Thermochemistry is the study of heat change in chemical reactions. Chapter 5 Energy & Rates of Reaction Unit 3
Thermochemistry is the study of heat change in chemical
reactions.
Chapter 5
Energy & Rates of Reaction
Unit 3
Chapter 5 Key Concepts: After completing this chapter you will be able to…..
• Compare the energy changes of different substances, perform calorimetry calculations, and describe the relationship between reactions using enthalpy terms.
• represent thermochemical equations using different methods, and determine whether a reaction is exothermic or endothermic
• Calculate and use bond energies to estimate the enthalpy change of a reaction.
• Define and solve problems using Hess’s law.
• Write formation reactions and calculate enthalpy changes using standard enthalpy of formation values.
• Describe current and future energy sources and explain their advantages and disadvantages
Energy: is the capacity to do work (units Joules, J)
• Work transferred energy by a force causing an object to move a distance (unit Joules)
• Thermal energy is the total quantity of potential and kinetic energy within a substance.
• Potential energy is the energy available due to an object’s position or composition
• Kinetic energy is the energy of the object in motion.
5.1
Heat is the transfer of thermal energy between two bodies that are at different temperatures. Heat moves from warm objects to cooler objects
Energy Changes in Chemical Reactions
Temperature is a measure of the average kinetic energy of the particles in a substance.
Temperature = Thermal Energy
900C 400C
greater thermal energy 5.1
Law of Conservation of Energy:
Energy cannot be created or destroyed
The system is the specific part of the universe that is of interest in the study.
open
mass & energy Exchange:
closed
energy isolated nothing
SYSTEM
SURROUNDINGS
5.1
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
2H2 (g) + O2 (g) 2H2O (l) + energy
H2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)
5.1
energy + H2O (s) H2O (l)
Nuclear Energy
• All nuclear reactions are exothermic. • Nuclear reactions release much more energy than exothermic
chemical reactions – per unit of mass. • Two main types of Reactions: Fusion and Fission • A fusion reaction occurs when two or more nuclei of small atomic
mass combine to form larger molecule. 21H + 31H → 42He + 10n
In Fission, large nuclei with high atomic mass are split to smaller,
lighter mass by collision with a neutron:
23592U + 10n → 92
36Kr + 14156Ba + 31
0n + 1.9 x 10 10 kJ/mol 5.1
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.
5.2: Calorimetry and Enthalpy
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 = mc Heat (q) absorbed or released:
q = mcΔT
q = CΔT
ΔT = Tfinal - Tinitial
5.2
Calorimetry: the measurement of heat into or out of a system for chemical and physical processes.
the heat released = the heat absorbed The device used to measure the absorption or release of heat in chemical or physical processes is called a “Calorimeter”
5.2
a “Calorimeter”
Calorimetry Calculations When analyzing data obtained using a calorimeter, make these assumptions:
• Any thermal energy transferred from the calorimeter to the outside environment is negligible.
• Any thermal energy absorbed by the calorimeter itself is negligible.
• All dilute, aqueous solutions have the same density (1.00 g/mL) and specific heat capacity (4.18 J/(g·°C)) as water
• Thermal energy absorbed or released by a chemical system is given the symbol q .
qwater = mcΔT
5.2
Calculations Involving Thermal Energy Transfer Thermal Energy transfer by metal
No heat enters or leaves! 5.2
How much heat is given off when an 869 g iron bar cools from 940C to 50C? Solution: Given: c of Fe = 0.444 J/g • 0C m = 869 g ΔT = Tfinal – Tinitial = 50C – 940C = -890C q = mcΔt = 869 g x 0.444 J/g • 0C x (–890C) = -34,000 J
5.2
A student places 50.0 mL of liquid water at 21.00 °C into a coffee-cup calorimeter. She places a sample of gold at 100 °C into the calorimeter. The final temperature of the water is 21.33 °C. The specific heat capacity of water is 4.18 J/g·°C and the density of water, d, is 1.00 g/mL. Calculate the quantity of thermal energy, q , absorbed by the water in the calorimeter
Calculations of metal in water
SOLUTION: Given: VH2O = 50.0 mL; c = 4.18 J/(g·°C) dH2O = 1.00 g/mL; Tinitial=21.00 °C; Tfinal=21.33 °C Unknown q = ? Equations d =m/v m=vd ∆T = Tfinal – Tinitial
1. Determine Mass m= vd, m =50mL x 1.00g/mL = 50 g 2. Determine Temperature change ∆T = 21.33oC – 21.00oC Calculate Themal energy q = mc∆t =50 g x 4.18 J x 0.33oC g. oC = 69 J The thermal energy absorbed by the water is 69 J.
