Energy of Reactions – Thermochemistry Vocabulary

Energy in Reactions

• Syllabus overview: chemical reactions and phase changes involve enthalpy changes, observable as changes in the temperature of the surroundings and/or the emission of light.
• Endothermic and exothermic reactions can be explained in terms of the Law of Conservation of Energy and the breaking of existing bonds & forming of new bonds; heat energy released or absorbed by the system to or from the surroundings can be represented in thermochemical equations.
• Enthalpy changes (ΔH) describe the heat content change of a system at constant pressure during a reaction.
• Practical observation: temperature change of surroundings and light emission are common indicators of enthalpy change.

Exothermic vs Endothermic Reactions

• Exothermic reactions: reactions that lose heat to the surroundings (e.g., combustion).
• Endothermic reactions: reactions that gain heat from the surroundings (absorb heat).
• These concepts align with energy conservation: heat flows between system and surroundings depending on enthalpy changes.
• Core idea: heat energy transferred corresponds to changes in enthalpy of the reacting system.
• Activate Prior Knowledge (from transcript):
  • A) What is energy? Energy is the ability to do work; measured in Joules.
  • The total amount of energy in a system is constant; if energy decreases in the system, it appears as increased energy in the surroundings.

Energy and Enthalpy: Foundational Concepts

• Energy definition: Energy is the ability to do work; measured in joules (J).
• Conservation principle: the total energy present in a closed system is constant.
• If the energy in the system falls, there is a corresponding rise in the energy of the surroundings.
• System vs surroundings: the reaction mixture is the system; the test tube and everything around it constitutes the surroundings. (CFU2 from Page 3)
• Enthalpy is a state function representing the total energy content of a system, including:
  • Chemical potential energy stored in bonds
  • Kinetic energy due to particle motion
• As temperature increases, kinetic energy increases.
• Quick definitions:
  • System: the part of the universe being studied (e.g., the reaction mixture).
  • Surroundings: everything else around the system (e.g., the container, atmosphere).

Enthalpy Change in Chemical Reactions

• During a chemical reaction, bonds are broken and new bonds are formed, changing stored chemical potential energy.
• Energy must be conserved; any change in chemical potential energy is balanced by an opposite change in another form of energy (usually kinetic energy/heat).
• This exchange manifests as a temperature change and/or heat flow into or out of the reaction system.
• Definition: \Delta H = H(\text{products}) - H(\text{reactants}) where H = enthalpy and ΔH = change in enthalpy.
• Reminder: Heat flows spontaneously from hotter regions to colder regions.

Sign Convention for ΔH

• Exothermic reactions: ΔH is negative (heat released to surroundings).
• Endothermic reactions: ΔH is positive (heat absorbed from surroundings).
• Graphical and symbolic representations:
  • Exothermic example (enthalpy decreases):
    \Delta H < 0\,&\; \text{(heat released)}
  • Endothermic example (enthalpy increases):
    \Delta H > 0\,&\; \text{(heat absorbed)}

Representations of Energy Changes

• Energy profile diagrams can visualize ΔH as the difference in energy between products and reactants.
• Chemical equations can also represent energy changes by including ΔH.
• Practical notation:
  • Exothermic: Reactants → Products, \Delta H = -\text{(magnitude)}\;\text{kJ}
  • Endothermic: Reactants → Products, \Delta H = +\text{(magnitude)}\;\text{kJ}

Worked Exothermic Reactions (examples from transcript)

