Download Ch 5 HEAT IN CHEMICAL REACTIONS Chemical reactions and the

Ch 5 HEAT IN CHEMICAL REACTIONS
Chemical reactions and the associated energy transfers are
extremely important to our lives.
 Our bodies metabolism, photosynthesis, fuels,
batteries
Energy: the ability to do work or produce heat
Kinetic energy: energy of motion
KE = ½ mv2 KE = kinetic energy, Joules
m = mass, kg
v = velocity, m/s
As velocity increases, KE increases
Particles in motion have kinetic energy
Potential energy: stored energy (or energy an object has
due to its position relative to another object)
Includes energy stored in chemical bonds.
Chemical potential energy: energy stored in a substance
electrostatic potential energy: is important in
chemistry.
It is the energy resulting from the
interactions of charged particles.
Thermochemistry: the study of changes in heat in chemical
reactions.
Heat: the flow of energy from a warmer object to a cooler
object
When studying chemical reactions we consider the energy in
the reactants and the products. This is our “system”. The
container and everything else around is the “surroundings”.
a. Closed systems (what we will be considering): in which
the system can exchange energy but not matter with the
surroundings.
 The energy exchanged with the surroundings can be
in the form of heat or work.
b. Open systems: can exchange matter and energy with the
surroundings.
c. Isolated systems: cannot exchange matter or energy
with the surroundings.
First Law of Thermodynamics
Law of Conservation of Energy: energy can be converted
from one form to another or from one object to
another, but it cannot be created or destroyed.
Internal energy, E, of a system includes motion of the particles
(kinetic energy) and the potential energy stored in
bonds.
∆E = Efinal - Einitial
Einitial = energy of reactants
Efinal = energy of the products
If ∆E is positive, the system has gained energy from the
surroundings
If ∆E is negative, the system has lost energy to the
surroundings
A closed system can exchange energy with its surroundings as
either heat or work.
∆Esystem = q + w
q = heat (energy added or lost)
W = work
Signs for ∆E, q, and w and the meaning
q
w
∆E
+ system gains heat
+ work done on system
+ net gain of E by system
- (neg) system loses heat
- work done by system
- net loss of energy by system
Problems:
A. In a sample reaction, A and B are gases that combine
to form C, a solid. A(g) + B(g)  C(s).
The system loses 1150 J heat to the surroundings
The surroundings do 480 J work on system.
What is the change in the internal energy of the
system?
∆Esystem = q + w
= -1150 J + 480 J
= -670J The system has lost energy to the
surroundings
B. Calculate the ∆E for the process in which a system
absorbs 140J from the surroundings and does 85J
work on the surroundings.
∆Esystem = q + w
= 140J -85J
= 55J
Heat units:
1. calorie (cal) – the amount of heat needed to increase
the temp of 1 g of water 1 0C.
2. Calorie – a kilocalorie, 1000 calories, used for
nutrition
3. Joule (J) – SI unit of energy and heat
1kJ = 1000J
1 J = 0.02390 cal
1 cal = 4.184 J
Almost all chemical reactions absorb or release energy in the
form of heat.
Energy must be input to break bonds and energy is
often released when new bonds are formed (due
to more favorable electron configurations)
1. Exothermic reactions:
 release heat to surroundings
 container feels warm
 heat is a product and is written into the chemical
equation as a quantity of Joules (J), or kilojoules (kJ).
C3H8 + 5O2  3CO2 + 4H2O + 2043kJ
In exothermic reactions the energy required to break
bonds in the reactants is less than the energy released
when new bonds form in the products. This extra energy is
released to the surroundings.
2. Endothermic reactions:
 absorb heat from the surroundings
 container will feel cold
 input heat is written as a reactant
C + 2H2O + 113 kJ  CO2 + 2H2
In endothermic reactions the energy needed to break
bonds is greater than the energy released when bonds
are formed in the products.
Summary:
Energy input
to break bonds
vs
Exothermic lesser
Endothemic greater
Energy released
when bonds are formed in products
greater
lesser
State functions: a property of the system that is determined by
the systems conditions or state, for example by its
pressure or temperature.
