CHEM 3310 Thermodynamics

CHEM 3310

Thermodynamics The study of energy changes accompanying physical and chemical processes. From the laws of thermodynamics, one can: Predict the results of chemical reactions Ascertain whether or not a reaction is possible Predict quantitatively the effect of changes in the experimental conditions CHEM 3310

What is Energy? 1. Kinetic Energy (KE) – the energy of motion due to the motion of the system's particles (translations, rotations, vibrations) 2. Potential Energy (PE) – the stored energy in an object by virtue of its position or composition (chemical bonds, electric energy of the atoms within the molecules) Energy cannot be created nor destroyed. The total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred. CHEM 3310

The total energy in any isolated system is constant. Terminology: System is the portion of the physical world that is selected for a thermodynamic study. Examples: A beaker of water A polluted lake Surroundings System Surroundings is the portion of the universe with which the system interacts. Isolated System = System + Surroundings The total energy in any isolated system is constant. CHEM 3310

E = KE + PE Internal Energy, E All substance possess internal energy, because the molecules of all substances all have kinetic energy and potential energy. Internal energy, represented by E, is essentially the thermal energy of the particles making up the system (all the atoms or molecules of the body due to their random motion and position). It is the sum of the kinetic and potential energies of the particles that form the system. E = KE + PE CHEM 3310

E = KE + PE Internal Energy, E T1 < T2 E = 3/2nRT In an ideal gas the molecules are so far apart that intermolecular forces can be ignored. Therefore, there is no potential energy. So all the energy is kinetic. E = KE + PE The simplest system is an ideal gas. Kinetic Molecular Theory assumes that the temperature of a gas is directly proportional to the average kinetic energy of its particles. At T1 At T2 T1 < T2 The internal energy of an ideal monatomic gas is therefore directly proportional to the temperature of the gas. R = gas constant T = temperature in Kelvin E = 3/2nRT CHEM 3310

Internal Energy, E Calculate the internal energy of a pressurized bottle of Helium gas with volume 0.0498 m3 at 1.01x107 Pa. You can use the gas law  PV=nRT  to express the internal energy in terms of pressure and volume: 0.368 m x 0.368 m x 0.368 m = 0.0498 m3 E = 3/2nRT = 3/2PV = 3/2 (1.01x107 Pa)(0.0498 m³) = 7.54 x 105 Pa m³ = 7.54 x 105 J (1 atm = 101.325 kPa) CHEM 3310

E = f(T) Internal Energy, E The internal energy of other systems that are more complex than the ideal gas are harder to measure. But the internal energy of the system is still has a unique value for each temperature. E = f(T) Where f is monotonically increasing function. As the temperature of the system increases, the internal energy of the system also increases. CHEM 3310

Temperature is a state function. Temperature, T Use a thermometer to measure a beaker of water at room temperature. T = 21.5 °C This measurement describes the state of the system at a particular moment in time. It doesn’t tell us how the water got to room temperature. Temperature is a state function. CHEM 3310

Internal Energy, E Because the internal energy of the system has a fixed value for any temperature, internal energy is also a state function. (E depends on the state of the system in terms of P, V and T.) Any change in the internal energy of the system is equal to the difference between its initial and final values. Esystem = Efinal - Einitial CHEM 3310

The First Law of Thermodynamics The total energy of the universe is constant. Energy can be transferred from the system to its surroundings and vice versa, but it cannot be created nor destroyed. The energy lost by the system is gained by the surroundings such that - Esystem = Esurroundings Euniverse = Esystem + Esurroundings = 0 Surroundings System CHEM 3310

There are two ways to change the internal energy of a system: By flow of heat, q Heat is the transfer of thermal energy between the system and the surroundings 2. By doing work, w Work can be converted into heat and vice versa. q and w are process dependent, and are not state functions. CHEM 3310

Conversion: 1 calorie = 4.184 joules Units of Energy Calorie: One calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 degree celsius. Joule: One joule is the work done when a force of one Newton is used to move an object one meter. 1 J = 1 Nm Conversion: 1 calorie = 4.184 joules CHEM 3310

Heat, q A form of energy Passes from one body to another as a result of a difference in temperature Heat flows from a higher temperature to a lower temperature Heat cannot be measured, but the effects which it produces is measurable. CHEM 3310

Measurement of the Effects of Heat The measurements of heat effects is known as calorimetry. The instrument used is the calorimeter. System : the molecules inside the reaction chamber Surroundings : the container holding the sample and the water bath Isolated System : the calorimeter The condition under which the measurement is made is constant volume. This is a schematic of a bomb calorimeter CHEM 3310

