Chapter 17: Free Energy and

Thermodynamics (again)

Mrs. Brayfield

17.2: Spontaneous and Nonspontaneous

Processes

A spontaneous process is one that occurs without

outside intervention

For example, when you drop a book it falls to the floor

In chemistry, we want to know the chemical potential

that will predict the direction of a chemical system

Spontaneous vs Nonspontaneous

DO NOT confuse spontaneity of a reaction with its

speed

Thermodynamics is the direction and the extent to which a

chemical reaction will proceed

Kinetics is the speed – how fast the reaction takes place

Thermodynamic ≠ Kinetic

Spontaneous vs Nonspontaneous

The conversion of diamond to graphite is

thermodynamically spontaneous, but the process is VERY

slow – so diamonds won’t turn to graphite quickly

Also, the rate of a reaction can be increased with a catalyst, but

a nonspontaneous process still will not go with one

A nonspontaneous process is not impossible, it can

happen

You just need to supply enough energy to it

17.3: Entropy and the 2nd Law

Remember entropy back from chapter 6…

Entropy is a function that increases with the number of

energetically equivalent ways to arrange the components

of a system to achieve a particular state (aka disorder)

This is melting ice (endothermic) happens spontaneously…

As the ice melts, the molecules become more “disordered” (or they

have more energy states), so entropy has increased (∴ spontaneous)

Entropy

The definition of entropy can be mathematically defined:

𝑆 = 𝑘𝑙𝑛𝑊 [Ludwig Boltzmann]

Entropy, like enthalpy (H) is a state function:

∆𝑆 = 𝑆𝑓𝑖𝑛𝑎𝑙 − 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙

A chemical system proceeds in the direction that

increases the entropy of the universe (or towards the

largest number of ways to arrange the components)

Entropy

How many ways can we arrange the following gas

molecules?

2nd Law of Thermodynamics

The second law of thermodynamics states:

For any spontaneous process, the entropy of the universe

increases (∆Suniv > 0)

Entropy Changes with States

Prediction of ∆S

In general, entropy increases (∆S > 0) for each of the

following:

Phase transition from solid to liquid

Phase transition from solid to gas

Phase transition from liquid to gas

An increase in the number of moles of gas during a chemical

reaction

Predicting

Predict the sign of ∆S for each of the following processes:

1. The boiling of water

2. I2(g) → I2(s)

3. CaCO3(s) → CaO s + CO2(g)

positive

positive

negative

Homework Problems: #2, 3, 6, 8, 10, 12

17.4: Heat Transfer and Changes in Entropy

If going from a liquid to a solid is spontaneous, why does

entropy decrease?

Well it doesn’t, even though you would think that.

Why not?

∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟

So the entropy of the system CAN decrease, as long as the

entropy of the surroundings increases by a greater amount

So that the entropy of the universe still increases

Spontaneity of Water Freezing

If we look at water freezing at a low temperature:

But if we looked at a high temperature:

This is like giving a rich verses a poor man $1000

Entropy Changes

In general…

A process that emits heat into the surroundings (qsys negative)

increases the entropy of the surroundings (positive ∆Ssurr)

A process that absorbs heat from the surroundings (qsys

positive) decreases the entropy of the surroundings (negative

∆Ssurr)

The magnitude of the change in entropy of the surroundings is

proportional to the magnitude of qsys

We then get the relationship:

∆𝑆𝑠𝑢𝑟𝑟 =−∆𝐻𝑠𝑦𝑠

𝑇 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃, 𝑇)

If at constant pressure qsys = ∆Hsys

Calculating Entropy Changes Example

Consider the following reaction:

2𝑁2 𝑔 + 𝑂2 𝑔 → 2𝑁2𝑂 𝑔 ∆𝐻𝑟𝑥𝑛 = +163.2𝑘𝐽

a. Calculate the entropy change in the surroundings

associated with this reaction occurring at 25˚C.

∆𝑆𝑠𝑢𝑟𝑟 =−∆𝐻

𝑇=−(163.2𝑘𝐽)

298𝐾= −548𝐽/𝐾

b. Determine the sign of the entropy change for the

system.

