Wednesday, Oct. 31st: “A” DayThursday, Nov. 1st: “B” Day

AgendaHomework questions/quick reviewSec. 10.3 quiz: “Changes in Enthalpy During

Chemical Reactions”Section 10.4: “Order and Spontaneity”

Entropy, Standard Entropy, Gibbs energyHomework:

Sec. 10.4 review, pg. 367: #3-5, 7-11Sec 10.4 concept review*Quiz next time over section 10.4*

Homework

Pg. 15 practice worksheetSec. 10.3 review, pg. 357: #1-5

Calculating a Reaction’s Change in Enthalpy Sample Prob. E, pg.356

Calculate the change in enthalpy for the reaction below using data from Table 2 on pg 355.

2 H2(g) + 2 CO2(g) 2 H2O(g) + 2 CO(g)

State whether the reaction is exothermic or endothermic.

ΔHreaction = ΔH f0

products - ΔH f0

reactants

ΔHf0

prod = [(2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol)]

= -704.6 kJΔHf

0reactants = [(2 mol)(0 kJ/mol) + (2 mol)(-393.5 kJ/mol)]

= -787 kJΔHreaction= (-704.6 kJ) – (-787 kJ) = 82.4 kJ

*Reaction is endothermic because ΔH is positive.*

Sec. 10.3 Quiz: “Changes in Enthalpy During Chemical

Reactions”You can use your notes and your book to

complete the quiz with a partner of your choice…

Good Luck!

EntropyEntropy: a measure of the randomness

or disorder of a system

Symbol: SUnits: J/K mol∙

A process is more likely to occur if it is accompanied by an increase in entropy( ΔS is positive)

Factors that Affect EntropyEntropy increases as molecules or ions become

dispersed. (Diffusion)Entropy increases as solutions become more dilute

or when the pressure of a gas is reduced.Mixtures of gases have more entropy than a single

gas.Entropy increases when

total # moles product > total # moles reactantEntropy increases when a reaction produces more

gas particles, because gases are more disordered than liquid or solids.

Hess’s Law Also Applies to EntropyStandard Entropy, So: the entropy of 1

mole of a substance at a standard temperature, 298.15 K.

The entropy change of a reaction can be calculated by:

ΔSreaction = S˚products - S˚reactants

Elements can have standard entropies of formation that have values other than zero.

Practice #1, pg. 361

Find the change in entropy for the reaction below by using Table 4 and that S˚ for CH3OH(l) is 126.8 J/K·mol

CO(g) + 2 H2(g) CH3OH (l)

ΔSreaction = S˚product - S˚reactants

S˚product = (1mol)(126.8 J/K·mol)

= 126.8 J/KS˚reactant=[(1mol)(197.6 J/K·mol)+(2mol)(130.7J/K·mol)]

= 459 J/K

ΔSreaction = 126.8 J/K – 459 J/K = -332.2 J/K

ExampleCalculate the change in entropy for the following

reaction using Table A-11 on pg. 833.2 Na(s) + 2 HCl(g) 2 NaCl(s) + H2(g)

ΔSreaction = S˚product - S˚reactants

S˚product= [(2 mol)(72.1 J/mol K)+(1 mol)(130.7 J/mol K)]∙ ∙

= 274.9 J/KS˚reactants= [(2 mol)(51.5 J/mol K)+(2 mol)(186.8 J/mol K)]∙ ∙

= 476.6 J/K

Δsreaction= 274.9 J/K – 476.6 J/K = -201.7 J/K

Gibbs Energy

If ΔH is negative and ΔS is positive for a reaction, the reaction will likely occur.

If ΔH is positive and ΔS is negative for a reaction, the reaction will NOT occur.

How can you predict what will happen if ΔH and ΔS are both positive or negative?

Gibbs EnergyGibbs Energy: the energy in a system

that is available for work. (also called free energy)

Symbol: GG = H – TS

ORΔG = ΔH – TΔS

H = enthalpy (kJ or J)S = entropy (J/K)T = temperature (K)

Gibbs Energy Determines Spontaneity

Spontaneous reaction: a reaction that does occur or is likely to occur without continuous outside assistance, such as the input of energy.

Non-spontaneous reaction: a reaction that will never occur without assistance.

Spontaneous vs. non-spontaneous On a snow-covered mountain in winter, an

avalanche is a spontaneous process because it may or may not occur, but it always CAN occur.

The return of snow from the bottom of the mountain to the mountaintop is a non-spontaneous process, because it can NEVER happen without assistance.

Gibbs Energy Determines Spontaneity

If ΔG is negative, reaction is spontaneous

If ΔG is greater than 0, reaction is non-spontaneous

If ΔG is exactly 0, reaction is at equilibrium

Entropy and Enthalpy Determine Gibbs Energy

Standard Gibbs energy of formation: the change in energy that accompanies the formation of 1 mole of the substance from its elements at 298.15 K.

Symbol: ΔGfo

Unit: kJ/mol

ΔGreaction = ΔGf˚products – ΔGf˚reactants

Sample Problem G, pg. 364Given that the change in enthalpy and entropy are -139 kJ and 277 J/K respectively for the reaction given

below, calculate the change in Gibbs energy. Then, state whether the reaction is spontaneous at 25˚C.

C6H12O6(aq) 2 C2H5OH(aq) + 2 CO2(g)

ΔG = ΔH – TΔSΔH = -139 kJΔS = 277 J/K (Change to kJ 0.277 kJ)T = 25°C + 273 = 298 KΔG = (-139 kJ) – [(298K) (0.277 kJ/K)] = -222 kJ

Reaction is spontaneous because ΔG is negative.

Sample Problem H, pg. 365Use Table 5 to calculate ΔG for the following water-gas reaction with graphite.

C(s) + H2O(g) CO(g) + H2 (g)ΔGreaction = ΔGf

˚ products - ΔGf

˚ reactants

ΔGf˚

products = [(1mol)(-137.2 kJ/mol) + (1mol)(0)]= -137.2 kJ

ΔGf˚

reactants= [(1mol)(0) + (1mol)(-228.6 kJ/mol)]= -228.6 kJ

ΔGreaction = -137.2 kJ – (- 228.6 kJ) = 91.4 kJ

Predicting SpontaneityΔH ΔS ΔG Spontaneous

?

Negative Positive Negative Yes, at all Temps

Negative Negative Either Positive or Negative

Only if T < ΔH/ΔS

Positive Positive Either Positive or Negative

Only if T > ΔH/ΔS

Positive Negative Positive Never

Predicting Spontaneity

Since ΔG = ΔH – TΔS, temperature may greatly affect ΔG.

Increasing the temperature of a reaction can make a non-spontaneous reaction spontaneous.

Homework

Sec 10.4 review, pg. 367: #3-5, 7-11Concept Review: “Order and Spontaneity”

Next Time:Quiz over section 10.4/Lab Write-

upTues/Wed: Calorimetry Lab

Chapter 10 test/concept review due: Friday, 11-16: “A” Day

Monday, 11-19: “B”: Day