Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 19: 4 March


Calorimetry, cont.

Example: 1.40 g of vegetable oil is placed in a bomb calorimeter with excess oxygen and ignited with a spark. If the calorimeter temperature changes from 20.000 °C to 21.195 °C, find the energy released per gram of oil . The calorimeter contains 2.50 kg of water. The calorimeter without water has a heat capacity of 1.00 kJ°C-1.

q = nCpDeltaT, where Cp is the molar heat capacity at constant pressure (= 75.3 J C-1mol-1 for water).

q= qwater + qcalorimeter

qwater= {(2.50 kg H2O)(1000g/kg) / (18.02 g H2O/mole)}{75.3 J C-1mol-1}{1.195 °C} = 1.25 x 104J

qcalorimeter = CDeltaT = (1.00 x 103J°C-1)(1.195 °C) = 1.195 x 103J

qtot= 1.25 x 104J + 1.195 x 103J = 1.369 x 104J

E/g = (1.369 x 104J) / 1.40 g = 9.78 kJ/g

Notice that this is now the energy released, and it will also be the energy you could potentially get from consuming this much oil, since we are working with state functions, and the pathway (fire or metabolism) doesn't matter.

Hess's Law

Hess's Law states that changes in enthalpy in any process depends only on the nature of the reactants and products, and is independent of the number of steps in the process or the pathway taken. Hess's Law is thus a result of the fact that enthalpy is a state function.

Hess's Law turns out to be extremely useful for determining the energy of various processes, and thus the conditions necessary for reactions to proceed. The pathway independence is particularly nice, because we can look at processes that have never been observed to occur in a laboratory, and reasonably discuss the thermochemistry of processes that might occur at the Earth's core or the heart of a comet etc.

In order to use Hess's Law we need to keep some properties of enthalpy in mind.

Standard Enthalpies (Heats) of Formation

Formation Reactions and Standard Enthalpies of Formation: If we consider a reaction in which compounds are formed from elements in their standard states then the value of DeltaH is the standard enthalpy (heat) of formation.

Standard States:

With this information we can now find DeltaH for any chemical reaction!

Example: Find the value of DeltaH for the reaction:

2 CO2 + 7 H2 right arrow C2H6 + 4 H2O(g)

From Table find DeltaH values:

Often just list compounds, since known to be from elements.
 C + O2 right arrow CO2  DeltaH = -393.5 kJ mol-1
 2 C + 3 H2 right arrow C2H6  DeltaH = -84.6 kJ mol-1
H2 + 1/2 O2 right arrow H2O(g)  DeltaH = -241.8 kJ mol-1

now we can put the reactions together, adding the enthalpies for products, and subtracting enthalpies for reactants (since the reaction directions are reversed), and multiplying enthalpy values by the coefficients of the balanced equation (since all of the formation reactions were based on coefficients of one).

-2 (DeltaHCO2) -7 (DeltaHH2) + (DeltaHC2H6) + 4 (DeltaHH2O) = DeltaHrxn

-2 (-393.5 kJ mol-1) - 7 (0) + (-84.6 kJ mol-1) + 4 (-241.8 kJ mol-1) = -2.648 x 102kJ mol-1

Atomic Structure & Electromagnetic Radiation (Light)

Electromagnetic Radiation comprises the various types of forms of radiation which propagate through space not associated with mass. The visible spectrum encompasses a very narrow region of the overall electromagnetic spectrum as seen below and on figure 7.2 on p 276 of your text.

diagram of electromagnetic spectrum

Electromagnetic radiation behaves in most circumstances as waves [Figure 7.1 p 276] and can thus be characterized as waves.

Labeled sine wave diagram


Three parameters determine a wave:

These parameters are related by the the expression:

v = f•lambda

For electromagnetic radiation (light) the speed is defined in a vacuum: v = c = 2.9979 x 108m/s

c = f• lambda in vacuo



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© R A Paselk

Last modified 4 March 2011