# Thermochemistry.

## Presentation on theme: "Thermochemistry."— Presentation transcript:

Thermochemistry

Specific Heat Formula Q = Energy (heat) lost or gained
cp = Specific Heat T = Temperature change m = Mass

Two Types of Thermal Reactions
Exothermic: Releases Thermal Energy (heat) sign is – Endothermic: Absorbs Thermal Energy (heat) sign is +

Exothermic Processes Processes in which energy is released as it proceeds, and surroundings become warmer Reactants  Products + energy

Endothermic Processes
Processes in which energy is absorbed as it proceeds, and surroundings become colder Reactants + energy  Products

Enthalpy = H In a chemical reaction, Enthalpy (H) is equal to the energy that flows as heat. (at a constant pressure) Example: When 1 mol of methane gas is burned it releases 890 kJ of energy CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat H = -890kj = exothermic reaction

Solution: Molar mass of CH4 = 16.04 g / mol
Calculate H for a process in which 5.8 g of methane are burned. H = -890kJ per mol CH4 Rxn: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) +heat Solution: Molar mass of CH4 = g / mol 5.8 g CH4 1 mol CH4 = 0.36 mol CH4 16.04 g CH4 0.36 mol CH kJ = -320 kJ 1 mol CH4

Solution: Molar mass of SO2 = 64.07 g / mol
Calculate H when 12.8 g of sulphur dioxide reacts with excess oxygen to form sulphur trioxide. H = kJ per mol SO2 Solution: Molar mass of SO2 = g / mol 12.8 g SO2 1 mol SO2 = mol SO2 64.07 g SO2 mol SO kj = kJ 1 mol SO2

Hydrogen peroxide decomposes according to the following thermochemical reaction:
H2O2(l) → H2O(l) + 1/2 O2(g) ΔH = kJ Calculate the change in enthalpy (ΔH) when 4.00 grams of hydrogen peroxide decomposes. Calculate the change in enthalpy (ΔH) when grams of water are created…

The reaction that occurs in hand warmers is:
4Fe(s) + 3O2(g)  2Fe2O3(s) + heat H of reaction = -1652kJ How much heat is released when 2.5 grams of Fe(s) is reacted with excess oxygen?

Law of Concert Tickets! Miller Stitts  LadyGaga \$600 2 Winters + Miller  Justin Bieber + \$300 Stitt  2 Elvis Winter \$500 Lady Gaga  Justin Bieber Elvis \$_____ ???

Law of Concert Tickets = Hess’s Law
You can add KNOWN equations to solve UNKOWN equations Two Rules: FLIP (reverse) the equation – you must FLIP the sign (- +) MULTIPY / DIVIDE equation – must Multiply / Divide the ΔH!

Example #1: Calculate the enthalpy for this reaction:
2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermo chemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l) ΔH° = kJ C(s) + O2(g) ---> CO2(g) ΔH° = kJ H2(g) + (1/2)O2(g) ---> H2O(l) ΔH° = kJ a) first eq: flip it so as to put C2H2 on the product side  b) second eq: multiply it by two to get 2C  c) third eq: do nothing.  kJ + (-787 kJ) + ( kJ) = kJ

2C(s) + H2(g) ---> C2H2(g)
ΔH° = ??? kJ C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l) ΔH° = kJ C(s) + O2(g) ---> CO2(g) ΔH° = kJ H2(g) + (1/2)O2(g) ---> H2O(l) ΔH° = kJ

+234 + (+592) + (-1220) = -394 Example #2: Given the following data:
Find the ΔH of the following reaction: C(s) + O2(g) ---> CO2(g) SrO(s) + CO2(g) ---> SrCO3(s) ΔH = -234 kJ 2SrO(s) ---> 2Sr(s) + O2(g) ΔH = kJ 2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g) ΔH = kJ a) first equation - flip it b) second equation - divide by two c) third equation - flip it, divide by two (+592) + (-1220) = -394

+234 + (+592) + (-1220) = -394 Example #2: Given the following data:
Find the ΔH of the following reaction: C(s) + O2(g) ---> CO2(g) SrO(s) + CO2(g) ---> SrCO3(s) ΔH = -234 kJ 2SrO(s) ---> 2Sr(s) + O2(g) ΔH = kJ 2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g) ΔH = kJ a) first equation - flip it b) second equation - divide by two c) third equation - flip it, divide by two (+592) + (-1220) = -394

Hess's Law In a reaction, the change in enthalpy (ΔH) is the same - regardless if the reaction occurs in a single step or in several steps. If a series of reactions are added together, the net change in ΔH is the sum of the enthalpy changes for each step.

Rules for using Hess's Law
If the reaction is multiplied (or divided) by some factor, Δ H must also be multiplied (or divided) by that same factor. If the reaction is reversed (flipped), the sign of Δ H must also be reversed.

Water phase changes & Energy
constant Temperature remains __________ during a phase change.

Phase Change Diagram Processes occur by addition of energy 
 Processes occur by removal of energy

Phase Diagram Represents phases as a function of temperature and pressure. Critical temperature: temperature above which the vapor can not be liquefied. Critical pressure: pressure required to liquefy AT the critical temperature. Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).

