Transcript Document

• What do you visualize when thinking about
chemical reactions? You probably imagine…
o Heat, Light, Sound
• These are energy, and chemical reactions involve
energy because chemical bonds are being broken
and made!
• Does the reaction in the picture release
energy or absorb energy?
Chemical Reactions That
Involve Heat
• Energy: the capacity to do work.
• What kinds of energy are there?
Radiant (solar)
Thermal (motion of atoms & molecules)
Chemical (stored within chemical bonds)
Potential (available because of an object’s
position)
o Kinetic (energy of motion)
o
o
o
o
• Heat is the transfer of thermal energy
between two objects at different temperature.
• Thermochemistry is the study of heat
changes that occur in chemical reactions.
o
‘Thermes’ is Greek for “heat.”
• Thermodynamics is the broader study of
energy and work changes during such
processes.
Learning Target:
Be able to calculate the amount of energy absorbed or
released during a chemical or physical change.
Learning Outcomes:
Complete the endothermic exothermic minilab
and calculations for ∆H.
CHAP. 12 OUTLINE DUE TODAY!!!
• Remember the “Law of Thermodynamics”
o “Energy cannot be created or destroyed, but may be
converted from one form to another.”
• To study thermochemistry we must be careful about
the ‘system’ involved.
Open System: allows exchange of both energy and mass.
Closed System: allows the exchange of energy, but not
mass.
o Isolated System: allows the exchange of neither energy nor
mass.
o
o
Mass Loss
or Gain
CLOSED
OPEN
Energy Loss
or Gain
Energy Loss
or Gain
ISOLATED
Lab Activity
 Break out into lab groups to collect data on the
reaction that is given to you at your lab station.
 When you have collected your data we will
compile it to investigate endothermic versus
exothermic reactions.
• Exothermic reactions release heat energy.
o
‘Exo’ means outside.
 CH4(g) + 2O2(g) →CO2(g) + 2H2O(g) + Heat
• Endothermic reactions absorb heat energy.
o
‘Endo’ means inside.
2HgO(s) + Heat →2Hg(l) + O2 (g)
• Most chemical processes we use occur at normal, constant
atmospheric pressure, and heat energy is either gained or
lost during the chemical reaction.
• Chemists use a special term, enthalpy, to describe heat
changes for chemical reactions at constant pressure.
• Enthalpy is the energy of a substance plus a small
adjustment for pressure-volume work.
o
Enthalpy (H) = E + PV
= Internal Energy + Pressure-Volume (Work)
= Heat absorbed or released by a reaction
• U.S. Army “MREs” (“Meals ready to eat”)
Magnesium reacts with water to release heat and
hydrogen gas. Write the reaction:
o Mg + 2H2O →Mg(OH)2 + H2 + Heat
o MREs heat the food to 60° C in 15 minutes!
o EXOTHERMIC Reaction
• Ammonium Nitrate (A common fertilizer)
o NH4NO3 dissolves in water, causing a drop in
temperature. Write the reaction:
o NH4NO3 + H2O + Heat →NH41+ (aq) + NO31- (aq)
o NH4NO3 is used in cold packs for sports injuries.
o ENDOTHERMIC Reaction
o Note: NH4NO3 can also undergo an EXOTHERMIC
reaction when it explodes, hence its use as a blasting
agent in mining and (unfortunately) in terrorists’ weapons,
such as that used in the Oklahoma City attack.
o
Learning Target:
Be able to calculate the amount of energy absorbed or
released during a chemical reaction.
Learning Outcomes:
Complete the Heat Calculation Problems.
•
Enthalpy change, ΔH, for a chemical reaction that occurs at
constant pressure is the heat released or absorbed in the
reaction.
ΔH is just the difference between the enthalpy of the products minus
the enthalpy of the reactants:
ΔH = ∑Hproducts – ∑Hreactants
Think of enthalpy in terms of a savings account balance. You are just
keeping track of money going into the account and the money spent
vs. the net balance.
• We use diagrams to show enthalpy changes.
• For exothermic reactions, Hproducts is lower than
Hreactants so ΔH is negative.
o
Heat is released (lost by the system).
• For endothermic reactions, Hproducts is higher
than Hreactants so ΔH is positive.
o
Heat is absorbed (gained by the system).
• See diagrams on the next slide.
