Thermochemistry Unit theme: Energy

Thermochemistry Potential energy diagrams Heat capacity Endothermic and exothermic reactions enthalpy Specific heat Heats of changes of state Units of heat calorimetry Thermochemical equations Hess’ Law calorie joule Gibb’s Free energy entropy

What evidence can you see that energy is involved?

Energy Defined as the capacity to do work or supply heat Chemical potential energy: aka chemical energy Energy stored in chemicals because of their compositions Different substances store different amount of energies

Heat A form of energy that flows from a warmer object to a cooler object Q Can’t be measured directly Can only measure its effect on temperature When heat is added to a system, its temperature rises Temperature is not heat! Temperature is a measure of average kinetic energy

Thermochemistry The study of heat changes that occur during chemical reactions and physical changes of state

Units of Heat calorie joule The quantity of heat needed to raise the temperature of 1 g of pure water by 1 degree Celsius 1 kcal = 1000 calories = 1 Calorie (nutrition) joule SI unit, named after British physicist newtonmeter 1 cal = 4.186 joules 1000 J = 1 kJ Commonly used because joules are so small

Heat Capacity The amount of heat it takes to change an object’s temperature by 1 Celsius degree Depends partly on mass More mass  greater heat capacity

Problem How many calories are required to heat 32.0 g of water from 25.0oC to 80.0oC? How many joules is this? Answer: 1760 calories 7360 joules = 7.36 kJ

Specific Heat Not all substances respond the same way to the input of heat Some get hot much more quickly than others!

Specific Heat The amount of heat required to raise the temperature of 1 g of a substance by 1oC A measure of how well a substance stores heat energy Substances with low specific heat (ex. metals) heat quickly, cool quickly Substances with high specific heat (ex. water) take a long time to heat and cool

Using Specific Heat C or Cp Units: J/goC or cal/goC Q = m C T where Q = total heat change m = mass T = temperature change

Phase Changes Melting, freezing Boiling, condensing Sublimation, deposition Phase changes occur without temperature changes

Calculating Energy Changes in Phase Changes Can’t use specific heat, because no temperature change is involved Instead, use this formula Q = m  Heat of fusion (or heat of vaporization) use heat of fusion for freezing, melting use heat of vaporization for boiling, condensation

Heating/Cooling Curves Regions A, C, E have temperature change Q = mCDT Choose specific heat to match state of matter Regions B, D are phase changes B: Q = mHfus D: Q = mHvap

Enthalpy of the reaction The total energy change associated with a chemical or physical change Given the symbol DHrxn DHrxn = (energy of products) – (energy of reactants)

Classifying Heat Changes Thermite reaction Produces molten iron Formerly used in welding railroad tracks, shipbuilding Gives off light and heat! Highly exothermic

Exothermic reactions Energy is released to the surroundings The temperature of the surroundings increases The products have less energy than the reactants

Endothermic Reactions Energy is absorbed from the surroundings The temperature of the surroundings decreases The products have MORE energy than the reactants

Thermochemical Equations Exothermic reactions Energy released by system Can treat energy as a product DH is negative 2 equivalent ways to write equation: CaO + H2O  Ca(OH)2 + 65.2 kJ CaO + H2O  Ca(OH)2 DH = - 65.2 kJ

Thermochemical Equations Endothermic reactions Energy absorbed by system Can treat energy as a reactant DH is positive 2 equivalent ways to write equation: 2 NaHCO3 + 129 kJ  Na2CO3 + H2O + CO2 2 NaHCO3  Na2CO3 + H2O + CO2 DH = + 129 kJ

2 ways to manipulate thermochemical equations 1) Write equation backwards Sign of DH must change A + B  C DH = + 123 kJ C  A + B DH = - 123 kJ 2) Multiply everything by a coefficient Must multiply DH by coefficient, too! 3 A + 3B  3C DH = 3 (+123 kJ) = + 369 kJ

Problems with thermochemical equations Consider the equation: 2 NaHCO3  Na2CO3 + H2O + CO2 DH = + 129 kJ How many kJ would be released if 4.5 moles NaHCO3 reacted?

Hess’ Law If you add two or more thermochemical equations to give a final equation, then you can also add the heat changes to give the final enthalpy of reaction.

Example What is the enthalpy change, DHrxn, for the decomposition of hydrogen peroxide? Target: 2 H2O2(l)  2 H2O(l) + O2(g) Given: H2(g) + O2(g)  H2O2(l) DH = -187.9 kJ H2(g) + ½ O2(g)  H2O(l) DH = -285.8 kJ

Standard Heats of Formation, DHof The standard heat of formation of a compound is the change in enthalpy that accompanies the formation of one mole of a substance from its elements in their standard states. The heat of formation of elements in their standard states is arbitrarily set to zero.

Using heats of formation Thermodynamic stability: a measure of the energy required to decompose the compound Compounds with large, negative enthalpies of formation are thermodynamically stable Many heats of formation have been measured. (see Appendix A-6 in textbook) Another way to do Hess’ Law! DHrxn = SnDHf(products) – SmDHf(reactants) where n represents the coefficients for the products m represents the coefficients for the reactants

Example Problems Compute DHrxn for the following reaction. (Refer to Appendix A-6) 2NO(g) + O2(g)  2 NO2(g) 4 FeO(cr) + O2(g)  2 Fe2O3(cr) Ans: -144.14 kJ, -560.0 kJ

Entropy Symbol: S A quantitative measure of the degree of disorder in a system The greater the disorder, the larger the value of S Solids have a high degree of order (low entropy) Gases have a low degree of order (high entropy) More particles (moles) results in higher entropy

Entropy, cont. Systems tend to proceed to higher disorder Examples Stirring sugar into your coffee The neatness of your locker The order of cards in a pack of playing cards after shuffling

J. Willard Gibbs 1839-1903 First American to earn a Ph.D. in science from a US university Yale, 1863 One of nation’s best scientists

Will a reaction occur spontaneously? The answer depends on the balance between enthalpy (heat changes) and entropy Gibb’s Free Energy The energy available from the system to do useful work

Gibb’s Free Energy DG = DH – TDS If DG is negative, the reaction will occur spontaneously and can proceed on its own. If DG is positive, the reaction is nonspontaneous and needs a sustained energy input to proceed. If DG is zero, the reaction is at equilibrium (both the forward and reverse reactions take place!)