Chemistry 103A; Sections 5, 6, 7, 8; Lecture 16, 29 Sep 00

Review - Using heats of formation

Here's the general formula for a hypothetical reaction that we used:

a A + b B ® c C + d D, (where a, b, c, and d are small integers which balance the equation and A, B, C, and D are compounds).

Then the heat of the reaction is given by,

DHorxn = c DHof,C + d DHof,D - a DHof,A - b DHof,B .


Why does it work?

It works because of Hess's law.

When we first talked about Hess's law we said that you could add or subtract chemical reaction equations. What does it mean to "subtract" a chemical reaction equation?

Consider the reaction equation for the formation of H2O(l)

H2(g) + 1/2 O2(g) ® H2O(l) DHfo = - 285.8 kJ/mol.
To subtract a chemical reaction equation is equivalent to writing the equation in reverse and then adding it. But when we write the reaction in reverse we have to change the sign of DHfo. So we get,
H2O(l) ® H2(g) + 1/2 O2(g) DHfo = + 285.8 kJ/mol.
This makes sense. If the reaction is exothermic - gives off heat - then its reverse must be endothermic - absorbs heat.

Go back to our first example.

Reaction 11: 2 NH3(g) ® N2H4(l) + H2(g)
The formation reactions used here are:
Reaction 9: 1/2 N2(g) + 3/2 H2(g) ® NH3(g) - 46.11 kJ/mol

Reaction 10: N2(g) + 2 H2(g) ® N2H4(l) + 50.63 kJ/mol

In order to obtain Reaction 11 from Reactions 9 and 10 we must take
1 mol ´ Reaction 10 + 2 mol ´ (- Reaction 9) = Reaction 11
(Since taking the negative of a reaction is the same as writing that reaction backwards, - Reaction 9 would be,
- Reaction 9: NH3(g) ® 1/2 N2(g) + 3/2 H2(g) + 46.11 kJ/mol.
Double this (- Reaction 9) to get,
2 mol ´ (- Reaction 9): 2 NH3(g) ® N2(g) + 3 H2(g) + 92.22 kJ.
Now add this last one to 1 mol ´ Reaction 10
1 mol ´ Reaction 10: N2(g) + 2 H2(g) ® N2H4(l) + 50.63 kJ

+

2 mol ´ (- Reaction 9) : 2 NH3(g) ® N2(g) + 3 H2(g) + 92.22 kJ

to get,
Reaction 11: 2 NH3(g) ® N2H4(l) + H2(g) 142.85 kJ
Notice that the N2(g) cancelled because it appears on both sides and two of the H2(g) cancelled for the same reason.

I will do one more example of using heats of formation to find the heat of a reaction using data from your text.

Consider the reaction:

H2S(g) + 4 H2O2(l) ® SO3(g) + 5 H2O(l)
The table of heats of formation gives the following data:
H2(g) + 1/8 S8(s) ® H2S(g) DHfo = - 20.63 kJ/mol

H2(g) + O2(g) ® H2O2(l) DHfo = - 187.78 kJ/mol

1/8 S8 + 3/2 O2(g) ® SO3(g) DHfo = - 395.72 kJ/mol

H2(g) + 1/2 O2(g) ® H2O(l) DHfo = - 285.83 kJ/mol

We know how find DHo for the reaction using our general formula, which in this case would look like,
DHo =  1 mol ´ DHfo(SO3(g)) + 5 mol ´ DHfo(H2O(l))
- 1 mol ´DHfo(H2S(g)) - 4 mol ´ Hfo(H2O2(l))
 
= 1 mol ´ (- 395.72 kJ/mol) + 5 mol ´ (- 285.83 kJ/mol) - 1 mol ´ (- 20.63 kJ/mol) - 4 mol ´ (- 187.78 kJ/mol)) = - 1053.12 kJ


However, If we want to see that this is a direct consequence of Hess's law we can re write the stack of reaction equations above leaving the product equations forward and reversing the reactant equations.

1/8 S8 + 3/2 O2(g) ® SO3(g) DHfo = - 395.72 kJ/mol

5 (H2(g) + 1/2 O2(g) ® H2O(l) DHfo = - 285.83 kJ/mol)

H2S(g) ® H2(g) + 1/8 S8(s) DHfo = + 20.63 kJ/mol

4 (H2O2(l) ® H2(g) + O2(g) DHfo = + 187.78 kJ/mol)

If we add this stack of reaction equations we get our initial reaction and if we add the heats of formation (after multiplying by the appropriate coefficient) we get the same heat of reaction. Notice that in the addition the S8, O2, and H2 all cancel out.
 