• Example 1 (gas-phase water formation):
  • Reaction: 2\mathrm{H}2(\mathrm{g}) + \mathrm{O}2(\mathrm{g}) \rightarrow 2\mathrm{H}_2\mathrm{O}(\mathrm{g})
  • Enthalpy change: \Delta H = -572\ \mathrm{kJ}
  • Interpretation: heat released to surroundings (exothermic).
• Example 2 (condensed notation for heat release): the same reaction can be described as heat released of magnitude +572\ \mathrm{kJ}, with ΔH negative.
• Example 3 (water formation in liquid product):
  • Reaction: \mathrm{H}2(\mathrm{g}) + \mathrm{O}2(\mathrm{g}) \rightarrow 2\mathrm{H}_2\mathrm{O}(\mathrm{l})
  • Enthalpy change: \Delta H = -574\ \mathrm{kJ}
• Example 4 (exothermic enthalpy from oxidation of acetylene in a welding context):
  • Steps (as in transcript):
  1. Type: Exothermic – causes temperature to rise; heat released to surroundings.
  2. Balanced equation: 2\mathrm{C}2\mathrm{H}2(\mathrm{g}) + 5\mathrm{O}2(\mathrm{g}) \rightarrow 2\mathrm{H}2\mathrm{O}(\mathrm{g}) + 4\mathrm{CO}_2(\mathrm{g})
  3. Include heat: 2\mathrm{C}2\mathrm{H}2(\mathrm{g}) + 5\mathrm{O}2(\mathrm{g}) \rightarrow 2\mathrm{H}2\mathrm{O}(\mathrm{g}) + 4\mathrm{CO}_2(\mathrm{g}) + 2.510 \times 10^6\ \mathrm{J}
  4. Equation with ΔH: 2\mathrm{C}2\mathrm{H}2(\mathrm{g}) + 5\mathrm{O}2(\mathrm{g}) \rightarrow 2\mathrm{H}2\mathrm{O}(\mathrm{g}) + 4\mathrm{CO}_2(\mathrm{g})\quad \Delta H = -2.510 \times 10^6\ \mathrm{J}
  5. Enthalpy diagram: labeled axes with reactants, products, and ΔH (note: diagram drawing not included here).
• Example 5 (cold-pack dissolution):
  • Reaction: \mathrm{NH}4\mathrm{Cl}(s) \rightarrow \mathrm{NH}4^+(aq) + \mathrm{Cl}^-(aq)
  • Heat absorbed: \Delta H = +1.4\ \mathrm{kJ\ per\ mole\ NH_4Cl}
  • This is endothermic; enthalpy is positive; enthalpy diagram would show products at higher enthalpy than reactants.
• Example 6 (acid-base neutralization – hydrolysis of salt formation):
  • Reaction: \mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H_2O}(l)
  • Heat change: \Delta H = -8.360\ \mathrm{kJ} per mole of reaction (i.e., per mole of HCl and NaOH reacting in the stated quantities).
  • Endothermic or exothermic designation: Exothermic (heat released).

Practical Calculations of Heat (ΔH and moles)

• General steps for calculating heat produced by a reaction:
  1) Determine the moles of the reactant involved.
  2) Multiply the heat of reaction by the moles (n × ΔH per mole base).
  3) If required, determine how much of the substance is present first to find the limiting amount.
• Cetane cracking example (from transcript):
  • Reaction: \mathrm{C}{16}\mathrm{H}{34}(\mathrm{l}) + 283.11\ \mathrm{kJ} \rightarrow \mathrm{C}7\mathrm{H}{16}(\mathrm{l}) + 3\mathrm{C}2\mathrm{H}4(\mathrm{g}) + \mathrm{C}3\mathrm{H}6(\mathrm{g})
  • Mass of crude oil sample: 2.89 tonnes; cetane mass percentage: 7.22% by mass.
  • Steps shown in transcript (paraphrased):
  • m(cetane) = (2.89\ \text{t}) \times (7.22\% ) = 0.208658\ \text{t} = 208,658\ g
  • n(cetane) = \dfrac{m}{M} = \dfrac{208,658}{226.432} \approx 2.62\times 10^2\ \text{mol}
  • Energy produced: Energy = n × 283.11 = 7.4223\times 10^4\ \text{kJ}
• Another example: Methane combustion
  • Reaction: \mathrm{CH}4(\mathrm{g}) + 2\mathrm{O}2(\mathrm{g}) \rightarrow \mathrm{CO}2(\mathrm{g}) + 2\mathrm{H}2\mathrm{O}(\mathrm{g})
  • Given: standard enthalpy of combustion for methane is \Delta H^\circ = -890\ \mathrm{kJ\,mol^{-1}}
  • Mass of methane sample: 3.50\ \text{kg}
  • Molar mass: M(\mathrm{CH}_4) = 16.042\ \mathrm{g\,mol^{-1}}
  • Calculate moles: n = \dfrac{m}{M} = \dfrac{3500\ \mathrm{g}}{16.042\ \mathrm{g\,mol^{-1}}} \approx 218.18\ \mathrm{mol}
  • Heat produced: \text{Energy} = n \times |\Delta H^\circ| = 218.18 \times 890 \approx 1.94 \times 10^5\ \mathrm{kJ}
  • Note: sign remains negative for exothermic reactions (heat released).

Guided Practice: Photosynthesis (Endothermic)

• CFU1 Type of reaction: Endothermic – absorbs heat from surroundings.
• CFU2 Balanced equation with heat included:
  • Step 2: 6\mathrm{CO}2(\mathrm{g}) + 6\mathrm{H}2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{C}6\mathrm{H}{12}\mathrm{O}6(\mathrm{s}) + 6\mathrm{O}2(\mathrm{aq})
  • Step 3 (including heat): 6\mathrm{CO}2(\mathrm{g}) + 6\mathrm{H}2\mathrm{O}(\mathrm{g}) + 2.4\ \mathrm{kJ\,mol^{-1}} \rightarrow \mathrm{C}6\mathrm{H}{12}\mathrm{O}6(\mathrm{s}) + 6\mathrm{O}2(\mathrm{aq})
  • Step 4 (ΔH included): 6\mathrm{CO}2(\mathrm{g}) + 6\mathrm{H}2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{C}6\mathrm{H}{12}\mathrm{O}6(\mathrm{s}) + 6\mathrm{O}2(\mathrm{aq})\quad \Delta H = +14.4\ \mathrm{kJ}\ (\text{since } 2.4\ \text{kJ} \times 6)
• Step 5: Enthalpy diagram labeling (Hreactants vs Hproducts).
• Final equation (illustrative): 6\mathrm{CO}2(\mathrm{g}) + 6\mathrm{H}2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{C}6\mathrm{H}{12}\mathrm{O}6(\mathrm{s}) + 6\mathrm{O}2(\mathrm{aq}) with \Delta H = +14.4\ \mathrm{kJ} per mole of reaction as stated.

