01 · Explore
Thermodynamic Terms and Systems
To study energy changes in chemical reactions, we must define the boundaries of our observation.
A system refers to the specific part of the universe where observations are made, such as a reaction mixture in a beaker. Everything else in the universe constitutes the surroundings. In practice, surroundings are the immediate neighborhood that can interact with the system. The system is separated from its surroundings by a boundary, which may be real or imaginary.
Systems are classified based on the exchange of matter and energy. An open system can exchange both matter and energy with its surroundings. A closed system can exchange energy but not matter. An isolated system can exchange neither matter nor energy; a reactants in a thermos flask is a close approximation of an isolated system.
The state of a system is described by measurable macroscopic properties like pressure (p), volume (V), temperature (T), and amount (n). These are called state functions because their values depend only on the current state of the system, not on how that state was reached.
| System Type | Exchange of Matter | Exchange of Energy |
|---|---|---|
| Open | Yes | Yes |
| Closed | No | Yes |
| Isolated | No | No |
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Why are pressure and temperature called state functions?
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02 · Explore
Internal Energy and the First Law
Internal energy represents the total energy of a system, which can change through work or heat transfer.
Internal energy (U) is the sum of all forms of energy in a system (chemical, electrical, mechanical, etc.). While we cannot measure the absolute value of U, we can measure its change (ΔU). U changes when heat (q) passes into or out of the system, work (w) is done on or by the system, or matter enters or leaves.
The First Law of Thermodynamics states that energy can neither be created nor destroyed; the energy of an isolated system is constant. Mathematically, this is expressed as ΔU = q + w. This equation shows that the change in internal energy is the sum of heat added to the system and work done on the system.
IUPAC sign conventions are critical: q is positive when heat is transferred from surroundings to the system (endothermic), and w is positive when work is done on the system (compression). Conversely, q and w are negative when heat is lost by the system or work is done by the system (expansion).
Calculating Internal Energy Change
ΔU = q + w
If a system absorbs 701 J of heat and does 394 J of work, q = +701 J and w = -394 J. ΔU = 701 - 394 = 307 J.
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What is the value of ΔU for an isolated system?
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03 · Explore
Pressure-Volume Work
Mechanical work in chemical systems often involves changes in volume against an external pressure.
Consider a gas in a cylinder with a frictionless piston. When the gas expands or is compressed, it performs work. For a constant external pressure (pex), the work done is w = -pex ΔV, where ΔV = Vf - Vi. In compression, Vf < Vi, making ΔV negative and work (w) positive, meaning work is done on the system.
A process is 'reversible' if it proceeds so slowly through a series of equilibrium states that it can be reversed by an infinitesimal change. For a reversible isothermal expansion of an ideal gas, work is calculated using the integral of pressure over volume, resulting in the formula: w_rev = -2.303 nRT log(Vf / Vi).
Free expansion occurs when a gas expands into a vacuum (pex = 0). In this case, no work is done (w = 0), regardless of whether the process is reversible or irreversible.
Reversible Isothermal Expansion
w = -2.303 nRT log(Vf / Vi)
For 1 mol of an ideal gas expanding from 2L to 10L at 298K, using R = 0.08206 L atm / (K mol): w = -2.303 * 1 * 0.08206 * 298 * log(10/2) = -39.366 L atm. To convert to Joules, 1 L atm ≈ 101.325 J, giving approximately -3.99 kJ.
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Why is work zero during expansion into a vacuum?
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04 · Explore
Enthalpy: Heat at Constant Pressure
Most chemical reactions occur at constant atmospheric pressure, making enthalpy a more convenient state function than internal energy.
Enthalpy (H) is defined as H = U + pV. At constant pressure, the change in enthalpy (ΔH) is equal to the heat absorbed or evolved by the system (qp). Thus, ΔH = ΔU + pΔV. For reactions involving gases, we can substitute pΔV using the ideal gas law to get ΔH = ΔU + ΔngRT, where Δng is the change in the number of moles of gaseous products and reactants.
ΔH is negative for exothermic reactions (heat released) and positive for endothermic reactions (heat absorbed). For systems containing only solids or liquids, volume changes are negligible, so ΔH is approximately equal to ΔU.
Enthalpy is a state function because it depends on U, p, and V, all of which are state functions. This means ΔH for a reaction is independent of the path taken.
Relating ΔH and ΔU
ΔH = ΔU + ΔngRT
In the vaporization of 1 mol of water at 100°C (373K), ΔH = 41 kJ/mol. Δng = 1 (gas) - 0 (liquid) = 1. ΔU = ΔH - ΔngRT = 41000 - (1 * 8.314 * 373) = 37904 J/mol or 37.9 kJ/mol.