Experiment 8 – Measure E of a combustion reaction Parr Bomb

Experiment 8 – Measure E of a combustion reaction Combustion of Cyclohexanol

q  nAT q = nACAT Measurement of the Effects of Heat In the isolated system containing Substance A q  nAT where nA is the number of moles of substance A T = T2 – T1 (T2 = final temperature, T1 = initial temperature) Introduce a proportionality constant, CA, such that q = nACAT CA is the molar heat capacity; it is characteristic of the substance. CHEM 3310

q = mAcAT Molar Heat Capacity Specific Heat is the quantity of heat necessary to raise the temperature of one mole of the material by one degree Celsius. Specific Heat is the quantity of heat necessary to raise the temperature of one gram of the material by one degree Celsius. q = mAcAT where mA is the mass of substance A T = T2 – T1 cA is the specific heat of substance A CHEM 3310

Conversion: 1 calorie = 4.184 joules Units of Heat Calorie: One calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 degree Celsius. Specific heat of water = 1.00 cal g-1 °C-1 Molar heat of water = 18.0 cal mole-1 °C-1 Conversion: 1 calorie = 4.184 joules Specific heat of water = 4.18 J g-1 °C-1 CHEM 3310

Heat Capacity varies as a function of temperature CA @ 100 K is not the same as CA @ 300 K is different if the heating is done under constant pressure or under constant volume conditions Heat capacity measured under constant pressure is Cp Heat capacity measured under constant volume is Cv For liquids and solids, there is little difference between Cp and Cv. For gas, Cp > Cv. CHEM 3310

Sign Convention The internal energy and temperature of a system decrease(E < 0) when the system loses heat to its surroundings. The internal energy and temperature of a system increase (E > 0) when the system gains heat from its surroundings. q > 0 when heat is added to the substance q > 0 when T2 > T1 (i.e. T>0) q < 0 when heat is lost by the substance q < 0 when T2 < T1 (i.e. T<0) CHEM 3310

Specific heat, Cp, of Pb is 0.0308 cal g-1 °C-1 Example: 100.0 g of Pb heated to 100 °C is dropped into 10.00 g of water at 0oC. What is the final temperature? Assume that heat transfer is between the two substances. Specific heat, Cp, of Pb is 0.0308 cal g-1 °C-1 Specific heat, Cp, of water is 1.00 cal g-1 °C-1 Answer: Heat lost by the Pb = Heat gained by the water - qPb= qwater - mPbcPb (T2-T1)Pb= mwatercwater (T2-T1)water T2 = ? T1 = 100 °C T2 = ? T1 = 0 °C T2 = 23.6 °C CHEM 3310

Latent Heat Effects The heat effects associated with changes in physical state such as fusion and vapourization are known as latent heat effects. Heat of Fusion - the heat required to melt a substance at its normal melting temperature - the energy required to reorganize the intermolecular structure of a substance. Let’s heat up 1 g of water! CHEM 3310

Latent Heat Effects Heat of Vaporization – the heat required to vaporize a substance at its normal boiling temperature Heat of fusion of ice is 79.71 cal/g at 0oC and 1 atm. Heat of vaporization of water is 539.6 cal/g 100oC and 1 atm. CHEM 3310

Sign Convention The internal energy and temperature of a system decrease (E < 0) when the system loses heat. The internal energy and temperature increase (E > 0) when the system gains heat from its surroundings. Surroundings HEAT System loses heat System E < 0, q < 0 E > 0, q > 0 System gains heat CHEM 3310

Calculation : Example 1: A 1.00 g sample of the rocket fuel hydrazine, N2H4, is burned in a bomb calorimeter containing 1200. g of water. The temperature rises From 24.62 °C to 27.96 °C. Taking the heat capacity of the calorimeter to be 200. cal °C-1, calculate: q for the combustion of the 1 g sample The molar heat of combustion of hydrazine. Answer: (a) -4680 calories (b) -150. kcal/mole

Calculation: Example 2: When 5.00 g of sodium hydroxide is added to 100. g of water, the temperature rises from 25.0oC to 37.5 °C. Calculate the molar heat of reaction for the process NaOH (s)  Na+ (aq) + OH- (aq) taking the specific heat of water to be 1.00 cal g-1 °C-1 and that of NaOH to be 0.48 cal g-1 °C-1. Answer: -10.2 kcal/mole

Calculation: Example 3: An experiment is designed to measure the heat of fusion of ice. 25 g of ice at 0 °C was dropped into 195 g of water at 30 °C. The water is contained in a copper calorimeter of mass 100 g. The final temperature was 18 °C. Given the specific heat of copper is 0.093 cal g-1 °C-1, find the heat of fusion of ice. Answer: Heat of fusion of ice = 80.1 cal/g

Calculation: Example 4: Steam at 100. °C is condensed in a large calorimeter. The heat capacity of the calorimeter is expressed as “water equivalent” to 272 g. The calorimeter contains 2.82 kg of water at 5.0 °C. The final temperature of 27.8 °C is reached after 115 g of steam has been condensed. Find the latent heat of vaporization of water as given by these data. Answer: Heat of vaporization of water = 540 cal/g

Next we will talk about work! HEAT WORK System E < 0, q < 0 Surroundings HEAT WORK Work is done by the system System loses heat System E < 0, q < 0 E < 0, q < 0 E > 0, q > 0 E > 0, q > 0 Work is done on the system System gains heat CHEM 3310