Sparks CH301 THERMODYNAMICS and ENTROPY UNIT 4 Day 5 What are we - PowerPoint PPT Presentation

Sparks CH301 THERMODYNAMICS and ENTROPY UNIT 4 Day 5

What are we going to learn today? Heats of Formation, Hess’s Law, and Bond Energies Second Law of Thermodynamics Concept of Entropy

QUIZ: iClicker Question Which of the following is not a “formation” reaction? (For which would Δ H rxn NOT equal Δ H f ?) A) Mg (s) + ½ O 2(g)  MgO (s) B) ½ N 2(g) + 3/2 H 2(g)  NH 3(g) C) NaF (s) + Li (s)  LiF (s) + Na (s) D) Li (s) + ½ F 2(g)  LiF (s)

Standard Enthalpy of Formation, Δ H f ° Standard Enthalpy of Formation Δ H for the formation of 1 mole of a compound from its elements in their most stable form at standard conditions

• Consider combustion of methane, CH 4 . Calculate  H o for the reaction below from  H o f  values. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)  H f C (s, graph) + 2H 2 (g)  CH 4 (g) ∆ H f = -75 kJ o = -75 kJ CH4  H f C (s, graph) + O 2 (g)  CO 2 (g) ∆ H f = -394 kJ o = -394 kJ CO2  H f ½ O 2 (g) + H 2 (g)  H 2 O (g) ∆ H f = -286 kJ o = -286 kJ H2O

Schematic diagram of the energy changes for combustion of methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) “FORM” PRODUCTS “UN - form” REACTANTS from ELEMENTS into ELEMENTS

CALCULATING Δ H FROM BOND ENERGIES

One way to think about energy changes in reactions: • Chemical reactions require the rearrangement of atoms – Breaking bonds ___________ energy – Forming bonds ___________ energy • Net result of bond breaking and bond forming will give overall enthalpy of reaction

POLL: iClicker Question Consider a reaction for which: E required break bonds > E released form bonds This reaction would be: A. endothermic B. exothermic

Bond Enthalpies Bond Enthalpy The heat required to break a mole of bonds at constant pressure. ° = Σ BE reactants - Σ BE products Δ H r

• To calculate the total energy change, we assume all bonds in reactant molecules are broken and then the atoms are reassembled into product molecules.

Spontaneity Almost every process in the world happens in only one direction (in isolation = “on its own”) Imagine the following situations. Are they spontaneous? Dropping an object Burning logs A gas expanding into the room Heat flow from high T to low T Ice melting in a glass of water Food dye dropped into water

Spontaneity We will refer to any process that happens in isolation as spontaneous . The forward reaction will happen but the reverse reaction will never happen on its own. How might these processes be reversed? Dropping an object Burning logs A gas expanding into the room Heat flow from high T to low T Ice melting in a glass of water

The Second Law of Thermodynamics The Second Law of Thermodynamics states that any process that happens spontaneously will lead to an increase in the entropy of the universe

Entropy The entropy of the universe is the total entropy of the system and surroundings. Spontaneous

Entropy What is Entropy? What words or ideas pop into your head with respect to Entropy?

Entropy Entropy is related to the dispersal of energy at a given temperature. • The more energy dispersed, the greater the entropy change. • The wider the energy dispersal, the greater the entropy change. • The lower the temperature, the greater the entropy change for a given amount of energy.

Entropy Examples When a gas expands in a vacuum, identify the System Surrounding Initial State Final State

POLL: iClicker Question When a gas expands in a vacuum, ∆ S total is A.> 0 B. = 0 C. < 0 D.No way to know

Spontaneity For a process that is spontaneous

POLL: iClicker Question When a gas expands in a vacuum, ∆S system is A.> 0 B. = 0 C. < 0 D.No way to know

Entropy Examples A container of gas was opened and the gas was allowed to fill the room. In this example, the system is the gas and the surroundings is the room. Increasing volume leads to an increase in entropy. The process was spontaneous The surroundings are unchanged The expansion led to an increase in the entropy of the system

Entropy Why does the increase in volume lead to an increase in entropy? We must use a microscopic view of dispersal of energy. Unfortunately, it is difficult to “visualize” energy, but it is easy to visualize molecules. States of highest entropy are simply the most likely to happen.

Microstates Let’s imagine the gas in our previous example, where the gas is in a container with a left-hand side and a right-hand side What if we only had one gas particle? There are two possibilities, both of which are equally likely

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Microstates What if we only had Avogadro’s number of particles? It is extremely unlikely that we will find all the molecules entirely on the left or right side. The most likely situation will have half of the particles on each side. If there are two molecules in the two-bulbed flask, there is one chance in four that both molecules will be in the left bulb.

Ω = 1 Ω = 4 Ω = 6 29

Entropy and Microstates Entropy is measure of the number of equivalent microstates. More volume more microstates more entropy More molecules more microstates more entropy Higher temperature more microstates more entropy Higher Energy more microstates more entropy

Entropy and Microstates It is harder to visualize microstates for energy, but it is the same idea, where more microstates means higher entropy Macroscopically, we can quantify this with heat flow The heat will always be the reversible heat for the processes we investigate in this course

What have we learned today? The change in Enthalpy can be calculated based on a variety of tabulated data: Heats of formation/Other Heats of Reaction/Bond Energies Understand the concept of entropy, S, and change in entropy ΔS. Understand the concept of change in entropy of a system, surroundings and universe.

Important Information HW, LM31, & LM32 Due Wednesday