AP CHEMISTRY NOTES 4-1 THERMOCHEMISTRY: ENTHALPY AND ENTROPY

Reaction Rate – how fast a chemical reaction occurs

Collision Theory In order for a chemical reaction to occur, the following conditions must be met:

1) The reacting particles must collide – the more collisions, the faster the reaction occurs

2) The reacting particles must collide with enough energy to cause bonds to break and new bonds to form

3) The reacting particles must collide with the correct orientation

Factors Affecting the Rate of a Chemical Reaction

*Temperature –higher To = faster the motion of particles = more collisions

*Concentration – higher concentration = greater number of particles = more collisions

*Particle Size (Surface Area) – smaller particle size = greater surface area = more room for collisions

*Catalyst – increases the rate of a chemical reaction by lowering the activation energy “hump” (see below); catalysts are not themselves changed in the chemical reaction

SPONTANEITY OF CHEMICAL REACTIONS

Just because substances are mixed together does not mean they will react. Whether or not a reaction is spontaneous (ie. occurring on its own without the the continual addition of energy) is determined by two factors:

1. Enthalpy (H) – the heat gained or lost during a chemical reaction

*endothermic reaction – a chemical reaction in which heat energy is absorbed; products are higher in heat content than the reactants (although the container, etc., surrounding the reaction system is colder since heat has been removed from them to feed the reaction)

*exothermic reaction - a chemical reaction in which heat energy is given off; products are lower in heat content than the reactants (although the container, etc., surrounding the reaction system is warmer since heat has been given to them)

*Reactions tend to favor the lowest heat or energy state, so exothermic reactions are thermodynamically favored POTENTIAL ENERGY DIAGRAMS:

Note that both types of reactions require a certain amount of activation energy (Ea) – the amount of energy needed to begin a chemical reaction. The activation energy for an endothermic reaction is considerably larger than the activation energy for an exothermic reaction.

The activated complex is a transition state consisting of unstable “intermediate” forms of the atoms/molecules involved in the chemical reaction as reactants react to form the products.

The heat of reaction (Hrxn) is the energy change involved when a chemical reaction occurs. Note that the heat of reaction will be negative (-Hrxn) for an exothermic reaction (since heat is given off) and positive (+Hrxn) for an endothermic reaction (since heat is taken in or absorbed). 2. Entropy (S) – the amount of disorder (energy and positional microstates) in a system

*The Second Law of Thermodynamics states that everything in the universe moves toward greater disorder

*Reactions are “entropically” favored when the products have a higher state of entropy than the reactants.

In chemical reactions: *Entropy of a gas > Entropy of a liquid > Entropy of a solid (more energy and positional microstates) (fewer energy and positional microstates) (more disorder) (less disorder)

*Entropy increases when a substance is divided into parts (the more parts / particles, the greater the entropy)

*Entropy favors an increase in temperature

EXAMPLES:

______Cu(s) > Cu(l) Entropy ______

______MgCO3(s) > MgO(s) + CO2(g) Entropy ______

2+ 2- ______Ba (aq) + SO4 (aq) > BaSO4(s) Entropy ______

AP CHEMISTRY NOTES 4-2 THERMOCHEMISTRY: ENTHALPY AND ENTROPY CALCULATIONS

The “change in enthalpy” (H) of a reaction can be calculated using the following equation:

o o H = Hf (products) - Hf (reactants)

o *NOTE: 1. Hf is the standard heat of formation (the enthalpy change involved when a compound is formed from its elements )

o 2. Hf of an element in its free state (including diatomic elements) is “0”

EXAMPLE: Calculate the enthalpy change for the following reaction, determine whether it is endothermic or exothermic, and determine whether it is thermodynamically favored or thermodynamically un-favored:

______CH4(g) + 3/2 O2(g) > 2H2O(l) + CO(g)

The “change in entropy” S) can be determined using the following equation:

S = So (products) - So (reactants)

EXAMPLE: Calculate the entropy change for the following reaction, and determine whether it is entropically favored or not.

