Transcript Energy of Reactions

ENERGY OF
REACTIONS
Entropy, Enthalpy, and
Gibb’s Free Energy
ENERGY SO FAR
 Review:




Energy is the ability to do work or produce
heat
There is potential and kinetic energy
Energy can neither be created or
destroyed
Every compound needs energy to increase
temperature or to change from one state
of matter to another
ENERGY SO FAR
 Units
of energy:
Joule
(J)
Calorie (cal)
1 cal = 4.184J
OTHER ASPECTS OF ENERGY
 Energy
is also an important
component of chemical reactions
 Example:
 4Fe(s)
 In
+ 3O2(g)  2Fe2O3(s) + 1625kJ
this example, you combine iron
and oxygen to produce iron (III)
oxide. This reaction also produces
1625kJ of energy/heat.
ENERGY AND REACTIONS
 To
better explain the energy changes in
reactions, chemists have come up with:
 Enthalpy (H): the heat content of
substances under constant pressure
 For a chemical reaction, we describe the
change in enthalpy. This is called:
 Enthalpy (or heat) of reaction (δHrxn):
the change in heat or energy in a
chemical reaction
HOW DO WE MEASURE δHrxn
 To
measure the heat produced or
used by a reaction, scientists again
use calorimeters. (REVIEW: What is
a calorimeter?)
 To calculate, we have the following
formula:
δHrxn
= Hproducts - Hreactants
PREVIOUS EXAMPLE
 Previously,
we looked at the following:
 4Fe(s) + 3O2(g)  2Fe2O3(s) + 1625kJ
 According
to this equation, we produced
(or lost) 1625kJ of energy to the
environment (you feel this as hot) EXOTHERMIC
 Therefore Hreactants > Hproducts
 You would have to add 1625kJ of
energy to the products to equal the
reactants
WHAT THIS MEANS
 For



the reaction:
The reactants have 1625kJ more than the
products – you have to add energy to the
product side
δHrxn = Hreactants - Hproducts
δHrxn = -1625kJ
 Therefore,
exothermic reactions are
always a negative “-” δHrxn
ENDOTHERMIC
 Since
we know that exothermic
reactions have a negative heat of
reaction:
 WHAT IS THE SIGN FOR AN
ENDOTHERMIC REACTION?
WHAT THIS MEANS
 Endothermic
reactions are always a
positive “+” δHrxn
 You have to add energy to the reactant
side to equal the products
 Therefore we always have to add energy
to an endothermic reaction:
 Example:
 27kJ + NH4NO3 (s)  NH4+(aq) + NO3-(aq)
 δHrxn = Hproducts - Hreactants
 δHrxn = +27kJ
CALCULATING HEATS OF
REACTION
 Chemists
have measured different heats
of reaction for combustion reactions.
 In each case, they do this under a
condition of standard pressure and
temperature.
 They call these heats of combustion:
δHcomb
REVIEW
What
is the
general format for
a combustion
reaction?
COMBUSTION REACTION
 Molecule
+ O2  CO2 + H2O
 Therefore, these are values for the
combustion of 1 MOLE of the molecule
 Example:



We will look at the combustion of 1 mole of
glucose (sugar)
C6H12O6 + 6O2  6CO2 + 6H2O + 2808kJ
δHcomb = -2808kJ/mole
CALCULATING HEAT OF
COMBUSTION
 Example:
You start with 3.55x103 g of
glucose (C6H12O6). How much energy is
released when glucose goes through a
complete combustion reaction?
 STEPS:


