unknown organic compound, observing a temperature increase of 2.18°C. What additional information is essential for calculating the enthalpy of combustion of the unknown compound?
Answer: The molecular formula of the unknown compound to determine the number of moles burned
To calculate enthalpy of combustion in kJ/mol, the student needs to know how many moles were burned. Heat released = 8.25 kJ°C⁻¹ × 2.18°C = 18.0 kJ from 0.895 g. Without the molecular formula (or molar mass), the number of moles cannot be determined, making the per-mole calculation impossible. The calorimeter constant accounts for all heat capacity effects, and formation enthalpies aren't needed for direct calorimetric measurement.
- A student uses Hess's law to calculate the enthalpy of formation of methane by combining the following thermochemical equations:
(1) C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5 kJ/mol
(2) H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = -285.8 kJ/mol
(3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH₃ = -890.3 kJ/mol
If the student accidentally uses ΔH₂ = -241.8 kJ/mol (formation of water vapor instead of liquid water), what error is introduced in the calculated ΔH°f for CH₄(g)?
Answer: The calculated value will be 88.0 kJ/mol too positive because two moles of water are involved
Using Hess's law: ΔH°f(CH₄) = ΔH₁ + 2ΔH₂ - ΔH₃. The difference between liquid and vapor formation enthalpies is 285.8 - 241.8 = 44.0 kJ/mol per mole of water. Since 2 moles of water are formed, the total error is 2 × 44.0 = 88.0 kJ/mol. Using the less negative value makes the calculated ΔH°f more positive (less negative) than the correct value.
- 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?
Answer: Assuming the specific heat capacity of the solution is the same as that of pure water.
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.
- 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)?
Answer: -188
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.
- 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.
Answer: 1100 J
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.
- 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?
Answer: Extrapolate the cooling curve back to the time of mixing (t = 60 s) to find a corrected maximum temperature.
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.
- Which of the following processes has a positive enthalpy change (ΔH > 0)?
Answer: The thermal decomposition of calcium carbonate.
A positive enthalpy change signifies an endothermic process, where energy is absorbed from the surroundings. Thermal decomposition reactions, such as CaCO₃(s) → CaO(s) + CO₂(g), require a continuous input of heat energy to break strong ionic and covalent bonds. Condensation, neutralization, and combustion are all exothermic processes (ΔH < 0).
- During a solution calorimetry experiment, a student dissolves 5.85 g of NaCl in 100.0 mL of water at 25.0°C. The final temperature is 24.1°C. The student then performs the same experiment using distilled water that was pre-cooled to 15.0°C. Assuming identical heat transfer conditions, what would be the most likely final temperature in the second experiment?
Answer: 14.1°C, because the enthalpy of solution is temperature-independent under these conditions
The enthalpy of solution for NaCl is essentially temperature-independent over this small range. The same amount of heat (q) is absorbed in both cases. Since q = mcΔT and m and c are approximately the same, ΔT ≈ 0.9°C in both experiments. Starting from 15.0°C, the final temperature would be 15.0°C - 0.9°C = 14.1°C. Option A incorrectly applies the absolute temperature change from the first experiment.
- Which statement correctly describes the system and surroundings for the combustion of methane in a Bunsen burner?
Answer: The reacting chemicals (methane and oxygen) are the system; the burner and the laboratory air are the surroundings.
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.
- Which statement correctly describes energy changes during an endothermic reaction?
Answer: The potential energy of the products is higher than that of the reactants, and heat is absorbed from the surroundings.
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.
- 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)
Answer: -42.1
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.
- 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?
Answer: 2 ΔT
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.
- 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)
Answer: +27.6
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.
