Laws of Thermodynamics and Thermochemistry

Laws of Thermodynamics

Zeroeth Law

• Temperature: Defines thermal equilibrium.

• Two systems in equilibrium with a third system are in thermal equilibrium with each other.

First Law

• Conservation of Energy: Energy can change forms, but is neither created nor destroyed.

• $total energy = equal$

• Mathematically represented as: $\Delta E = q + w$ where:

  • $\Delta E$ is the change in internal energy.

  • $q$ is heat.

  • $w$ is work.

• Pathway: Refers to the division of energy between work ($w$) and heat ($q$).

• The energy of the universe (system + surroundings) is constant. Any energy transferred from a system must be transferred to the surroundings and vice versa.

Second Law

• Entropy of an isolated system always increases.

• In an isolated system, processes are spontaneous when they lead to an increase in disorder or entropy.

• Entropy increases when a reaction increases the number of molecules or the amount of unusable energy in the universe.

Third Law

• Entropy of a system approaches a constant as temperature approaches absolute zero (0K).

• The entropy of a perfect crystal is zero when the crystal is at a temperature of absolute zero (0K).

Entropy

• Entropy is molecular randomness or disorder; breaking things apart, spreading them out, making them unusable, and achieving the most likely distribution of microstates.

• Entropy is a measure of the “dilution” of thermal energy.

• Spontaneous processes must increase the entropy of the universe.

Enthalpy (Sec 18-0)

Heat of Formation and Hess' Law

Heats of Formation

• $\Delta H_f^\circ$ – standard enthalpy of formation

• The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states.

• Example: $2 C(s) + 3 H<em>2(g) + \frac{1}{2} O</em>2(g) \rightarrow C<em>2H</em>5OH(l)$

  • Elements are combined in exactly the amount of each atom required to form a single product molecule.

• The standard enthalpy of formation of any element in its most stable form is zero.

• $\Delta H<em>f^\circ (O</em>2) = 0$

• $\Delta H<em>f^\circ (O</em>3) = 142 \text{ kJ/mol}$

• $\Delta H_f^\circ (C, \text{graphite}) = 0$

• $\Delta H_f^\circ (C, \text{diamond}) = 1.90 \text{ kJ/mol}$

• Elemental vs. Molecular examples illustrated using an Oreo Cookie analogy (12g diamond).

Standard States

• Compounds:

  • Gases: 1 atm

  • Condensed State: Pure solid or liquid

  • Aqueous or other solution: 1M

• Elements: Whatever state at 1 atm and 298K.

Tables for $\Delta H^\circ$

• Pure elements have $\Delta H^\circ = 0 \text{ kJ}$

• Values are typically in kilojoules ($1 \text{ kJ} = 1000 \text{ J}$).

Basic Rules for Enthalpy Calculations

• Stoichiometric coefficients always refer to the number of moles of a substance (phase is VERY important).

• If you reverse a reaction, the sign of $\Delta H$ changes.

• If you multiply both sides of the equation by a factor $n$, then $\Delta H$ must change by the same factor $n$.

Calculating $\Delta H_{rxn}$

• For chemical reactions (not just formation), using the balanced reaction:

  • $\Delta H<em>{rxn} = \sum n \Delta H</em>{f \text{ prod}}^\circ - \sum n \Delta H_{f \text{ react}}^\circ$

  • $\sum$ means add up all the $\Delta H_f^\circ$ values.

  • $n$ stands for the coefficients from the balanced reaction.

  • $\Delta = (\text{final} – \text{initial}) = (\text{products} – \text{reactants})$

Enthalpy Calculation: Heats of Reaction

• Example: Benzene ($C<em>6H</em>6$) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

Hess’s Law

• $\Delta H$ for the overall reaction will equal the sum of the enthalpy changes for the individual steps.

• If a reaction is reversed, the sign of $\Delta H$ is also reversed.

• The magnitude of $\Delta H$ is directly proportional to the balanced equation; if the equation is multiplied or divided by an integer, so is the enthalpy.

• Hess’s Law Definition: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Enthalpy of Reaction Example

• Determine the enthalpy of reaction for the reaction: $2 N<em>2(g) + 5 O</em>2(g) \rightarrow 2 N<em>2O</em>5(g) \qquad \Delta H = ?$

• Using the following heats of reaction:

• $Eqn 1: \quad 2H<em>2(g) + O</em>2(g) \rightarrow 2H_2O(l) \qquad \Delta H = -571.7 \text{kJ}$

• $Eqn 2: \quad N<em>2O</em>5(g) + H<em>2O(l) \rightarrow 2HNO</em>3(l) \qquad \Delta H= -92 \text{kJ}$

• $Eqn 3: \quad N<em>2(g) + 3O</em>2(g) + H<em>2(g) \rightarrow 2HNO</em>3(l) \qquad \Delta H= -348.2 \text{kJ}$

• Solution involves reversing equations, altering stoichiometry, and adding the equations to arrive at the target reaction.

• Reverse Eqn 1: $2H<em>2O(l) \rightarrow 2H</em>2(g) + O_2(g) \qquad \Delta H = +571.7 \text{kJ}$

• Reverse and multiply Eqn 2 by 2: $4HNO<em>3(l) \rightarrow 2N</em>2O<em>5(g) + 2H</em>2O(l) \qquad \Delta H= +184 \text{kJ}$

• Multiply Eqn 3 by 2: $2N<em>2(g) + 6O</em>2(g) + 2H<em>2(g) \rightarrow 4HNO</em>3(l) \qquad \Delta H= -696.4 \text{kJ}$

• $\Delta H = 59.3$

Gibbs Free Energy

• The maximum amount of work that a system can do, or the maximum energy change that any reaction can achieve.

• Gibbs free energy is a simple formula that incorporates enthalpy, entropy, and temperature.

  • $\Delta G = \Delta H - T \Delta S$

• The sign of the free energy indicates whether a reaction is spontaneous.

• When Gibbs free energy is negative, the reaction is spontaneous.

• $\Delta S$ is often reported in Joules.

Entropy and Gibbs Free Energy

• $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$

• Gibbs free energy is the energy that is available to do useful work.

• A reaction will spontaneously occur if \Delta G < 0 (exergonic reaction).

• A reaction will NOT spontaneously occur if \Delta G > 0 (endergonic reaction).