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Chemical Thermodynamics

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Title: Chemical Thermodynamics


1
Chemical Thermodynamics
2
First Law of Thermodynamics
  • The total energy of the universe is a constant.
  • Energy cannot be created and destroyed, however,
    it can be converted from one form to another.

3
  • Thermodynamics is concerned with the flow of heat
    from the system to its surroundings, and
    vice-versa.
  • In studying thermodynamics, try to identify these
    two parts
  • the system - the part on which you focus your
    attention, where the change is occuring.
  • the surroundings - includes everything else in
    the universe.
  • Together, the system and its surroundings
    represent everything in the universe.

4
Endo or Exo?
  • Heat flowing into a system from its
    surroundings
  • Object is gaining energy
  • q has a positive value
  • called endothermic
  • system gains heat (gets warmer) as the
    surroundings cool down

5
Endo or Exo?
  • Heat flowing out of a system into its
    surroundings
  • Object is losing energy
  • q has a negative value
  • called exothermic
  • system loses heat (gets cooler) as the
    surroundings heat up

6
Humor
  • A small piece of ice lived in a test tube, and
    fell in love with a Bunsen burner. One day it
    decided to profess its love
  • Bunsen! Oh Bunsen my flame! I melt whenever I
    see you said the ice.
  • The Bunsen burner replied , Dont worry, its
    just a phase youre going through.

7
  • Heat transfers between the system and the
    surrounding from hot objects to cooler objects.
  • Eventually they will reach a point of
    equilibrium where there is a balance between the
    two parts.

8
Calorimetry
  • Calorimetry - the measurement of the transfer of
    heat into or out of a system for chemical and
    physical processes.
  • Based on the fact that heat released by an system
    is equal to the heat absorbed by the
    surrounding!
  • Two different parts are involved, but the
    energy(Q) will be the same for both parts.(1st
    Law of Thermodynamics)

9
Calorimeter
  • The device used to measure the absorption or
    release of heat in chemical or physical processes
    is called a Calorimeter
  • The material inside the cup will be the System
    and the water will be the Surroundings

10
  • A 40.0 gram block is heated to 100.0 OC. It is
    then inserted into an insulated calorimeter
    containing 25.0 grams of water at 26.6 OC.
  • After stirring, the water equalizes at a
    temperature of 30.0 OC. What is the specific heat
    of the metal?
  • What are we solving for?
  • Who is losing energy?
  • Who is gaining energy?
  • What do we know about those values????

Surroundings
SYSTEM
H2O
??
m 25.0 g Ti 26.6oC
m 40.0 g Ti 100.0oC
11
-QMetal(s) QH2O(L)
  • -QMetal (M)(CMetal(s))(?T)
  • -QMetal (40.0 g) (CMetal(s))(30.0 OC -100 OC)
  • I have two unknowns! Lets look at the water.
  • QH2O (M)(CH20(L))(?T)
  • QH2O (25.0 g)(4.186 J/g OC)(30.0OC 26.6OC)
  • QH2O 355.81 J
  • Energy Gained is equal to the Energy Lost
    soQH2O 355.81 J -QMetal(s) - 355.81 J

12
Plug that back into the first equation
  • -355.81 J (40.0 g) (CMetal(s))(30.0OC -100OC)
  • Why is the joules a negative value???
  • CMetal(s) 0.127075
  • Sig Figs 0.127 J/g OC

13
Calorimetry Problem 2
  • Problem A 30.0 gram sample of Silver is heated
    to 100.0 OC. It is placed in 75.0 grams of water
    at 25.0 OC. Determine what the final temperature
    will be.
  • Who is losing energy?
  • Who is gaining energy?
  • What do we know aboutthe amount of energy
    forboth of them?

Surroundings
SYSTEM
H2O
Ag
m 75 g T 25oC
m 30 g T 100oC
14
A point of clarification
  • Youre not solving for ?T, instead we need the
    final temperature, Tf.
  • The warmer silver is going down from 100.0 to X.
    So its ?T (X 100)
  • The cooler water is going up starting at 25 and
    going X. So its ?T (x 25.0)
  • 25.0 OC X 100.0 OC
  • To ensure we get an answer between these values,
    you must make sure the Qs have the correct sign.
    (Silver loses, Water gains)

15
-QAg(s) QH2O(L)
  • -QAg (M)(CAg(s))(?T)
  • -QAg (30.0 g)(0.236 J/g OC)(X100.0OC)
  • I have two unknowns!
  • QH2O (M)(CH20(L))(?T)
  • QH2O (75.0 g)(4.18 J/g OC)(X 25.0OC)
  • Still have two unknowns, but what do we know
    about the two Qs?

16
Calorimeter
  • - (30.0 g) (0.236 J/g OC) (x -100OC)
  • (75.0 g) (4.186 J/g OC) (x 25.0OC)
  • - 7.08X 708 313.5x 7848.75 (Combine Terms)
  • 8556 320.58x
  • X 26.7 OC Final Temperature

17
240.0 g of water (initially at 20.0oC) are mixed
with an unknown mass of iron (initially at
500.0oC). When thermal equilibrium is reached,
the system has a temperature of 42.0oC. Find the
mass of the iron. C Fe(s) 0.4495 J/g OC C
H2O(l) 4.18 J/g OC
mass ? grams
-
LOSE heat GAIN heat
- (Cp,Fe) (mass) (DT) (Cp,H2O) (mass) (DT)
- (0.4495 J/goC) (X g) (42oC - 500oC)
(4.184 J/goC) (240 g) (42oC - 20oC)
- (0.4495) (X) (-458) (4.184) (240 g) (22)
Drop Units
205.9 X 22091
X 107.3 g Fe
Calorimetry Problems 2 question 5
18
A sample of ice at 12oC is placed into 68 g of
water at 85oC. If the final temperature of the
system is 24oC, what was the mass of the ice?
T -12oC
mass ? g
H2O
ice
Tf 24oC
GAIN heat - LOSE heat
qA (Cp,H2O) (mass) (DT)
mass x qA qB qC - (Cp,H2O)
(mass) (DT)
qA (2.077 J/goC) (mass) (12oC)
24.9 m
mass x qA qB qC - (4.184 J/goC)
(68 g) (-61oC)
qB (Cf,H2O) (mass)
333 m
qB (333 J/g) (mass)
458.2 mass 17339
qC (Cp,H2O) (mass) (DT)
m 37.8 g
qC (4.184 J/goC) (mass) (24oC)
100.3 m
qTotal qA qB qC
458.2 m
Calorimetry Problems 2 question 13
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