CH4(g) + 2O2(g) →CO2(g) + 2H2O(g) + Heat
CH4(g) + 2O2(g)
E
N
T
H
A
L
P
Y
ΔH = -803kJ
(EXOTHERMIC; HEAT
RELEASED FROM THE
‘SYSTEM’ TO THE
SURROUNDINGS)
CO2(g) + 2H2O(g)
REACTION PROGRESS (TIME)
NOTE: Hproducts IS
LOWER THAN
Hreactants
2HgO(s) + Heat →2Hg(l) + O2 (g)
2Hg(l) + O2(g)
E
N
T
H
A
L
P
Y
ΔH = +181kJ
(ENDOTHERMIC; HEAT
ABSORBED BY THE
‘SYSTEM’ FROM THE
SURROUNDINGS)
2HgO(s)
REACTION PROGRESS (TIME)
NOTE: Hproducts IS
HIGHER THAN
Hreactants
What is the enthalpy change for
the following reaction?
Mg(s) + 2H2O(l) → Mg2+(aq) + 2OH-(aq) + H2(g)
The ΔH for Mg(s) and H2(g) is 0 kJ/mol
a. Calculate the ΔH for the above reaction.
b. Is the reaction exothermic or endothermic?
c. Create an enthalpy diagram.
Learning Target:
Be able to use thermochemical equations to determine
enthalpy change (ΔH) associated with a given amount of
reactant.
Learning Outcomes:
1. Know how to calculate ΔH, and produce an enthalpy
diagram.
2. Know how to calculate enthalpy for a given amount
of reactant.
Learning Target:
Be able to calculate the amount of energy absorbed or
released during a chemical or physical change.
Learning Outcomes:
Complete the Heat Calculation Problems.
Warm Up
 Calculate the enthalpy change (∆H) for the following
reaction:
b. C2H5OH(l) + 3O2 (g) → 2CO2(g) + 3H2O(g)
How much heat will be released when 8.76g of oxygen gas
reacts with excess sulfur by the reaction:
2S(s) + 3O2 (g) →2SO3 (g)
Solution:
.
ΔH° = -791.4kJ
Calculate the enthalpy (∆H°) for the following
reaction:
2NH3(g) + 4H2O(l) → 2NO2(g) + 7H2(g)
How much energy would be released if 50.0g of
water were reacted?
How much energy is released if 10.0g of
ethanol, C2H5OH , were burned?
C2H5OH(l) + 3O2 (g) → 2CO2(g) + 3H2O(g)
∆H° = -1234.8 kJ
Warm Up
 Using the standard enthalpies of formation table,
calculate standard enthalpy change (ΔH°) for the
following reactions (Balance first):
CH3OH(l) + O2(g) → CO2(g) + H2O(g)
Using the above balanced equation and the ΔH° you
calculated, how much heat will be released when 24.0
g of methanol (CH3OH) reacts with excess oxygen?
Learning Target:
Be able to determine the amount of standard enthalpy
change (ΔH°) for a series of reactions by applying Hess’
Law.
Learning Outcomes:
Understand how to calculate ΔH° for a series of
reactions.
• To be exact when comparing enthalpy changes chemists
must define the conditions, such as pressure,
temperature and phase.
• The “Standard State” of a substance is its pure form at
the standard pressure of 1 atmosphere and 25° C.
o
o
o
Standard state of oxygen is as the diatomic gas.
Standard state of carbon is graphite.
Standard state of phosphorus is P4 (white phosphorus).
• Enthalpy changes measured under “standard state”
conditions are called “Standard Enthalpy Changes”
(ΔH°).
• Hess’s Law: If a series of chemical
reactions are added together, the enthalpy
change for the net reaction will be the sum of
the enthalpy changes for the individual steps.
Hess’ Law Example
 Example
N2 (g) + O2 (g)
2NO(g)
ΔH1 = 181 kJ
2NO (g) + O2 (g)
2NO2(g)
ΔH2 = -113 kJ
Net equation:
N2 (g) + 2 O2 (g) + 2NO (g)
2NO(g) + 2NO2(g)
To write the net equation, the sum of the reactants is
placed on the left hand side, and the sum of the
products is placed on the right side. Substances on
both sides cancel out – like algebra. In example above
the next reaction is:
N2 (g) + 2 O2 (g)
2NO2(g)
ΔH = ΔH1 + ΔH2 = 181kJ + -113 kJ = 68kJ (endo)
Applying Hess’s Law
 If the coefficients of an equation are multiplied by a
factor, the enthalpy change the reaction is multiplied
by the same factor.
 If the equation is reversed, the sign of the ΔH changes
also.
 For example
C (s) + H2O(g)
Then
CO(g) + H2(s)
CO(g) + H2(s)
ΔH = 113kJ
C (s) + H2O(g) ΔH = -113kJ
Hess’s Law Example
• Calculate the ∆H for the reaction:
2C(s) + O2(g) →
2CO(g) from
C(s) + O2(g) → CO2(g)
ΔH = -393.5 kJ
2CO(g) + O2(g) →
ΔH = -566.0 kJ
2CO2(g)
• The trick is to simply combine equations so that only
the relevant parts are present (everything else cancels
out). Remember, whatever you do to the equations,
you must do to the value of ΔH for the reaction.
