science ss1 thermochemistry 4 th class entropy and free energy

Science SS1Thermochemistry

4th ClassEntropy and Free Energy

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Thermodynamics vs. Kinetics

• Domain of Kinetics– Rate of a reaction depends

on the pathway from reactants to products.

• Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.


Spontaneous Processes and Entropy

• Thermodynamics lets us predict the direction in which a process will occur but gives no information about the speed of the process.

• A spontaneous process is one that occurs without outside intervention.


Entropy, ΔS

• The driving force for a spontaneous process is an increase in the entropy of the universe.

• A measure of molecular randomness or disorder.


The Expansion of An Ideal Gas Into an Evacuated Bulb

Positional Entropy

• A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system.

• Therefore: Ssolid < Sliquid << Sgas

Entropy

Entropy

ΔS°reaction = ΣnpS°products – ΣnrS°reactants

ΔG=ΔH-tΔS

Entropy ΔSJ/t×mole of Reaction

(Watch Units)

• Negative ----Going to order (Products more ordered than reactants)

• Positive -----Going to disorder (Product more disorder than reactants)

ΔSsurr

• ΔSsurr = +; entropy of the universe increases

• ΔSsurr = -; process is spontaneous in opposite direction

• ΔSsurr = 0; process has no tendency to occur



Predict the sign of ΔS for each of the following, and explain:

a) The evaporation of alcoholb) The freezing of waterc) Compressing an ideal gas at constant

temperatured) Heating an ideal gas at constant

pressuree) Dissolving NaCl in water

+––

+

+

CONCEPT CHECK!

ΔSsurr

• The sign of ΔSsurr depends on the direction of the heat flow.

• The magnitude of ΔSsurr depends on the temperature.


ΔSsurr

Heat flow (constant P) = change in enthalpy = ΔH


surr = H

ST

Free Energy

Free Energy

ΔG°reaction = ΣnpG°products

– ΣnrG°reactants

ΔG° = ΔH° – TΔS°

Equil ΔG° = -RTlnk

Electro ΔG° = -nfe

Free Energy ΔGKJ/mole of Reaction

• Negative = spontaneous reaction• Positive = non-spontaneous reaction (will not

occur at current conditions)• Zero = System is in equilibrium

Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is:

a) Exothermic and becomes more positionally randomSpontaneous

b) Exothermic and becomes less positionally randomCannot tell

c) Endothermic and becomes more positionally randomCannot tell

d) Endothermic and becomes less positionally randomNot spontaneous

Explain how temperature affects your answers.

CONCEPT CHECK!

Free Energy (G)

• A process (at constant T and P) is spontaneous in the direction in which the free energy decreases.– Negative ΔG means positive ΔSuniv.


univ = (at constant and ) G

S T PT

Relation between ΔG, ΔH, and ΔS

∆S

∆H

SpontaneousHigh temp

Always spontaneous

Never spontaneousSpontaneous Low temp

-∆H

-∆S

Based on the formula ΔG=ΔH-TΔS Δ G Δ H ΔS

Spontaneous at high temp

positve positive

Always spontaneous

negative positive

Spontaneous at low temp

negative negative

Never spontaneous

Positive negative

A liquid is vaporized at its boiling point. Predict the signs of:wqΔHΔSΔSsurr

ΔG

Explain your answers.


–+++–0

CONCEPT CHECK!

Third Law of Thermodynamics

• The entropy of a perfect crystal at 0 K is zero.• The entropy of a substance increases with

temperature.


A stable diatomic molecule spontaneously forms from its atoms.

Predict the signs of:ΔH° ΔS° ΔG°

Explain.


– – –

CONCEPT CHECK!

The Meaning of ΔG for a Chemical Reaction

• A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.


• The equilibrium point occurs at the lowest value of free energy available to the reaction system.

ΔG = 0 = ΔG° + RT ln(K)ΔG° = –RT ln(K)


Change in Free Energy to Reach Equilibrium


Question 1

Which of the following is a graph that describes the pathway of reaction that is exothermic and has high activation energy? A. B.

C. D.

Question 1 Breakdown

• Correct answer is C• What can you say about A?• What can you say about B?• What can you say about D?• Label Activation energy on each graph• Label ∆H for each graph

Question 2

When solid NH4SCN is mixed with solid Ba(OH)2 in a closed container, the temperature drops and a gas is produced. Which of the following indicates the correct signs for ΔG, ΔH, and ΔS for the process? ΔG ΔH ΔS A) – – –

B) – + –C) – + +D) + – +

Question 2 break down

• Correct answer is C• G is negative because it happens

spontaneously• S is positive since going from solid to gas• H is positive because it feels cold

(endothermic)

Question 3

N2(g) + 3 H2(g) → 2 NH3(g)

The reaction indicated above is thermodynamically spontaneous at 298 K, but becomes nonspontaneous at higher temperatures. Which of the following is true at 298 K?A) ΔG, ΔH, and ΔS are all positive.B) ΔG, ΔH, and ΔS are all negative.C) ΔG and ΔH are negative, but ΔS is positive.D) ΔG and ΔS are negative, but ΔH is positive.E) ΔG and ΔH are positive, but ΔS is negative.

Question 3 breakdown

• G = H - T S• Spontaneous means negative G.• S is negative since you are going from 4 gases

to 2 gases (disorder is leaving)• H must be negative if G is negative.• Correct answer is B.

