Transcript Unit 10

Unit 12
Thermochemistry
1. Energy is…
 The capacity for doing work or
supplying heat.
a) Energy has neither
volume
_______
mass nor _________
b) List the types of energy---
Thermal/heat
potential
chemical kinetic
electrical Solar/light/electromagnetic
2. Thermochemistry is
Study of energy changes during chemical
reaction and change in state.
FYI
Temperature is
The average kinetic energy.
Units: K (kelvin), oC (celsius),
3. Heat is …
Energy that transfers from one object to
another.
Heat flows: from a
Warmer object
 a colder object
Conversion factors FYI
1 Calorie = 1000 calorie
4.184 J = 1 cal
Practice
1.262 cal

301.1 cal
4. The law of conservation of energy
In any chemical or physical process, energy is
neither created or destroyed.
5. Endothermic process:
In an endothermic process, the system
absorbs heat as the surroundings cool down.
6. Exothermic process:
In an exothermic process, the system releases
heat as the surrounding warm up.
q surr
=mC∆T
q sys
7. Enthalpy
a) Enthalpy (H) is
The heat content of a system
b) The change in enthalpy (∆H)=
The heat released or absorbed by a
reaction
c) ∆H = heat
FYI: Energy Diagram
Exothermic Reaction
Endothermic Reaction
8. Endo or Exo during state change
a. Solidliquidgas (absorb heat) endo
b. Gasliquidsolid (release heat) exo
Gas
Endothermic reaction
Melting and
vaporization
Liquid
Exothermic reaction
Freezing and
Solid
Condensation
16. Heating curve of water
9. Thermochemical equation
a. Heat of reaction: the enthalpy change
for the chemical equation.
b. Heats of reaction reported as ∆H
c. ∆H is negative 
Exothermic reaction, heat as a product
d. ∆H is positive 
Endothermic reaction, heat as a reactant
9. Thermochemical equation
e. Ex: CaO (s) + H2O (l)  Ca(OH)2 (s)
∆H= -65.2kJ (endo or exo?) Exo, heat as
a product
f. Example of heat of combustion:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
∆H= -890kJ (endo or exo?)
exo
g. Thermochemical equation is same as
stoichiometry
Practice
Consider the following chemical equation
2NaHCO3 (s) + 129kJ  Na2CO3 (s) + H2O
(g) + CO2 (g)
i.
Is this an exothermic or endothermic
reaction? Explain.
Endo, heat as a reactant.
Practice
mole to heat conv., use stoich

65 kJ
65 kJ of heat
are absorbed
Practice
mole to heat conv., use stoich
Consider the following chemical equation
2NaHCO3 (s) + 129kJ  Na2CO3 (s) + H2O (g) + CO2 (g)
iii. How many kJ of heat are absorbed when
25.0 gram of sodium hydrogen carbonate is
decomposed?
19.2 kJ
19.2 kJ of heat
are absorbed
9. Specific heat capacity
a) The heat capacity is
the amount of heat needed to increase the
temperature of an object exactly 1oC
b) The heat capacity depends on both its mass
and its chemical composition.
c) The specific heat capacity is
the amount of heat it takes to raise the
temperature of 1g of the substance 1oC.
9. Specific heat capacity
d. Low specific heat
capacity
 easy to raise
temperature
e. High specific heat
capacity
 need more heat
to raise temperature
Substance
Specific heat
capacity (J/g oC)
H2O (l)
4.18
H2O (s)
2.03
Al (s)
0.89
Fe (s)
0.45
Hg (l)
0.14
C (s)
0.71
10.Calorimetry
a. Calorimetry is
the measurement of the heat flow into
or out of a system for chemical and
physical process.
b. Heat released by the system
= Heat absorbed by its surrounding
c. Heat absorbed by the system
= Heat released by its surrounding
q surr
=mC∆T
q sys
10.Calorimetry
d.
Calorimeter is
the insulated devise used to measure
the absorption or release of heat in
chemical or physical process.
i.
qsurr = mC∆T (We can measure this)
ii. qsys = - qsurr = - mC∆T (We can’t
measure this, but put opposite sign)
Important formula
q = mCΔT
q = heat absorbed/heat released,
enthalpy change(J)
m = mass (g)
C = The specific heat capacity (specific
heat) (J/g・oC)
ΔT = final temperature - initial
temperature (oC)
IF you have specific heat, use this formula
Practice
e. How many joules of heat are needed to
raise the temperature of 238 grams of water
from 20.0oC to 50.0oC? (Cwater= 4.18 J/g oC)
29800J
Practice
Convert from cal to joule
f. A 20.0 gram sample of iron at 25.0oC was
given 50.0 calories of heat. What is its final
temperature? Ciron=0.45 J/g oC)
Tf -25.0oC= 23.2
Tf = 48.2oC
12. Application
A student performed a
calorimetry lab to
determine the specific
heat of an unknown
metal. Consider the
following data and
answer the questions
from a) to e). Assume
Cwater = 4.18 J/goC
Mass of unknown
metal
12.50g
Mass of water
75.00g
Initial temperature of
water in calorimeter
15.8oC
Final temperature of
water + unknown
metal in calorimeter
25.1 oC
Initial temperature of
unknown metal in
calorimeter
88.3OC
a) What is the temperature difference (∆T) of
water?
25.1 oC– 15.8 oC = 9.3 oC
b) Calculate the heat absorbed by the water.
Q = (75.00g)(4.18J/goC)(9.3)=2920J
c) What is the heat released by unknown metal?
Q = -2920J
Change the sign!!
d) What is the temperature difference (∆T) of
unknown metal?
25.1 oC- 88.3 oC = -63.2 oC
e) Calculate the specific heat capacity of this
metal.
(12.50g)(C)(-63.2 oC)=-2920J
C = 3.70J/goC
11. Molar heats of solution: (∆Hsoln)
a.
b.
c.
d.
e.
f.
The heat is either released or absorbed during
formation of a solution.
Example of thermochemical equation:
+ (aq) + OH- (aq)
2O ( l )
NaOH (s) H
Na

