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Ellen Fraser
Study Guide: Chapter 6
Energy and Chemical Change
What is thermochemistry?
Thermochemistry is a branch of chemistry, which deals
with energy that is absorbed or released during a chemical
reaction. Heat (thermal energy) can be transferred between
objects having different temperatures. When this occurs,
the heat continues to flow until the objects reach the same
temperature (thermal equilibrium). Energy cannot be
created or destroyed, it can only be changed from one form
to another.
Endothermic and Exothermic Reactions
A reaction that has heat as a product is exothermic. This type of reaction
causes heat to be released.
Reactant
Product + Heat
A reaction that has heat as a reactant is endothermic. This type of reaction
causes heat to be absorbed.
Reactant + Heat
Product
Description of Energy Transfer
An example of energy transfer from a warmer to a cooler object: When heat from a hot cup of coffee flows from the cup into its
colder surroundings (i.e. a thermos). The molecules within the coffee are moving at a rapid pace as they collide with the slower moving
molecules of the thermos. This causes the coffee molecules to slow down and the thermos molecules to speed up. Eventually as the
temperature of the coffee decreases and the temperature of the thermos increases, the two establish thermal equilibrium. In this case,
thermal energy was transferred from the coffee to the thermos. The coffee is exothermic (releasing heat) while the thermos is endothermic
(absorbing heat).
*Remember: Heat and temperature are not the same thing!*
3 Types of Systems
In thermochemistry, a boundary can be visable (i.e. walls of a beaker), or invisible (i.e. the separation of warm and cold air along a weather
front). There are 3 possible types of systems depending on whether matter or energy can cross a certain boundary.
 Open System: can gain or loose mass and energy across their boundaries.
 Closed System: can absorb or release energy, but not mass across the boundary. The mass of a closed system always remains
constant.
 Isolated System: cannot exchange matter or energy with their surroundings. The energy of an isolated system always remains
constant.
Heat Capacity
There are 2 equations in relation to heat capacity. Heat capacity
depends on two factors , being size of sample and what the
sample is made of. Specific heat is an intensive property related
to heat capacity.
1) q=C t [heat capacity]
2) q=mc
t [specific heat capacity]
You can plug in known values in order to find an unknown value,
and in some situations, you can make these equations equal each
other in order to find an unknown value.
Specific Heats (Constants)
1 calorie = 4.18 J
Substance
Specific Heat
(J/g °C)
Carbon (graphite)
0.711
Copper
0.387
Ethyl alcohol
2.45
Gold
0.129
Granite
0.803
Iron
0.4498
Lead
0.128
Olive Oil
2.0
Silver
0.235
Water
4.18
*Plug in for “c” value in specific heat capacity equation depending on
which compounds/ elements you are dealing with.
Examples for Measurement of Heat
Question #1: You have a 250g sample of H20 and over the course of a reaction, the temperature
into the water (measured in calories)?
Answer:
Change in temp. = +5°C
Constant specific heat capacity for H20 (see chart above) = 4.18
Mass = 250g
q = (250)(4.18)(5)
q = 5225 J / 4.18 J
q = 1250 cal.
changed from 25°C to 30°C. How much energy was transferred
Question #2: A ball bearing at 220°C is dropped into a cup containing 250g of water at 20°C. The water and ball bearing come to a temperature of 30°C. What is the
heat capacity of the ball bearing?
Answer:
q1 = -q2
change in temp = tf – ti
C
t = -mc
t
C(-190) = -(250)(4.18)(10)
(-190)
(-190)
q1: 30°C - 220°C = -190°C
q2: 30°C - 20°C = +10°C
C (heat capacity of ball bearing) = 55 J/°C
*All “C” values should be positive
Enthalpy & Hess’s Law
Enthalpy is a measure of the total energy of a thermodynamic system. This is considered to be an internal energy.
Enthalpy is defined by the equation: H = E + PV
Hess’s Law states that the value of
H° for any reaction that can be written in steps equals the sum of the values of
H° of each of the individual steps.
Rules for Manipulating Thermochemical Equations
1)
2)
When an equation is reversed (written in the opposite direction), the sign of
H° must also be reversed.
Formulas canceled from both sides of an equation must be for the substance in identical physical states.
3)
If all the coefficients of an equation are multiplied or divided by the same factor, the value of
factor.
4)
For an endothermic change,
5)
For an exothermic change,
H° must likewise be multiplied or divided by that
H is positive.
H is negative.
Example for Hess’s Law
Question:
Coal gasification converts coal into a combustible mixture of carbon monoxide and hydrogen (called coal gas) in a gasifier.
H20(l) + C(s)
CO(g) + H2(g)
H° = ?
Calculate the standard enthalpy change for this reaction from the following half chemical equations:
2C(s) + O2(g)
2CO(g)
2H2(g) + O2(g)
H20(l)
2H20(g)
H20(g)
H° = -222 kJ
H° = -484 kJ
H° = +44 kJ
Steps Taken:
1)
2)
3)
4)
Divide 1st and 2nd half equations by 2 (do this to
H° values too)
nd
Flip 2 half equation (remember to reverse sign of
H° )
Remember to take into account physical states of compounds/ elements
Cancel & add
C(s) + 1/2O2(g)
CO(g)
H° = -111 kJ
H20(g)
H2(g) + 1/2O2(g)
H° = +242 kJ
H20(l)
H20(g)
H° = +44 kJ
H20(l) + C(s)
CO(g) + H2(g)
H° = +175 kJ
* you can check your answer by making sure final equation is the same as original
Enthalpy Calculations
At constant pressure, the energy of a system is called its enthalpy. At constant pressure,
is the enthalpy change (
H° =
E+P
V. The heat of reaction at constant pressure,
H°)
Product – Reactant
Example:
Calculate the enthalpy of formation (kJ/mol) of CO2(g). The enthalpy of reaction for the equation as written is -540.66 kJ/mol. If the answer is negative,
enter the sign and then the magnitude.
2 HCOOH(l) + O2(g) → 2 CO2(g) + 2 H2O(l) .
ΔHof (kJ/mol)
HCOOH(l)
-409
O2(g)
0
CO2(g)
?
H2O(l)
-285.83
P–R
(2 x CO2(g) + 2 x -285.83) – (2 x -409 + 0) = -540.66
2 x CO2(g) - 571.66 - - 818 = -540.66
2 x CO2(g) + 246.34= -540.66
2 x CO2(g) = -787
ΔHo CO2(g)= - 393.5 kJ/mol
Calorimetry
E=q+w
No matter how the change in energy, it is equal to the sum of q and w. The sign for q is negative when the system gives off heat or does work on the surroundings.
The sign is positive when the system absorbs heat or receives work energy done to it. When the volume of a system cannot change, as in a bomb calorimeter, w is
zero and q is the heat of reaction at a constant volume.