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1. Guiding question: What can be deduced from the temperature change that accompanies chemical or physical change?
When a chemical or physical change occurs, the accompanying temperature change tells us about the direction of energy transfer between the system and its surroundings. Temperature is a measure of the average kinetic energy of particles. If the surroundings become warmer, the system has released energy as heat: the reaction is exothermic and the enthalpy change (ΔH) is negative. If the surroundings become cooler, the system has absorbed heat: the reaction is endothermic and ΔH is positive.
For physical changes, such as melting or boiling, a fall in temperature of the surroundings indicates that energy is being absorbed to overcome intermolecular forces. Conversely, when gases condense or liquids freeze, energy is released into the surroundings.
Consequence: By carefully measuring temperature changes, we can deduce both the nature of the reaction (endo/exothermic) and calculate the enthalpy change using calorimetry.
2. Understand the difference between heat and temperature.
Temperature is a measure of the average kinetic energy of the particles in a substance. It tells us how fast the particles are moving, but not how many particles there are. Heat, in contrast, is the total energy content of a substance — it depends on both the temperature and the number of particles.
Why this matters: A pipette of boiling water and a beaker of boiling water have the same temperature, but the beaker contains much more heat because it has many more particles moving at that average speed.
Consequence: This distinction explains why temperature alone cannot describe energy changes in reactions — calorimetry must consider heat, not just temperature.
Structure 1.1—What is the relationship between temperature and kinetic energy of particles?
Temperature is directly proportional to the average kinetic energy of the particles in a system. As particles move faster, their average kinetic energy increases, and so does the temperature. Conversely, slower particle motion means lower kinetic energy and a lower temperature.
Consequence: This relationship is the reason why heating a system increases the likelihood of successful collisions in a chemical reaction — higher temperature means particles collide more frequently and with greater energy.
4. Understand the temperature change (decrease or increase) that accompanies endothermic and exothermic reactions.
All chemical reactions involve bond breaking (endothermic) and bond making (exothermic). Whether a reaction is overall endothermic or exothermic depends on which energy effect dominates.
In an exothermic reaction, the energy released when new bonds form is greater than the energy absorbed to break existing bonds. The system loses heat to the surroundings, the temperature of the surroundings increases, and ΔH is negative.
In an endothermic reaction, more energy is required to break bonds than is released when new ones form. The system absorbs heat from the surroundings, the temperature of the surroundings decreases, and ΔH is positive.
Why it matters: Exothermic reactions are thermodynamically favourable (more stable products), but may still require activation energy. Endothermic reactions can occur if driven by entropy or high temperatures.
5. Tool 1, Inquiry 2—What observations would you expect to make during an endothermic and an exothermic reaction?
In an exothermic reaction, we observe an increase in temperature of the surroundings. There may also be visible signs of heat release, such as flames in combustion, condensation of water vapour, or glowing.
In an endothermic reaction, the surroundings feel colder as heat is absorbed. Examples include the dissolving of certain salts (like ammonium nitrate) or photosynthesis, where light energy is absorbed.
Consequence: Observing temperature change and other visible energy effects provides qualitative evidence for classifying a reaction as endo- or exothermic.
6. Structure 2.2—Most combustion reactions are exothermic; how does the bonding in N₂ explain the fact that its combustion is endothermic?
Most combustion reactions are exothermic because the bonds in the reactants are relatively weak and the bonds formed in the products (such as C=O in CO₂ and O–H in H₂O) are much stronger.
However, in the case of nitrogen (N₂), the molecule has a very strong triple bond (N≡N) with a bond enthalpy of about 945 kJ mol⁻¹. Breaking this bond requires a very large energy input. When nitrogen combusts to form nitrogen oxides (NO, NO₂), the energy released by forming N–O bonds is not sufficient to compensate for the large energy absorbed in breaking the N≡N bond.
Consequence: The overall enthalpy change for nitrogen combustion is positive (endothermic). This explains why nitrogen gas is very stable and why high-energy conditions (like lightning or combustion engines) are needed to form nitrogen oxides.
7. Tool 1, Inquiry 1, 2, 3—How can the enthalpy change for combustion reactions, such as for alcohols or food, be investigated experimentally?