ΔSsys is negative (less moles on product side)

c. Determine the sign of the entropy change for the

universe. Will the reaction be spontaneous?

∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟

∆𝑆𝑢𝑛𝑖𝑣 = − + −

Therefore ΔSuniv is negative -> process is not spontaneous

17.5: Gibbs Free Energy

Using the equations for this chapter, we can solve for a

new thermodynamic function: (see page 653)

∆𝐺 = ∆𝐻 − 𝑇∆𝑆

Where G is Gibbs free energy

If you know the sign of G (positive or negative), you can

tell if the process is spontaneous or not

If ∆G is negative, the process is spontaneous

If ∆G is positive, the process is nonspontaneous

Gibbs Free Energy

Gibbs Free Energy

Case 1: ∆H negative, ∆S positive

G will be negative at all temperatures (∴ spontaneous)

Case 2: ∆H positive, ∆S negative

G will be positive at all temperature (∴ nonspontaneous)

Gibbs Free Energy

Case 3: ∆H negative, ∆S negative

Case 4: ∆H negative, ∆S positive

Predicting Spontaneity Example

Consider the following reaction:

𝐶2𝐻4 𝑔 + 𝐻2 𝑔 → 𝐶2𝐻6 𝑔 ∆𝐻 = −137.5𝑘𝐽; ∆𝑆 = −120.5𝐽/𝐾

Calculate ∆G at 25˚C and determine whether the

reaction is spontaneous. Does ∆G become more negative

or more positive as the temperature increases?

∆𝐺 = ∆𝐻 − 𝑇∆𝑆 = −317 × 103𝐽 − 298𝐾 −120.5𝐽

𝐾

∆𝐺 = −101.6𝑘𝐽

Reaction is spontaneous. As temperature increase, G

becomes more positive

Homework Problems: #14, 15, 17, 22, 24, 26, 28, 32, 38

17.6: Entropy Changes in Reactions

To find the change in entropy for any reaction, use an

equation that is similar to one we’ve already seen:

∆𝑆𝑟𝑥𝑛° = ∆𝑆𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠

° + ∆𝑆𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠°

Just use the standard molar entropy values (pg. 657 and

appendix IIB)

The third law of thermodynamics states:

The entropy of a perfect crystal at absolute zero (0K) is zero.

Entropy Changes Example

Calculate ∆Srxn˚ for the following reaction:

2𝐻2𝑆 𝑔 + 3𝑂2 𝑔 → 2𝐻2𝑂 𝑔 + 2𝑆𝑂2(𝑔)

∆𝑆𝑟𝑥𝑛°= 𝑛𝑝𝑆

°(𝑝𝑟𝑜𝑑) − 𝑛𝑟𝑆° 𝑟𝑒𝑎𝑐𝑡

∆𝑆𝑟𝑥𝑛°= 2𝑚𝑜𝑙 188.8

𝐽

𝑚𝑜𝑙𝐾+ 2𝑚𝑜𝑙 248.2

𝐽

𝑚𝑜𝑙𝐾

− 2𝑚𝑜𝑙 205.8𝐽

𝑚𝑜𝑙𝐾+ 3𝑚𝑜𝑙 205.2

𝐽

𝑚𝑜𝑙𝐾

∆𝑆𝑟𝑥𝑛°= 874

𝐽

𝐾− 1027

𝐽

𝐾= −153.2

𝐽

𝐾

17.7: Free Energy Changes

Using all of the thermodynamics information (and tables),

we can calculate the free energy change of a reaction:

∆𝐺𝑟𝑥𝑛° = ∆𝐻𝑟𝑥𝑛

° − 𝑇∆𝑆𝑟𝑥𝑛°

Example:

For the following reaction:

𝑁𝑂 𝑔 +1

2𝑂2(𝑔) → 𝑁𝑂2(𝑔)

Compute ∆Grxn˚ at 25˚C and determine if the process is

spontaneous.