 Energy of system is a state function. It has a fixed
value depending on the temperature and/or
pressure.
 ∆E depends only on the final overall change in
energy, not any of the up and down changes that
occurred between the initial and the final conditions.
CALORIMETRY – the study of heat flow and heat measurement
Heat capacity: the amount of heat needed to raise
the temp. of an object 1oC (or K)
 The greater the heat capacity, the greater the heat
required to change the temperature of the object.
Heat capacity depends both on the type and quantity
of matter in an object.
Molar heat capacity: the heat required to increase the
temperature of 1 mole of a substance 1 oC (or K)
Specific Heat - the amount of heat needed to raise
the temperature of 1 g of any substance by 1oC (or K)
 varies with substance, see table pg 176
Water has a high specific heat (it takes a lot of energy to
raise the temp of water compared to other types
of matter)
q = m C ∆T
q = heat absorbed or released by a substance
m = mass of substance, grams
C = specific heat of substance
T = Tfinal - Tinitial (either oC or K)
a –q value means heat is released
a + q value means heat is absorbed
(exothermic)
(endothermic)
Problems:
A. How much heat is absorbed by 5000g concrete heats up
from 19oC to 25pC, given Cconcrete = 0.84J/goC?
m = 5000g
∆T = 25-19 = 6oC
C = 0.84 J/goC
need to solve for q in q = mC∆T
q = 5000 x 0.84 x 6
q = 25200 J or 25.2 kJ
B. How much heat is released by the 5000g piece of
concrete when it cools from 74oC to 40oC at night?
m = 5000g
C = 0.84 J/goC
∆T = 40-74 = -34oC
need to solve for q in q = mC∆T
q = 5000 x 0.84 x -34
q = -142800 J or -142.8 kJ
(the neg sign reminds you the heat is released)
C. How much heat is needed to warm 250g water from
22oC to 98oC? What is the molar heat capacity of water?
m = 250 g
C = 4.184 J/goC
∆T = 98-22=76
q = mC∆T
= 250 * 4.184* 76 = 79J
Molar heat capacity is for 1 mole of water or 18 g water
We know Cwater = 4.184 J/gOC so re-write as a
fraction
4.184 J * 18 g
g oC 1 mole
= 75.2J/mole oC
HW Calorimetry problems 4 – 7, 9,10,12
Constant Pressure Calorimeter: a well- insulated container
filled with a known mass of water, used to measure
heat absorbed or released by a “reacting system”.
 Also known as a “coffee cup calorimeter”
 The water temp changes as the reacting system
absorbs or releases energy
 used by food scientists to determine Calories in food
qrxn = heat transfer of the reaction
qsur = heat change by the surroundings (the water)
qrxn = - qsur they are equal in magnitude, just
opposite in sign because the heat lost in a rxn = heat
gain of the water OR the heat gained in the rxn = heat
loss of the water)
Since q = mC ∆T,
mrxn Crxn ∆T rxn = - msur Csur ∆T sur
qrxn = - msur Csur ∆T sur
Use this equation when we are solving for heat change
of a substance by measuring heat change of its surroundings
(usually water in a calorimeter).
Note that if you mix two quantities of solutions or dissolve a
substance in water, the mass you use must be the mass of the
total solution.
Example: A calorimeter contains 125 g water at 25.6oC. A 50.0
g piece of metal at 115oC is put into the water. The temp. of the
water stabilizes at 29.3oC. Calculate the specific heat of the
metal, and identify the metal using the table of specific heats.
qrxn = - qsur think qmetal = - qwater
substitute mC∆T for each side: mC∆T metal = - mC∆T water
Look up specific heat of water = 4.184 J/goC
Write in values for variables and solve for Cmetal
(50) C (29.3 – 115) = - (125) (4.184) (29.3-25.6)
-4285C = -1935.1
C = 0.452 J/goC
The metal must be iron.
Problems
A. When a student mixes 50 mL of 1.0M HCl and 50mL of
1.0 M NaOH in a constant pressure calorimeter, the
temperature of the resultant solution increases from
21oC to 27.3oC. Calculate the enthalpy (heat) change in
J/mol HCl? In kJ/mol HCl assuming the total volume of
the solution is 100mL and the density is 1.0g/mL. The
Cwater = 4.184 J/g oC.
