Download Mr Alasdair Ross at Southpointe Academy: Math and Chemistry Pages

Southpointe Academy
AP Chemistry
Topic 2 – Thermochemistry
Thermochemistry is the study of the energy changes that occur during physical
processes and chemical reactions.
System – the part of the universe we are studying.
Surroundings – the rest of the universe.
Open System – Can exchange matter and energy with the surroundings.
Example:
Closed System – Can exchange energy but not matter with the surroundings.
Example:
Isolated System – Exchanges neither matter nor energy with its surroundings.
Example:
1
Internal Energy (U) – The internal energy of a system is the total energy contained within
a system, partly as kinetic energy and partly as potential energy (See page 218, figure
6.4).
Heat (q) – A quantity of energy transferred between a system and its surroundings as a
result of a temperature difference between them.
Work (w) – In this section, we will consider the work done when gases expand or are
compressed.
W  F d
(work = force x distance)
and:
P
F
A
(pressure = force/area)
so:
W  P  A d
now:
Area x distance = Volume
So:
w  P  V
(work = pressure X change in volume)
In fact
w  P  V
Because when a gas expands, V is positive and work is negative.
2
U  U final  U initial
In a physical or chemical change, energy can be exchanged between a system and its
surroundings, but no energy can be created or destroyed.
The First Law of Thermodynamics
U  q  w
The change in internal energy of a system U , must be related to the energy exchanges
that occur as heat (q) and work (w).
Note:
Energy entering system carries a positive sign. If heat is absorbed by the system, q>0. If
work in done on the system, w>0.
Energy leaving a system carries a negative sign. If heat is given off by the system, q<0.
If work in done by the system, w<0.
Example
A gas does 125J of work expanding, while at the same time it absorbs 156J of heat. What
is the change in Internal Energy?
(remember when a gas expands, work is negative)
3
Enthalpy Change - ∆H
A chemical reaction is a process where the reactants (initial state) change into the
products (final state).
Exothermic Reaction
Chemical energy in a system is converted into thermal energy.
Isolated System – temperature increase in the system.
Open or Closed System – heat given off to surroundings (q<0).
Endothermic Reaction
Thermal energy in a system is converted to chemical energy.
Isolated System – temperature decrease in the system.
Open or closed System – heat is absorbed from the surroundings (q>0).
4
Representing ∆H for a Chemical Reaction
Consider:
CH 4 ( g )  2O2 ( g )  CO2 ( g )  2H 2O(l )
H  890kJ
This means that the Enthalpy change when 1 mol of Methane is combusted is -890kJ.
We can represent this reaction on an Energy Diagram
The H value of a reaction is dependent on the direction of the reaction and the number
of mole of reactant/product present. The “standard H ” of a reaction is the H for the
balanced reaction equation.
Example:
Given:
H 2 ( g )  F2 ( g )  2HF ( g )
H  72kJ
Calculate H for the following reaction:
HF ( g ) 
1
1
H 2 ( g )  F2 ( g )
2
2
5
Example:
The complete combustion of liquid butane given off 21.5kJ of heat per gram of butane.
a) Write a balanced reaction equation.
b) Calculate a ∆H value for the reaction equation.
∆H in Stoichiometric Calculations
Consider:
H 2 ( g )  Cl2 ( g )  2HCl ( g )
H  184.6kJ
a) What is the Enthalpy change associated with the formation of 6.34mol of HCl(g)?
6
b) What is the Enthalpy change when 15.6g of Hydrogen gas reacts with excess
chlorine gas?
Calorimetry
The process of measuring quantities of heat is called calorimetry. The device in
which measurements are made is called a calorimeter. Calorimetry is based on the
law of conservation of energy. Whatever heat is lost by a system is gained by its
surroundings and vice-versa.
Heat Capacity
The heat capacity (C) of a system is the quantity of heat required to change the
temperature of the system by 1°C.
C
q
T
The units for Heat Capacity are J /  C
Molar Heat Capacity and Specific Heat
Molar Heat Capacity – The heat capacity of one mole of a substance
Specific Heat – The heat capacity of a one gram sample
Specific Heat =
C
m
7
Because: C 
q
,
T
Specific Heat =
q
m  T
In a typical calorimetric calculation, we are trying to find the amount of heat released
or absorbed. Rearranging the previous equation gives:
q  SpecificHe at  mass  T
Refer to table 6.1 on page 231 for the specific heats of some common substances.
Examples
(1)
Calculate the heat capacity of a iron block that must absorb 734J of heat from
its surroundings in order for its temperature to rise from 22°C to 153°C.
(2)
How much heat in joules and kilojoules does it take to raise the temperature of
345g of water from 35.0°C to 95.0°C?
8
(3)
What mass of water, in kilograms, can be heated from 6°C to75°C by
7.56  1010 J of heat?