Alternate process Question 3

• G is negative since it is spontaneous• G is becoming more positive as Temp

increases. Only S is affected by the temp. S must be negative since TS is positive.

• Since TS is positive H must be negative if G is negative.

Question 4

Of the following reactions, which involves the largest decrease in entropy?

A) 2 CO(g) + O2(g) → 2 CO2(g)

B) Pb(NO3)3(s) + 2 KI(s) → PbI2(s) + 2 KNO3(s)

C) C3H8(g) + O2(g) → 3 CO2(g) + 4 H2O(g)

D) 4 La(s) + 3 O2(g) → 2 La2O3(s)


• Correct answer D. Going from gas to solid• Which choice is largest increase? Why?

Question 5

Assume the data graphed was collected at aconstant pressure of 0.97 atm and represents four different temperature samples of pure neon gas. Which of the following temperatures most likely corresponds to the data graphed for sample “D”?

A) 273 KB) 298 KC) 305 KD) 338 K


• KE= ½ mv2

• Small molecules move faster than larger molecules. The high point of each curve hits the x axis in order from A to D. A is the slowest and biggest. D is the smallest and fastest.

• Temperature means molecules are moving faster therefore D is the highest temperature

Question 6

• At 298 K, as the salt MX dissolves spontaneously to form an aqueous solution, ∆S and ∆H are positive. Which describes the value of ∆G and the absolute values of its components, T∆S and ∆H?

• A) ∆G < 0; |T∆S| > |∆H|• B)∆G < 0; |T∆S| < |∆H|• C) ∆G > 0; |T∆S| > |∆H|• D) ∆G > 0; |T∆S| < |∆H|

Breakdown 6

• Spontaneous means G is negative or less than zero

• If both H and S are positive then S is the driving force of the spontaneity therefore if the reaction is spontaneous it must mean that TS is greater than H.

• Correct answer is A.

Free response 1

• C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)• When a 2.000-gram sample of pure phenol,

C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow.

substance Delta H KJ/mol@25

Delta S Joules/mol

CO2 −393.5 213.6

H2 0.00 130.6

H2O −285.85 69.91

O2 0.00 205.0

C6H5OH ? 144.0

• (a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.

• 2.00/94.12=.021 moles of phenol• .021moles/64.98 KJ= 1mole/X• X= -3058 KJ/mole = delta H• (b) Calculate the standard heat of formation,

ΔH°f, of phenol in kilojoules per mole at 25°C.• H= prod- react = -3058=(6(−393.5)+3(−285.85))-X• X=-161 J/K*mol

• (c) Calculate the value of the standard free-energy change, ΔG° for the combustion of phenol at 25°C.

• ∆ S= prod-react=(6(213.6)+3(69.91)) – (7(205.0) + 144.0)

• S= -87.67 J/mol• G= H-TS = -3058 – 298( -.08767 )• G = -3032 KJ/mol

Free Response 2• O3(g) + NO(g) → O2(g) + NO2(g)• Consider the reaction represented above.• (a) Referring to the data in the table below,

calculate the standard enthalpy change, ΔH, for the reaction atv25°C. Be sure to show your work.

KJ/mol O3 NO NO2

Std. enthalpy of formation∆Hf at 25◦C

143 90 33

• A. H= 33- (90+143)= -200 KJ/mol• (b) Make a qualitative prediction about the

magnitude of the standard entropy change, ΔS°, for the reaction at25°C. Justify your answer.

• 2 moles of gas to 2 moles of gas very small magnitude (very close to zero)

• (c) On the basis of your answers to parts (a) and (b), predict the sign of the standard free-energy change, ΔG°, for the reaction at 25°C. Explain your reasoning.

• H is negative and TS is very small after converting to KJ therefore G is negative.

Free Response 3• C2H2(g) + 2 H2(g) ---> C2H6(g)

• a) If the value of the standard entropy change, ΔS°, for the reaction is -232.7 joules per mole Kelvin, calculate the standard molar entropy, S°, of C2H6 gas.

• (b) Calculate the value of the standard free-energy change, ΔG°, for the reaction. What does the sign of ΔG° indicate about the reaction above?

• (c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K.

Substance S° (J/mol K) ΔH°f (kJ/mol)

C2H2(g) 200.9 226.7

H2(g) 130.7 0

C2H6(g) -------- -84.7

• S = prod –react = -237.2= X – (200.9 + 2(130.7))• X = 229.6 J/mol• B. H = prod –react= -84.7–226.7= -311.4 KJ/mol• G = H – TS = -311.4 – 298(-.2372)• G = -242.1 KJ/mol

• C. G = -RTlnK• -242000= -(8.31)(298)lnK• ln K = 97.7• K = 3x 1042

Formulas tell you what to do.

• ∆E = q (heat) + w(work)• q=mC∆T (water’s specific heat is 4.184 J/g °C)

• ∆H, S, G = (sum of products) – (sum of reactants)• ∆H bond energy = (sum of reactants) – (sum of products)

• ∆G = H-TS (watch for units on S)• ∆G=-RTlnK (watch for units on G)• ∆G=-nFE • Hess’s law match the equation and add up ∆Hrxn

(changes affect equation and ∆H)

science ss1 thermochemistry 4 th class entropy and free energy

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