Thermochemistry

ThermochemistryBy: Ms. Buroker1Thermochemistry is the study of the relationship between chemical reactions and energy changes.2The Nature of EnergyWe define energy as the capacity to do work or to transfer heat.

HEAT is the form of energy that flows between 2 objects because of their difference in temperature.

The Law of Conservation of Energy states that energy can be converted from one form to another but can be neither created nor destroyed.3The Nature of Energy

4Some Definitions .Thermal Energy: energy associated with the random motion of molecules

Chemical Energy: energy stored due to the structure of chemical substances

Kinetic Energy: energy of motion

Potential Energy: available energy due to an objects position

Radiant Energy: energy in the form of electromagnetic radiation

5Potential EnergyPotential Energy: energy due to position or composition. Stored energy resulting from attractions and repulsions an object experiences in relation to other objects.1.) Electrostatic Forces between charged particles2.) Chemical Energy3.) Thermal Energy6Kinetic EnergyKinetic Energy: energy in motion

Ek = mv2

MassVelocity7Potential and Kinetic Enegry

Potential Energy can be converted to kinetic energy and do work in the process.

Units of Energy:SI Unit: Joule, (J)Cgs Unit: Erg1J = 1kg-m2/s21Erg = 1g-cm2/s2

8Units of Energy The joule is not very large, so energies associated with chemical reactions are commonly expressed in kJ.

Non- SI Base Unit commonly used calorie (cal)

Calorie = the amount of energy required to raise the temperature of 1g of water 1C.1 cal = 4.184J (exactly)

9Internal Energy of the System

Kinetic (thermal) energy:Potential Energy:TranslationalNuclear ForcesRotationalVibrational

Electrostatic attractions (Chemical energy):Intramolecular - Bonds between atomsIntermolecular - Attractions between molecules10Temperature vs. Heat Temperature: a property that reflects the random motions of the particles in a particular substance.

Heat: involves the transfer of energy between two objects due to a temperature difference it is NOT a substance contained in the object!

11The more thermal energy a substance has, the greater the motion of its atoms and molecules.

The total thermal energy in an object is the sum of the individual energies of all the atoms, molecules, or ions in that object.

So The thermal energy of a given substance depends not only on temperature but also on the amount of substance. (E.g. Warm bath vs. hot coffee)Temperature vs. Heat Continued

12State Function A State Function or State Property is a property of a system that depends only on its present state. * It does not depend on how the system arrived at the present state, it depends only on the characteristics of the present state.* A change in this function or property in going from one state to another state is independent of the particular pathway taken between the two states.13An Example of a State Function Elevation on the earths surface vs. distance between to points.

Chicago (elevation 674 ft)Denver (elevation 5280 ft)

Elevation = State FunctionDistance = Not a State FunctionEnergy is a state function while, heat and work are NOT!!!!Energy, Pressure, Volume, and Temperature are State Functions.14System and SurroundingSystem: the portion we single out for our study

Surrounding: everything else

When studying chemical reactions the chemicals are the system and the containers and everything else are the surroundings.CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy(heat)15Three Types of Systems1. Open system is one in which both matter and energy (usually in the form of heat) are flowing into and out of the system from the surroundings, i.e., there is exchange of mass and energy with its surroundings.

2. Closed system is one in which matter does NOT flow into or out of the system but energy does flow, i.e., there can be an exchange of energy but not of mass.

3. Isolated system is one in which neither mass nor energy can flow from surroundings to the system or from the system to the surroundings, i.e., there is no exchange of either mass or energy .

16Thermal EquilibriumHeat transfer occurs when two objects at different temperaturesare brought into contact.Heat energy is transferred until the system comes to thermalequilibrium.

Within a system, the heat lost by the hotter object is equal to theheat gained by the cooler object.

When heat transfer occurs across the boundary between asystem and the surrounding, the directionality of the heat transferis described as:Exothermic: heat is transferred from a system to thesurroundings means out of (negative)

Endothermic: Heat is transferred from the surroundings to thesystem means in to (positive)

17The First Law of Thermodynamics The energy of the universe is constant.

Internal Energy, E, of a system can be defined most precisely as the sum of the kinetic and potential energies of all the particles in the system.