 

Chapter 7: Atomic Structure

Before we can get into the details of atomic structure we have to spend some time talking about light. This is because almost everything we know about atomic structure was learned through the interaction of atoms with light.

In addition, the three major experiments or phenomena that lead to our present understanding of atomic structure involved the interaction of light with matter.

Light is produced by oscillating electric charges. ("Oscillating" is a fancy word for vibrating.)

Conversely, light will interact with charged particles by trying to make them vibrate.

A charge produces an electric field.

If we move the charge the effect of having moved the charge moves through space with a speed, c. c is called the speed of light or the velocity of light.

c = 2.99792458 ´ 108 m/s (exactly).
(To within 0.07 % this is approximately 3.00 ´ 108 m/s. You will be expected to know the speed of light to this accuracy.)

If we oscillate the charge (vibrate it or "wiggle" it) the effect of that oscillation moves through space with the speed, c.

Suppose we find a way to make the charge vibrate at a frequency, n (lower case Greek "nu"). n has units 1/s because it repeats itself n times per second.

This vibration will create a pattern in the electric field due to the oscillating charge called an electromagnetic wave.

As the charge oscillates it is up and then down and then up again. The distance the electromagnetic wave travels between the time the charge is up, down, and then up again is called the wavelength. The wavelength is usually given the symbol, l (lower case Greek "lambda").

It is not hard to prove that

c = ln .
(You will be expected to know this equation.)

Since c is one of the fundamental constants of nature, if we know n we can find l , and vice versa.

(There are two other types of wave motion that are more familiar to us than light waves. These are sound waves and waves moving on the surface of a liquid. Sound waves can only exist in a gas, liquid, or solid. That is, they need some "medium" in which to propagate. Ocean waves are waves moving on the surface of a liquid. The medium in which they propagate is the liquid/air interface.

Both sound waves and ocean waves are good examples of wave motion, but they differ in several fundamental ways. Suppose you are watching a ping pong ball floating on the smooth surface of a pool and someone drops a rock in the pool at the other end. You can see the waves generated by the rock moving toward you. (They also have a wavelength and a frequency and a speed.) But when the waves hit the ping pong ball the ball does not move in the same direction the wave is moving. The ball moves up and down on the liquid surface. This motion is perpendicular to the direction the wave is travelling.

When the effect or action of a wave motion is perpendicular to the direction the wave is moving the wave is called a transverse wave.

A sound wave in a gas or liquid is a "pressure wave." Suppose you are listening to someone play a musical instrument. The instrument is producing an oscillating pressure in the air and the effect of this oscillation travels toward your ear. However, as the sound wave passes through the air the molecules in the air are moving back and forth along a line the runs from the instrument to your ear. That is, the effect or action of the wave is parallel to the direction of propagation.

When the effect or action of a wave motion is perpendicular to the direction of propagation the wave is called a longitudinal wave.)

So what kind of a wave is a light wave?

Polarization

We already know that light is an oscillating electric field. But an electric field is a "vector" quantity, it has a direction. In a light wave the oscillating part of the electric field points in a direction perpendicular to the direction the wave is travelling. So light is a transverse wave.

However, the direction that the electric field points in a light wave is the same as the direction that the oscillating charge is moving.

The direction the electric field is pointing in the light wave is called the polarization of the light wave.

The motion of the charges in incandescent and fluorescent lights is random - moving in all possible directions. So the light from incandescent and fluorescent light fixtures is "unpolarized" because the electric fields from these many different oscillating charges are pointing in all possible directions.

Polaroid filters will allow light with electric fields pointing in one direction to pass, but will block light with electric fields perpendicular to that direction.

The Electromagnetic Spectrum

The frequency of electromagnetic radiation ranges from essentially zero to over 1024 s-1. Conversely, the wavelength ranges from millions of miles to smaller than the nucleus of atoms.

The full range of electromagnetic radiation is technically light, but our eyes are only sensitive to a very small fraction of the total range of frequencies and wavelengths. There is an instructive diagram of the electromagnetic spectrum on page 295 of your text.