The enthalpy change of combustion can be measured using a simple calorimetry experiment. A known mass of the fuel (such as an alcohol) is burned in a spirit burner. The flame heats a known mass of water in a calorimeter (often a copper can), and the temperature rise of the water is measured.
How:
Record the initial mass of the fuel and the initial temperature of the water.
Burn the fuel until the water temperature rises by a measurable amount.
Record the final mass of the fuel and the final water temperature.
Use q=mcΔT to calculate the heat gained by the water, then divide by moles of fuel burned to get ΔHc.
Consequence: Experimental values are typically less exothermic than theoretical values, due to heat loss to the surroundings and incomplete combustion of the fuel.
8. Tool 1, Inquiry 3—Why do calorimetry experiments typically measure a smaller change in temperature than is expected from theoretical values?
In calorimetry experiments, several sources of error reduce the measured temperature change compared to the theoretical value.
Heat loss to the surroundings: not all the heat from the reaction is transferred to the water; some escapes into the air.
Heat absorbed by the apparatus: energy warms the calorimeter itself rather than only the water.
Incomplete combustion (for fuels): soot or carbon monoxide may form instead of full conversion to CO₂, releasing less energy.
Evaporation of fuel: mass readings may overestimate fuel actually burned.
Consequence: These factors mean the calculated ΔHc is less exothermic than the data-booklet value. Improved insulation, lids, and more efficient calorimeter designs (like a bomb calorimeter) reduce these errors.
Q1. Explain the difference between heat and temperature. Use the boiling water example.
Temperature is the measure of the average kinetic energy of particles in a substance. It reflects how fast the particles are moving, but not how many particles there are. Heat, on the other hand, is the total energy content of a system. It depends both on the speed of the particles (temperature) and on the number of particles present.
For example, a drop of boiling water and a beaker of boiling water both have the same temperature, 100 °C, because their particles move at the same average speed. However, the beaker contains many more particles, so its heat content is much greater.
Consequence: A drop of boiling water transfers little energy when spilled, but a beaker transfers a dangerous amount of energy, despite both being at the same temperature.
Q2. State and explain the law of conservation of energy in relation to chemical reactions.
The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed.
In chemical reactions, this means the total energy of the system and surroundings remains constant. Breaking bonds always requires energy (endothermic), while forming bonds releases energy (exothermic). The overall enthalpy change, ΔH, is the balance of these processes.
Consequence: Energy is never lost; it simply changes form or location. In combustion, for example, energy stored in chemical bonds is transferred as heat to the surroundings, raising their temperature, while the total energy remains constant.
Q3. Define open, closed, and isolated systems, with chemical examples.
A system is the part of the universe being studied, usually the reacting substances, while the surroundings are everything else.
Open system: Exchanges both matter and energy with the surroundings. Example: combustion of wood in air — gases and heat are released.
Closed system: Exchanges energy but not matter. Example: a sealed flask in which a chemical reaction occurs — heat can escape, but no reactants or products leave.
Isolated system: Exchanges neither matter nor energy. True isolated systems are theoretical, but a thermos flask approximates one.
Consequence: Calorimetry experiments aim to mimic a closed or isolated system so that energy changes can be measured accurately.
Q1. Explain the difference between endothermic and exothermic reactions in terms of:
(a) System and surroundings:
Endothermic: The system absorbs energy from the surroundings, so the surroundings cool down.
Exothermic: The system releases energy into the surroundings, so the surroundings heat up.
(b) Bond breaking and bond forming:
Endothermic: More energy is absorbed to break bonds than is released when new ones form.
Exothermic: More energy is released in forming bonds than is absorbed in breaking them.
(c) ΔH sign:
Endothermic: ΔH is positive.
Exothermic: ΔH is negative.
Consequence: Knowing whether ΔH is positive or negative allows chemists to predict how a reaction will affect temperature and whether it will need continuous energy input.
Q2. Describe the observable changes in temperature that would indicate whether a reaction is exothermic or endothermic.
In an exothermic reaction, the surroundings become warmer because heat is released into the environment; the thermometer reading increases. In an endothermic reaction, the surroundings become cooler because heat is absorbed from the environment; the thermometer reading decreases.
