Thermodynamics

Heat and Temperature Heat represents energy flow between two substances at different temperatures. Temperature represents the average amount of kinetic energy present in a given substance. Heat ≠ Temperature

Heat and Temperature Energy cannot be created nor can it be destroyed; the amount of energy in the universe is constant. This is known as the First Law of Thermodynamics. Heat, being a form of energy, cannot be destroyed nor created, therefore, it must be transferred between systems. Heat transfer occurs when atoms/molecules collide with each other transferring their kinetic energy. Exothermic processes transfer heat to its surroundings. Endothermic processes absorb heat from its surroundings.

State Functions Enthalpy change (ΔH), entropy change (ΔS), and free-energy change (ΔG) are state functions. These all depend only on the change between the initial and final states of a system, not on the process by which the change occurs. For chemical reactions, these state functions are independent of the reaction pathway. Ex. The addition of a catalyst to a reaction will have no effect on the overall energy or entropy change of the reaction.

Standard State Conditions A thermodynamic quantity under standard state conditions looks as follows: ΔH = ΔHo ΔS = ΔSo ΔG = ΔGo Standard State Conditions: All gases are at 1 atm pressure. All liquids are pure. All solids are pure. All solutions are at 1 M concentration. The energy of formation of an element in its normal state is defined as zero. The temperature used for standard state values is most commonly room temperature: 25 oC (298 K).

Enthalpy of Formation, ΔHof Enthalpy of formation (AKA heat of formation) is the change in energy that takes place when one mole of a compound is formed from its component pure elements under standard state conditions. EX. C(s) + 𝟏 𝟐 O2(g) + 2H2(g)  CH3OH In enthalpy of formation, there is always one mole of product. To have one mole of product, fractions can be used to balance equations. Fractions can only be used with diatomic molecules and only 1 2 is permitted.

Enthalpy of Formation, ΔHof ΔHof for a pure element is defined as zero. If ΔHof for a compound is negative, energy is released when the compound is formed from pure elements, and the product is more stable than its constituent elements. EXOTHERMIC If ΔHof for a compound is positive, energy is absorbed when the compound is formed from pure elements, and the product is less stable than its constituent elements. ENDOTHERMIC If the ΔHof values of the products and reactants are known, ΔH for a reaction can be calculated. ΔHo = Σ ΔHof products – Σ ΔHof reactants

Example 2CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g) ΔHo = Σ ΔHof products – Σ ΔHof reactants Compound ΔHof (kJ/mol) CH3OH(g) -201 O2(g) CO2(g) -394 H2O(g) -242

Enthalpy of Combustion The enthalpy of combustion describes the amount of energy released when ONE mole of hydrocarbon combusts, and is always exothermic. EX. How much heat is released when 5.00 g of CH3OH is combusted in excess oxygen?

Bond Energy Bond energy is the energy required to break a bond. Bond energy is always a positive value. When a bond is formed, energy equal to the bond energy is released. ΔHo = Σ Bond energies of bonds broken – Σ Bond energies of bonds formed The bonds broken will be the reactant bonds, and the bonds formed will be the product bonds.

Bond Energy The number of bonds broken and formed are affected by: The number of a particular bond type within a molecule The number of molecules that are in a balanced reaction Find the ΔHo for the following reaction: 2H2(g) + O2(g)  2H2O(g) Bond Bond Energy (kJ/mol) H-H 436 O=O 499 O-H 463

Hess’s Law Hess’s Law states that if a reaction can be described as a series of steps, then ΔH for the overall reaction is simply the sum of the ΔH values for all the steps. Rules for enthalpy calculations: If you flip the equation, flip the sign on the enthalpy value. If you multiply or divide an equation by an integer, also multiply/divide the enthalpy value by that same integer. If several equations, when summed up, create a new equation, you can also add the enthalpy values of those component equations to get the enthalpy value for the new equation.

4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Example Calculate the enthalpy for the following reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Given: Equation A: N2(g) + O2(g)  2NO(g) ΔH = 180.6 kJ/mol Equation B: N2(g) + 3H2(g)  2NH3(g) ΔH = -91.8 kJ/mol Equation C: 2H2(g) + O2(g)  2H2O(g) ΔH = -483.7 kJ/mol

Enthalpy of Solution When an ionic substance dissolves in water, a certain amount of heat is emitted or absorbed. The bond between the cation and anion breaks, which requires energy, but energy is released when those ions form new attractions to the dipoles of the water molecules.

Steps for Enthalpy of Solution Step 1: Breaking the solute bonds The bonds between the Na+ and Cl- ions must be broken. The amount of energy needed to break this bond is equal to the lattice energy. This step always has a positive ΔH since energy is required to break the bonds.

Steps for Enthalpy of Solution Step 2: Separating the Solvent Molecules The water molecules must spread out to make room for the Na+ and Cl- ions. This step always has a positive ΔH since energy is needed to weaken the IMFs between the water molecules.

Steps for Enthalpy of Solution Step 3: Creating new Attractions The free-floating ions are attracted to the dipoles of the water molecules. Energy is released during this process, even though no bonds are being formed, so this step always has a negative ΔH.

Enthalpy of Solution The energy values from step 2 and step 3 combined are often called the hydration energy. This value is always negative. Hydration energy is a Coulombic energy and thus increases as the ions either increase in charge or decrease in size. If you add the energy values for all 3 steps, the enthalpy of solution for that compound can be determined. If hydration energy > lattice energy, the enthalpy of solution is negative. If hydration energy < lattice energy, the enthalpy of solution is positive.