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When does ΔH equal ΔU?
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05 · Explore
Properties and Heat Capacity
Thermodynamic properties are categorized by their dependence on the amount of matter, and heat capacity quantifies how temperature changes with energy input.
An extensive property depends on the quantity or size of matter in the system (e.g., mass, volume, enthalpy). An intensive property is independent of the amount of matter (e.g., temperature, density, pressure). Molar properties, like molar volume (Vm), are intensive because they are defined per mole.
Heat capacity (C) is the coefficient that relates heat transfer to temperature rise: q = C ΔT. Specific heat capacity (c) is the heat required to raise the temperature of one unit mass by one degree, while molar heat capacity (Cm) is the heat required for one mole.
For an ideal gas, the relationship between heat capacity at constant pressure (Cp) and constant volume (CV) is Cp - CV = R. This difference arises because at constant pressure, some energy is used to do expansion work.
Extensive vs. Intensive Properties
- 1
System (V, T)
A gas at volume V and temperature T.
- 2
Partition
Divide the system into two equal halves.
- 3
Result (V/2, T)
Volume is halved (extensive), but temperature remains the same (intensive).
This sequence demonstrates how extensive properties change with system size while intensive properties remain constant.
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Is molar heat capacity an intensive or extensive property?
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06 · Explore
Calorimetry: Measuring ΔU and ΔH
Calorimetry is the experimental technique used to measure energy changes in physical or chemical processes.
ΔU is measured at constant volume using a bomb calorimeter. A known mass of a combustible substance is burnt in a steel 'bomb' immersed in a water bath. Since the volume is fixed, no work is done, and the heat evolved (qv) equals ΔU. The temperature rise of the water and calorimeter is used to calculate the energy change.
ΔH is measured at constant pressure, typically in a simpler coffee-cup calorimeter open to the atmosphere. The heat absorbed or evolved (qp) at constant pressure is the enthalpy of reaction (ΔrH).
In these experiments, the heat lost by the reaction is equal to the heat gained by the calorimeter: q = -C_calorimeter * ΔT.
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What condition is maintained in a bomb calorimeter?
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07 · Explore
Enthalpy of Reaction and Hess's Law
The enthalpy change of a reaction can be calculated from the enthalpies of its individual steps.
The standard enthalpy of reaction (ΔrH°) is the enthalpy change when all participating substances are in their standard states (pure form at 1 bar). It is calculated as the sum of standard enthalpies of products minus the sum of standard enthalpies of reactants, multiplied by their stoichiometric coefficients.
Hess's Law of Constant Heat Summation states that if a reaction takes place in several steps, the overall standard reaction enthalpy is the sum of the standard enthalpies of the intermediate steps. This is possible because enthalpy is a state function and independent of the path.
Hess's Law is extremely useful for calculating enthalpies of reactions that are difficult to measure directly, such as the partial combustion of carbon to carbon monoxide.
Applying Hess's Law
ΔrH° = ΔrH1 + ΔrH2
To find ΔH for C + 1/2 O2 -> CO: (1) C + O2 -> CO2 (ΔH = -393.5 kJ); (2) CO + 1/2 O2 -> CO2 (ΔH = -283.0 kJ). Reverse (2) and add to (1): -393.5 + 283.0 = -110.5 kJ/mol.
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What is the standard state of an element?
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08 · Explore
Standard Enthalpies of Various Processes
Enthalpy changes are specifically named based on the type of process occurring.
Standard Enthalpy of Combustion (ΔcH°) is the enthalpy change when one mole of a substance is completely burnt in oxygen. These are always exothermic (negative).
Standard Enthalpy of Formation (ΔfH°) is the enthalpy change for the formation of one mole of a compound from its constituent elements in their most stable states. By convention, ΔfH° for any element in its reference state is zero.
Enthalpy of Atomization (ΔaH°) is the enthalpy change required to break one mole of bonds completely to obtain atoms in the gas phase. For diatomic molecules like H2, this is the same as the bond dissociation enthalpy.
Phase transformations also have specific enthalpies: Fusion (melting), Vaporization (boiling), and Sublimation (solid to gas). These processes are endothermic (positive) as they require energy to overcome intermolecular forces.
| Process | Symbol | Definition |
|---|---|---|
| Formation | ΔfH° | 1 mole compound from elements |
| Combustion | ΔcH° | 1 mole substance burnt in O2 |
| Fusion | ΔfusH° | 1 mole solid to liquid |
| Vaporization | ΔvapH° | 1 mole liquid to gas |
| Atomization | ΔaH° | 1 mole bonds broken to atoms |
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Why is ΔfH° of O2(g) taken as zero?
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09 · Explore
Bond and Lattice Enthalpies
Energy changes at the molecular and ionic level involve breaking and forming attractions.
