Unit I – Energy & Thermochemistry (full study notes)

Energy: Basic Principles

• Energy = capacity to do work or supply heat.
• Work
  • Mechanical energy that causes a mass to move.
• Heat
  • Energy that causes the temperature of an object to rise.
• Two common energy units
  • Joule (J)
  – 1\,\text{J}=1\,\text{kg}\,\text{m}^2\text{/s}^2 (derived from kinetic energy of a 2\,\text{kg} mass moving at 1\,\text{m\,s}^{-1}).
  • Calorie (cal)
  – Heat that raises 1\,\text{g} H$_2$O by 1\,^{\circ}\text{C}.
  • Conversion: 1\,\text{cal}=4.184\,\text{J}.

Thermodynamics & Thermochemistry

• Thermodynamics = study of energy transformations.
• Thermochemistry = branch that tracks heat flow during chemical reactions/physical changes.

Law of Conservation of Energy

• Energy cannot be created nor destroyed.
• It can be converted from one form to another.

System vs. Surroundings

• System = portion of the universe chosen for study (reaction mixture, piston, etc.).
• Surroundings = everything else.

First Law of Thermodynamics

• Applies conservation of energy to the system.
• Internal Energy change: \Delta U \;\text{(or) }\;\Delta E
  • Sum of all P.E. + K.E. inside the system.
  • \boxed{\Delta U = q + w}
  – q = heat into (+) or out of (–) system.
  – w = work done on (+) or by (–) the system.

Sign conventions (chemistry):

• q>0 : heat absorbed by system (endothermic).
• q<0 : heat released by system (exothermic).
• w>0 : work done on system by surroundings (e.g.
  compression).
• w<0 : work done by system on surroundings (e.g.
  expansion).

Worked example

• System absorbs +140\,\text{J} heat; performs -85\,\text{J} work (work on surroundings).
• \Delta U = (+140\,\text{J}) + (-85\,\text{J}) = +55\,\text{J}.

Thermal Equilibrium & Direction of Heat Flow

• Heat spontaneously flows from higher T object to lower T object until T{\text{high}} = T{\text{low}}.

Endothermic vs. Exothermic

• Endothermic
  • Heat transferred surroundings → system.
  • Energy of system ↑ ; surroundings ↓.
• Exothermic
  • Heat transferred system → surroundings.
  • Energy of system ↓ ; surroundings ↑.

Heat Capacity Concepts

• Heat required depends on
  • Mass (m)
  • Temperature change (\Delta T)
  • Identity / composition.

• Heat Capacity (general)
  • Heat to raise object by 1\,\text{K} (or 1\,^{\circ}\text{C}).
  • Symbol C (or K_c).

• Specific Heat Capacity (Cp or Cv)
  • Heat to raise 1 g by 1\,\text{K}.

• Molar Heat Capacity (C_{\text{molar}})
  • Heat to raise 1 mol by 1\,\text{K}.

• Trend: Larger C ⇒ more heat needed for same \Delta T.

Representative values (25 °C)

• Metals (≈ 25\,\text{J mol}^{-1}\text{K}^{-1}):
  • Al 0.897\,\text{J g}^{-1}\text{K}^{-1} (24.2 J mol$^{-1}$K$^{-1}$)
  • Fe 0.449 (25.1)
  • Cu 0.385 (24.5)
  • Au 0.129 (25.4)
• Water phases
  • liquid 4.184 (75.4)
  • ice 2.06 (37.1)
  • steam 1.86 (33.6)
• Ethylene glycol 2.39\,\text{J g}^{-1}\text{K}^{-1} (14.8 J mol$^{-1}$K$^{-1}$)
• Wood ≈ 1.8; Glass ≈ 0.8.

Fundamental Heat Equation

\boxed{q = m\, C_p \, \Delta T}

• m = mass.
• \Delta T = T{\text{final}} - T{\text{initial}}.
• Sign: q>0 when heat enters the object, q<0 when heat leaves.

Quick qualitative comparison

• 100 g water (C$p$ = 4.184) vs 100 g olive oil (C$p$ ≈ 2.0): same heat → oil’s temperature rises twice as much.

