THERMOCHEMISTRY

Energy The ability to do work or transfer heat. Work: Energy used to cause an object that has mass to move. Heat: Energy used to cause the temperature of an object to rise.

Definitions #1 Energy: The capacity to do work or produce heat Potential Energy: Energy due to position or composition Kinetic Energy: Energy due to the motion of the object

Definitions #2 Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms The First Law of Thermodynamics: The total energy content of the universe is constant

Definitions #3 Internal energy: The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes potential and kinetic energy. The sample is at a height from the ground etc).  Symbol for Internal Energy Change: ΔU

A) is the sum of the kinetic energy of all of its components B) is the sum of the rotational, vibrational, and translational energies of all of its components C) refers only to the energies of the nuclei of the atoms of the component molecules D) is the sum of the potential and kinetic energies of the components E) none of the above 1) The internal energy of a system __________.

Definitions #4 Enthalpy: Enthalpy is the amount of heat content used or released in a system at constant pressure

E = q + w E = change in internal energy of a system q = heat flowing into or out of the system -q if energy is leaving to the surroundings +q if energy is entering from the surroundings w = work done by, or on, the system -w if work is done by the system on the surroundings +w if work is done on the system by the surroundings

2) Which one of the following conditions would always result in an decrease in the internal energy of a system? A) The system loses heat and does work on the surroundings. B) The system gains heat and does work on the surroundings. C) The system loses heat and has work done on it by the surroundings. D) The system gains heat and has work done on it by the surroundings. E) None of the above is correct.

3) The change in the internal energy of a system that releases 2,500 J of heat and that does 7,655 J of work on the surroundings is __________ J. A) -10,155 B) -5,155 C) 1.9x107 D) 10,155 E) 5,155

Work problems Chapter 5 5.25 5.27 A and B 5.31 All

Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

1 joule = 4.184 calories The Joule The unit of heat used in modern thermochemistry is the Joule 1 joule = 4.184 calories

A Bomb Calorimeter

A Cheaper Calorimeter

Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·K) Water (liquid) 4.18 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) 0.897 Carbon (graphite,solid) 0.709 Iron (solid) 0.449 Copper (solid) 0.385 Mercury (liquid) 0.140 Lead (solid) 0.129 Gold (solid)

Calculations Involving Specific Heat OR s = Specific Heat Capacity q = Heat lost or gained T = Temperature change

4) The temperature of a 15-g sample of lead metal increases from 22 °C to 37 °C upon the addition of 29.0 J of heat. The specific heat capacity of the lead is __________ J/g-K. 7.8 1.9 29 0.13

Problems 5.53 a and B

State Functions depend ONLY on the present state of the system ENERGY IS A STATE FUNCTION A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane! WORK IS NOT A STATE FUNCTION WHY NOT???

State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached room temperature from either direction.

State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so, E depends only on Einitial and Efinal.

State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its E is the same. But q and w are different in the two cases.

Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −PV

Work, Pressure, and Volume Expansion Compression +V (increase) -V (decrease) -w results +w results Esystem decreases Esystem increases Work has been done by the system on the surroundings Work has been done on the system by the surroundings

Energy Change in Chemical Processes Endothermic: Reactions in which energy flows into the system as the reaction proceeds. + qsystem - qsurroundings Exothermic: Reactions in which energy flows out of the system as the reaction proceeds. - qsystem + qsurroundings

Endothermic Reactions

Exothermic Reactions

5) Which one of the following is an exothermic process? A) ice melting B) water evaporating C) boiling soup D) condensation of water vapor E) Ammonium thiocyanate and barium hydroxide are mixed at 25 °C: the temperature drops

Enthalpy H = E + PV At constant pressure and volume the change in enthalpy is the heat gained or lost H = q

Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants

Enthalpies of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Hess’s Law “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 a series of steps.”

Hess’s Law

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ Step #3: Multiply reaction #3 by 2

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ CH4 + 2O2  CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O Hrxn =  Hf(products) -   Hf(reactants)     Substance Hf   CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

  13) Given the following reactions the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) +O2 (g)  2SO3(g) is __________ kJ. 2SO2  2S + 2O2 ΔH = 594 kJ 2SO3  2S + 3 O2 ΔH = 790 kJ

6. 2SO2  2S + 2O2 ΔH = 594 kJ keep 2SO3  2S + 3 O2 ΔH = 790 kJ reverse 2S + 3 O2  2SO3 ΔH = -790 kJ Ans: -196kJ

Problems 5.63