Chemistry 123: Physical and Organic Chemistry · 2/9/09 3 Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry Winter 2009 Page 5 …

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Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry http://people.ok.ubc.ca/orcac/chem123out.html

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Topic 2: Thermochemistry Text: Chapter 7 and 19 (~ 3 weeks) 2.0 Introduction, terminology and scope 2.1 Enthalapy and Energy Change in a chemical process; 1st law of Thermodynamics 2.2 Enthalapy of Formation, Calorimetry, Heat capacity, Bond Energies, and Standard State 2.3 Entropy; 2nd and 3rd laws of thermodynamics 2.4 Free Energy; Spontaneity and Equilibrium. 2.5 Relationship between dH°,dS°, dG° and K 2.6 Change of dG with Temperature adn Concentration; Effect of Temperature on K (van't Hoff)

Topic 2: Introduction,

From: http://www.dilbert.com


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2.0 Introduction, terminology and scope

• A System is the part of the universe under study. •  The Surroundings is anything outside of the System. •  An open system allows for exchange of both material and energy •  A closed system can exchange only energy • An isolated system, does not permit the transfer of mass or Energy. •  Energy is the ability to do work •  Work occurs when a force acts through a distance. •  Kinetic Energy is the energy of a moving object. •  Thermal Energy is the kinetic energy associated with random molecular motion

Topic 2.0: Introduction,

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Topic 2.0: Introduction, Kinetic Energy

ek = 1 2

mv2 [ek ] = m s

= J kg 2

Work

w = Fd [w ] = kg m s2

= J m



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Understanding

Calorie (cal): The quantity of heat required to change the temperature of one gram of water by one degree Celsius.

Joule (J): International System of Units (SI) unit for heat

1 cal = 4.184 J


James Joule (1818-1889)

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The quantity of heat (q) 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 times c = C specific heat.

◦  Water: c = 4.184 g j-1 °C-1

q = m c ΔT

q = C ΔT

Heat Capacity


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Conservation of Energy Topic 2.0: Introduction,


In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed.

qsystem + qsurroundings = 0

qsystem = -qsurroundings


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Determination of Specific Heat Topic 2.0: Introduction,


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EXAMPLE 7-2

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qlead = -qwater

qwater = mcΔT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C

qwater = 1.4×103 J

qlead = -1.4×103 J = mcΔT = (150.0 g)(clead)(28.8 - 100.0)°C

clead = 0.13 Jg-1°C-1

Determining Specific Heat from Experimental Data. Using the data presented on the last slide to calculate the specific heat of lead. Water is the surroundings…


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♦ Chemical energy. • Contributes to the internal energy of a

system. ♦ Heat of reaction, qrxn.

•  The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature.

Heats of Reaction and Calorimetry

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Topic 2.0: Introduction, Heats of Reaction and Calorimetry ♦  Exothermic reactions.

•  Produces heat, qrxn < 0. ♦  Endothermic reactions.

•  Consumes heat, qrxn > 0.

♦  Calorimeter •  A device for measuring quantities

of heat.

Add water


qrxn < 0 qrxn > 0 Winter 2009 Page 12


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qrxn = -qcal

qcal = q bomb + q water + q wires +…

Define the heat capacity of the calorimeter: qcal = ΣmiciΔT = CΔT

all i

heat

C = heat capacity of the calorimeter


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Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of 1.010 g sucrose, in a bomb calorimeter, causes the temperature to rise from 24.92 to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C. (a) What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11? (b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.

1 food Calorie ( 1 Calorie with a capital C) is actually 1000 cal or Kcal

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Topic 2.0: Introduction, Calculate qcalorimeter:

qcal = CΔT = (4.90 kJ/°C)(28.33-24.92)°C = (4.90)(3.41) kJ = 16.7 kJ

Calculate qrxn: qrxn = -qcal = -16.7 kJ per 1.010 g

qrxn = -qcal = -16.7 kJ

1.010 g

= -16.5 kJ/g

343.3 g

1.00 mol = -16.5 kJ/g

= -5.65 × 103 kJ/mol

qrxn

(a)

Calculate qrxn for one teaspoon:

4.8 g

1 tsp

= (-16.5 kJ/g)( qrxn (b) )( )= -19 kcal/tsp 1.00 cal 4.184 J


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Coffee cup Calorimeter Topic 2.0: Introduction,

A simple calorimeter. •  Well insulated and therefore isolated. •  Measure temperature change.

qrxn = -qcal

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In addition to heat effects chemical reactions may also do work.

Gas formed pushes against the atmosphere.

Volume changes.

Pressure-volume work.

Pressure- Volume Work of a reaction

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Topic 2.0: Introduction, Pressure- Volume Work of a reaction


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Topic 2.0: Introduction, Pressure- Volume Work of a reaction

Assume an ideal gas and calculate the volume change: Vi = nRT/P = (0.100 mol)(0.08201 L atm mol-1 K-1)(298K)/(2.40 atm) = 1.02 L Vf = 2.04 L

Calculating Pressure-Volume Work. Suppose the gas in the previous figure is 0.100 mol He at 298 K and the each mass in the figure corresponds to an external pressure of 1.20 atm. How much work, in Joules, is associated with its expansion at constant pressure.

ΔV = 2.04 L-1.02 L = 1.02 L

w = -PΔV = -(1.20 atm)(1.02 L)( = -1.24 × 102 J

) -101 J

1 L atm

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Chemistry 123: Physical and Organic Chemistry · 2/9/09 3 Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry Winter 2009 Page 5 …