Question of the Day:1. Is an endo or exo reaction more likely to be spontaneous?

2. Use the diagram below: If 1 mol of water is produced in the combustion, how much heat is involved?

Day 1 2-3

CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)H

+ –802

How would you diagram the reaction below?

CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ

CaO(s) + H2O(l)

H + –65.2

Ca(OH)2(s)

http://wps.prenhall.com/esm_brown_chemistry_9/2/660/169060.cw/index.html

Homework # 2 Chapter 5 – w/ discussion partner – show me successful screen (100%)

4

Predicting Spontaneity

• H2O(l) → H2O(g)

∆H = + 40.7 kJ; 1 atm, 100°C

5

Predicting Spontaneity

• CaCO3(s) → CaO(s) + CO2(g)

∆H = +178.0 kJ;

298 K vs. 1100 K

6

Predicting Spontaneity

7

Entropy

• Entropy is the measure of the disorder or randomness of a system.• Entropy is a state function.• Entropy changes follow Hess’ Law.• A system has high entropy if it

• _______________________________________• _______________________________________

Is Very dsorigniaez

d

(Very disorganiz

ed)

Has freedom of motion

8

Entropy & Spontaneity

• Reactions will be spontaneous if ________ is increased.

• Examples• ice cubes melting vs. putting water in

freezer to make ice cubes

• room gets messy vs. doing work to put things in order

etnopry

9

Standards & Constants

• Conditions & Units• T = 25oC = 298K• P = 1 atm• Units: J/K = Joules/Kelvin

• What is the relationship between entropy and temperature?• As T 0 K, S 0 • T↓, S↓ and T↑, S↑

10

Standards & Constants

• Why are the entropy values for elements and compounds always positive?• T always > 0K • Therefore everything is always moving

• How do entropies compare among the phases of matter?• So solid < So liquid < So gas

11

Standard Entropy Changes

• The equation

∆So = Soproducts – So

reactants

• Example: 2HCI(aq) + 2Ag(s) → 2AgCl(s) + H2(g)1 atm, 25oC

• Predict if this reaction will have + ∆So or – ∆So.

• Calculate ∆So.

Question of the Day:1. Use the diagram below: If 36 grams of water is produced in the combustion, how much heat is involved?

Day 2 2-6

CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)H

+ –802

Question of the Day:1. If 32.6 kJ of heat are given off, how many grams of calcium hydroxide are produced?

Day 3 2-7

CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ

37 grams

14

Standard Entropy Changes

• The equation

So = nSoproducts – mSo

reactants

• Example: 2 H2(g) + O2(g) → 2 H2O(g) 1 atm, 25oC

• Predict if this reaction will have + ∆So or – ∆So.

• Calculate ∆So.

15

Standard Entropy Changes

Example: 2HCI(aq) + 2Ag(s) → 2AgCl(s) + H2(g) 1 atm, 25°C

a. Predict if this rxn will have + ∆S° or – ∆S°.

b.Calculate ∆S°.

16

Entropy and Molecular Motion

• Molecules always have positive entropy values because molecules are always moving.

• How do molecules move?• Rotation

• Translation

• Vibration

-an atom spins around a single bond-entire molecule spins

-entire molecule moves in one direction

-atoms in molecule get closer together or further apart along a bond

ASSIGNMENT:

READ section 18.5 and complete #s 46-53

C(s, diamond) → C(s, graphite)Hess’s Law

Although the enthalpy change for this reaction cannot be measured directly, you can use Hess’s law to find the enthalpy change for the conversion of diamond to graphite by using the following combustion reactions.

a. C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

Hess’s Law

Write equation a in reverse to give:

c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ

When you reverse a reaction, you must also change the sign of ΔH.

a. C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

C(s, diamond) → C(s, graphite)

Hess’s Law

If you add equations b and c, you get the equation for the conversion of diamond to graphite.

C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ

C(s, diamond) → C(s, graphite)

b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ

C(s, diamond) → C(s, graphite)

Hess’s Law

If you also add the values of ΔH for equations b and c, you get the heat of reaction for this conversion.