Bond Enthalpy (ΔbondH°) is the energy required to break one mole of a specific bond in a gaseous molecule. In polyatomic molecules like CH4, we use 'mean bond enthalpy' because the energy to break each successive C-H bond differs slightly. The reaction enthalpy can be estimated as: ΔrH° = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products).
Lattice Enthalpy (ΔlatticeH°) is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions. It cannot be measured directly.
The Born-Haber Cycle is an indirect method to calculate lattice enthalpy using Hess's Law. It links the enthalpy of formation of an ionic solid to steps like sublimation, ionization, bond dissociation, and electron gain.
- 1
Sublimation of metal: Na(s) -> Na(g)
- 2
Ionization of metal: Na(g) -> Na+(g) + e-
- 3
Dissociation of non-metal: 1/2 Cl2(g) -> Cl(g)
- 4
Electron gain by non-metal: Cl(g) + e- -> Cl-(g)
- 5
Lattice formation: Na+(g) + Cl-(g) -> NaCl(s)
The sum of these steps equals the Enthalpy of Formation (ΔfH°). Lattice enthalpy is found by rearranging this sum.
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How is enthalpy of solution related to lattice enthalpy?
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10 · Explore
Entropy and the Second Law
Spontaneity is not determined by energy alone; disorder also plays a vital role.
Entropy (S) is a measure of the degree of randomness or disorder in a system. Spontaneous processes tend toward increased disorder. For example, a gas expanding to fill a volume or two gases diffusing into each other increases entropy.
The Second Law of Thermodynamics states that for a spontaneous process in an isolated system, the entropy increases (ΔS > 0). For non-isolated systems, the total entropy change (ΔStotal = ΔSsys + ΔSsurr) must be positive for spontaneity.
Entropy is a state function. For a reversible process, ΔS = q_rev / T. Entropy increases with temperature because molecular motions (translational, rotational, vibrational) become more vigorous. Gases have higher entropy than liquids, which have higher entropy than solids.
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Predict the sign of ΔS when a liquid crystallizes into a solid.
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11 · Explore
Gibbs Energy and Spontaneity
Gibbs energy combines enthalpy and entropy into a single criterion for spontaneity at constant temperature and pressure.
Gibbs energy (G) is defined as G = H - TS. The change in Gibbs energy at constant temperature is ΔG = ΔH - TΔS. This is known as the Gibbs equation.
ΔG provides a definitive criterion for spontaneity: (1) If ΔG is negative, the process is spontaneous. (2) If ΔG is positive, the process is non-spontaneous. (3) If ΔG is zero, the system is at equilibrium.
A reaction with positive ΔH (endothermic) can still be spontaneous if ΔS is positive and the temperature (T) is high enough such that TΔS outweighs ΔH. This explains why some endothermic reactions occur at high temperatures.
| ΔH | ΔS | ΔG | Spontaneity |
|---|---|---|---|
| - | + | Always - | Spontaneous at all T |
| + | + | - at high T | Spontaneous at high T |
| - | - | - at low T | Spontaneous at low T |
| + | - | Always + | Non-spontaneous at all T |
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What is the physical meaning of ΔG?
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12 · Explore
Gibbs Energy and Equilibrium
The standard Gibbs energy change is directly related to the equilibrium constant of a reaction.
At equilibrium, ΔG = 0. The standard Gibbs energy change (ΔrG°) is related to the equilibrium constant (K) by the equation: ΔrG° = -RT ln K or ΔrG° = -2.303 RT log K.
If ΔrG° is strongly negative, K will be much larger than 1, meaning the reaction goes nearly to completion. If ΔrG° is strongly positive, K will be very small, and the reaction will hardly proceed.
The Third Law of Thermodynamics states that the entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero (0 K). This allows for the calculation of absolute entropy values.
Calculating Equilibrium Constant
log K = -ΔrG° / (2.303 RT)
If ΔrG° = -13.6 kJ/mol at 298 K: log K = -(-13600 J/mol) / (2.303 * 8.314 J/K mol * 298 K) = 2.38. K = antilog(2.38) = 2.4 * 10^2.
Dissociation of Dinitrogen Tetroxide
ΔrG° = -RT ln Kp
At 60°C (333 K), N2O4 is 50% dissociated at 1 atm. The mole fractions are x(N2O4) = 0.5/1.5 and x(NO2) = 1/1.5. Kp = (pNO2)^2 / pN2O4 = (0.667)^2 / 0.333 = 1.33. ΔrG° = -8.314 * 333 * 2.303 * log(1.33) = -763.8 J/mol or -0.7638 kJ/mol.
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What does a ΔrG° value of zero imply for the equilibrium constant?
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