Thermochemistry Unit 7. thermochemistry the study of heat changes in chemical reactions and physical changes.

ThermochemistryUnit 7

thermochemistry

the study of heat changes in chemical reactions and physical changes

open

mass & energy**Exchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

system—specific part being analyzedsurroundings—everything outside the system (usually the immediate area)universe = system + surroundings

Energy is the capacity to do work

• Thermal energy is the energy associated with the random motion of atoms and molecules

• Chemical energy is the energy stored within the bonds of chemical substances

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Electrical energy is the energy associated with the flow of electrons

• Potential energy is the energy available by virtue of an object’s position

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

Temperature = Thermal Energy

900C400C

greater thermal energy

Units for Measuring Heat Flow

• calorie - the amount of heat needed to raise the temperature of 1 g of H2O by 1 oC.

-Used except when referring to food

• Calorie (capital C) refers to the energy in food

Units for Measuring Heat Flow

1 Calorie = 1 kilocalorie = 1000 cal

SI unit of heat and energy = Joule 4.184 J = 1 cal1 J = 0.2390 cal   

Energy Transformations

• Heat – (variable is q) – energy that transfers from one object to another

– only changes can be detected!– flows from warmer to cooler

object

Exothermic - process that gives off heat- energy goes from system to surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

Energy is written on the product (right) side of the equation

q and ΔH are negative (-)

Δ is a Greek letter – means change in

The products are lower in energy than the reactantsThus, energy is released.ΔH = -395 kJ• The negative sign does not mean negative energy, but instead that energy is lost.

exothermic

Endothermic - process in which heat has to be supplied – energy goes from surroundings to system

energy is written on the reactant (left) side of the equation

q and ΔH are positive (+)

energy + 2HgO (s) 2Hg (l) + O2 (g)

The products are higher in energy than the reactants Thus, energy is absorbed.ΔH = +176 kJ• The positive sign means energy is absorbed

endothermic

When a substance changes state the temperature remains constant during the actual phase change.- this is because the heat energy is being used up by the phase change process.

Changes of state

The specific heat (C) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

q = heatm = mass ( in grams)t = tfinal – tinitial

C = specific heat

q = H

Specific heat formulaq = m X t X C

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

C of Fe = 0.444 J/g • 0C m = 869 g

t = tfinal – tinitial = 50C – 940C = - 890C

q = mCt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

6.4

The specific heat of water is high due to hydrogen bonding

Calorimetry the measurement of the heat into or out of a system

heat released = the heat absorbed

• enthalpy changes are measured with a calorimeter

No heat enters or leaves!

The Law of Conservation ofEnergy

• in any chemical or physical process, energy is neither created nor destroyed.

- All the energy is accounted for as work, stored energy, or heat.

thermo chemical equations

thermo chemical equations— equations that show heat changes

enthalpy (H)—heat content of a substance

Cannot measure directly but we CAN measure

• change in enthalpy = ΔH; – heat change for a process; usually

measured in kJ (kilojoules)

H2O (s) H2O (l)H = 6.01 kJ

• The physical states of all reactants and products must be specified

Heat of reaction

H2O (l) H2O (g)H = 44.0 kJ

Equation – = given mol x H

1 # mol

Thermo chemical Equations

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P41 mol P4

123.9 g P4

x 3013 kJ1 mol P4

x = 6470 kJ

• ΔHf = HEAT OF FORMATION = heat absorbed or released to make 1 mol of a compound from its elements

•ΔHsoln = Heat of Solution = heat change caused by dissolving of one mole of substance

Hf for the formation of rust (Fe2O3) is –826 kJ/mol. How much energy is involved in the formation of 5

grams of rust

5.0 g Fe2O3

1

1 mol Fe2O3

160 g Fe2O3

x 826 kJ1 mol Fe2O3

x = 25.9 kJ

When 1.0 g of solid NaOH (Hsoln = – 445.1 kJ/mol) dissolves in 10 L of water, how much heat is released?

1.0 g NaOH 1

1 mol NaOH

40.0 g NaOHx

445.1 kJ1 mol NaOH

x = 11.1 kJ

Molar Heat of Combustion = ΔHcomb = heat released in combustion of 1 mol of substance

Molar Heat of Fusion (Hfus.) = the heat absorbed by 1 mol of a substance in melting from a solid to a liquid

Molar Heat of Solidification (Hsolid.) = the heat lost when 1 mol of liquid solidifies (or freezes) to a solid

Molar Heat of Vaporization (Hvap.) = the amount of heat necessary to vaporize 1 mol of a given liquid.

Molar Heat of Condensation (Hcond.) = amount of heat released when 1 mol of vapor condenses to a liquid

Standard heat of reaction

• Hess’s law of heat summation states that if you add two or more thermochemical equations you can add the heats of reaction to give the final heat of reaction

Standard Heats of Formation

• The standard heat of formation (ΔHf0) of a compound is the change in enthalpy that accompanies the formation of one mole of a compound

Standard heat of formation for elements = 0

Standard Heat of Reaction

H = H (products) – H (reactants)

1. Multiply the standard heat of formation by the number of moles for each reactant and product.

2. Add the reactants together3. Add the products together4. Subtract the sum of the reactants from

the sum of the products

• Calculate H for the following reaction.

C2H4(g) + H2(g) C2H6(g)

H for C2H4(g) = 52.5 kJ/mol;

H for H2(g) = 0 kJ/mol; (free element)

H for C2H6(g) = –84.7 kJ/mol

- 84.7 - 52.5 = - 137.2 kJ