chm 108 suroviec spring 2014 chapter 6 energy transfer
TRANSCRIPT
CHM 108Suroviec
Spring 2014
Chapter 6Energy Transfer
I. Principles of Heat Flow
Energy is the capacity to do workWork is the result of a force acting through a distance
A. Different Types of Energy
Kinetic energy – energy associated with motion of an object
Thermal energy – energy associated with temperature of an object
Potential energy – energy associated with position of an object
Chemical energy – energy associated with position of an electron around an atom
B. Law of Conservation of Energy
Energy cannot be created or destroyedEnergy can be transferred from one kind to another
II. First Law of Thermodynamics
Total energy of the universe is constant and energy is neither created or destroyed
A.Internal Energy (IE) Sum of kinetic and potential energies of all the particles that compose the system
Is a state function
A. Internal Energy
Internal energy change = E
B. Exchanging Energy
A system can exchange energy with the surroundings through heat and work
B. Exchanging Energy
III. Quantifying Heat and Work
A. Heat Heat = exchange of thermal energy
between a system and its suroundings
B. Heat Capacity
When a system absorbs heat (q) its temperature changes by T.
Heat capacity = quantity of heat required to raise T by 1oC. (J/oC)
Specific heat capacity = quantity of heat required to raise T of 1 gram of material by 1oC. (J/goC)
Example
Suppose that you have the idea to start making stained glass. Given that glass turns to a pliable liquid around 1500oC and you are starting at 25oC, how much heat does a 2.50 g piece of glass absorb?
C. Pressure – Volume work
How does work change with the volume changes?
Example
Inflating a balloon requires P-V work1. If I inflate a balloon from 0.200L to 0.985L at an external pressure of 1.00 atm, how much work in Latm is done?
2. Given that 101.3 J = 1 Latm, what is this value of work in Joules?
IV. Measuring ΔE
The way to measure this is with constant volume calorimetry
A. Calorimetry
Calorimetry measures change in thermal energy between a system and its surroundings.At constant volume the change in temperature is related to heat absorbed by entire calorimeter
Example
When 2.09g of sucrose (C12H22O11) is combusted the temperature rose from 23.89 oC to 28.45 oC. What is the rxn for this in kJ and then in kJ/mole?
V. Enthalpy
In most cases we cannot hold the volume constant
We are interested with how much energy is given off at constant pressure
V. Enthalpy
The signs of H and E are similar in meaning
Example
If we burn 1 mole of fuel and constant pressure it produces 3452 kJ of heat and does 11 kJ or work. What are the values of H and E?
A. Stoichiometry involving H
Enthalpy for chemical reactions = Hrxn
For the reaction below, if you start with 13.2 kg of C3H8 what is q in kJ?
C3H8 (l) + 5O2 (g) 3CO2 (g) + 4H2O (l) Hrxn = -2044 kJ
VI. Measuring Hrxn
Coffee cup calorimetry is a way to measure Hrxn.
Knowing the mass of solution, the heat evolved will cause a T. Combing that with Cs of solution you can determine q.
Example
We want to know the Hrxn of the following reaction:
Ca(s) + 2HCl (aq) CaCl2 (aq) + H2 (g)
Given 0.204 g of Ca(s) in 100.0 mL of HCl (aq), the temperature rose from 24.8 °C to 33.9 °C. The density of the solution is 1.00 g/mL and the Cs,soln is 4.18 J/g°C
H2O (s) H2O (l)H = 6.01 kJ
• The stoichiometric coefficients always refer to the number of moles of a substance
• If you reverse a reaction, the sign of H changes
H2O (l) H2O (s)H = -6.01 kJ
• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.
2H2O (s) 2H2O (l)H = 2 x 6.01 = 12.0 kJ
IV. Thermochemical Equations
H2O (s) H2O (l)H = 6.01 kJ
• The physical states of all reactants and products must be specified in thermochemical equations.
H2O (l) H2O (g)H = 44.0 kJ
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn
S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn
The final reaction for formation of CS2 is:
C(graphite) + 2S(rhombic) CS2 (l)
Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
f
The standard enthalpy of formation of any element in its most stable form is zero.
Hf0 (O2) = 0
Hf0 (O3) = 142 kJ/mol
H0 (C, graphite) = 0f
H f0 (C, diamond) = 1.90 kJ/mol
V. Enthalpy of formation
VII. Standard Heats of Formation
Example
Write an equation for the formation of C6H12O6 from elements. The Hf
0 for C6H12O6 is -1273.3 kJ/mole
B. Calculating Hf0 for a
reaction
To calculate H rxn0 subtract the heats
of formations of the reactant multiplied by their stoichiometric coefficient from heats of formation of the products multiplied by their stoichiometric coefficents.
The standard enthalpy of reaction (H0rxn
) is the enthalpy of a reaction carried out at 1 atm.
aA + bB cC + dD
H0
rxndH0 (D)
fcH0 (C)
f= [ + ] - bH0 (B)faH0 (A)f[ + ]
H0
rxnnH0 (products)
f= mH0 (reactants)
f-
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
A. Hess’s Law
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Example
Compound Hf0 (kJ/mole)
C6H6 (l) +49.04
O2 (g) 0
CO2 (g) -393.5
H2O (l) -187.6