E = Eproducts Ereactants E = q + wHeat Transfer work18

Specific Heat Capacity and Energy Transfer

Specific Heat: The amount of energy required to increase one gram of a substance by 1K (or oC)19E = q + w The sign reflects the systems point of view.

If energy flows into the system via heat q is positive and is thus an endothermic process.

If energy flows out of the system via heat q is negative and is thus an exothermic process.

20Specific Heat and Energy Transfer

21Endothermic vs. Exothermic

22Sample Problems

1. In an experiment it was determined that 59.8J was required tochange the temperature of 25.0g of ethylene glycol (a compoundused as antifreeze in automobile engines) by 1.00K. Calculate the specific heat capacity of ethylene glycol from these data.

2. A 15.5g piece of chromium, heated to 100.0oC, is dropped into55.5g of water at 16.5oC. The final temperature of the metal andthe water is 18.9oC. What is the specific heat capacity ofchromium? (Assume no heat is lost to the container or to thesurrounding air.)

3. A piece of iron (400.g) is heated in a flame and then droppedinto a beaker containing 1,000.g of water. The originaltemperature of the water was 20.0oC, and the final temperature of the water and iron is 32.8oC after thermal equilibrium has been attained. What was the original temperature of the hot iron bar? (Assume no heat is lost to the beaker or to the surrounding air.)You need the specific heat capacity of iron = 0.449J/gK

23Answer: Sample Problem #1Sample Problems

1. In an experiment it was determined that 59.8J was required to change the temperature of 25.0g of ethylene glycol (a compound used as antifreeze in automobile engines) by 1.00K. Calculate thespecific heat capacity of ethylene glycol from these data.

Answer:q= c x m x DTSpecific Heat Capacity Units = J/gK59.8J / [(25.0g)(1.00K)] = 2.39 J/gK24Answer: Sample Problem #2Sample Problems

2. A 15.5g piece of chromium, heated to 100.0oC, is dropped into 55.5g of water at 16.5oC. The final temperature of the metal and the water is 18.9oC. What is the specific heat capacity of chromium? (Assume no heat is lost to the container or to the surrounding air.)

Answer:Heat lost by chromium:-q = C(15.5g)(18.9oC 100.0oC) = -1257.05(C)Heat gained by water:q = (55.5g)(4.184J/goC)(18.9oC 16.5oC) = 557.309J1257.05g(C) = 557.309JC = 557.309 / 1257.05 = .443J/goC25Answer: Sample Problem #3Sample Problems

3. A piece of iron (400.g) is heated in a flame and then droppedinto a beaker containing 1,000.g of water. The originaltemperature of the water was 20.0oC, and the final temperature ofthe water and iron is 32.8oC after thermal equilibrium has beenattained. What was the original temperature of the hot iron bar?(Assume no heat is lost to the beaker or to the surrounding air.)You need the specific heat capacity of iron = 0.449J/gK

Answer:Ti = 330.992oC = 331oC26Phase ChangesSolid Liquid Gas

Heat of Fusion: Heat required to convert a solid at its melting point to a liquid

Heat of Vaporization: Heat required to convert a liquid at its boiling point to a gas.

Heat of Sublimation: Heat required to take a substance directly from its solid to its gas state.27

Note that the temperature of a substance does not change as it isgoing through a phase change.28Example ProblemHow much heat must be absorbed to warm 25.0g of liquidmethanol, CH3OH, from 25.0oC to its boiling point (64.6oC) and thento evaporate the methanol completely at that temperature? Thespecific heat capacity of liquid methanol is 2.53J/gK. The heat ofvaporization of methanol is 2.00kJ/g

Answer:Step 1q1 =c x m x DT q1 = (2.53J/gK) x (25.0g) x (64.6oC 25.0oC) q1 = 2504.7J

Step 2q2 = 25.0g (2000.0J/g) q2= 50,000J

29Work Associated with Chemical Processes

Work done by a gas expansionWork done to a gas compression

W = -PVP= external pressure-V= compression+V= expansionNOTE: For an ideal gas, work can occur ONLY when its volume changes. Thus, if a gas is heated at constant volume, the pressure increases but no work occurs.30Enthalpy Thermodynamics defines a new function: ENTHALPY(H) (q p) as the heat flow part of the change in internal energy when the pressure is kept constant as it is in an open container or in an expandable container.