Quantitative examples

1. Heating 250 g H$_2$O from 22.0 °C → 98.0 °C.
   – \Delta T = 76.0\,^{\circ}\text{C}
   – q = (250\,\text{g})(4.184)(76.0) = 7.95 \times 10^4\,\text{J}.
2. Solar-rock storage (50.0 kg rock, C$_p$ = 0.082 J g$^{-1}$K$^{-1}$, ΔT = 12 °C)
   – Heat absorbed q = (5.00\times10^4\,\text{g})(0.082)(12) = 4.9\times10^4\,\text{J}.
   – Reverse: emitting 450 kJ ⇒ \Delta T = \frac{450\times10^3}{(5.00\times10^4)(0.082)} \approx 110\,^{\circ}\text{C} drop.

Calorimetry—Heat Transfer Between Two Substances

• Closed system (metal + water): q{\text{metal}} + q{\text{water}} = 0.
• Signs: q{\text{metal}} negative (cools), q{\text{water}} positive (warms).

Metal-in-water example

• 55.0 g Fe at 99.8 °C into 225 g H$2$O at 21.0 °C; equilibrium 23.1 °C. – q{\text{water}} = (225)(4.184)(2.1) = 1.98\times10^3\,\text{J}.
  – q{\text{metal}} = -1.98\times10^3\,\text{J}. – Cp(\text{metal}) = \frac{|q|}{m\,\Delta T} = \frac{1.98\times10^3}{55.0\times(99.8-23.1)} \approx 0.45\,\text{J g}^{-1}\text{K}^{-1} → matches iron.

Constant-Pressure Calorimetry (Coffee-cup)

• The calorimeter insulates: q{\text{rxn}} + q{\text{solution}} (+ q{\text{cal}})=0. • If calorimeter heat capacity Kc given: q{\text{cal}} = Kc\,\Delta T.

Worked acid/base example

• 50.0 mL 1.0 M HCl + 50.0 mL 1.0 M NaOH.
  – Total mass ≈ 100 g, Cp\text{(soln)} = 4.184. – \Delta T = 300.5\text{K}-294.0\text{K}=6.5\text{K}. – q{\text{soln}}=(100)(4.184)(6.5)=2.72\times10^3\,\text{J}.
  – q_{\text{rxn}} = -2.72\times10^3\,\text{J}.
  – Moles HCl reacted =0.0500; \Delta H = \frac{-2.72\,\text{kJ}}{0.0500\,\text{mol}} = -54\,\text{kJ mol}^{-1}.

Unknown-metal example with calorimeter

• 451 g metal @ 99.9 °C into 100 mL H$2$O (22.3 °C → 26.8 °C), Kc = 10\,\text{cal K}^{-1}.
  – Convert: Cp(\text{H}2\text{O})=1.0\,\text{cal g}^{-1}\text{K}^{-1}.
  – Solve q{\text{metal}} + q{\text{water}} + q{\text{cal}} = 0 for Cp(\text{metal}) (procedure mirrors earlier example).

Constant-Volume Calorimetry (Bomb)

• Relationship: q{\text{bomb}} + q{\text{water}} + q_{\text{rxn}} = 0 (volume fixed ⇒ q = \Delta U not \Delta H).

Octane combustion example

• 1.00 g C$8$H${18}$, m{\text{water}}=1.20\times10^3\,\text{g}, Kb=837\,\text{J K}^{-1}, \Delta T = 8.2\,\text{K}.
  • q{\text{bomb}} = 837\times8.2 = 6.87\times10^3\,\text{J}. • q{\text{water}} = 1.20\times10^3 \times 4.184 \times 8.2 = 4.12\times10^4\,\text{J}.
  • q{\text{rxn}} = -(4.12\times10^4 + 6.87\times10^3) = -4.80\times10^4\,\text{J}. • Heat of combustion: -4.80\times10^4\,\text{J g}^{-1} or -5.48\times10^6\,\text{J mol}^{-1} (M$r$ = 114.26 g mol$^{-1}$).

Enthalpy (H)

• State function for constant-pressure heat flow.
• Definition: H = U + PV.
• Differential form (only P–V work):
  \Delta H = \Delta U + P\,\Delta V = (qp + w) - w = qp.
  • Thus, at constant P, heat exchanged = \Delta H.
• Signs
  • \Delta H

Thermochemical Equations

• Chemical equation + associated \Delta H under specified conditions.
  Example: \text{H}2(g) + \tfrac12\text{O}2(g) \rightarrow \text{H}2\text{O}(l) \quad \Delta H{\text{rxn}} = -241.8\,\text{kJ}.
• Units often written kJ per equation (kJ mol$^{-1}$rxn).

