Free Energy and Temperature

•Free energy decreases (becomes more negative) as temperature

•At low T, Gm for solid phase is lower than that of liquid or vapour, so the solid phase is prevalent

•As we increase T to Tfus and higher, the liquid state has a lower Gm, so it is the phase that prevails

•As we increase T further to Tb, the gas phase has the lowest value of Gm

7.13: Gibbs Free Energy of Reaction

• To determine the spontaneity of a reaction, we use the change in the Gibbs Free Energy, G, or the Gibbs Free Energy of Reaction

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G = nGm Products∑ − nGm Reactants∑ We’ve seen something like this before somewhere…

Standard Gibbs Free Energy of Formation, Gf°

Gf° = The standard Gibbs Free Energy of reaction per mole for the formation of a compound from its elements in their most stable form.

• Most stable form?– Hydrogen = ?– Oxygen = ?– Iodine = ?– Sodium = ?

Gf°is for the formation of 1 mole of product– Different amounts of reactants may be used…Be vigilant!

Gf°: What Does it Mean?

• Compounds with Gf° > 0 are Thermodynamically Unstable

• Compounds with Gf° < 0 are Thermodynamically Stable

Gibbs Free Energy and Nonexpansion Work

• we = ‘Extra work’– Nonexpansion work is any kind of work other than that done

against an opposing pressure

• Stretching a spring, moving a rope, importing a sugar molecule into a cell are all examples of nonexpansion work

• All cellular processes are examples of nonexpansion work

• How are the Gibbs Free Energy and we related?

Gibbs Free Energy and we

G = we

• If we know the change in free energy, we know how much nonexpansion work can be done

• What does this mean?– Let’s look at the combustion of glucose.

G° and the Combustion of Glucose

• The G° of the reaction is -2879 kJFor 1 mole of glucose, we get 2879 kJ of energy

OrFor 180 g of glucose, we get 2879 kJ of energy

• To make one mole of peptide bonds, 17 kJ of work must be done.– If we get 2879 kJ of energy from one mole of glucose, we

should be able to make 170 moles of peptide bondsOne molecule of glucose will provide enough

energy to add 170 amino acids to a growing protein (in actuality, you can only add 10 amino acids)

C6H12O6 (s) + 6 O2 (g) --> 6 CO2 (g) + 6 H2O (l)

The Effect of Temperature on G°

• Remember that H° or S° is the sum of the individual enthalpies or entropies of the products minus those of the reactants

• If we change the temperature, both are affected to the same extent, so the H° and S° values don’t significantly change

• This is not the case with G°. Why?

G° = H° - TS°

The Effect of Temperature on G°

The Effect of Temperature on G°

The Effect of Temperature on G°

The Effect of Temperature on G°

Thermodynamics Review

Let’s Look at some of the most important equations we’ve covered over the past 2 chapters…

The First Law of Thermodynamics

• Up until now, we have only considered the changes in the internal energy of a system as functions of a single change: either work or heat

• However, these changes rarely occur singly, so we can describe the change in internal energy as:

U = q + w (The 1st Law)

• The change in internal energy is dependent upon the work done by the system and the heat gained or lost by the system

Heat Capacity

q = CT = mCsT

nCmT

System: Solution and chemicals that reactSystem: Solution and chemicals that react

Surroundings: Cup and the world around it!Surroundings: Cup and the world around it!

Assumptions: We use 2 cups to prevent energy Assumptions: We use 2 cups to prevent energy transfer to the surroundings (we assume that it transfer to the surroundings (we assume that it works as designed)works as designed)

Expected Changes:Expected Changes:

i)i) As the chemical reaction occurs, the potential As the chemical reaction occurs, the potential energy in the reactants will be released as heat or energy in the reactants will be released as heat or the solution can supply heat to allow formation of a the solution can supply heat to allow formation of a product with a higher potential energyproduct with a higher potential energy

ii)ii) The solution will absorb or release energy during The solution will absorb or release energy during the reaction. We will see this as a temperature the reaction. We will see this as a temperature changechange

qqrr + q + qsolutionsolution = 0 = 0

Coffee Cup CalorimeterCoffee Cup Calorimeter

We place 0.05g of Mg chips in a coffee cup We place 0.05g of Mg chips in a coffee cup calorimeter and add 100 mL of 1.0M HCl, and calorimeter and add 100 mL of 1.0M HCl, and observe the temperature increase from 22.21observe the temperature increase from 22.21°C °C to 24.46°C. What is the to 24.46°C. What is the ΔΔH for the reaction?H for the reaction?