Convert to moles
Use the δHcomb for glucose to calculate
energy
ANSWER
 Molar

mass of glucose: 180g/mole
δHcomb = -2808kJ/mole
 3.55x103
g | 1 mole | -2808kJ
| 180g | 1 mole
 -5.54x104 kJ
TRY THESE
1.
2.
The heat of combustion for octane
(C8H18) is -5471kJ/mole. If you start with
1550g of octane, how much energy is
released?
Sucrose (C12H22O11), or table sugar, has a
heat of combustion of -5644kJ/mole. If
you add 2.5g of sucrose to your cereal,
how much energy will be added to your
cereal?
ANSWERS
4
1.-7.44X10
2.-41
kJ
kJ
TRY THIS
 You
have a cup filled with 125mL of
glucose (C6H12O6). If the density of
glucose is 1.54g/mL, how many moles of
glucose do you have?
 If the heat of combustion for glucose is
-2805 kJ/mole, what is the heat produced
from the cup of glucose?
ANSWER
1.07
moles
glucose
3
-3.00 x10 kJ
TRY THIS
A
combustion reaction with octane
(C8H18) releases a total of -5.55x104
kJ of energy. If the heat of
combustion for octane is
-250kJ/mole, how many grams of
octane did you start with?
ANSWER
4
2.53X10
g
HEATS OF REACTION
 As
we said earlier, most heats of reaction
or heats of combustion are measured
using a calorimeter. Sometimes (because
some reactions are toxic or unstable), we
cannot use a calorimeter and must find
an alternative way to measure heats of
reaction.
HEATS OF REACTION
 Hess’s
law: If you add two or more
thermochemical equations to produce a
final equation for a reaction, then the sum
of the enthalpy changes for the individual
reactions is the enthalpy change for the
final reaction
EXAMPLE: FIND THE HEAT OF
REACTION
 2S
+ 3O2  2SO3
 This reaction can occurs in 2 steps:


S + O2  SO2 ;
2SO3  2SO2 + O2 ;
 How
δH = -297kJ
δH = 198kJ
do these reactions go together to
form the final reaction?
2S + 3O2  2SO3
 Step
1: reverse the second reaction
because SO3 is a product. Therefore, you
have to change the sign for the heat of
reaction


S + O2  SO2 ;
2SO2 + O2  2SO3 ;
 NOTE:
δH = -297kJ
δH =-198kJ
Now the second reaction is
exothermic
2S + 3O2  2SO3
 Step
2: there are 2 moles of S in the first
reaction. This means we have to double
everything in the first reaction (including
the heat of reaction)

2(S + O2  SO2 ;
δH = -297kJ)

2S + 2O2  2SO2;
2SO2 + O2  2SO3 ;
δH = -594kJ
δH =-198kJ

2S + 3O2  2SO3
 Step
3: you add the two reactions
together to get the final product

2S + 2O2  2SO2;
2SO2 + O2  2SO3 ;
δH = -594kJ
δH = -198kJ__

2S + 3O2  2SO3;
δH =-792kJ__

 NOTE:
If a compound is on opposite sides
of a reaction, they cancel out
SUMMARY
1.
2.
3.
4.
5.
You need to have known chemical
reactions that you can combine to form the
final chemical reaction
You need the δH for each reaction
If you need to reverse a reaction, you must
also change the sign of δH
If you need to multiply a reaction by a
number, you must also multiply the δH
When all reactions are completed, you then
add them together to get the final δH
TRY THE FOLLOWING
 Combine
the two reactions to find the
heat of reaction when hydrogen peroxide
(H2O2) breaks apart to form water and
oxygen)
 2H2O2


 2H2O + O2;
2H2 + O2  2H2O;
H2 + O2  H2O2;
ΔH = ?
δH = -572kJ
δH = -188kJ
2H2O2  2H2O + O2
 Step
1: We need to reverse the second
reaction


2H2 + O2  2H2O;
H2O2  H2 + O2;
 Step


δH = -572kJ
δH = 188kJ
2: You need 2 moles of H2O2
2H2 + O2  2H2O;
2H2O2  2H2 + 2O2;
δH = -572kJ
δH = 376kJ
2H2O2  2H2O + O2