______CH4(g) + 3/2 O2(g) > 2H2O(l) + CO(g)

BOTH enthalpy and entropy act together to determine whether the reaction is thermodynamically favored. Unfortunately (depending on your point of view!) these two don’t always work together.

A reaction that has a ____ H and a ____ S will always be thermodynamically favored. (spontaneous)

A reaction that has a ____ H and a ____ S will always be thermodynamically unfavored. (non-spontaneous)

However, in situations in which H and S have the same signs (either both negative or both positive), it is necessary to calculate the change in free energy (G) in order to determine whether the reaction is thermodynamically favored (spontaneous) or not.

Gibb’s Free Energy – the maximum amount of energy that can be coupled to another process to do useful work (ie. the total energy available in the reaction system)

*Gibb’s Free Energy depends upon the size and direction of enthalpy and entropy changes

Exergonic reactions – thermodynamically favored (spontaneous) reactions which release free energy (-G)

Endergonic reactions – thermodynamically “un-favored” (non-spontaneous) reactions which absorb free energy (+G)

Gibb’s Free energy can be calculated using the following equation:

Go = Ho - TSo

(Note that temperature is in Kelvins)

EXAMPLE: Determine whether the following reaction is thermodynamically favored at 25oC:

______CH4(g) + 3/2O2(g) > 2H2O(l) + CO(g)

AP CHEMISTRY NOTES 4-3 THERMOCHEMISTRY – WORK AND HEAT

BASIC TERMS:

1. Thermodynamics – the study of energy and its interconversions

2. Energy – the capacity to do work

3. Kinetic Energy – the energy of motion

4. Potential Energy – energy that can be converted into useful work

5. Heat – the transfer of energy between two objects

6. Work – force times distance

7. State Function – a property independent of pathway (P, V, T, H, S, G, your bank account)

8. System – that which we focus on

9. Surroundings – everything else in the universe

10. Exothermic – energy (as heat) flows out of the system (-H) *number value shown on the products side of a chemical equation

11. Endothermic – energy (as heat) flows into the system (+H) *number value shown on the reactants side of a chemical equation

12. Internal Energy (E) – the sum of all of the potential and kinetic energy of the system

13. Enthalpy (H) – the heat content of the system

14. Entropy (S) – a measure of disorder (or chaos)

15. Free Energy (G) – the energy available in a system to do work

*THREE LAWS OF THERMODYNAMICS

1. The First Law of Thermodynamics: The energy of the universe is constant

2. The Second Law of Thermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe

3. The Third Law of Thermodynamics: The entropy of a perfect crystal at 0 Kelvins is zero *ENERGY

Internal energy can be changed by a) a flow of heat b) a flow of work c) a flow of both work and heat

This change in internal energy (E) can be calculated by using the following equation:

E = q + w

q = heat (in “cal” or “J”) w = work (in “cal” or “J”)

+q = heat absorbed +w = work done on the system - q = heat released - w = work done by the system

+E = system gains energy - E = system loses energy

EXAMPLES:

Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J.

Calculate the change in internal energy for a system undergoing an exothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

*GASES AND WORK

A common type of work accomplished with chemical processes is that of “work done to a gas” (through compression of the gas) or “work done by a gas” (through expansion of the gas).

To determine the amount of work involved when a gas expands or is compressed, the following equation is used:

w = - PV where P = pressure (in atmospheres) V = change in volume (in liters)

Also recall the following equation:

q = mCp∆T where q = heat (in Joules or calories) m = mass (in grams) or moles .o .o Cp = specific heat (J/g C or cal/g C) or molar heat capacity (J/mol.oC) ∆T = change in temperature (oC)

Note – “Kelvins” may be used in place of oC with no change in numerical value

EXAMPLES:

Calculate the work associated with the expansion of a gas from 46.0L to 64.0 L at a constant external pressure of 15atm. Express the answer in both L.atm and Joules. (1 L.atm = 101.3 J)