** Under constant pressure we can assume enthalpy = q
qrxn = - mC∆T
qrxn = ?
m = we can solve for using
density and volume
Cwater = 4.184 J/g oC (given)
∆T = 27.3 – 21 = 6.5
Solve for mass : density = mass/vol
1 = mass/100 ml, so mass = 100g
Plug values into equation: qrxn = - 100 * 4.184* 6.5
= - 2717 J
Now divide by 1000 for kJ
-2717J * 1kJ / 1000 J = 2.715 kJ
The last thing is to solve for a per mole basis. How many
moles of HCl do we start with?
50 mL * 1L * 1 mole HCl = 0.05 moles
1
1000mL 1 L
Now divide kJ by moles - 2.715 kJ = -54 kJ/mol
0.05 mol
Note that the negative sign tells you the rxn is exothermic.
Bomb Calorimeter: a constant volume calorimeter often used
in combustion reactions b/c it can withstand high
pressure.
 The sample is put in a sealed container the ( “bomb”)
which is then placed in a precise amount of water.
 The heat released is absorbed by the water and the
calorimeter, so we must include a value for the heat
capacity of this calorimeter in calculations.
 More precise than a “coffee cup calorimeter”.
Equation for constant volume (Bomb) calorimetry problems.
qrxn = - Ccal ∆T
qrxn = heat change of rxn
Ccal = specific heat of the calorimeter
∆T = Tfinal - Tinitial in oC or K
Problem : The combustion of liquid rocket fuel is shown in the
following equation. When 4.00 g of CH6N2 is combusted in a
bomb calorimeter, the temp increases from 25 oC to 39.50oC.
The Ccal is 7.794 kJ/oC. Calculate the heat of reaction for the
combustion of 1 mole of CH6N2.
2CH6N2 + 5 O2  2N2 + 2CO2 + 6H2O
qrxn
= - Ccal ∆T
= - (7.794) ( 39.50 – 25)
= -113 kJ
This is per 4 grams of CH6N2 , so we need to convert it to
per mole by writing kJ / g, then using conversion factors
to get it per mole.
- 113 kJ x 46
4g
1 mol
= -1299.5 kJ/mol
Group practice: A 0.5865 g sample of lactic acid, HC3H5O3 is
burned in a bomb calorimeter whose heat capacity is 4.812
kJ/°C. The temp increases from 23.10 °C to 24.95°C. Calculate
the heat of combustion for lactic acid
a. per gram
b. per mole
HW Problems 19 - 21
Enthalpy , H, is the energy (heat) in a substance at a
constant pressure.
 H = E + PV
H = Enthalpy
E = internal energy
PV = small component accounting for
pressure and volume
 Enthalpy is a state function, and depends on
temperature, state , and composition of the
substance, so we must include states in chemical
equations used with enthalpy and enthalpy changes.
 Enthalpy is an extensive property it is proportional to
the quantity of reactants.
We can easily measure the change in enthalpy of a chemical
reaction. This is called:
∆Hrxn, or enthalpy (heat) of reaction and is equal to heat
absorbed or released in a chemical rxn (assuming
constant pressure)
∆ Hrxn = Hproducts - Hreactants
Endothermic reactions have a positive ∆ H, meaning
the H products > H reactants
Exothermic reactions have a negative ∆ H,meaning
the H reactants > H products
We will be dealing with reactions at constant pressure so
Heat, q = ∆ Hrxn
Enthalpy Diagrams: Show the enthalpy change of a reaction.
 The y axis is increasing enthalpy, ∆H
 The reactants are written on a line then an arrow
points to a line with the products.
Standard Enthalpy Change, ∆Ho are enthalpy changes that
occur under standard conditions of 1 atm, 25o C AND
assuming the reactants and products are in their
standard states.
(not to be confused with STP, 1 atm, 0OC of gases)
 the standard state of a substance is
its natural state in pure form at 1 atm
 if it has 2 natural states (such as graphite and
diamond) the most stable form will be considered the
standard state (graphite)
∆Ho values are used to compare values because the heat
absorbed or released varies with temp, pressure and
state.
