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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