A balloon is being inflated to its full extent by heating the 89.3 kg of helium inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L under a constant pressure of 620 torr. The temperature of the balloon rises from 23.3oC to 79.4oC. Determine “q”, “w”, and E (in kJ) for this process. (1 L.atm = 101.3 J; the molar heat capacity of helium is 20.8 J/mol.oC)

AP CHEMISTRY NOTES 4-4

THERMOCHEMISTRY: ENTHALPY & CHEMICAL REACTIONS

When the amount of heat involved in a chemical process is calculated, it is often necessary to use the specific heat capacity of a substance in that calculation. Depending upon the information available to us, however, it may be necessary to convert specific heat capacities to “molar heat capacities” (the amount of heat required to change the temperature of 1 mole of a substance by 1oC) and vice vesa.

EXAMPLE: The specific heat capacity of water is 4.18 J/g.oC. Determine the molar heat capacity of water.

The change in enthalpy (H) can be calculated for specific amounts of reactants and products involved in a chemical reactions.

EXAMPLE:

The equation for the fermentation of glucose to alcohol and carbon dioxide is:

______C6H12O6 > 2C2H5OH + 2CO2

The enthalpy change for the reaction is -67 kJ. Is this reaction endothermic or exothermic? Is energy, in the form of heat, absorbed or released during the reaction? Put the heat term in the proper position in the equation above. Calculate the enthalpy change that occurs with the production of 100.0 grams of ethanol.

EXAMPLE:

Consider the combustion of propane:

______C3H8 + 5O2 > 3CO2 + 4H2O

If 0.500 g of propane produces 25.2 kJ of energy when it reacts, what is the molar heat of combustion (H) of propane?

AP CHEMISTRY NOTES 4-5

THERMOCHEMISTRY: ENTHALPY & PHASE CHANGES

When substances move from one phase to another, enthalpy changes are involved in the process:

HEATING CURVE

A → B ______

B → C ______

C → D ______

D → E ______

E → F ______

EXAMPLE: Determine the amount of heat required to change the temperature of 15 grams of H2O2 from ice at -5.0oC to steam at 122oC.

Specific Heat of Water = 4.18 J/g.oC Specific Heat of Ice = 2.0 J/g.oC Specific Heat of Steam = 2.0 J/g.oC Heat of Vaporization = 49.8 kJ/mol Heat of Fusion = 6.0 kJ/mol

Specific Heat Capacity Substance at 25oC in J/goC

H2 gas 14.267 He gas 5.300

H2O(l) 4.184 lithium 3.56 ethyl alcohol 2.460 ethylene glycol 2.200 ice @ 0oC 2.010 steam @ 100oC 2.010 vegetable oil 2.000 sodium 1.23 air 1.020 magnesium 1.020 aluminum 0.900 Concrete 0.880 glass 0.840 potassium 0.75 sulphur 0.73 calcium 0.650 iron 0.444 nickel 0.440 zinc 0.39 copper 0.385 brass 0.380 sand 0.290 silver 0.240 tin 0.21 lead 0.160 mercury 0.14 gold 0.129

AP CHEMISTRY NOTES 4-6

THERMOCHEMISTRY: CALORIMETRY

CALORIMETRY – the science of measuring heat flow, based on observing temperature change when a quantity of matter absorbs or releases heat.

There are two types of calorimetry:

*Coffee-Cup Calorimetry (“constant pressure” calorimetry) – the type normally done in labs

* Bomb Calorimetry (“constant volume” calorimetry) – weighed reactants are placed inside of a rigid steel container and ignited

Coffee Cup Calorimeter Bomb Calorimeter

In a thermochemical reaction carried out in a calorimeter:

The heat lost by one part of the system will equal the heat gained by another part of the system

the amount of heat = the amount of heat + amount of heat produced by the reaction absorbed by the calorimeter absorbed by the solution

- q total = q calorimeter + q solution

- q total = Kcal (T) + mCT

o Kcal = heat capacity of calorimeter (“calorimeter constant”) in J/ C C = specific heat capacity of water in J/g.oC EXAMPLES:

3.358 kJ of heat are added to a calorimeter that contains 50.00 g of water. The temperature of the water and calorimeter, originally at 22.34oC, increases to 36.74oC. Calculate the heat capacity of the calorimeter (the o calorimeter constant, Kcal) in J/ C.