When writing thermochemical equations to show
∆ Ho, you pull energy out of the equation and write it to
the right using a + or – sign to designate endothermic
or exothermic reactions
(recall + ∆Ho = endothermic and - ∆ Ho = exothermic)
so writing the same equations as before in this way….
C3H8 + 5O2  3CO2 + 4H2O
C + 2H2O
 CO2 + 2H2
∆ Ho = - 2043 kJ
∆Ho = + 113 kJ
Stoichiometry Calculations involving Enthalpy (heat)
We can solve for ∆ Ho by using a conversion factor relating
moles of any substance in a balanced equation to the ∆ Ho for
that chemical reaction.
Ex: 2H2O2 (l)  2H2O(l) + O2 (g)
∆Ho = -190 kJ
1. Conversion factors you can make from the
relationships in this equation are:
2 mole H2O2 and
-190 kJ
2 mole H2O
-190 kJ
and
1 mole O2
-190 kJ
2. these conversion factors can be used to covert from
moles  kJ or kJ  mole
3. use other stoichiometry relationships to make other
conversions….such as
moles A from the coefficients in the balanced eq
moles B
__1 mole
molar mass,g
1 mole
6.02 x 1023parts
1 mole
22.4 Lgas at STP
STEPS FOR SOLVING ∆Ho ENTHALPY PROBLEMS
You will usually be given a quantity of a substance and will be
asked for the change in enthalpy.
1. Start with a balanced chemical equation with ∆Ho.
2. Write given quantity as a fraction over 1.
3. Convert given quantity to moles by multiplying by
1 mole
molar mass, g
(omit this step if you are given moles)
4. Then convert from moles of chemical to ∆Ho (heat)
x
∆Ho
moles substance (from coefficients in eq)
Example: How much heat will be released if 1.0g hydrogen
peroxide decomposes in a Bombedier beetle to produce a
steam spray?
2H2O2(l)  2H2O(l) + O2(g) ∆Ho = -190 kJ
Balanced chemical equation
Write given over one and convert to moles
1gH2O2 x 1 mole H2O2
1
34g H2O2
Then multiply by moles H2O2 : kJ from equation
1gH2O2 x 1 mole H2O2 x -190 kJ
1
34g H2O2
2 mole H2O2
= -2.79 kJ
Ex2: How much heat is transferred when 9.22g glucose in your
body reacts with O2?
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) Ho = -2803kJ
Ex 3: How much heat is transferred when 147g NO2 gas is
dissolved in 100g H2O?
3NO2(g) + H2O(l)  2HNO3(aq) + NO(g) Ho = -138 kJ
You must do 2 conversions, because we do not know which
compound is limiting the amount of product, the 147 g NO2 or
the 100g H2O.
HW –# 22 – 26 and Enthalpy Stoichiometry Additional
problems sheet
Hess’s Law: If a series of reactions are added together, the
enthalpy changes for the overall (net) reaction will be
the sum of the enthalpy changes for the individual steps.
Ex. The formation of smog occurs in two steps:
Step 1:
N2(g) + O2(g)  2NO(g)
∆Ho = +181 kJ
Step 2:
2NO(g) + O2(g)  2NO2(g)
∆Ho = -113 kJ
The overall (net) reaction is
N2(g) + 2O2(g)  2NO2 (g)
∆Ho = 181kJ + (-113kJ)
= +68 kJ
(Notice that the overall reaction is endothermic although one of the
steps is exothermic)
This illustration is an enthalpy diagram showing the two step
reactions and the overall reaction for the combustion of
methane.
Hess’s Law allows us to calculate energy changes that are
difficult to measure directly , by measuring the ∆H of the
individual step reactions.
Steps for applying Hess’s Law to solve enthalpy problems:
When you are given individual (step) reactions and ∆Ho for
each and asked to solve for the ∆Ho of the overall or net
reactions, which is also usually given, you must:
1. Compare the step reactions to the net reaction you are
asked to solve for. Arrange the equations so you can
add them together to give you the overall reaction.