Consequence: Simple temperature observations can classify a reaction as endothermic or exothermic even without detailed calculations.
Q3. Explain why some reactions that are exothermic do not occur spontaneously.
Even though an exothermic reaction has a negative enthalpy change (ΔH), it still requires activation energy (Ea) to get started. Activation energy is the minimum energy required for reactant particles to break their bonds and form new ones.
If the reactants cannot overcome this barrier under normal conditions, the reaction will not proceed, despite being energetically favourable.
Example: The combustion of methane in oxygen is highly exothermic, but methane and oxygen can coexist indefinitely without reacting — until a spark provides the activation energy.
Consequence: This distinction highlights that thermodynamic feasibility (ΔH) is not the same as kinetic feasibility (Ea).
Q1. Sketch and annotate energy profile diagrams for exothermic and endothermic reactions.
For an exothermic reaction: draw products lower than reactants; label ΔH as negative; show the activation energy hump with the transition state at the top.
For an endothermic reaction: draw products higher than reactants; label ΔH as positive; show a higher activation energy hump.
Consequence: Energy profile diagrams clearly display the difference between bond-breaking and bond-forming processes and allow comparison of energy requirements.
Q2. Explain why the activation energy is essential for both endothermic and exothermic reactions.
Activation energy is essential because it provides the minimum energy required to break bonds in the reactants. Even in exothermic reactions, some energy input is needed to weaken existing bonds before new, more stable bonds can form.
Consequence: Without sufficient activation energy, reactant particles will collide but not react. This explains why catalysts, which lower activation energy, speed up both endothermic and exothermic reactions.
Q3. Compare the activation energies of exothermic and endothermic reactions.
Exothermic reactions: Generally have lower activation energies because the energy released in bond formation outweighs the energy absorbed in bond breaking.
Endothermic reactions: Require higher activation energies because more energy must be absorbed to break bonds than is released when new bonds form.
Consequence: Endothermic reactions are less likely to occur without continuous energy input, whereas exothermic reactions often proceed once initiated.
Apply ΔH = –Q/n to determine ΔH when 0.25 mol of reactant causes this temperature change.
Q3. Explain the difference between Q = mcΔT and the standard enthalpy change ΔH°.
Q = mcΔT gives the heat transferred in a specific experiment to a specific mass of material, under the experimental conditions.
ΔH° is the standard enthalpy change per mole of reaction, measured under standard conditions (298 K, 1 atm, 1 mol dm⁻³ solutions, substances in standard states).
Consequence: Q is an experimental measurement, while ΔH° is a standardized value that allows comparisons across different reactions.
Q1. Define the following standard enthalpy changes (conditions: 298 K, 100 kPa, 1 mol dm⁻³, substances in standard states):
Reaction (ΔHr°): Enthalpy change when a reaction occurs according to the balanced equation, under standard conditions.
Formation (ΔHf°): Enthalpy change when 1 mol of a compound is formed from its elements in their standard states.
Combustion (ΔHc°): Enthalpy change when 1 mol of a substance is completely burned in oxygen, under standard conditions.
Neutralisation (ΔHneut°): Enthalpy change when 1 mol of water is formed by the reaction of an acid with a base, under standard conditions
Q2. Explain why the enthalpy change of combustion and neutralisation are always exothermic.
Combustion: Breaking C–H and C–C bonds in fuels requires less energy than is released when strong C=O and O–H bonds form in CO₂ and H₂O. The large bond enthalpies of these products ensure ΔHc° is always negative.
Neutralisation: Neutralisation forms strong O–H bonds in water from H⁺ and OH⁻ ions. This process always releases energy, so ΔHneut° is always negative.
Consequence: Both processes release energy to the surroundings, making them consistently exothermic.
Q1. Describe how a simple calorimeter is set up to measure the enthalpy change of a reaction in solution.
A simple calorimeter uses a polystyrene cup with a lid to reduce heat loss. A thermometer measures the temperature change of the solution. Known quantities of reactants are mixed in the cup, and the temperature is recorded before and after the reaction.
Q2. State the assumptions made in calorimetry experiments in solution, and explain their impact.