DE = q - P (DV)DE = q p - P (DV) (q p is given the name Enthalpy and the symbol (H))DE = DH - P (DV) or re-arrange to express it:DH = DE + P (DV) = Change in Enthalpy

31Enthalpy ContinuedAt constant pressure, the change in enthalpy of the system is equal to the energy flow as heat.DH = qp

DH = Hproducts - Hreatctants

Note: DH = qp ONLY at constant pressure the change in enthalpy of a system has no easily interpreted meaning except at constant pressure, where DH = heat.

At constant pressure, exothermic means DH is negative ; endothermic means DH is positive.32Enthalpy Changes For Chemical Reactions

Enthalpy (heat) of ReactionsEnthalpy changes accompany chemical Reactions:H2O(g) H2(g) + O2(g) DH = +241.8kJ (Endothermic)33For the opposite reaction

H2(g) + O2(g) H2O(g)H = -241.8kJ (Exothermic)34Calorimetry The devise used to determine the heat associated with a chemical reaction is called a calorimeter.

Calorimetry is the science of measuring heat, based on the temperature change a body undergoes when it absorbs or discharges energy as heat.35Heat CapacityThe heat capacity of a substance, is defined as:

C = heat absorbed increase in temperature

Specific Heat Capacity: the energy required to raise the temperature of one gram of a substance by one degree Celsius.Molar Heat Capacity: the energy required to raise the temperature of one mole of a substance by one degree Celsius.36Coffee Cup Calorimetry Simple way of measuring heat. Two Styrofoam cups are nested together where the inner cup holds the solution in which the reaction takes place.

An example of constant-pressure calorimetry. Used in determining the changes in enthalpy for reactions occurring in solution.Remember!! Under the conditions of constant pressure enthalpy equals heat!

37Bomb CalorimetryConstant- Volume Calorimetry

qreaction = -Ccal x DTCalorimeter Constant

38Bomb Calorimetry . E = qv = qrxn = -(qbomb + qwater)

The energy transferred to the bomb is given by the heat capacity (usually in J/K)

The energy transferred to the water is given by the specific heat equation:qwater = m x C x T

39Example Problems

1. Assume you mix 200.mL of 0.400M HCl with 200.mL of 0.400M NaOH in a coffee-cup calorimeter. The temperature of the solutions before mixing was 25.10oC; after mixing and allowing the reaction to occur, the temperature is 27.78oC. What is the molar enthalpy of neutralization of the acid? (Assume that the densities of all solutions are 1.00g/mL and their specific heat capacities are 4.20 J/gK)

2. A 1.00g sample of ordinary table sugar (sucrose C12H22O11) is burned in a bomb calorimeter. The temperature of 1.50x103g of water in the calorimeter rises from 25.00oC to 27.32oC. The heat capacity of the bomb is 837J/K, and the specific heat of the water is 4.20J/gK.a) Calculate the heat evolved per gram of sucrose.b) Calculate the heat evolved per mole of sucrose.40Answer: Sample Problem #11. Assume you mix 200. mL of 0.400M HCl with 200. mL of 0.400MNaOH in a coffee-cup calorimeter. The temperature of thesolutions before mixing was 25.10oC; after mixing and allowing thereaction to occur, the temperature is 27.78oC. What is the molarenthalpy of neutralization of the acid? (Assume that the densitiesof all solutions are 1.00g/mL and their specific heat capacities are4.20 J/gK)

HCl + NaOH NaCl + H2O

Total volume of solutions = 200.mL + 200.mL = 400.mLTotal mass = 400.mL (1g/mL) = 400.gMoles of HCl and NaOH (each) = .200L (0.400mol/L) = .0800molq = (400.g)(4.20J/gK)(27.78oC 25.10oC) = 4502.4JMolar enthalpy = 4502.4J / .0800mol = 56,280J/mol = 56.3kJ/mol41Answer: Sample Problem #22. A 1.00g sample of ordinary table sugar (sucrose C12H22O11) isburned in a bomb calorimeter. The temperature of 1.50x103 g ofwater in the calorimeter rises from 25.00oC to 27.32oC. The heatcapacity of the bomb is 837J/K, and the specific heat of the wateris 4.20J/gK.a) Calculate the heat evolved per gram of sucrose.b) Calculate the heat evolved per mole of sucrose.

