OB: Introduction to Thermochemistry; energy
units and energy unit conversion math,
review of Table B with descriptions, review of
Table I and descriptions, and the
inclusion of heat to Stoichiometry of balanced
chemical equations.
Get out reference tables,
get a calculator
Thermo = heat, the energy changes that happen in a chemical reaction.
Sometimes heat is given off which is exothermic,
sometimes energy is absorbed to make the reaction happen, this is
endothermic.
Energy Units: there are four, some you “know”, some will sound odd, that’s okay.
Calories are “food” calories.
A loose science word that means 1000 calories (with a small “c”)
Calories are really kilo-calories, or 1000 cal.
In our class we will use CAPITAL C for the food or kilo calories.
With a lower case “c”, calories are the “science” calories, a smaller amount of
energy. There are 1000 calories in 1 Calorie.
A calorie is the amount of energy required to
raise the temperature of one gram of water by
exactly 1 Kelvin (or 1.0°C)
1000 cal = 1 C = 1 kcal
1 Hershey bar has about 230 Calories, how many calories is that?
This is a one step unit conversion problem (do it now…)
1 Hershey bar has about 230 Calories, how many calories is that?
This is a one step unit conversion problem (do it now…)
230 C x 1000 cal
= 230,000 cal
1
1C
Another unit of energy is the JOULE, named after
J.P. Joule, the English physicist who studied energy and
work. His ideas led to the development of the theory
(now a law) of conservation of energy in a chemical
reaction. (wowza!)
Joules and calories are related with this equality:
4.18 Joules = 1.0 calories
The last unit is kilo-Joule or kJ
1000 Joules = 1 kilo-Joule
1.0 Calorie = 1000 cal = 4180 joules = 4.180 kJ
Converting back and forth between units will take some practice, but not much.
An idea that sometimes escapes students is this:
Food is the equivalent of energy. You eat food, which has its energy content
measured in Calories.
Too much food eaten = too many Calories of energy eaten = too much Charlie
Eat less food means taking in less energy. You get tired quicker, and begin to
use the stored energy in your body (fat becomes energy, you lose mass)
That’s called going on a diet.
Food is stored by the body (this is basic biology) as fat. Food (energy) is
stored to be used when the body needs energy. If you never need more
energy because you live now, and there are supermarkets and lunchrooms, you
don’t need to worry about your hunting skills, or the weather, then the
season of less food never comes. Your body stores the excess energy you
eat (in the form of cupcakes or salmon, food is energy), and you grow.
Food is energy. Calories turn into usable energy, or stored energy.
The more food = the more energy
Table B (let’s take a lookie-look at what gifts we have there!)
The Physical Constants of Water. These 3 constants are for water. Every
substance will have these 3 constants, although their numbers will be
different.
The Heat of Fusion
The heat of fusion for water means the amount of energy required to be added
to one gram of ice at the melting point, to melt it into one gram of water,
without changing the temperature. This means, 273 K solid to 273 K liquid.
It’s the cold phase change energy for water.
It also means the reverse: how much energy needs to be removed from
one gram of liquid water at the freezing point to turn it into a solid.
Heat of fusion for water is 334 Joules/gram
This is ONE GRAM of ICE CUBE.
1 gram of H2O(S)
To MELT this gram of ice it takes a CERTAIN AMOUNT of
energy added in, called the HEAT OF FUSION for water.
+ 334 Joules of energy
Energy must be added
This is ONE GRAM of WATER.
1 gram of H2O(L)
To FREEZE this gram of ice it takes a CERTAIN AMOUNT of
energy removed, called the HEAT OF FUSION for water.
– 334 Joules of energy
Energy must be removed
The Heat of Vaporization
The heat of vaporization for water means the amount of energy required to
be added to one gram of water at the boiling point, to vaporize it into one
gram of steam gas without changing the temperature. This means, 373 K
liquid to 373 K gas.
It’s the hot phase change energy for water.
It also means the reverse: how much energy needs to be removed from
one gram of steam at the condensing point to turn it into a liquid.
Heat of vaporization for water is 2260 Joules/gram
----------------------------------------------For water…
HF = 334 J/g
HV = 2260 J/g
The last constant for water here is the specific heat capacity. This is the amount of
energy required to change the temperature of one gram of water by one Kelvin (or
one degree centigrade). This change can be heating, or cooling. Energy added will
heat the water, energy removed will cool the water.
This constant can only be used when H2O is liquid phase, and there is a change in
temperature.
for water, the specific heat capacity (C) is
C = 4.18 Joules/gram x Kelvin
C = 4.18 J/g·K
----------------------------------------------Go to table B now. Under it, in small letters, add:
1 C = 1000 cal
4.18 J = 1 cal
1000 J = 1 kJ
Table I shows us the HEATS OF REACTION for
25 different balanced chemical reactions.
This means how much energy is released (exo) or absorbed (endo) by
these reactions when they occur at room temperature and normal
pressure.
Let’s look over this first reaction, the combustion of methane gas.
CH4 + 2O2 ---> CO2 + 2H2O
we can read this as:
1 mole CH4 combusts with 2 moles O2, into 1 mole CO2 + 2 moles H2O
The ΔH for this reaction is -890.4 kJ/mole
which means that when one mole of methane combusts, it
releases 890.4 kJ of energy.
(the - sign only means exothermic!)
There’s NO SUCH THING as negative energy. All energy is energy.
So, this becomes...
CH4 + 2O2 ---> CO2 + 2H2O + 890.4 kJ
we can read this as:
1 mole CH4 combusts with 2 moles oxygen,
into 1 mole CO2 and 2 moles water + 890.4 kJ energy
CH4 + 2O2 ---> CO2 + 2H2O + 890.4 kJ
Remember how to do stoich? You can determine the answer to this
(right) If 23.4 moles of methane combust, how many molecules of
water form?
Well, now I’m telling you that ENERGY is also in the mole ratio as well.
the mole ratio here is 1:2:1:2:890.4 kJ
You could do this problem now too… If 23.4 moles of methane
combust, how many kJ of energy are released?
If 23.4 moles of methane combust, how much energy is released?
MR
CH4
energy
1
890.4 kJ
23.4
X kJ
X = (890.4)(23.4) = 20,835.36 kJ
X = 20,800 kJ with 3 SF
Or, if I were feeling devilish on a Wednesday:
If 23.4 moles of methane combust,
how many Calories of energy is released (or joules, or calories)?