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