The solution has the same density and specific heat capacity as water (1 g cm⁻³ and 4.18 J g⁻¹ K⁻¹).
All the heat released or absorbed by the reaction is transferred to the solution, with none lost to the surroundings or the calorimeter.
Impact: If these assumptions are false, calculated enthalpy changes may be less exothermic than the true values.
Q3. Describe the experimental setup for measuring enthalpy of combustion. Identify two errors and solutions.
A spirit burner containing the fuel is weighed, lit, and used to heat a known mass of water in a calorimeter (often a copper can). The temperature change of the water is recorded, and the mass of fuel burned is measured.
Error 1: Heat loss to the surroundings → reduce by insulating the calorimeter and using a lid.
Error 2: Incomplete combustion of the fuel → reduce by ensuring sufficient oxygen supply and using a draught shield.
Q4. Explain why the calculated enthalpy of combustion is often less exothermic than accepted values.
Because of heat loss to the surroundings, incomplete combustion, and heat absorbed by the calorimeter itself, the measured temperature rise is smaller than expected.
Consequence: Experimental ΔHc values are less negative than data-booklet values
Q1. Explain why heat loss to the surroundings is more significant in slow/weak exothermic reactions.
In slow or weak reactions, energy is released gradually and in small amounts, so a greater proportion escapes to the surroundings before being measured. In fast or vigorous reactions, heat is released quickly and in large amounts, so the relative loss is less significant.
Q2. Describe how to construct a temperature correction graph for a displacement reaction (Zn + CuSO₄).
Record the temperature of the solution at regular intervals before and after adding zinc. Plot temperature vs. time. Extrapolate the cooling curve back to the time of mixing to find the true maximum temperature.
Q3. Explain why extrapolating the cooling curve provides a more accurate enthalpy change.
Extrapolation corrects for heat loss to the surroundings during the experiment. The maximum temperature predicted from the graph represents the temperature change if no heat had escaped.
Consequence: This gives a more reliable value for ΔH of the reaction.
Q1. Discuss how enthalpy changes in solution are measured with calorimetry. Explain the role of Q = mcΔT.
In calorimetry, known amounts of reactants are mixed in a polystyrene cup and the temperature change is measured. The heat transferred is calculated using Q = mcΔT, where m is the mass of the solution, c is the specific heat capacity, and ΔT is the temperature change. Dividing Q by the number of moles of reactant gives ΔH.
Consequence: This allows enthalpy changes of neutralisation, displacement, or dissolution to be determined experimentally.
Q2. Describe how calorimetry experiments link conservation of energy, heat transfer, and enthalpy changes.
Calorimetry relies on the conservation of energy: energy lost by the reaction is gained by the solution. By measuring the heat transfer as a temperature change, we can calculate the enthalpy change of the reaction using Q = mcΔT and ΔH = –Q/n.
Example: In the neutralisation of HCl with NaOH, the energy released in forming O–H bonds in water is transferred as heat to the solution, raising its temperature.
Consequence: This demonstrates the fundamental link between thermodynamics (energy conservation) and practical experimental measurements.
Q1. Define enthalpy (H) and explain why only enthalpy changes (ΔH) can be measured experimentally, not absolute enthalpy values.
Enthalpy (H) is the total energy content of a system, including both the internal energy of the particles and the energy associated with pressure and volume. It represents the stored chemical energy of a substance.
Why only ΔH: Absolute enthalpy values cannot be measured because we cannot determine the total energy of all bonds, electrons, and interactions in a system. However, when a reaction occurs, we can measure the difference in energy between reactants and products, expressed as the enthalpy change ΔH.
How: In practice, enthalpy changes are determined by measuring temperature changes in calorimetry experiments and relating them to heat transfer.
Consequence: This limitation means that data tables record only enthalpy changes (e.g., enthalpy of combustion, neutralisation, or formation) rather than absolute enthalpy values.
Q2. Explain why the enthalpy change of neutralisation has a very similar value (≈ –57 kJ mol⁻¹) for many strong acid–strong base reactions.
The enthalpy change of neutralisation is the energy released when one mole of water is formed from the reaction between H⁺ ions from an acid and OH⁻ ions from a base under standard conditions.