Stoichiometric use

• Mg combustion: 2\,\text{Mg}(s)+\text{O}_2(g) \rightarrow 2\,\text{MgO}(s) \; \Delta H=-1204\,\text{kJ}.
  • 2.4 g Mg ⇒ heat q = \frac{2.4\text{ g}}{24.31\text{ g mol}^{-1}} \times \frac{-1204\,\text{kJ}}{2\,\text{mol}} \approx -59\,\text{kJ}.
  • Given \Delta H=-96.0\,\text{kJ}, produced MgO mass:
  \text{mol MgO}=\frac{|-96.0|}{1204/2}=0.159\,\text{mol} ⇒ 6.41\,\text{g}.

Hess’s Law

• Because H is a state function, \Delta H is path-independent.
  • If reaction proceeds via series, \Delta H{\text{overall}} = \sum \Delta H{\text{steps}}.

Manipulating thermochemical equations

1. Multiply entire equation by n ⇒ multiply \Delta H by n.
2. Reverse equation ⇒ change sign of \Delta H.
3. Add equations ⇒ add their \Delta H values.

Example (Making acetylene)

• Target: 2\,\text{C}(s)+\text{H}2(g) \rightarrow \text{C}2\text{H}_2(g).
• Using
  • \text{C}(s)+\text{O}2 \rightarrow \text{CO}2 \quad \Delta H=-393.5
  • \text{H}2+\tfrac12\text{O}2 \rightarrow \text{H}2\text{O} \quad \Delta H=-285.8 • \text{C}2\text{H}2+\tfrac52\text{O}2 \rightarrow 2\text{CO}2+\text{H}2\text{O} \quad \Delta H=-1299.6
• After scaling/reversing, \Delta H_{\text{target}} = -226.8\,\text{kJ}.

Standard Enthalpy of Formation (\Delta H_f^{\circ})

• \Delta H for creating 1 mol of compound from elements in their standard states (1 atm, 25 °C).
• Elements in standard state have \Delta Hf^{\circ}=0 (e.g. \text{C}(s,\text{graphite}),\,\text{O}2(g),\,\text{N}_2(g)).
• Common diatomic elements: \text{H}2,\text{N}2,\text{O}2,\text{F}2,\text{Cl}2,\text{Br}2,\text{I}_2.

Reaction enthalpy from formation data

\boxed{\Delta H{\text{rxn}}^{\circ}=\sum n\, \Delta Hf^{\circ}(\text{products}) - \sum n\, \Delta H_f^{\circ}(\text{reactants})}

Ethanol combustion example

• Reaction: \text{CH}3\text{CH}2\text{OH}(l)+3\,\text{O}2(g)\rightarrow2\,\text{CO}2(g)+3\,\text{H}_2\text{O}(l)
• Data (kJ mol$^{-1}$): Ethanol –277.7; CO$2$ –393.5; H$2$O(l) –285.8.
• \Delta H_{\text{rxn}}^{\circ} = [2(-393.5)+3(-285.8)] - [(-277.7)+0] = -1366.7\,\text{kJ} per mol ethanol.
• If 2.00\times10^5\,\text{J} released:
  • Moles ethanol = \frac{2.00\times10^5\,\text{J}}{1.3667\times10^6\,\text{J mol}^{-1}} = 0.146\,\text{mol}.
  • Mass =0.146\times46.08=6.73\,\text{g}.
  • Volume =\frac{6.73\,\text{g}}{0.789\,\text{g mL}^{-1}}\approx 8.5\,\text{mL}.

Ethical, Practical & Real-world Connections

• Energy conservation principles underpin green chemistry and engineering efficiency.
• Calorimetric measurements critical for
  • Fuel evaluation (combustion data).
  • Food caloric values (bomb calorimeters for nutrition).
  • Material selection in thermal management (spacecraft, electronics).
• Endo-/exothermic awareness informs industrial safety (heat-release hazards).