Guided Practice: Neutralization and Salt Formation (Exothermic)

• Mixing HCl and NaOH yields table salt (NaCl) and water, with heat release of +8,360\ \mathrm{J} (i.e., -8,360 J for ΔH per reaction conventions).
• Steps (paraphrased):
  • Step 1: Exothermic – heat released to surroundings.
  • Step 2: Balanced equation: \mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H}_2\mathrm{O}(l)
  • Step 3: Include heat of reaction: \mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H}_2\mathrm{O}(l) + 8360\ \mathrm{J}
  • Step 4: ΔH expression: \mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H}_2\mathrm{O}(l)\quad \Delta H = -8360\ \mathrm{J}
  • Step 5: Enthalpy diagram sketch (Hreactants vs Hproducts).

Guided Practice: Methane and Hydrocarbon Combustion (Exothermic)

• Complete the following reaction: Methane combustion with heat given as 1.0 MJ released per reaction.
  • Step 1: Exothermic – heat released to surroundings.
  • Step 2: Balanced equation: \mathrm{CH}4(\mathrm{g}) + 2\mathrm{O}2(\mathrm{g}) \rightarrow \mathrm{CO}2(\mathrm{g}) + 2\mathrm{H}2\mathrm{O}(\mathrm{g})
  • Step 3: Include heat of reaction: \mathrm{CH}4(\mathrm{g}) + 2\mathrm{O}2(\mathrm{g}) \rightarrow \mathrm{CO}2(\mathrm{g}) + 2\mathrm{H}2\mathrm{O}(\mathrm{g}) + 1.0\times 10^6\ \mathrm{J}
  • Step 4: ΔH: \Delta H = -1.0\times 10^6\ \mathrm{J} (negative for exothermic).
  • Step 5: Enthalpy diagram labeling.

Guided Practice: Calcium Carbonate Decomposition (Endothermic)

• Thermal decomposition of CaCO3: CaCO3(s) → CaO(s) + CO2(g)
• Given heat of reaction (ΔH) = +776 kJ per mole CaCO3.
• Steps:
  • Step 1: Endothermic – temperature falls; heat absorbed from surroundings.
  • Step 2: Balanced equation: \mathrm{CaCO}3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}2(\mathrm{g})
  • Step 3: Include heat: \mathrm{CaCO}3(\mathrm{s}) + 776\ \mathrm{kJ} \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}2(\mathrm{g})
  • Step 4: ΔH: \Delta H = +776\ \mathrm{kJ} per mole of CaCO3.
  • Step 5: Enthalpy diagram sketch.

Guided Practice: Enthalpy Calculations with Crude Oil Cracking

• Reaction: \mathrm{C}{16}\mathrm{H}{34}(\mathrm{l}) + 283.11\ \mathrm{kJ} \rightarrow \mathrm{C}7\mathrm{H}{16}(\mathrm{l}) + 3\mathrm{C}2\mathrm{H}4(\mathrm{g}) + \mathrm{C}3\mathrm{H}6(\mathrm{g})
• Practice scenario: A 8.43 tonne sample contains 5.99% cetane by mass. Calculate the heat produced.
• Steps (as in transcript):
  • m(cetane) = (8.43\ \text{t}) \times (5.99\%) = ?\ \text{kg} \n - n(cetane) = m/M; M = 226.432\ \mathrm{g\,mol^{-1}}
  • Energy produced = n × 283.11\ \mathrm{kJ}
• Note: The transcript lists placeholders to fill in with numbers; the approach remains the same: determine moles of cetane then multiply by the per-mole heat of reaction.

Guided Practice: Calculating Heat for Cetane Cracking (8.9 t sample)

• Given: 8.9 tonne sample with cetane content 10.39% by mass.
• Steps (as in transcript):
  • m(cetane) = (8.9\ \text{t}) \times (10.39\%) = \text{(compute)}\n - n(cetane) = m/M = \text{(compute)}\n - Energy = n × 283.11\ \mathrm{kJ} = \text{(compute)}\n

Guided Practice: Heat Produced for Hydrocarbon Cracking (Summary)

• The