A 28.2-gram sample of iron is heated to 99.8oC and placed in a coffee-cup calorimeter containing 150.0 g of water at a temperature of 23.5oC. What is the final temperature of the system? (Specific heat of iron = 0.45 J/g.oC)

EXAMPLE:

A 50.0-mL sample of 0.400 M copper(II) sulfate solution at 23.35oC is mixed with 50.0 mL of 0.600 M sodium hydroxide solution, also at 23.35oC, in a coffee-cup calorimeter. After the reaction occurs, the temperature of the resulting mixture is measured to be 25.23oC. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved. Assume that the specific heat of the solution is the same as that of pure water. (Note the calorimeter constant from the first example.) In addition, calculate the amount of heat produced per mol NaOH.

AP CHEMISTRY NOTES 4-7

THERMOCHEMISTRY: ENTHALPY & HEAT OF FORMATION

o HEAT OF FORMATION (∆Hf ) – the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states

*The superscript “o” indicates that the reaction was carried out under standard conditions: Pressure = 1 atm Concentration = 1 M Temperature = 25oC

*Remember, ∆H = 0 for any element in its standard state (including diatomic elements)

EXAMPLE:

o Calculate ∆Hf for ammonium perchlorate in the following reaction

3Al(s) + 3NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3NO(g) + 6H2O(g) + 2677 kJ

given the following values:

o _Substance ∆Hf value (kJ/mol)_ Al2O3 - 1676 AlCl3 - 705 NO 90 H2O - 242

The heat of formation of acetic acid (CH3COOH) is -876.1 kJ/mol. Show a balanced reaction for the formation of this substance from its elements.

Is this reaction endothermic or exothermic?

AP CHEMISTRY NOTES 4-8

THERMOCHEMISTRY: ENTHALPY & HESS’S LAW

HESS’S LAW OF CONSTANT HEAT SUMMATION - if a reaction is the sum of two or more other reactions, then ∆Hrxn for the overall process must be the sum of the ∆H values of the constituent reactions

*Steps: 1. Based on the overall reaction, change the constituent reactions so that the substances involved match in both number and side of the reaction.

a. Multiply or divide a constituent equation to match the coefficients in the overall equation; in addition, multiply or divide the ∆H value by the same factor

b. Reverse a constituent equation in order to place the substances on the same side as those in the overall equation; in addition, change the sign of the ∆H value for the constituent equation that was reversed

2. Cancel out substances that appear on opposite sides of two constituent equations

EXAMPLE:

Find ∆H for the following reaction:

2H3BO3(g) → B2O3(g) + 3H2O(g)

given the constituent reactions below:

H3BO3(g) → HBO2(g) + H2O(g) ∆H = - 0 020 kJ

H2B4O7(g) + H2O(g) → 4HBO2(g) ∆H = - 11.3 kJ

H2B4O7(g) → 2B2O3(g) + H2O(g) ∆H = 17.5 kJ

AP CHEMISTRY NOTES 4-9

THERMOCHEMISTRY: ENTHALPY & BOND ENERGY

BOND ENERGY – the amount of energy necessary to break one mole of bonds in a gaseous covalent substance to form products in the gaseous state at constant temperature and pressure (also called “bond enthalpy”)

*When bonds are broken, energy must be added to the system (endothermic)

*When bonds are formed, energy is released (exothermic)

∆H = bonds broken - bonds formed

*NOTE: This results in an equation of “reactants minus products” (the only such equation involved on the AP test – everything else is “products minus reactants”!!!)

EXAMPLE:

Using the bond energies given below, calculate the change in energy that accompanies the following reaction:

H2(g) + F2(g) → 2HF(g)

Bond Energies:

H-H 432 kJ/mol F-F 154 kJ/mol H-F 565 kJ/mol