∆𝐻𝑟𝑥𝑛°= 33.2

𝑘𝐽𝑚𝑜𝑙 − 91.3

𝑘𝐽𝑚𝑜𝑙 +

1

20

∆𝐻𝑟𝑥𝑛°= −58.1𝑘𝐽

∆𝑆𝑟𝑥𝑛°= 240.1

𝐽𝑚𝑜𝑙𝐾

− 210.8𝐽𝑚𝑜𝑙𝐾 +

1

2205.2𝐽𝑚𝑜𝑙𝐾

∆𝑆𝑟𝑥𝑛°= −73.2

𝐽𝐾

∆𝐺𝑟𝑥𝑛°= ∆𝐻𝑟𝑥𝑛

° − 𝑇∆𝑆𝑟𝑥𝑛°

∆𝐺𝑟𝑥𝑛°= −58.1 × 103𝐽 − 298𝐾 −73.2

𝐽𝐾

= −36.3𝑘𝐽 ∴ 𝑟𝑥𝑛 𝑖𝑠 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠

17.8: Free Energy Changes with

Nonstandard States

Just use the equation:

∆𝐺𝑟𝑥𝑛 = ∆𝐺𝑟𝑥𝑛° + 𝑅𝑇𝑙𝑛𝑄

Where Q is the reaction quotient (equilibrium)

T in K

R is gas constant

8.314 J/mol•K

Example

Consider the following reaction at 298K:

2𝐻2𝑆 𝑔 + 𝑆𝑂2 𝑔 → 2𝑆 𝑠 + 2𝐻2𝑂 𝑔 ∆𝐺𝑟𝑥𝑛

°= −102𝑘𝐽 Compute ∆Grxn under the following conditions:

𝑃𝐻2𝑆 = 2.00𝑎𝑡𝑚 𝑃𝑆𝑂

2= 1.50𝑎𝑡𝑚 𝑃𝐻

2𝑂 = 0.0100𝑎𝑡𝑚

Is the reaction more or less spontaneous under these conditions than under standard states?

𝑄 =𝑃𝐻2𝑂

2

𝑃𝐻2𝑆2𝑃𝑆𝑂2=(0.1)2

2 2(1.5)= 1.67 × 10−5

∆𝐺𝑟𝑥𝑛 = ∆𝐺° + 𝑅𝑇𝑙𝑛𝑄

= −102kJ + (8.314𝐽

𝑚𝑜𝑙𝐾)(298𝐾)(ln 1.67 × 10−5

= −129𝑘𝐽 𝑡ℎ𝑒 𝑟𝑥𝑛 𝑖𝑠 𝑚𝑜𝑟𝑒 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠

17.9: Free Energy and Equilibrium

If we know that at equilibrium, ∆Grxn = 0 and Q = K,

∆𝐺𝑟𝑥𝑛° = −𝑅𝑇𝑙𝑛𝐾

When K < 1, lnK is negative, ∆Grxn˚ is positive

Reaction is spontaneous in reverse direction

When K > 1, lnK is positive, ∆Grxn˚ is negative

Reaction is spontaneous in forward direction

When K = 1, lnK is zero, ∆Grxn˚ is zero

Reaction is at equilibrium

Example

Compute ∆Grxn˚ at 298K for the following reaction:

𝐼2 𝑔 + 𝐶𝑙2 𝑔 ↔ 2𝐼𝐶𝑙 𝑔 𝐾𝑝 = 81.9

∆𝐺𝑟𝑥𝑛°= −𝑅𝑇𝑙𝑛𝐾

∆𝐺𝑟𝑥𝑛°= −(8.314

𝐽

𝑚𝑜𝑙𝐾)(298𝐾)(ln 81.9

= −10.9𝑘𝐽𝑚𝑜𝑙

Homework Problems: #40, 42, 46

Review: #50, 51, 53

Entropy:

http://www.youtube.com/watch?v=ZsY4WcQOrfk&index

=21&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr

Electrochemistry:

http://www.youtube.com/watch?v=IV4IUsholjg&index=38

&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr