Thermochemistry Chapter 5
Intro Vocabulary ●
Thermodynamics - Study of energy and its transformations
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Thermochemistry - Relationship between chemical reactions and energy changes
5.1 - Energy Energy: Capacity to do work
Work: Energy used to cause an object to move
Heat: energy used to cause the temperature of an object to increase
5.1 - Energy Potential Energy vs Kinetic Energy
Kinetic Energy Examples 1.
Determine the kinetic energy of a 500. kg motorcycle moving at 100. km/hr.
2.
Determine the kinetic energy of an argon atom moving at a speed of 650 m/s. Note: 1 amu = 1.66 x 10-27 kg
5.1 - Energy Units of Energy ●
joule
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calorie
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1 cal = 4.184 J
5.1 - Energy System vs. Surroundings
5.1 - Energy Transferring Energy: Work and Heat ● ● ●
work: energy transferred when a force moves an object force: push or pull w=Fxd
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heat: moves from hotter to colder object
5.1 - Energy Important Aspects of Heat [Thermal Energy] 1. 2. 3.
NOT THE SAME AS TEMPERATURE More thermal energy = more motion of particles Thermal energy depends on number of molecules present
Work Examples 1.
Determine the work, in joules, done to lift a 124 kg object a distance of 1.45 m.
2.
Determine the work, in joules, to lift a 1500 lb box 15.2 m. Note: 1 kg = 2.20 lb
5.2 - The First Law of Thermodynamics States that energy is conserved ●
Energy can be neither created nor destroyed
5.2 - The First Law of Thermodynamics Internal Energy ● ●
Sum of ALL kinetic and potential energy Concerned with the change in energy
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ΔE =
5.2 - The First Law of Thermodynamics Three parts of thermodynamic quantities:
5.2 - The First Law of Thermodynamics Equation Examples - N2(g) + O2(g) → 2 NO(g) + heat
- heat + 2 H2O(g) → 2 H2(g) + O2(g)
Energy Diagrams
5.2 - The First Law of Thermodynamics Relating ΔE to Heat and Work ●
ΔE = q + w
Calculating ΔE Examples 1.
2.
Calculate the change in the internal energy for a process in which a system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. As a reaction occurs, the system loses 1150 J of heat to the surroundings. The surroundings do 480 J of work on the system; what is the change in the internal energy of the system?
5.2 - The 1st Law of Thermodynamics Endothermic and Exothermic Processes
5.2 - The 1st Law of Thermodynamics State Functions
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Value depends only on the present state of the system, not on the path the system took to reach that state Examples:
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Non-Examples:
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5.3 - Enthalpy Internal energy plus the product of the pressure and volume:
All are state functions Symbol:
5.3 - Enthalpy Pressure-Volume Work ● ● ● ●
Work involved in the expansion or compression of gases At constant pressure Only work is mechanical work associated with a change in volume w = -PΔV
w = -PΔV Examples 1.
A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 0.250 L and the final volume is 0.980 L. If the piston expands against a constant pressure of 1.35 atm, how much work, in J, is done? Note: 1 L・atm = 101.3 J. Was work done by the system or on the system?
2.
Calculate the work, in J, if the volume of a system contracts from 1.55 to 0.85 L at a constant pressure of 0.985 atm. Was work done by the system or on the system?
5.3 - Enthalpy Enthalpy Change
5.4 - Enthalpies of Reaction Enthalpy change for a reaction:
Called: enthalpy of reaction or heat of reaction
5.4 - Enthalpies of Reaction Two ways to show it with a reaction:
5.4 - Enthalpies of Reaction Energy Diagrams
5.4 - Enthalpies of Reaction Key points of thermochemical equations: 1.
Magnitude is proportional to the amount of reactants
5.4 - Enthalpies of Reaction Key points of thermochemical equations: 2.
Reverse reaction has the opposite sign.
5.4 - Enthalpies of Reaction Key points of thermochemical equations: 3.
ΔH depends on the state of matter of reactants and products
Enthalpies of Reaction Examples 1.
How much heat is released when 3.5 moles of methane is burned in a constant-pressure system? CH4(g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
ΔH = -890 kJ
Enthalpies of Reaction Examples 2.
What is the enthalpy change for the oxidation of 5.00 g of sugar? C12H22O11(s) + 12 O2(g) → 12 CO2(g) +11 H2O(l) ΔH = -5645 kJ
5.5 - Calorimetry Can determine the value of ΔH experimentally: Calorimetry - measurement of heat flow using a calorimeter
5.5 - Calorimetry Heat Capacity - amount of heat required to raise the temperature by 1 K (or 1 oC) - higher heat capacity = more heat required to raise temp
Specific Heat Capacity Example 1.
Determine the quantity of heat that must be added to raise the temperature of 250. g of water from 20.5°C to 95.6°C. The specific heat capacity of water is 4.184 J/g・oC .
2.
Determine the enthalpy change per mole of water in the above problem.
5.5 - Calorimetry Constant-Pressure Calorimetry - Coffee-cup calorimetry - Need to know three things: -
mass of solution specific heat of solution amount of reactants
- qrxn = -qsoln
Constant-Pressure Calorimetry Examples 1.
You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from 25.10°C to 27.78° C. What is the molar enthalpy of neutralization of the acid in kJ/mol? Assume all densities are 1.00 g/mL and the specific heat capacities are 4.184 J/g*K.
Constant-Pressure Calorimetry Examples 1.
You place 0.500 g of Mg chips in a coffee cup calorimeter and then add 100.0 mL of 0.100 M HCl. The temperature increases from 295.4K to 318.0K. What is the enthalpy for the reaction per one mole of Mg? Assume the specific heat capacity of the solution is 4.184 J/g*K and the density of the HCl solution is 1.00 g/mL. Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq)
5.6 - Hess’s Law Can’t always measure the value of ΔH of a reaction in a calorimeter Can use the values of ΔH from known reactions to find another reaction’s value Hess’s Law - If a reaction is carried out in a series of steps, ΔH for the overall reaction equals the sum of the enthalpy changes for the individual steps.
5.6 - Hess’s Law Example: +
A+2B→2C+D ____ 2C → 2 E_____ A + 2B → D + 2 E
ΔH = -802 kJ ΔH = -88 kJ ΔH = -890 kJ
Hess’s Law Examples 1.
Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) → CO2(g) + ½ N2(g)
ΔH? = ?
Given the following information, calculate the ΔH?: Eqn A: CO(g) + ½ O2(g) → CO2(g) ΔHA = -283.0 kJ Eqn B: N2(g) + O2(g) → 2 NO(g) ΔHB = 180.6 kJ
Hess’s Law Examples 2.
Nitrogen oxides undergo many interesting reactions. Calculate ΔH for the following overall equation: 2 NO2(g) + ½ O2(g) → N2O5(s) ΔH = ? Useful Reactions: N2O5(s) → 2 NO(g) + 1.5 O2(g)
ΔH = 223.7 kJ
NO(g) + ½ O2(g) → NO2(g)
ΔH = -57.1 kJ
Hess’s Law Examples 3.
Calculate ΔH for the reaction: NO(g) + O(g) → NO2(g). Helpful equations: NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -198.9 kJ O3(g) → 3/2 O2(g) ΔH = -142.3 kJ O2(g) → 2 O(g) ΔH = 495.0 kJ
5.7 - Enthalpies of Formation Definition: Enthalpy change associated with the formation of a compound from its elements AKA: heat of formation Symbol: Dependent on temperature, pressure, and state of matter of reactants and products
5.7 Enthalpies of Formation To compare values between different substances, need standards. Standard State:
Standard Enthalpy Change: Symbol:
5.7 - Enthalpies of Formation Standard Enthalpy of Formation Definition: Change in enthalpy for the reaction that forms one mole for the compound from its elements in their standard states. Symbol: Example:
5.7 - Enthalpies of Formation Another Examples of Standard Enthalpy of Formation:
STANDARD ENTHALPY OF FORMATION FOR MOST STABLE FORM OF AN ELEMENT IS ZERO!
5.7 - Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction Equation:
Enthalpies of Formation Examples 1.
CaCO3(s) → CaO(s) + CO2(g) Use the values below to determine the ΔH for the above reaction. Is this reaction endothermic or exothermic? Draw an Energy Diagram.
Compound
ΔHfo (kJ/mol)
CaCO3(s)
-1207.6
CaO(s)
-635.1
CO2(g)
-393.5
Enthalpies of Formation Examples 2.
Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizers, dyes and explosives. The first step in the industrial process is the oxidation of ammonia: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Calculate ΔHrxn° from ΔHf° values given: ΔHf° [NH3(g)]: -45.9 kJ/mol ΔHf° [NO(g)]: 90.3 kJ/mol ° ΔHf [H2O(g)]: -241.8 kJ/mol
