Chapter 5 Thermochemistry

Energy of objects Objects can possess 2 kinds of energy. KE= energy of motion E k = ½ mv 2 PE= stored energy (energy of position)

Units of heat Joule= 1 kg-m 2 /s 2 1 cal= Joule

The universe is divided into two halves. the system and the surroundings. The system is the part you are concerned with. The surroundings are the rest. Exothermic reactions release energy to the surroundings. Endo thermic reactions absorb energy from the surroundings.

Energy is... The ability to do work or transfer heat. Energy can be transferred by doing work Force- any kind of push or pull exerted on an object Work= any energy used to move an object against a force W=f x d

Energy can be transferred as heat. Always in the direction from hotter object to a colder one. When you apply an ice pack your body part feels cooler because your body is transferring heat to the ice pack.

Energy is conserved! First law of thermodynamics- any energy that is lost by a system must be gained by its surroundings, and vice versa. The change in energy that occurs during a reaction can be calculated

Potential energy Heat

Potential energy Heat

Energy changes Every energy measurement has three parts. 1. A unit 2. A number (indicating magnitude) 3. and a sign to tell direction. negative – exothermic (system loses energy) positive- endothermic (system gains energy)

System Surroundings Energy  E <0

System Surroundings Energy  E >0

LOOK AT FIGURE 5.5 on page 150.

Calculating energy changes System can change energy by work or heat Heat given off is negative. Heat absorbed is positive. Work done by system on surroundings is positive.

Work done on system by surroundings is negative. Thermodynamics- The study of energy and the changes it undergoes.

First Law of Thermodynamics Law of conservation of energy. q = heat w = work  E = q + w Take the systems point of view to decide signs. Look at Table 5.1

The internal energy of a system is a state function, it is independent of it past history. ΔE= state function (q and w are not though)

Practice Hydrogen and oxygen gases are in a cylinder. As a reaction occurs between them, the system loses 1150 joules of heat to the surroundings. The reaction also causes a piston to rise as the hot gas expands. The expanding gas does 480 J of work on the surroundings. What is the change in internal energy of the system?

Enthalpy Most reactions occur at constant pressure and so energy is transferred mainly in the form of heat. Enthalpy is the heat absorbed or released by a system under constant pressure abbreviated H

More about enthalpy State function Cannot be measured, but the change can. ΔH equals the heat (q p ) gained or lost by the system when the process occurs under constant pressure. ΔH = H final – H initial = q p

Since H is a state function it depends only on the starting and ending states of the system, not how they got there.

Enthalpies of reactions ΔH rxn = H(products)- H(reactants) Thermochemical equations 2H 2 + O 2 = 2H 2 O ΔH= kj Enthalpy is an extensive property (dependent on amount)

Calorimetry Measuring heat. Use a calorimeter. Two kinds Constant pressure calorimeter (called a coffee cup calorimeter) heat capacity for a material, C is calculated C= heat absorbed/  T =  H/  T specific heat capacity = C/mass

Calorimetry molar heat capacity = C/moles heat = specific heat x m x  T heat = molar heat x moles x  T Make the units work and you’ve done the problem right. A coffee cup calorimeter measures  H. An insulated cup, full of water. The specific heat of water is 1 cal/gºC Heat of reaction=  H = sh x mass x  T

Examples How much heat is released when 4.5 g of methane is burned in a constant pressure system. CH  CO 2 + H 2 O ΔH=-890 kJ

Calorimetry Constant volume calorimeter is called a bomb calorimeter. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. The heat capacity of the calorimeter is known and tested. Since  V = 0, P  V = 0,  E = q

Bomb Calorimeter thermomete r stirrer full of water ignition wire Steel bomb sample

Properties intensive properties not related to the amount of substance. density, specific heat, temperature. Extensive property - does depend on the amount of stuff. Heat capacity, mass, heat from a reaction.

Hess’s Law Enthalpy is a state function. It is independent of the path. We can add equations to to come up with the desired final product, and add the  H Two rules If the reaction is reversed the sign of  H is changed If the reaction is multiplied, so is  H

N2N2 2O 2 O2O2 NO 2 68 kJ NO kJ -112 kJ H (kJ)

Standard Enthalpy The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions) Symbol  Hº When using Hess’s Law, work by adding the equations up to make it look like the answer. The other parts will cancel out.

Use the following data to calculate the enthalpy of combustion of C  CO ( C + ½ O 2  CO ) 1. C + O 2  CO 2 ΔH= kJ/mol 2. CO + ½ O 2  CO 2 ΔH= kJ/mol

Standard Enthalpies of Formation Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. Standard states are 1 atm, 1M and 25ºC

ΔHrxn= ΣnΔHf (products) – ΣmΔH f (reactants) Use appendix C to calculate the standard enthalpy change for the combustion of 1 mol of benzene (C 6 H 6 ).

Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O which leads us to this rule.

Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O which leads us to this rule.

You are responsible for all of chapter this includes the part we did not talk about on foods and fuels. Make sure you study it. It will be on the test!