Unit 14 – ThermodynamicsThe study of energy

relationshipsChapter 19

Warm-Ups

• Complete ONE Page of the Chemical Equations Review per day

• ONE minute per question!!!

Spontaneous Reactions• Naturally occurring reactions (happens on

its own) that favor the formation of products

• Do not favor product formation nor provide substantial product yield.

*Does NOT refer to Rate! Reactions can have parts of both.

*Temperature and Pressure have great impacts on spontaneous reactions

Nonspontaneous Reactions

Free Energy, ∆G• Energy available to do work• No reaction is 100% efficient• Energy can only be obtained if the

reaction occurs (spontaneous)CO2 (g) C(s) + O2 (g)

• Measure of the disorder of a system• Law of Disorder: Processes will

spontaneously move in the direction of maximum disorder or randomness

Entropy, ∆S : Chemical Chaos

Entropy and Phase of Matter• Entropy of gases is greater than the

entropy of a liquid or solid• Liquid entropy is greater than a solid’s

entropy• Entropy for a reaction will increase when

a solid changes into a liquid or a gas.• Entropy also increases when a substance

is divided into parts.• Entropy increases when the total # of

product molecules is larger than that of the # of reactant molecules

• Temp Increase = Entropy Increase

The Meaning of the Signs of Thermodynamic Properties

Property Positive (+) Negative (-)∆H Endothermic

Energy INExothermicEnergy OUT

∆S More Disordered More Ordered = Less Disordered

∆G NOT Spontaneous

Spontaneous

Heat Change, ∆H

Entropy, ∆S Reaction Type

- (exothermic)

+ (more disorder)

SpontaneousHeat Release + More disorder =favorable

+ (endothermic)

+ SpontaneousGreater increase in entropy than heat absorbed

+ + Non-SpontaneousMore heat is absorbed than increase in entropy

- - (less disorder)

SpontaneousMore heat is released than the decrease in disorder

- - Non-SpontaneousNot enough heat is released compared to entropy decrease

+ - Non-SpontaneousAdditional Heat and less disorder = unfavorable

Example H2O (s) H2O (l)

• When is this process spontaneous?• At temperatures greater than 0oC

• When is this process non-spontaneous?

• At temperatures less than 0oC

• It all depends on the TEMPERATURE!

Calculating Entropy, ∆S• Quantitative measurement of Disorder

of a system• Symbol = ∆S• Units = J/K or J/(K*mol)• Standard Entropy = ∆So

at 25 oC and 101.3kPa• A perfect crystal at 0 K would have

∆S=0• ∆So = ΣSo

products - ΣSoreactants

Example (use Appendix p. 558)

H2 (g) + Cl2 (g) 2HCl (g)∆So = (2x186.7) – (130.6+223.0)∆So = 373.4 – 353.6 = 19.8 J/(K*mol)

FORWARD: ∆So = (2*69.94) - (2*130.6 + 205.0) ∆So = 139.88 - 466.2 = -326.32

J/(K*mol)REVERSE: ∆So = (2*130.6 + 205.0) – (2*69.94)

∆So = 466.2 – 139.88 = 326.32 J/(K*mol)

Calculate the ∆S for both the forward and reverse reactions for the

synthesis of water.

Free Energy Calculations• Gibbs Free Energy Change, ∆G• Maximum amount of energy that can be coupled

to another process to do useful work• Relates to Entropy and Enthalpy changes

∆G = ∆H - T∆S enthalpy temp. Entropy in K change• If ∆G is negative, the process is Spontaneous• If ∆G is positive, the process is Nonspontaneous• If ∆G is 0, the process is at equilibrium

Example: Calculate the Free Energy for the following reaction to

determine if it is spontaneous at 25oCC(s) + O2(g) CO2(g)

Watch units!!! Convert J to kJ

Then Calculate ∆ H and ∆ S

∆ H of 0.0 0.0 -393.5 kJ/mol

S o 5.69 205 214 J/(K/mol)

∆ H of 0.0 0.0 -393.5 kJ/mol

So .00569 0.205 0.214 kJ/(K/mol)

∆So =∆So products - ∆So reactants

= 0.214 – (0.0057 + 0.205)= 0.003 kJ/(K*mol)

∆Ho = ∆Hoproducts - ∆Ho

reactants

= -393.5 kJ/mol – (0.00 + 0.00)kJ/mol

= -393.5 kJ/mol

∆Go = ∆Ho – T∆So

T = 25oC = 298.15 K∆Go = ∆Ho – T∆So

= -393.5 kJ/mol – (298.15 K * 0.003kJ/(K*mol)

= -394.4 kJ/mol

Reaction is Spontaneous

because ∆Go is negative!

Predict whether the equilibrium lies to the left or to the right

Free Energy and Equilibrium∆Go = ∆Ho – T∆ So can be used algebraically to solve for any unknown quantity ∆Ho, T, or ∆ So

∆G = ∆Go + RT lnQ Free energy at Reaction Quotient Non-standard (compare to Keq )

Conditions

At standard conditions, Q = 1∆Go = - RT lnKeq

∆Go = negative, Keq > 1

∆Go = 0, Keq = 1

∆Go = positive, Keq < 1

Calculating Enthalpy, ∆H1. Calorimetry: q = mC∆T 2. Hess’s Law : Add reactions3. Hess’s Law : Σproducts – Σ reactants4. Bond Energies : Bonds Broken – Bonds

Formed

5. ∆Go = ∆Ho – T∆ So 6. ∆Go = -nF€o

7. ∆Go = - RT lnKeq

Calculating Free Energy, ∆G

Example: Calculate ∆H, ∆S, and ∆Go for the following reaction at 298K: Use the

appendix in your book to find the individual values.2SO2(g) + O2(g) 2SO3(g)

Example: Calculate ∆H, ∆S, and ∆Go for the following reaction at 298K: Use the appendix in your book to find the individual values.

2SO2(g) + O2(g) 2SO3(g)

∆ H = 2(-396) – 2(297) = -198 kJ/mol∆ S = 2(257) – [2(248) + 205] = -187 j/(K*mol) = -0.187 kJ/(K*mol)∆ G = ∆ H – T∆S = -198 –(298)(-0.187)

=-142 kJ/mol

∆ Hf -297 0.0 -396 kJ/mol S o 248 205 257 J/(K/mol)

Example: Consider the ammonia synthesis reaction: N2(g) + 3H2(g) 2NH3(g)

• Where ∆ G = -33.3 kJ/mol of N2 consumed at 25oC

• Calculate the value for the equilibrium constant.

∆Go = - RT lnKeq

Example: Consider the ammonia synthesis reaction: N2(g) + 3H2(g) 2NH3(g)• Where ∆ G = -33.3 kJ/mol of N2 consumed

at 25oC• Calculate the value for the equilibrium

constant. ∆Go = - RT lnKeq

∆Go = -33.3 kJ/mol = -33,300 J/molR = 8.3145 J/(K*mol)-33,300 = -(8.3145)(298K) (lnKeq)

13.4 = ln K

e13.4 = elnK K = 686780 (no units for Keq)

Does this favor products or reactants?

Example: Calculate the value of ∆Go for the following reaction at 389K where the [NH3] = 2.0 M, [H2] = 1.25 M, and [N2] =

3.01 M

• N2(g) + 3H2(g) 2NH3(g)

Example: Calculate the value of ∆Go for the following reaction at 389K where the [NH3] = 2.0 M, [H2] = 1.25 M, and [N2] = 3.01 M

N2(g) + 3H2(g) 2NH3(g)

3.01 1.25 2.0

Keq = (2.0)2 .

(3.01)(1.25)3