 Section 1 – Thermochemistry  Section 2 – Driving Force of Reactions

 Define temperature and state the units in which it is measured.  Define heat and state its units.  Perform specific-heat calculations.  Explain enthalpy change, enthalpy of reaction, enthalpy of formation, and enthalpy of combustion.  Solve problems involving enthalpies of reaction, enthalpies of formation, and enthalpies of combustion.

 Virtually every chemical reaction is accompanied by a change in energy. ◦ What two ways can energy be involved?  Thermochemistry: study of the transfer of energy as heat that accompanies chemical reactions and physical changes.

 Calorimeter: used to measure heat absorbed or released.  Energy released from the reaction is measured from the temp change of the water

 measure of the average kinetic energy of the particles in a sample of matter. ◦ The greater the kinetic energy of the particles in a sample, the hotter it feels.  Remember: How do we convert Celsius to Kelvin?

 Heat: the energy transferred between samples of matter because of a difference in their temperatures. ◦ Joule: SI unit of heat and energy  Energy transferred as heat moves spontaneously  from matter at a higher temp  lower temp

 Energy transferred depends on: ◦ The material ◦ The mass  Specific Heat: specific heat of a substance is the amount of energy required to raise the temperature of one gram by one Celsius degree (1°C) or one kelvin (1 K). ◦ Units: J/(g°C) or J/(gK) ◦ The temperature difference as measured in either Celsius degrees or kelvins is the same.

Heat released Specific heat: or absorbed: c p = specific heat at a given pressure q = energy lost or gained m = mass of the sample ∆T = difference between the initial and final temperatures.

 A 4.0 g sample of glass was heated from 274 K to 314 K, and was found to have absorbed 32 J of energy as heat. a. What is the specific heat of this type of glass? b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

Given: m = 4.0 g ∆ T = 314 – 274 = 40. K q = 32 J Unknown:a. c p in J/(gK) b. q for ∆ T of 314 K → 344 K Solution: a.

b. **See Practice Worksheet for more examples

How much energy does it take heat water from ice to vapor?? (2 equations: 1 for change of temp./ 1 for phase change) Phase Change:  Heat of Fusion: H fus ◦ Heat energy needed to melt 1 mol (s to l)  Heat of Vaporation: H vap ◦ Heat energy need to boil 1 mol of a substance ( l – g)

 Enthalpy(∆H): “change in enthalpy” energy absorbed as heat during a chemical reaction at constant pressure

 The difference between the enthalpies of products and reactants. ∆H = H products – H reactants

 Equation that includes the quantity of energy released or absorbed as heat 2H 2 (g) + O 2 (g) → 2H 2 O(g) kJ  States of matter are ALWAYS shown because it can influence the energy of released or absorbed.  Coefficients are always viewed as number of moles in the reaction.  If 2x as many reactants were provide, 2x as many moles of water would be produced and 2x as many kJ of energy would be released.

 chemical reaction that releases energy the energy of the products is less than the energy of the reactants ◦ example: 2H 2 (g) + O 2 (g) → 2H 2 O(g) kJ

 chemical reaction that absorbs energy the products have a larger enthalpy than the reactants 2H 2 O(g) kJ → 2H 2 (g) + O 2 (g)

 Enthalpy is often not included in the reaction expression, instead as a ∆ H value ◦ Example: 2H 2 (g) + O 2 (g) → 2H 2 O(g) ∆ H = –483.6 kJ ◦ Negative ∆H is exothermic – releases heat ◦ Positive ∆H is endothermic – absorbs heat

Energy released, ∆ H is negative.

Energy is absorbed, ∆ H is positive.

 molar enthalpy of formation: enthalpy change that occurs when one mole of a compound is formed from its elements in their standard state at 25°C and 1 atm. ◦ Enthalpies of formation are given for a standard temperature and pressure so that comparisons between compounds are meaningful. standard states heat of FORMATION change in enthalpy

 In your textbook! (or handout)  The values are given as the enthalpy of formation for one mole of the compound from its elements in their standard states. ◦ **because Δ H f is reported for the formation of 1 mol of a substance, it may be necessary to write on ½ mol for one of the reactants ◦ Ex: 1/2 N 2 (g) + 1/2 O 2 (g)  NO (g)

 Compounds with a large negative enthalpy of formation are very stable.  Compounds with positive values of enthalpies of formation are typically unstable.

 ∆H c : enthalpy change that occurs during the complete combustion of one mole of a substance **Enthalpy of combustion is defined in terms of one mole of reactant **Enthalpy of formation is defined in terms of one mole of product

 Review: ∆ H rxn energy absorbed or released during a chemical reaction  Use Hess’s Law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process.  This means that the energy difference between reactants and products is independent of the route taken to get from one to the other.