• Calculate the ∆H for the reaction:
2S(s) + 2OF2(g) →
SO2(g) + SF4(g)
OF2(g) + H2O(l) → O2(g) + 2HF(g)
∆H = -277 kJ
SF4(g) + 2H2O(l) → SO2(g) + 4HF(g) ∆H = -828kJ
S(s) + O2(g) → SO2(g)
∆H = -297kJ
Exit Question
From the following enthalpy changes,
H2S(g) + 3/2O2(g) → H2O(l) + SO2(g)
CS2(l) + 3O2(g) → CO2(g) + 2 SO2(g)
∆H° = -563kJ
∆H° = -1075kJ
Calculate the value of ∆H° for
CS2(l) + 2H2O(l) → CO2(g) + 2H2S(g)
12-4 Calorimetry
• What happens during chemical reactions?
o During an exothermic reaction, the surroundings gain heat from
the ‘system.’
o During an endothermic reaction, the surroundings lose heat to
the ‘system.’
• Calorimetry is the study of heat flow and heat
measurement during chemical reactions.
• Calorimetry experiments use accurate measurements
of temperature changes to determine the enthalpy
changes (heat flow).
Calorimeter Experiment
Learning Target:
Be able to use the equation “q = m . C . ΔT” to solve for
heat absorbed or released, mass, specific heat or
temperature change of a substance during a chemical or
physical change.
Learning Outcomes:
Understand how to calculate for the variables in the
heat equation.
Finish calculations for Hess’s Law Lab
HEAT vs. TEMPERATURE
• Heat Capacity: the amount of heat needed to raise the
temperature of an object by 1 degree Celsius (1° C).
• Specific Heat: the heat capacity of one gram of a
substance.
o Specific Heat of water = 4.184 J/g.C°
o To raise the temperature of one gram of liquid water by 1
degree C requires 4.184 J of heat energy, or 1 calorie
(cal.)
• Note that heat and temperature are related, but
different, concepts.
• We detect heat changes by measuring temperature
changes.
o A small temperature change does not mean a small
amount of heat transfers! (Consider melting a lake!)
o Use specific heat and mass to measure heat changes.
o Heat equation:
q = m X C X (Tf – Ti)
Hess’ Law Lab
 On a separate sheet of paper answer the following:
1. Write 2 specific experimental errors that occurred in
your lab? (Don’t just give me I made a mistake!)
2. How did they impact your results?
A 15.75-g piece of iron absorbs 1086.75 joules of heat energy,
and its temperature changes from 25°C to 175°C. Calculate the
specific heat capacity of iron.
How many joules of heat are needed to raise the temperature of
10.0 g of aluminum from 22°C to 55°C, if the specific heat of
aluminum is 0.90 J/g°C?
Calorimeter Experiment
• Calorimeter: a well insulated container filled with
a known mass of water, a way to conduct a
chemical reaction, a stirrer and a thermometer.
•
•
•
•
•
Measure the initial temperature of the water (Ti).
Conduct the chemical reaction.
Measure the temperature periodically.
Determine the final temperature (Tf).
Note that all heat released by a reaction is gained
by the water in the calorimeter, and vice versa!
• Mathematically we may say:
(qrxn = -qsur)
• Calculate the quantity of heat absorbed by the water (qsur) from the
mass of water, the Specific Heat of water and the temperature
change (Tf – Ti). (And vice versa.)
qsur = m X C X (Tf – Ti)
Calorimeter Experiment
• Our body is like a chemical factory, taking in raw materials
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of various kinds to make new products that keep us alive
and help us grow.
These reactions involve bond-breaking and bond-making,
and they meet the energy needs of our body.
Carbohydrates and fats are major food sources, and they
release lots of energy (exothermic reactions).
Glucose releases -2803kJ/mol C6H12O6
See Table on page 399.
“Calories Count” Project
Recall 1000 cal = 1 Kcal = 1 Cal (the food kind!)
• “Caloric Theory”
o Heat was thought of as an invisible, weightless fluid capable
of flowing from a hot object to a colder one.
• Benjamin Thompson (Count Rumford) showed that
friction heat between objects was continuous as long
as two objects were rubbed together.
o
This was not consistent with ‘Caloric Theory.’
• James Joule used a paddlewheel experiment to
convert mechanical energy into heat, thus showing
there is a ‘mechanical equivalent of heat.’
• Modern “Kinetic Theory”
o
Heat is defined as the transfer of kinetic energy from a hotter
object to a colder one.