CDO Chemistry 2015
st
1
Law of Thermodynamics
 1st Law – energy cannot be
created or destroyed it can just
change forms
 Energy can be changed from
Potential to Kinetic or vice versa
System Vs. Surroundings
Two Parts to the Universe
System – is the part of the
universe being studied
Surroundings – the rest of the
universe that interacts with
the system
Directionality of Heat
 Heat – (q)
 Flow from the
surrounding to the
system is positive;
q>0
 Flow from the
system into the
surroundings is
negative; q < 0
Endothermic Processes
 Endothermic
process (endo =
in) is one that
absorbs heat
from the
surrounding (it
feels cold).
Exothermic Processes
 Exothermic
process (exo =
out) is one that
transfers heat to
the surrounding
(it feels hot)
 All combustion
and
neutralization
reactions
Heat and Temperature
 Temperature – is the measure of the
average kinetic energy of the
particles
 Heat energy does not necessarily
change the temperature of 2 objects
in the same way.
 Increase in temperature depends on
 Mass of the object
 The heat added
 Nature of the substance
Heat Capacity and Specific Heat
 The amount of energy required to raise the
temperature of a quantity of a substance
by 1 K (1C) is its heat capacity.
 We define specific heat capacity (or simply
specific heat - C) as the amount of energy
required to raise the temperature of 1 g of
a substance by 1 K.
 Specific Heat of water = 4.184 J/g oC
Finding Heat (q)
q = m C DT
q = quantity of heat in Joules
m = mass in grams
C = specific heat capacity = J/(g oC)
DT = change in Temp
Example 1
 How much heat is released when 10.0 g of Cu
with a specific heat capacity of 0.385 J/g oC is
cooled from 85.0 oC to 25.0 oC?
Example 2
If 500. J of heat is added to 100.0 g samples of
water (specific heat = 4.184 J/goC) and Silver
(specific heat = 0.237 J/goC) which substance will
have the biggest change in Temperature?
Calorimetry
 We measure the transfer of heat (at a constant
pressure) by a technique called calorimetry.
 In calorimetry ...
 the heat released by the system is equal to
the heat absorbed by its surroundings.
 the heat absorbed by the system is equal to
the heat released by its surroundings.
 The total heat of the system and the
surroundings remains constant.
Calorimetry
 We use an insulated device called a
calorimeter to measure this heat transfer.
 A typical device is a “coffee cup
calorimeter.”
 The calorimeter is often filled with
water and the change in
temperature of the water is
measured
Calorimetry
-q substance = +q water
So…….
-mCΔT = +mCΔT
Example
 If a 25.5 g sample of a metal at 95.0oC is
placed into a 100.0 g of water at 25.0oC and
the temperature of the water is raised to
32.6oC, what is the specific heat of the
metal?
Example
 An insulated cup contains 75.0g of water at
24.00oC. A 26.00g sample of metal at
82.25oC is added. The final temperature of
the water and metal is 28.34oC. What is the
specific heat of the metal?
Enthalpy

When a reaction is carried out a constant
pressure, the heat (q) that is transferred
in the reaction is given a special name
“Enthalpy”, ΔH

Enthalpy is measured in kilojoules per mole
(KJ/mol)
∆H
 Endothermic Reactions - Heat is transferred from
the surrounding to the system, so the change in
enthalpy is positive (ΔH > 0)
 Exothermic Reactions – Heat transferred from the
system to the surrounding, so the change in
enthalpy is negative (ΔH < 0)
Methods of Finding DH
 Measure it using a coffee cup calorimeter
 Calculate it using Hess’s Law
 Calculate it using average bond energies
Coffee Cup Calorimetry
 The ΔH for the
reaction is equal in
magnitude but
opposite in sign to the
q for the calorimeter
DH rxn
-q

1000n
 q is heat lost or gained
 n is moles of
substance
Important Assumptions
 Assume the volume of the water is equal to its mass in




grams.
If water is present use its mass in the q =mcΔT
Always use the specific heat of water 4.184 J/goC as the
c value
If two solutions are being reacted add volumes
together and assume it is equal to mass in grams.
Use the substance identified in the problem to
determine n.
Example
 Calculate the enthalpy change of the combustion of
ethanol (C2H5OH) from the following data. Assume
all of the heat from the reaction is absorbed by
the water..
Mass of water in
calorimeter
200.00 g
Temperature increase in
water
13.00 oC
Mass of ethanol burned
0.45 g
Practice
 1.56 grams of methanol (CH3OH) is burned in a
calorimeter. This causes the 150.Og of water to
increase its temperature from 20.0oC to 38.6oC.
What is enthalpy of combustion for methanol?
Practice
 100.0 mL of water was measured out and poured into a
polystyrene cup with an initial temp of 18.3 oC. 5.20 g
of NH4Cl was added to the cup and it dissolved. The
minimum temp. was recorded to be 15.1 oC. Calculate
the enthalpy of solution for NH4Cl per mole.
Practice
 200.0 mL of water was measured out and poured into a
polystyrene cup with an initial temp of 20.0 oC. 8.25 g
of CaCl2 was added to the cup and it dissolved. The
maximum temp. was recorded to be 35.0 oC. Calculate
the enthalpy of solution for CaCl2.
Example
 When 1.00 L of 1.00M Ba(NO3)2 solution at 25.0 °C
is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25
°C in a calorimeter, the white solid BaSO4 forms and
the temperature of the mixture increases to 28.1 °C .
What is the enthalpy of reaction?
Practice
 When 1.00 L of 1.50 M HCl solution at 30.0 °C is
mixed with 1.00 L of 1.50 M NaOH solution at 30.0
°C in a calorimeter, and the temperature of the
mixture increases to 41.5 °C . What is the enthalpy of
neutralization?
Hess’s Law
 If a chemical equation is the sum of multiples
of other equations, the DH of this equation
equals a similar sum of multiples of DH's for
the other equations.
Enthalpy Relationships
Given:
H2(g) + 1/2 O2(g)  H2O(g)
∆H˚ = -242 kJ
If multiplied:
2 H2(g) + O2(g)  2 H2O(g)
∆H˚ = -484 kJ
If Reversed the sign on the enthalpy value
switches:
H2O(g)  H2(g) + 1/2 O2(g)
∆H˚ = +242 kJ
If a different Phase the enthalpy value changes:
H2(g) + 1/2 O2(g)  H2O(liquid)
∆H˚ = -286 kJ
Example 8
CO2(g)  CO(g) + 1/2O2(g)
DH = +283.0 KJ
C(s) + O2(g)  CO2(g)
DH = -393.5 KJ
_________________________________________
C(s) + 1/2O2(g)  CO(g)
DH = ?
Example 9
SO2(g)  S(s) + O2 (g)
DH = +296.8 KJ
2SO3(g)  2S(s) + 3O2 (g) DH = +791.4 KJ
________________________________
2SO2 (g) + O2 (g)  2SO3(g) DH = ?
Average Bond Enthalpies
 This table lists the
average bond
enthalpies for many
different types of
bonds.
 Average bond
enthalpies are
positive, because
bond breaking is an
endothermic process.
© 2009, Prentice-Hall,
Inc.
Enthalpies of Reaction
DHrxn = (bond enthalpies of bonds broken) -
(bond enthalpies of bonds formed)
Prior to using this formula the Lewis Structure for
reactant and product must be drawn
Example
Estimate the enthalpy change for the
chlorination of ethylene:
CH2CH2(g) + Cl2(g) CH2ClCH2Cl
Example
 2H3COH + 3O2  2CO2 + 4H2O