Step 3: Combine the reactions

2H2 + O2  2H2O;
2H2O2  2H2 + 2O2;
δH = -572kJ
δH = 376kJ

2H2O2  2H2O + O2
δH = -196kJ

 NOTE:
When you cancel out reactants
and products, 2H2 cancels out 2H2, but
only 1O2 is cancelled out in the
products
TRY THIS
2 Ca + 2C + 3O2  2CaCO3
1.
2.
3.
Ca + 2C  CaC2;
dHrxn = -62.8kJ
CO2  C + O2;
dHrxn = -393kJ
2CaCO3 + 2CO2  2CaC2 + 5 O2; dHrxn =
+1538kJ
OTHER PRACTICE
4
NH3 + 5 O2  4 NO + 6 H2O
 N2 + O2  2 NO
 ∆H = -180.5 kJ
 N2 + 3 H2  2 NH3
 ∆H = -91.8 kJ
 2 H2 + O2  2 H2O
 ∆H = -483.6 kJ
PREDICTING SPONTANEOUS
CHEMICAL REACTIONS
 Energy
is an essential ingredient for a
chemical reaction
 Not all chemical reactions happen
spontaneously
 To determine if a reaction occurs
spontaneously, we need to look at
another concept: ENTROPY
ENTROPY
 Entropy
(S): a measure of the number of
possible ways that the energy of a system
can be distributed
 In other words, entropy is the tendency for
molecules to spread out as far as possible
from each other
 Since molecules spreading out is
dependent on temperature, the unit for
entropy is (J/K) – Joules/Kelvin
KELVIN TEMPERATURE
 Before
we go any further, we need to
review Kelvins:

Kelvin (K)
A
temperature scale based on absolute 0
(the coldest possible temperature)
 The Celsius temperature converts to
Kelvin:

K = °C + 273
THERMODYNAMICS
 Thermodynamics:
The laws of
thermodynamics, in principle, describe
the specifics for the transport of heat and
work.
 The
only law we are interested in is the
second law of thermodynamics
SECOND LAW OF
THERMODYNAMICS
Second
Law of
Thermodynamics:
spontaneous processes
always proceed in such a
way that the entropy of
the universe increases.
PREDICTING CHANGES IN
ENTROPY
 Reminder:

δHrxn = Hreactants – Hproducts
 We



have a similar equation for entropy
δSrxn = Sproducts - Sreactants
If the entropy increases during a reaction, then
Sproducts > Sreactants and δSrxn is positive
If the entropy decreases during a reaction, then
Sproducts < Sreactants and δSrxn is negative
PREDICTING CHANGES IN
ENTROPY

Changing states of matter: entropy
changes when you go between solid,
liquid or gas



H2O(l)  H2O (g); δSrxn > 0
Dissolving a gas in a solid or liquid
always decreases entropy
When you increase the number of
gas particles in a reaction, entropy
tends to increase

Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g); δSrxn > 0
PREDICTING CHANGES IN
ENTROPY

With some exceptions, entropy increases
when a solid dissolves in a liquid


The solid tends to break apart
Random motion of particles increases as
the temperature increases

As you increase temperature, the entropy
increases
PREDICT THE FOLLOWING
 Try
to determine whether entropy
increases or decreases for the following
reactions:
1.
2.
3.
4.
CF(g) + F2(g)  CF3 (g)
NH3 (g)  NH3 (aq)
C10H8 (s)  C10H8 (l)
H2O (l)  H2 (g) + O2 (g)
ANSWERS
1.
2.
3.
4.
Entropy decreases: You go from 2
molecules of gas to 1
Entropy decreases: You are dissolving a
gas into a liquid
Entropy increases: You have gone from
a solid to a liquid
Entropy increases: You are going from
no gas to 2 molecules of gas
TRY3THIS
CH4 + NH
 HCN + 3 H2
 N2
C
 H2
+ 3 H2  2 NH3;
+ 2 H2  CH4;
∆H = -91.8 kJ
∆H = -74.9 kJ
+ 2 C + N2  2 HCN; ∆H = +270.3 kJ
TRY THIS
N2H4 + H2 → 2NH3
N2H4 + CH4O  CH2O + N2 + 3H2
kJ
N2 + 3H2 → 2NH3
CH4O → CH2O + H2
ΔH = -37
ΔH = -46 kJ
ΔH = -65 kJ
TRY THIS
2C + 2H2O  CH4 + CO2
C
+ H2O  CO + H2; ∆H = 131.3kJ
 CO + H2O  CO2 + H2; ∆H = -41.2kJ
 CH4 + H2O  3H2 + CO; ∆H = 206.1kJ
PUTTING IT ALL TOGETHER
 Generally,
exothermic reactions are
spontaneous (you do not need to add
energy to make them work)
 Generally, spontaneous reactions will
increase in entropy
 Yet, there is a way to know for sure if a
reaction will occur spontaneously or not
GIBB’S FREE ENERGY
 In
1878, William Gibbs (an American)
combined enthalpy and entropy to
determine if a reaction was spontaneous
 He called his equation Gibbs Free Energy
equation