Why values are similar: Strong acids and strong bases are fully dissociated in aqueous solution. This means the reaction in all cases is effectively:
H⁺ (aq)+OH⁻ (aq)→H₂O (l)
The enthalpy change depends only on the formation of the O–H bond in water, which is the same in all such reactions.
Consequence: The value is consistently around –57 kJ mol⁻¹ for strong acid–strong base combinations. For weak acids or bases, the value differs because additional energy is required to ionise the weak species.
Q3. Describe two key differences between measuring the enthalpy change of combustion and the enthalpy change of neutralisation, and explain why combustion experiments typically give less accurate results.
WHAT is different:
In neutralisation, the reaction takes place in aqueous solution and heat is transferred directly to the solution.
In combustion, the reaction involves burning a fuel in air and transferring heat to water in a calorimeter.
WHY combustion is less accurate:
Heat is easily lost to the surroundings in combustion experiments, while solution reactions can be better insulated.
Combustion often suffers from incomplete burning of the fuel, so less heat is released than expected.
Consequence: Measured enthalpy of combustion values are usually less exothermic than literature values, whereas neutralisation experiments give more reliable results.
Q4. Explain why calorimetry experiments assume that the specific heat capacity and density of solutions are the same as pure water, and discuss the limitations of this assumption.
In calorimetry, the heat absorbed by the solution is calculated using Q = mcΔT, where c is the specific heat capacity and m depends on the density.
Why assumption is made: To simplify calculations, it is assumed that aqueous solutions behave like pure water, with a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹. This is reasonable because most dilute aqueous solutions are close to these values.
Limitation: Some solutions differ significantly from water in density or heat capacity. This leads to systematic errors in calculated enthalpy changes.
Consequence: The assumption improves practicality but reduces accuracy. Advanced calorimetry uses measured values of c and density for greater precision.
Q5. In an enthalpy of combustion experiment, identify two sources of systematic error and two sources of random error. Suggest improvements to minimise each.
Systematic errors:
Heat loss to surroundings → solution: insulate the calorimeter and use a lid.
Incomplete combustion of fuel → solution: supply excess oxygen or use a bomb calorimeter.
Random errors:
Fluctuations in flame size or position → solution: shield flame from draughts, ensure consistent setup.
Temperature measurement uncertainty → solution: use a digital thermometer with higher resolution.
Consequence: Reducing systematic errors improves accuracy, while reducing random errors improves precision. Both are essential for reliable experimental results.
Q6. Explain why bond enthalpies can be used to estimate enthalpy changes, but why such estimates may differ from experimental values.
Bond enthalpy is the average energy required to break one mole of a particular bond in the gas phase. By comparing the energy needed to break bonds in reactants with the energy released in forming bonds in products, we can estimate ΔH for a reaction:
ΔH≈Σ(bonds broken)−Σ(bonds formed)
Why values differ:
Bond enthalpies are averages taken from many compounds, so they may not match the exact bond energies in the specific molecules of the reaction.
Bond enthalpy calculations assume all species are in the gas phase, while many reactions involve liquids or solids.
Experimental ΔH values also include effects like solvation or lattice energies, which bond enthalpy estimates ignore.
Consequence: Bond enthalpies provide a useful approximation for enthalpy changes, but accurate values require experimental determination or Hess’s law cycles.
Q7. Compare the advantages and disadvantages of calorimetry experiments in solution versus bomb calorimetry for combustion enthalpy determinations.
Calorimetry in solution:
Advantages: Simple, inexpensive, quick; suitable for school laboratories.
Disadvantages: Significant heat loss to surroundings; incomplete combustion; fuel evaporation; lower accuracy.
Bomb calorimetry:
Advantages: Fuel is burned in a strong, sealed container (bomb) in excess oxygen, with excellent insulation. Nearly all heat is transferred to the water, giving highly accurate values.
Disadvantages: Expensive equipment, requires specialist setup, not always accessible.
Consequence: For precise measurement of combustion enthalpies (e.g., in industry), bomb calorimetry is used. For educational or approximate purposes, simple calorimetry in solution is sufficient.