a) Temperature change = 27.32oC 25.00oC = 2.32oCqwater = (1500g)(4.20J/gK)(2.32oC) = 14,616Jqbomb = 837J/K(2.32oC) = 1,941.84Jqrxn = -(14,616J + 1,941.84J) = -16,557.8J = -16.6kJ/g

b) Moles sucrose = 1.00g / (342.2965g/mol) = 2.921x10-3molqrxn = -16,557.8J / 2.921x10-3mol = 5.6685x106J/mol = -5.67x103kJ/mol42Hesss Law Since enthalpy is a state function, the change in enthalpy in going from some initial state to some final state is independent of the pathway..

So, in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.43Characteristics of Hesss Law1.) If a reaction is reversed, the sign of DH is also reversed.

2.) The magnitude of DH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of DH is multiplied by the same integer.44

Example: Given the reaction information to the left, determine theunknown enthalpy of reaction a.Answer: Ha + -283.0kJ = -393.5kJ Ha = -110.5kJ45Example Problem #1Use Hesss law to calculate the enthalpy change for the formationof CS2(l) from C(s) and S(s) from the following enthalpy values:

C(s) + O2(g) CO2(g)DH = -393.5kJ

S(s) + O2(g) SO2(g)DH = -296.8kJ

CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g)DH = -1103.9kJ

C(s) + 2 S(s) CS2(l)DH = ?

Hint: Remember you can reverse and multiply equations through.Whatever you do to the reaction equation, you do to the enthalpy.46Answer: Example Problem #1Use Hesss law to calculate the enthalpy change for the formationof CS2(l) from C(s) and S(s) from the following enthalpy values:C(s) + O2(g) CO2(g)DH = -393.5kJS(s) + O2(g) SO2(g)DH = -296.8kJCS2(l) + 3 O2(g) CO2(g) + 2 SO2(g)DH = -1103.9kJ

C(s) + 2 S(s) CS2(l)DH = ?

C(s) + O2(g) CO2(g)DH = -393.5kJ2S(s) + 2O2(g) 2SO2(g)DH = -593.6kJ (Double)CO2(g) + 2 SO2(g) CS2(l) + 3 O2(g)DH = 1103.9kJ (Reverse)

C(s) + 2 S(s) CS2(l)H = -393.5kJ + -593.6kJ + 1103.9kJ = 116.8kJ47Example #2Diborane (B2H6) is a highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program. Calculate DH for the synthesis of diborane from its elements, according to the equation:2B(s) + 3H2(g) B2H6(g)

Use the following data:2B(s) + 3/2O2(g) B2O3(s) DH= -1273kJ B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g)DH = -2035kJ H2(g) + 1/2O2(g) H2O(l) DH = -286kJ H2O(l) H2O(g) DH = 44kJ48Hints for using Hesss LawHint #1: Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reaction at your disposal.

Hint #2: Reverse any reactions needed to give the required reactants and products.

Hint #3: Multiply reactions to give the correct number of reactants and products.Standard Enthalpies of Formation Standard enthalpy of formation (DHf) for a compound is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.50 The degree symbol on a thermodynamic function, for example, DHf, indicates that the corresponding process has been carried out under standard conditions.

The standard state for a substance is a precisely defined reference state.

51CalculationDHreaction = SnpDHf(products) SnrDHf(reactants)

Elements in their standard states are not included in the DHreaction calculations. That is, DHf, for an element in its standard state is zero.52Sample ProblemCalculate the standard enthalpy of combustion for benzene, C6H6.C6H6(l) + 7.5 O2(g) 6 CO2(g) + 3 H2O(l)Horxn = ? Hof[C6H6(l)] = +49.0kJ/mol. Other values needed can be found in the appendix in the back of your book.

Horxn = [6(-393.51) + 3(-285.83)] [1(49.0) + 7.5(0)]= -3,267.55 = -3.27x103kJ/mol53