Free energy: the energy that is available to
do work (δG)
GIBB’S FREE ENERGY
 Gibbs
Free Energy Equation:
δGrxn
 δG
= δHrxn – TδSrxn
(kJ): represents free energy
 δH (kJ): represents change in enthalpy
 T (K): temperature in kelvins
 δS (J/K): represents change in entropy
SPONTANEOUS?
 Like
entropy or enthalpy, Gibb’s free
energy can be positive or negative


If positive (+): if the Gibb’s free energy is
positive, the reaction is not spontaneous
(you would need to add energy to make
the reaction work)
If negative (-): if the Gibb’s free energy is
negative, the reaction is spontaneous
EXAMPLE

N2(g) + 3H2(g)  2NH3(g)



1.
2.
3.
δHrxn = -91.8kJ
δSrxn = -197J/K
T = 25°C
Why does the entropy decrease
in this reaction?
What is the Gibb’s free energy?
Is this reaction spontaneous?
STEP 1: WRITE WHAT YOU
KNOW
 δG
=?
 δH = -91.8kJ
 T = 25°C +273 = 298K
 δS = -197J/K = -0.197kJ/K
 NOTE:
Temperature must be in Kelvin
and entropy is converted to kJ so that
it matches the units for δG and δH.
STEP 2: PUT THE NUMBERS IN
THE FORMULA
 δGrxn
= δHrxn – TδSrxn
 δGrxn = -91.8kJ – (298K)(-0.197kJ/K)
 δGrxn = -91.8kJ – (-58.7kJ)
 δGrxn = -91.8kJ + 58.7kJ
 δGrxn
= -33.1kJ
STEP 3: DETERMINE IF IT IS
SPONTANEOUS
1.
2.
3.
The reason why the entropy of this
reaction decreased is that you are going
from 4 moles of gas to 2 moles of gas.
The amount of gas is reduced and
lowers the entropy.
δGrxn = -33.1kJ
Since the Gibb’s free energy is negative,
this is a spontaneous reaction.
TRY THIS
A
Chemical reaction has the following
information:



1.
2.
δHrxn = 145 kJ
δSrxn = 322 J/K
T = 109°C
What is the Gibb’s free energy?
Is this reaction spontaneous?
ANSWER
1.
2.

δG = +22kJ
Since the Gibb’s free energy is
positive, this reaction is NOT
spontaneous
Challenge: At what
temperature would this reaction
become spontaneous?
ANSWER



 We
δHrxn = 145 kJ
δSrxn = 322 J/K
T = 109°C
need to find what temperature
δG<0




0kJ = 145kJ – T(0.322kJ/K)
T(0.322kJ/K) = 145kJ
T = 450K
the reaction has to be at least 450K to be
spontaneous
WHEN ARE REACTIONS
SPONTANEOUS?
 Reactions



δHrxn < 0; δSrxn > 0
δHrxn < 0; δSrxn < 0 (spontaneous at lower
temperatures only)
δHrxn > 0; δSrxn > 0 (spontaneous at higher
temperatures only
 Reactions

are spontaneous when:
are not spontaneous when:
δHrxn > 0; δSrxn < 0