Determining specific heat capacity of a substance Using the value of q above, calculate the specific heat capacity, c, of gold if its mass is 6.77 g. Final temperature of the gold is same as that of water in the calorimeter.
Given: mAu =6.77 g q = 69 J (+ or - ?) T initial =100 oC T final = 21.33oC ∆T =21.33 oC – 100oC = - 78.67 oC Required: cAu q = mc∆T Rearrage and solve for c
c = q/m∆T c = (-69J)/(6.77g x -78.67oC) c = 0.13 J/g oC The specific heat capacity of the sample of gold is 0.13 J/g oC
Enthalpy (H): Total amount of thermal energy in a substance
Enthalpy Change (∆H) :
Energy changes in Physical and Chemical Reactions The enthalpy change of the chemical system is equal to the flow of thermal energy in and out of the system,
∆H system = |q system| ∆H > 0 Endothermic reaction ∆H < 0 Exothermic reaction
5.2
ΔH = H (products) – H (reactants) ΔH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
ΔH < 0
Hproducts > Hreactants
ΔH > 0 5.2
Energy Changes during Exothermic & Endothermic Reaction in an Open System
Molar Enthalphy Change
Energy changes involving 1 mole of a substances… (∆H) ∆H = n∆Hr
Where n is the amount and ∆Hr is the molar enthalpy change of the reaction Types of Molar Enthalpies of Reaction Solution (∆Hsol) NaBr(s) →Na(aq) + Br(aq)
Combustion (∆Hc) CH4(g) + 2O2(aq) → CO2(aq) + H2O(l)
Vapourization (∆Hvap) CH3OH(l) → CH3OH(g)
Neutralization (∆Hneut) 2NaOH(aq) + H2SO4(aq) →Na2SO4(aq) + H2O(l)
Formation (∆Hf) C(s) +2H2(g) + 1/2O2(g) → CH3OH(l)
5.2
Molar Enthalpy Calculations
Given: methanol=1.00 g; ∆Hvap=38.6 kJ/mol ∆H = ? n = ? Equation: ∆H = n∆Hvap
5.2
Methanol= 24 +6 +16 = 46g/mol
n = m / M n = 1 g x 1 mol/46 g = 0.022 ∆H = 0.022 mol x 38kJ/mol = 0.836 KJ = 836 J
Calculate ∆H for Vaporization Reactions Ethanol, CH3CH2OH(l), is used to disinfect the skin prior to an injection. If a 1.00 g sample of ethanol is spread across the skin and evaporated, what is the expected enthalpy change? The molar enthalpy of vaporization of ethanol is 38.6 kJ/mol.
Representing Molar Enthalpy Changes
Energy changes can be communicated in four different ways. Three of them are thermochemical equations and one uses diagram. 1. By including an energy value as a term in the thermochemical equation e.g., CH3OH(l) + 3/2 O2 → CO2(g) + 2H2O(g) + 726 KJ(Exothermic Reaction) 2. By writing a chemical equation and stating its enthalpy change e.g., CH3OH(l) + 3/2O2(g)→CO2(g)+ 2H2O(g) ∆H = -726 KJ (Exothermic
Reaction)
3. By stating the molar enthalpy of a specific reaction e.g., ∆Hcombustion or ∆Hc = - 726 KJ/mol CH3OH
4. By drawing a chemical potential energy diagram Note: Fractions are convenient in many thermochemical equations. They apply to fractions of moles of substances e.g., 3/2 mol represents 1.5 mol rather than fractions of actual molecules
5.2
H2O (s) H2O (l) ΔH = 6.01 kJ
Is ΔH negative or positive?
System absorbs heat
Endothermic
ΔH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
5.2
Potential Energy Diagram
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ΔH = -890.4 kJ
Is ΔH negative or positive?
System gives off heat
Exothermic
ΔH < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
5.2
Thermochemical Equations