Free Energy and Temperature Free energy decreases (becomes more negative) as temperature At low T, G m for solid phase is lower than that of liquid or vapour, so the solid phase is prevalent As we increase T to T fus and higher, the liquid state has a lower G m, so it is the phase that prevails As we increase T further to T b, the gas phase has the lowest value of G m

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 We’ve seen something like this before somewhere…

Standard Gibbs Free Energy of Formation,  G f °  G f ° = 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 = ?  G f °is for the formation of 1 mole of product –Different amounts of reactants may be used…Be vigilant! 

 G f °: What Does it Mean? Compounds with  G f ° > 0 are Thermodynamically Unstable Compounds with  G f ° < 0 are Thermodynamically Stable

Gibbs Free Energy and Nonexpansion Work w e = ‘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 w e related?

Gibbs Free Energy and w e  G = w e 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 kJ For 1 mole of glucose, we get 2879 kJ of energy Or For 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 bonds One 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) C 6 H 12 O 6 (s) + 6 O 2 (g) --> 6 CO 2 (g) + 6 H 2 O (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° - T  S°

The Effect of Temperature on  G°

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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 = C  T = mC s  T nC m  T

System: Solution and chemicals that react Surroundings: Cup and the world around it! Assumptions: We use 2 cups to prevent energy transfer to the surroundings (we assume that it works as designed) Expected Changes: i)As the chemical reaction occurs, the potential energy in the reactants will be released as heat or the solution can supply heat to allow formation of a product with a higher potential energy ii)The solution will absorb or release energy during the reaction. We will see this as a temperature change q r + q solution = 0 q r + q solution = 0 Coffee Cup Calorimeter

We place 0.05g of Mg chips in a coffee cup calorimeter and add 100 mL of 1.0M HCl, and observe the temperature increase from 22.21°C to 24.46°C. What is the ΔH for the reaction? Mg(s) + 2HCl (aq) --> H 2 (g) + MgCl 2 (aq) Assume: C p of the solution = 4.20 J/gK Density of HCl is 1.00 g/mL Density of HCl is 1.00 g/mL Constant Pressure Calorimetry: An Example

To solve this: Δ T = (24.46°C – 22.21°C) = (297.61K – K)=2.25K Mass of solution = Now, let’s calculate q solution : q solution = mC m Δ T = (100.05g)(4.20 J/gK)(2.25K) = J = J Now, let’s calculate q r : q r = -q solution = J Constant Pressure Calorimetry: An Example

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 = q Enthalpy 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:

Entropy Change as a Function of Temperature at Constant Volume If T 2 > T 1, 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

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 V 2 > V 1, the entropy increases Note: Units are still J/K 

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: Entropy decreases for a samples that has been compressed isothermally (P 1 >P 2 ) 

Boiling Water and Entropy Let’s get 3 facts straight: 1.At a transition temperature (T f or T b ), 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 (at the boiling temperature)

Ice Melting and Entropy We use the same logic to determine the entropy of fusion,  S fus

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 = x 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 

The Surroundings If  S Tot is positive, the reaction is spontaneous If the  S system is negative, the reaction will still be spontaneous if  S Surr is that much more positive  S Tot =  S System +  S Surr

Gibbs Free Energy  G =  H - T  S (at constant T)  G = -T  S (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 T  S is large and positive, the value of  G may be large and negative