Measuring Specific Heats
We use a calorimeter to measure specific heats. We can use foam cups to make a simple
calorimeter (recall Chemistry 11). Foam cups work well because:
1) They have a low heat capacity (The foam only gains a small amount of heat)
2) They are good insulators (Very little heat escapes from the calorimeter)
From the mass of the solid and the temperature change of the water, we can determine
the amount of heat transferred.
Example:
A 16.5g sample of a metal is heated to 90.0°C and then dropped into 25.0g of water
in a calorimeter. The temperature of the water rises from 22.0°C to 26.2°C.
Calculate the specific heat of the metal.
9
Problem
Consider the following neutralization reaction:
HCl (aq)  NaOH (aq)  NaCl(aq)  H 2O(l )
A 50.0mL sample of 0.500M HCl at 21.0ºC is added to 50.0mL of 0.500M NaOH,
also at 21.0ºC. After mixing, the temperature of the solution rises to 24.5ºC.
a) Calculate the heat released.
b) Calculate the ∆H of this reaction (the H for the reaction of 1 mole of
Hydrochloric Acid with 1 mole of Sodium Hydroxide).
a)
b)
(Note: units for H can be kJ or kJ/mol though there is some argument about this)
10
Hess’ Law
The heat of a reaction is constant, whether the reaction is carried out directly in one step
or indirectly in a number of steps.
Example:
a)
1
C ( graphite)  O2 ( g )  CO( g )
2
H  110.5kJ
1
CO( g )  O2 ( g )  CO2 ( g ) H  283.0kJ
2
___________________________________________
c)
C( graphite)  O2 ( g )  CO2 ( g )
H  393.5kJ
b)
H a  H b  H c
11
Problem
Calculate the enthalpy change for reaction (a) given the data in equations (b), (c) and (d).
(a)
2C( graphite)  2H 2 ( g )  C2 H 4 ( g )
H  ?
(b)
C( graphite)  O2 ( g )  CO2 ( g )
H  393.5kJ
(c)
C2 H 4 ( g )  3O2 ( g )  2CO2 ( g )  2H 2O(l )
H  1410.9kJ
(d)
1
H 2 ( g )  O2 ( g )  H 2 O(l )
2
H  285.8kJ
12
Standard Enthalpies of Formation
We know:
We cannot, however, obtain absolute values of enthalpies. We use a scale of relative
enthalpies called enthalpies of formation to overcome this problem.
The standard state of a solid or liquid is defined as the pure element or compound at
1atm pressure and the temperature of interest (usually 25ºC).
The standard state of a gas is the pure gas behaving as an ideal gas at 1 atm pressure
and the temperature of interest (usually 25ºC).
The standard enthalpy of reaction ( H  ) is the enthalpy change for a reaction in which
the reactants in their standard states yield products in their standard states. (This is not
the same as DH ).
The standard enthalpy of formation ( H  f ) of a substance is the enthalpy change that
occurs as 1 mol of substance forms from its elements when both the products and the
reactants are in their standard states.
The standard enthalpy of formation ( H  f ) is often called the heat of formation.
The elements must be in their reference form (the most stable form of the element at
1atm and the given temperature).
The standard enthalpy of formation ( H  f ) of a pure element in its reference form is
0. This is because the formation of the reference form of an element from itself is not an
actual change.
13
Example:
Two forms of Carbon are readily attainable at 25ºC and 1 atm, Graphite and Diamond.
The two forms are different states of carbon and must have different enthalpies of
formation.
CO2 ( g )  C( graphite)  O2 ( g )
H   393.5kJ
CO2 ( g )  C(diamond )  O2 ( g )
H   395.4kJ
What is the H  for the formation of diamond from graphite? (Graphite is chosen as the
reference form of Carbon).
14
Calculations Based on Standard Enthalpies of Formation
Suppose we want to find the standard enthalpy change for the conversion of NO2 ( g ) to
N 2 O4 ( g ) at 25ºC and 1 atm.
(a) 2NO2 ( g )  N 2O4 ( g )
H   ?
The standard enthalpies of formation of these two gases can be found in Table 6.2 Page
240.
(b)
1
N 2 ( g )  O2 ( g )  NO2 ( g )
2
(c) N 2 ( g )  2O2 ( g )  N 2O4 ( g )
H   H  f NO2 ( g )
H   H  f N 2 O4 ( g )
Using Hess’ Law, we can combine these two equations and arrive at the one we want:
-2(b)+(c)=(a)
So H  =
15
Generally:




H    Coeff  H  f p   Coeff  H  f r
Example:
Find H  for the following reaction:
3CH 4 ( g )  2H 2O(l )  CO2 ( g )  4CO( g )  8H 2 ( g )
16
Example:
The combustion on isopropyl alcohol is represented by the following equation:
2(CH 3 ) 2 CHOH (l )  9O2 ( g )  6CO2 ( g )  8H 2 O(l )
Find the standard enthalpy of formation of isopropyl alcohol.
17
H   4011kJ