Phase Changes

Effect of Pressure on Boiling Point
Boiling Point of Water at Various Locations Location Feet above sea level Patm (kPa) Boiling Point (C) Top of Mt. Everest, Tibet 29,028 32 70 Top of Mt. Denali, Alaska 20,320 45.3 79 Top of Mt. Whitney, California 14,494 57.3 85 Leadville, Colorado 10,150 68 89 Top of Mt. Washington, N.H. 6,293 78.6 93 Boulder, Colorado 5,430 81.3 94 Madison, Wisconsin 900 97.3 99 New York City, New York 10 101.3 100 Death Valley, California -282 102.6 100.3

Phase Diagram Represents phases as a function of temperature and pressure. Critical temperature: temperature above which the vapor can not be liquefied. Critical pressure: pressure required to liquefy AT the critical temperature. Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).

Phase changes by Name

Water

Phase Diagram for Carbon
Carbon dioxide Phase Diagram for Carbon dioxide

Phase Diagram for Carbon

Phase Diagram for Sulfur

Reaction Pathway Shows the change in energy during a chemical reaction

Exothermic Reaction 2H2(l) + O2(l)  2H2O(g) + energy
reaction that releases energy products have lower PE than reactants energy released 2H2(l) + O2(l)  2H2O(g) + energy

Endothermic Reaction 2Al2O3 + energy  4Al + 3O2
reaction that absorbs energy reactants have lower PE than products energy absorbed 2Al2O3 + energy  4Al + 3O2

Latent Heat of Phase Change
Molar Heat of Fusion The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. Molar Heat of Solidification The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

Latent Heat of Phase Change #2
Molar Heat of Vaporization The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point. Molar Heat of Condensation The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C? Mass of ice Molar Mass of water Heat of fusion

Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent. Substance Heat of Solution (kJ/mol) NaOH -44.51 NH4NO3 +25.69 KNO3 +34.89 HCl -74.84

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 6.1

Thermochemistry is the study of heat change in chemical reactions.
The system is the specific part of the universe that is of interest in the study. SURROUNDINGS SYSTEM open closed isolated Exchange: mass & energy energy nothing 6.2

2H2 (g) + O2 (g) 2H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2 (g)
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) Hg (l) + O2 (g) 6.2

DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.3

Thermochemical Equations
Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ 6.3

Thermochemical Equations
Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = kJ 6.3

Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) H2O (l) DH = 2 x 6.01 = 12.0 kJ 6.3

Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.3

Heat (q) absorbed or released:
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.4

Dt = tfinal – tinitial = 50C – 940C = -890C
How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = J/g • 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x J/g • 0C x –890C = -34,000 J 6.4

Constant-Volume Calorimetry
qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = msDt qbomb = CbombDt Reaction at Constant V DH = qrxn DH ~ qrxn No heat enters or leaves! 6.4

Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.4

General Chemistry: Chapter 7
Chemistry 140 Fall 2002 Terminology Energy, U The capacity to do work. Work Force acting through a distance. Kinetic Energy The energy of motion. Energy is from the Greek “work within”. Moving objects do work when they slow down or are stopped. Kinetic means “motion” in greek. Prentice-Hall © 2002 General Chemistry: Chapter 7

General Chemistry: Chapter 7
Chemistry 140 Fall 2002 Energy Kinetic Energy 1 kg m2 ek = mv2 [ek ] = = J 2 s2 w = Fd [w ] = kg m s2 = J m Work [w] means UNITS OF WORK, not concentration in this case. Prentice-Hall © 2002 General Chemistry: Chapter 7

General Chemistry: Chapter 7
Chemistry 140 Fall 2002 Energy Potential Energy Energy due to condition, position, or composition. Associated with forces of attraction or repulsion between objects. Energy can change from potential to kinetic. Energy changes continuously from potential to kinetic. Energy is lost to the surroundings. Prentice-Hall © 2002 General Chemistry: Chapter 7

Energy and Temperature
Thermal Energy Kinetic energy associated with random molecular motion. In general proportional to temperature. An intensive property. Heat and Work q and w. Energy changes. Prentice-Hall © 2002 General Chemistry: Chapter 7

General Chemistry: Chapter 7
Chemistry 140 Fall 2002 Heat Energy transferred between a system and its surroundings as a result of a temperature difference. Heat flows from hotter to colder. Temperature may change. Phase may change (an isothermal process). Heat is transfer of energy. Bodies do NOT contain heat. Prentice-Hall © 2002 General Chemistry: Chapter 7

General Chemistry: Chapter 7
Units of Heat Calorie (cal) The quantity of heat required to change the temperature of one gram of water by one degree Celsius. Joule (J) SI unit for heat 1 cal = J Prentice-Hall © 2002 General Chemistry: Chapter 7

General Chemistry: Chapter 7
Heat Capacity The quantity of heat required to change the temperature of a system by one degree. Molar heat capacity. System is one mole of substance. Specific heat capacity, c. System is one gram of substance Heat capacity Mass  specific heat. q = mcT q = CT Prentice-Hall © 2002 General Chemistry: Chapter 7