 1. The sign ∆ H rxn of depends on the direction of the reaction  (If you flip the reaction around, the a positive enthalpy would become negative)  2. The magnitude ∆ H rxn of is proportional to the amount of the substance in a reaction.  (If you change of the coefficients then multiply the ∆ H, too) *from enthalpy packet

 If you know the reaction enthalpies of individual steps in an overall reaction, you can calculate the overall enthalpy without having to measure it experimentally. What is the heat of formation of methane? C(s) + 2H 2 (g) → CH 4 (g)

H 2 (g) + ½O 2 (g) → H 2 O(l) C(s) + O 2 (g) → CO 2 (g) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l)  The component reactions in this case are the combustion reactions of carbon, hydrogen, and methane: This reaction must be reversed! The Δ H is changed to positive! CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ∆ H 0 = kJ 2 [ H 2 (g) + ½O 2 (g) → H 2 O(l) ] This reaction must x2! The Δ H is x2! Goal Reaction: C(s) + 2H 2 (g) → CH 4 (g)

2H 2 (g) + O 2 (g) → 2H 2 O(l) C(s) + O 2 (g) → CO 2 (g) C(s) + 2H 2 (g) → CH 4 (g) CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g)

∆ H 0 = Σ [( ΔH 0 f of products) × (mol of products)] – Σ [( ΔH 0 f of reactants) × (mol of reactants)] *Need to use reference sheet for H f values **Keep in mind: elements in their standard state have an enthalpy of formation of zero ◦ Ex: Ag(s) = 0 ; N 2 (g) = 0  See Practice WKST (Calculating Enthalpy from Heats of Formation)

 Explain the relationship between enthalpy change and the tendency of a reaction to occur.  Explain the relationship between entropy change and the tendency of a reaction to occur.  Discuss the concept of free energy, and explain how the value of this quantity is calculated and interpreted.  Describe the use of free energy change to determine the tendency of a reaction to occur.

 Majority of spontaneous reactions are exothermic ◦ Reactions proceed in a direction that leads to a lower energy state  But, some spontaneous reactions are endothermic ◦ Example: melting (particles have more energy in the liquid state)  So there are 2 factors that drive a reaction

 2 Factors: Enthalpy, but also Entropy  Entropy (S) : the degree of randomness or disorder of the particles in a system ◦ Increases when a system goes from one state to another (even if there is no enthalpy change) 2NH 4 NO 3 (s)  2N 2 (g) + 4H 2 O(l) + O 2 (g)  The arrangement of particles on the right- hand side of the equation is more random than the arrangement on the left side and hence is less ordered.

 There is a tendency of nature to go from order to disorder ◦ Think about your bedroom, does it always stay neat and organized? Or does it get messier and messier? ◦ Increase in disorder  Increase in entropy ◦ The entropy of the universe is ALWAYS increasing

 How does entropy change with temperature? As you go from a solid  liquid  gas What happens to the entropy?  As you go from a gas  liquid  solid What happens to the entropy?  What about when you dissolve a solute?  Or what about when you dilute a substance? Entropy increases! Entropy decreases

 Unit: kJ/(mol K) or J/mol K  Change in Entropy ΔS ◦ Difference between the entropy of the products and the reactants  Increase in Entropy  +ΔS (more disorder)  Decrease in Entropy  -ΔS (less disorder)

 Calculate entropy like calculating enthalpy from heats of formation, but use S values instead  ∆S = S products - S reactants  See practice WKST, use thermodynamic data sheet for S f values

 Gibb’s Free Energy (G): ◦ Combined enthalpy-entropy function  Spontaneous Reactions: ◦ Least enthalpy (very negative #) ◦ Greatest entropy (large positive #)

 Only the change in free energy can be measured (KJ/mol)  At a constant pressure and temperature… ∆G 0 = ∆H 0 – T ∆ S 0 *T must be in Kelvin since ∆ S will be in units of KJ/mol k *∆S should also be in KJ

If -∆G  the reaction is spontaneous +∆G  the reaction is not spontaneous ∆H∆H ∆S∆S ∆G∆G -+- /spontaneous +-+ /not spontaneous --- /Low temps spontaneous ++- /High temps spontaneous See “Free Energy” worksheet to practice calculation See Practice Worksheet – “Pg. 78 Gibbs Free Energy” to review concepts

For the reaction NH 4 Cl (s) → NH 3(g) + HCl (g) ∆H 0 = 176 kJ/mol and ∆S 0 = kJ/(molK) at K. Calculate ∆G 0, and tell whether this reaction is spontaneous in the forward direction at K.

Given: ∆ H 0 = 176 kJ/mol at K ∆ S 0 = kJ/(molK) at K Unknown: ∆ G 0 at K Solution: The value of ∆ G 0 can be calculated according to the following equation: ∆G 0 = ∆ H 0 – T ∆ S 0 ∆G 0 = 176 kJ/mol – 298 K [0.285kJ/(molK)] ∆G 0 = 176 kJ/mol – 84.9 kJ/mol ∆G 0 = 91 kJ/mol