2. Rules for arranging equations in order to get the net
reaction:
 If you multiply or divide the coefficients of a
reaction, you must do the same to the ∆Ho value.
 If you reverse an equation, you must change the sign
on the ∆Ho. (Logically, if a reaction absorbs heat in
one direction, it will give off heat in the other
direction.)
 When adding equations together to get the net
(overall) reaction, you cross out equal quantities of
the same substances that appear on both sides of the
equation.
 Add the ∆Ho values of the re-arranged step reactions
to get the ∆Ho of the net reaction.
3. Check that your final equation matches the net reaction
you were asked to solve for. Sometimes you must
simplify the coefficients by dividing them AND the ∆Ho
value.
Ex: The combustion of sulfur can produce SO2 as well as SO3
depending on the supply of oxygen. From the following
reactions and Ho, calculate the standard enthalpy changes for
the combustion of sulfur to produce SO2.
Net rxn: S(s) + O2(g)  SO2(g)
Step rxns:
2SO2(g) + O2(g)  2SO3(g) ∆Ho = -196 kJ
2S(s) + 3O2(g)  2SO3(g) ∆Ho = -790 kJ
Solution:
We need SO2 as a product – it is a reactant in the given step
equation,, so we must reverse that equation AND change the
sign on the ∆Ho.
2SO3(g)  2SO2(g) + O2(g) ∆Ho = +196 kJ
Write the step reactions and their ∆Ho in columns so they can
be added together:
2SO3(g)
 2SO2(g) + O2(g) ∆Ho = +196 kJ
2S(s) + 3O2(g)  2SO3(g)
∆Ho = -790 kJ
2S(s) + 2O2(g)  2SO2(g)
∆Ho = -594 kJ
We must divide the coefficients and the ∆H of our resultant
equation by 2 to match the given net reaction.
S(s) + O2(g)  SO2(g) ∆Ho = -594/2 = -297 kJ
Ex. 2: From
H2S + 3/2 O2  H2O + SO2
∆Ho = -563 kJ
CS2 + 3O2  CO2 + 2SO2 ∆Ho = -1075 kJ
Calculate ∆Ho for CS2 + 2H2O  CO2 + 2H2S
Solution:
We need to reverse the first reaction to get H2S as a product –
remember to change the sign of ∆Ho.
H2O + SO2  H2S + 3/2 O2
∆Ho = +563 kJ
Must multiply coefficients and ∆Ho of the rxn 1 so that
quantities of O2 and SO2 are equal and can cancel out.
(1) 2H2O + 2SO2  2H2S + 3O2
∆Ho = +1126 kJ
(2) CS2 + 3O2
 CO2 + 2SO2
∆Ho = -1075 kJ
CS2 + 2H2O
 CO2 + 2H2S
∆Ho = 51 kJ
HW #27 – 31 and Hess’ Law Practice Problems
Enthalpy of formation, ∆Hfo , enthalpy changes associated
with the formation of one mole of a compound from its
constituent elements in their standard states.
 Aka “heat of formation”
 Under standard conditions of 1 atm, 25oC
 Equations are always written to show 1 mole of the
compound, so fraction coefficients , like ½ are
commonly used.
 ∆Hfo units are kJ/mole
Ex: ½ N2(g) + O2(g)  NO2(g) ∆Hfo = 34 kJ/mol
 The ∆Hfo for an element in its standard state is 0.
Using ∆Hfo values to calculate Enthalpies of Reactions
1. Look up ∆Hfo values for each reactant and product.
2. Write a step reaction showing the formation of 1 mole
of each reactant and product that is a compound. The
reactants in these formation reactions must be elements
(including diatomic gases). You may use fraction
coefficients when needed to form just one mole of the
compound.
3. Use Hess’s Law to combine step reactions and calculate
the ∆Ho for the overall reaction
Example: What is the ∆Ho for the combustion of propane?
C3H8(g) + 5O2 (g)  3CO2 (g) + 4H2O(l)
1. Write chemical equations to show formation of one mole
of each reactant and product and look up ∆Hfo values in
a table.