Mg(s) + 2HCl (aq) Mg(s) + 2HCl (aq) -->--> H H22(g) + MgCl(g) + MgCl22(aq)(aq)

Assume: CAssume: Cpp of the solution = 4.20 J/gK of the solution = 4.20 J/gK

Density of HCl is 1.00 g/mLDensity of HCl is 1.00 g/mL

Constant Pressure Constant Pressure Calorimetry: An ExampleCalorimetry: An Example

To solve this:To solve this:

ΔΔT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25KT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25K

Mass of solution = Mass of solution =

Now, let’s calculate qNow, let’s calculate qsolutionsolution::

qqsolutionsolution = mC = mCmmΔΔT = T = (100.05g)(4.20 J/gK)(2.25K)(100.05g)(4.20 J/gK)(2.25K)

= 945.5 J= 945.5 J

Now, let’s calculate qNow, let’s calculate qrr::

qqrr = -q = -qsolutionsolution = -945.5 J = -945.5 J

Constant Pressure Constant Pressure Calorimetry: An ExampleCalorimetry: An Example

g 100.05 0.05g ml

1g x ml100 =+⎟

⎠

⎞⎜⎝

⎛

Enthalpy

• In a constant volume system in which no work is done (neither expansion nor non-expansion), we can rearrange the first law to:

U = q + w (but w=0)

U = q• Most systems are constant pressure systems which can

expand and contract• When a chemical reaction takes place in such a system, if gas

is evolved, it has to push against the atmosphere in order to leave the liquid or solid phase

– Just because there’s no piston, it doesn’t mean that no work is done!

Enthalpy• Let’s look at an example:

• If we supply 100J of heat to a system at constant pressure and it does 20J of work during expansion, the U of the system is +80J (w=-20J)

– We can’t lose energy like this

• Enthalpy, H, is a state function that we use to track energy changes at constant pressure

H=U + PV

• The change in enthalpy of a system (H) is equal to the heat released or absorbed at constant pressure

Enthalpy• Another way to define enthalpy is at constant

pressure:

H = qEnthalpy is a tricky thing to grasp, but we can look at it this way:• Enthalpy is the macroscopic energy change (in the form of

heat) that accompanies changes at the atomic level (bond formation or breaking)

• Enthalpy has the same sign convention as work, q and U– If energy is released as heat during a chemical reaction

the enthalpy has a ‘-’ sign– If energy is absorbed as heat from the surrounding during

a reaction, the enthalpy has a ‘+’ sign

The 2nd Law of Thermodynamics

The entropy of an isolated system

increases in the course of any

spontaneous change

• We can summarize this law mathematically as:

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S =q

T

Entropy Change as a Function of Temperature at Constant Volume

• If T2 > T1, then the logarithm is ‘+’ and entropy increases– Makes sense since we are raising the temperature and

thermal motion will increase

• The greater (higher) the heat capacity, the higher the entropy change

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S = C lnT2

T1

⎛

⎝ ⎜

⎞

⎠ ⎟

Entropy Change as a Function of Changing Volume

• We can use a similar logic to derive the change in entropy when the volume changes:

• When V2 > V1, the entropy increases

• Note: Units are still J/K€

S = nR lnV2

V1

⎛

⎝ ⎜

⎞

⎠ ⎟

Entropy Change as a Function of Pressure

• Remember Boyle’s Law?

• We can substitute this relationship into the equation for entropy change as a function of volume to get:

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S = nR lnP1

P2

⎛

⎝ ⎜

⎞

⎠ ⎟

Entropy decreases for a samples that has been compressed isothermally (P1>P2)

Boiling Water and Entropy

Let’s get 3 facts straight:

1. At a transition temperature (Tf or Tb), the temperature remains constant until the phase change is complete

2. At the transition temperature, the transfer of heat is reversible

3. Because we are at constant pressure, the heat supplied is equal to the enthalpy

Water Boiling and Entropy

• We use the ‘ º ’ superscript to denote the standard entropy or the entropy at 1 bar of pressure

€

S =ΔHvap

Tb

(at the boiling temperature)

€

So =ΔHvap

o

Tb

Ice Melting and Entropy

• We use the same logic to determine the entropy of fusion, Sfus

€

Sfus =ΔH fus

Tb

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Sfuso =

ΔH fuso

Tb

The Boltzmann Formula

• W is a reflection of the ensemble, the collection of molecules in the system

• This entropy value is called the statistical entropy

S = k lnW

Where:

k = Boltzmann’s constant

= 1.381 x10-23 J/K

W=# of ways atoms or molecules in the system can be arranged and still give the same total energy

Calculating the Entropy of a Reaction

• Sometimes we can’t always use our judgement and we need to calculate the entropy

• In order to do this, we need the standard molar entropies of the products and the reactants as well as the number of moles of each

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S° = nSm°∑ (products) - nSm

°∑ (reactants)

The Surroundings

• If STot is positive, the reaction is spontaneous

• If the Ssystem is negative, the reaction will still be spontaneous if SSurr is that much more positive

STot = SSystem + SSurr

Gibbs Free Energy

G = H - TS (at constant T)

G = -TS (at constant T and P)

• A negative value of G indicates that a reaction will spontaneously occur

• Large negative H values (like we’d have in a combustion reaction) would probably give you a large negative G

• If TS is large and positive, the value of G may be large and negative