C(s, diamond) → C(s, graphite)

C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ

C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ

Hess’s LawC(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ

CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ

C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ

•How can you determine ΔH for the conversion of diamond to graphite without performing the reaction?

CHEMISTRY & YOUCHEMISTRY & YOU

•How can you determine ΔH for the conversion of diamond to graphite without performing the reaction?

CHEMISTRY & YOUCHEMISTRY & YOU

You can use Hess’s law by adding thermochemical equations in which the enthalpy changes are known and whose sum will result in an equation for the conversion of diamond to graphite.

• Suppose you want to determine the enthalpy change for the formation of carbon monoxide from its elements.

• Carrying out the reaction in the laboratory as written is virtually impossible.

Hess’s LawAnother case where Hess’s law is useful is when reactions yield products in addition to the product of interest.

C(s, graphite)+ O2(g) → CO(g) ΔH = ? 1 2

Hess’s LawYou can calculate the desired enthalpy change by using Hess’s law and the following two reactions that can be carried out in the laboratory:

C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

CO2(g) → CO(g) + O2(g) ΔH = 283.0 kJ

C(s, graphite)+ O2(g) → CO(g) ΔH = –110.5 kJ

1 2

1 2

Hess’s LawC(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ

CO2(g) → CO(g) + O2(g) ΔH = 283.0 kJ

C(s, graphite)+ O2(g) → CO(g) ΔH = –110.5 kJ

1 2

1 2

•According to Hess’s law, it is possible to calculate an unknown heat of reaction by using which of the following?

A. heats of fusion for each of the compounds in the reaction

B. two other reactions with known heats of reaction

C. specific heat capacities for each compound in the reaction

D. density for each compound in the reaction

•According to Hess’s law, it is possible to calculate an unknown heat of reaction by using which of the following?

A. heats of fusion for each of the compounds in the reaction

B. two other reactions with known heats of reaction

C. specific heat capacities for each compound in the reaction

D. density for each compound in the reaction

30

Gibbs Free Energy• Gibbs free energy is the definitive method

for determining the spontaneity of a reaction.• Gibbs free energy incorporates both enthalpy

and entropy.• ∆G, the change in free energy, is the driving

force of a chemical reaction.• Free energy is the energy that a system can

release through heat and disorder.

• The Gibbs-Helmholtz equation

G = H - TS

31

Why does the Gibbs-Helmholtz equation work?

• The role of ∆H• If -H : heat out & spontaneity indicated

with - sign• If H: no sign means +H

• The role of – T∆S• T given in K ! always +• “ – ” will change sign of ∆S so spontaneity

is indicated in like manner with ∆H

32

How does the value of ∆G determine spontaneity?

• – ∆G• Spontaneous• Net release of energy through heat & disorder• Becomes more stable

• + ∆G• Not spontaneous• Must do work or supply energy to make rxn go• Reverse reaction is spontaneous

• ∆G = 0• Reaction is at equilibrium• Forward rate = reverse rate

33

Standardize!

• Why?• Changes in pressure and concentration

affect entropy values.

• P = 1 atm for gases• Concentration = 1M for solutions

34

Calculating G

2 H2(g) + O2(g) → 2 H2O(l)

35

Effect of Temperature on Reaction Spontaneity

H S G = H - TS Remarks

– + – Spon. @ all temps

+ – + Not spon. @ any T

– – – OR +Spon. @ low T *Not spon. @ high T

+ + – OR +Spon. @ high T *Not spon. @ low T

* High & low temps are relative to equilibrium temp

36

Finding Equilibrium

• ∆G = ∆H – T∆S = 0• At equilibrium, the free energy

released by the spontaneous reaction is being absorbed by the reverse, nonspontaneous reaction.

• What is the formula for finding the temperature of equilibrium?

T = H / S

37

Finding Equilibrium

• What are the possibilities for the equilibrium temperature?• May be an ordinary temp• May be extremely high or low• May be negative

• !Impossible• System can NEVER be in equilibrium