3C(s) + 4H2 (g)  C3H8 (g)
C(s) + O2 (g)  CO2 (g)
H2 (g) + ½ O2 (g)  H2O (l)
∆Hfo = -103.85 kJ/mol
∆Hfo = -393.5 kJ/mol
∆Hfo = -285.8 kJ/mol
2. Use Hess’s Law to rearrange these equations to get the
reactants on the left and the products on the right.
Remember to change signs on the ∆Hfo values if you
reverse a reaction. Multiply coefficients and ∆Hfo values
if needed to enable you to cross out substances and add
sides to get the net reaction.
C3H8 (g) 3C(s) + 4H2 (g)
3C(s) +3 O2 (g) 3 CO2 (g)
4H2 (g) + 2 O2 (g)  4H2O (l)
∆Hfo = 103.85 kJ/mol
∆Hfo = 3 * -393.5 kJ/mol
∆Hfo = 4* -285.8 kJ/mol
C3H8(g) + 5O2 (g)  3CO2 (g) + 4H2O(l)∆Hfo = -2220 kJ/mol
Practice problem:
Using step reactions and Hess’s law, find the ∆H° for the
following reaction:
H2S (g) + 4F2 (g)  2HF(g) + SF6(g)
Another way….
The ∆Ho of a reaction is equal to the sum of the ∆Hfo of the
products minus the ∆Hfo of the reactants.
∆Horxn = ∑ n( ∆Hfo products) - ∑ n( ∆Hfo reactants)
∑ = sum of values for all products (or reactants)
n = coefficient of specific compound from balanced eq
Elements are NOT included in this equation, since their
∆Hof values are zero.
Example: Using standard enthalpies of formation, calculate
the ∆Ho for the net reaction below:
4NH3 (g) + 7O2 (g)  4NO2 (g) + 6H2O (l)
NH3 (g)
O2 (g)
NO2 (g)
H2O (l)
∆Hof = -46.2 kJ/mol
∆Hof = 0 (element)
∆Hof = 33.8 kj/mol
∆Hof = -285.8 kJ/mol
coeff = 4
coeff = 4
coeff = 6
∆Horxn = ∑ n( ∆Hfo products) - ∑ n( ∆Hfo reactants)
= [(4 * 33.8) + (6 * -285.8)] – (4 * -46.2)
= - 1396 kJ
When you use this equation, you do not switch signs on any
∆Hof values.
Problem: Calculate the ∆Ho for the combustion of 1 mol
benzene , C6H6:
C6H6 (g) + 15/2 O2 (g)  6 CO2 (g) + 3H2O (l)
Calculate the ∆Ho for the combustion of 1 mol of sulfuric acid
SO3(g) + H2O(l)  H2SO4(aq)
FOOD & FUEL
Fuel value: the energy released when 1 g of a substance is
combusted.
 The fuel values of food are measured in a
calorimeter.
Most energy for our bodies metabolism comes from the
combustion of carbohydrates and fats.
 Carbohydrates are broken down easily and provide
rapid source of energy for our cells
 Average food value for
Carbohydrate = 17 kg/mol
Fats
= 38 kJ/mol
Protein
= 17 kJ/mol
 Our bodies store extra calories as fat because it
stores more energy per gram (its lighter) and it is not
soluble in water.
 Proteins are used primarily for building materials, not
fuel, although they provide the same energy/mol as
carbohydrates.
Fuels : the complete combustion of fuels converts all Carbon to
CO2 and all hydrogen to H2O.
 The greater the proportion of C and H in a fuel, the
greater the fuel value.
The US has only 4.5% of the world’s population, yet accounts
for 20% of the world’s energy consumption.
Fossil Fuels – petroleum (oil), coal, natural gas
- All formed from the remains of plants and animals
deposited millions of years ago.
- Natural gas is a mixture made up of mainly methane,
CH4, with some ethane, C2H6 , some propane, C3H8,
and some butane, C4H10.
- Petroleum is a liquid combination of large and small
hydrocarbons with some S, N, and O
- Coal is solid hydrocarbons with some S, O, and N. We
have coal reserves to last about 100 years, but coal
use has concerns including
o SO2 pollution and CO2 pollution (contributing
to global warming)
o Hazards associated with mining