∆Hsoln = - 445.1 kJ/mol (- 445.1 kJ per 1 mol)
Is this endo or exo?
Is the temperature of the solution increasing or
decreasing?
Because system release heat, so surrounding
absorbs heat.
Practice
A student dissolved 2.500 mol NaOH (s)
in water
a. Write the thermochemical equation
(l )
NaOH (s) HO
Na+ (aq) + OH- (aq) + 445.1kJ
2
(∆Hsoln = -445.1 kJ/mol)
b. Is it an endo or exo?
exo
Practice
A student dissolved 2.500 mol NaOH (s) in water
(l )
 Na+ (aq) + OH- (aq) + 445.1kJ
NaOH (s) HO
2
(∆Hsoln = -445.1 kJ/mol)
c. How much heat (in kJ) is released when
2.500 mol NaOH (s) is dissolved in water?
2.500mol NaOH x -445.1 kJ = -1113kJ
1mol NaOH
Lab report
Data table II
Mass of water (g)
Use density formula: d = m/v
D(water)=1.0 g/mL
Lab report
2. Use qsys = - qsurr
Post lab question/Analysis
3. Calculate the Calories per gram of
marshmallow burned.
Energy released by the marshmallow
divided by the mass of marshmallow
burned
Post lab question/Analysis
4. Find the average Calculate the Calories
per gram of marshmallow burned.
Energy released by the marshmallow
divided by the mass of marshmallow
burned
Marshmallow lab
Introduction
Heat unit in science
joule (J)
kilojoule (kJ)
calorie (cal)
kilocalorie (kcal)
Heat unit in Food
Calorie (Cal)
Marshmallow lab
Procedure
1. Measure and record the mass of
1. Marshmallow
2. Paper clip
3. Watch glass together
Marshmallow lab
Procedure
2. Make your paper clip into a holder
for marshmallow
Marshmallow lab
Procedure
3. Measure 25.0 mL of cold water with
a graduated cylinder. Record the
volume of water used precisely. Pour
it into the aluminum can.
Marshmallow lab
Procedure
4. Place chopstick through the tab so
you can hang it from the ring support
of the ring stand.
Marshmallow lab
Procedure
5. Measure and record the initial
temperature of the water.
Measurements
Initial mass marshmallow,
paperclip, and watch glass (g)
Volume water (mL)
Initial temperature water (°C)
Final temperature water (°C)
Trial 1
Trial 2
Marshmallow lab
6. Light the marshmallow.
Keep the thermometer above
the bottom of the can.
Marshmallow lab
7. Allow the marshmallow to
stop burning on its own, It
does not have to completely
burn to black. Keep an eye
on the temperature to figure
out the highest temperature
reached. Record the temp.
Marshmallow lab
8. Measure and record the mass of the
burned marshmallow, paper clip, and watch
glass together.
 To find the mass of burned marshmallow,
subtract final mass from initial mass.
Marshmallow lab
Perform trial 2
Burned marshmallow should be thrown
away
 use the same paper clip but washed
 use a new can
 Measure all over again
 when finished trial 2, return 2 cans,
paper clip, chop stick
Marshmallow lab
Assumption
Heat released by marshmallow
is same as heat absorbed by
water.
q sys = q surr
Pre lab quiz
q sys
q surr