Calorimetry

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 = qwater + qcalorimeter

T = 21.195 °C – 20.000 °C = 1.195 °C

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

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

or, using specific heat for water, q = mCspT,

qwater= (2.50 kg H2O)(1000g/kg) (4.184J C-1g1)(1.195 °C) = 1.25 x 104J

qcalorimeter = CT = (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.

Real Gases

Know the van der Waals equations, its postulates, and how to use it: [Pobs + a(n/V)2] x (V – nb) = nRT. Note this simple model takes two factors into account, the small attraction of particles for each other, a(n/V)2 and the volume of the particles, nb, to more closely model actual gas behavior.

Material for this Friday's Exam ends here.

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.

• When a reaction is written in reverse, the sign of H is reversed.
• The magnitude of H is directly proportional to the amount of reactants. Thus if the coefficients of a reaction are multiplied, then H is multiplied by the same amount.

Hess's Law & 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 H is the standard enthalpy (heat) of formation.

Standard States:

• The form of a pure substance stable at one atm and 25°C. (Actually other temperatures are tabulated, so have to check when looking at tabulated values.)
• For a substance in solution the standard state is defined for a concentration of exactly one molar.
• For an element the standard state is the form stable at one atm and 25°C. Note that many elements have allotropes: different forms of the pure element. For example carbon has 3 allotropes, the most common of which are graphite and diamond. Graphite is the stable form (a diamond is not forever at 1 atm and 25°C!).

•The enthalpy of formation of a pure element in its standard state is defined to be 0.

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

Example: Find the value of H for the reaction:

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

From Table find H values:

 C + O2 CO2 H = -393.5 kJ mol-1 2 C + 3 H2 C2H6(g) H = -84.6 kJ mol-1 H2 + 1/2 O2 H2O(g) H = -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).

{(HC2H6) + 4 (HH2O)} – {2 (HCO2) + 7 (HH2)}= Hrxn

= {(-84.6 kJ mol-1) + 4 (-241.8 kJ mol-1)} – {2 (-393.5 kJ mol-1) + 7 (0)}

Hrxn == -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 297 of your text.

Electromagnetic Radiation comprises the various types or forms of radiation which propagate through space thar are not associated with mass.

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