Title: Chemical Thermodynamics
1Chemical Thermodynamics
2First 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.
4Endo 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
5Endo 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
6Humor
- 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.
8Calorimetry
- 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)
9Calorimeter
- 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 -
12Plug 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
13Calorimetry 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
14A 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? -
16Calorimeter
- - (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
17240.0 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
18A 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