chemistry xxi unit 5 how do we predict chemical change? in order to make predictions about the...
TRANSCRIPT
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IUnit 5
How do we predict chemical change?
In order to make predictions about the likelihood of a
chemical process, we need to explore four main features:
THERMODYNAMICS
DirectionalityExtent
KINETICS
RateMechanism
The central goal of this unit is to help you identify and apply the different factors that help
predict the likelihood of chemical reactions.
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IUnit 5
How do we predict chemical change?
M3. Measuring Rates Analyzing the factors that affect reaction rate.
M2. Comparing Free EnergiesDetermining the directionality and
extent of chemical reactions.
M1. Analyzing Structure Comparing the relative stability of different substances
M4. Understanding Mechanism Identifying the steps that determine reaction rates.
FOUR MAIN MODULES
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Context
To illustrate the power of chemical ideas and models in predicting whether a chemical change will occur
or not, we will focus our attention in some of the processes that seem to lie behind the origin of life.
Why would the analysis of directionality, extent, rate, and mechanism of chemical
reactions be important to understand the EMERGENCE OF LIFE in our planet?
Why do we
care?
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The Problem
One of the central questions of modern science is how life started in our planet.
The answer to that question depends on our understanding of how we went from simple
molecules, such as N2 , H2O, and CO2, to complex, such as proteins and DNA.
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IUnit 5
How do we predict chemical change?
Module 1: Analyzing Structure
Central goal:
To evaluate the relative stability of substances
based on relevant structural features.
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The Challenge
Imagine that you have information about the chemical composition
of the early Earth. How could you decide what reactions were likely
to occur among these “primordial” components?
What structural features of the reactants and products could help you make the prediction?
How could you qualitatively predict the directionality of a chemical reaction?
TransformationHow do I change it?
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IRelative Stability
The extent of the reaction is determined by the relative stability of the reactants and products.
Properly evaluating the relative stability of two or more substances is a crucial skill in making
predictions about reaction directionality.
Potential
Energy
Ea
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Two Crucial Factors
In making judgments about relative stability, we need to consider two types of factors:
ENERGETIC FACTORS ENTROPIC FACTORS
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Energetic Factors
In general, energetic stability increases with decreasing potential energy of a substance.
Let′s think!
Which features of a substance, or its structural units (ions, molecules), may
affect its potential energy?
These are some of the relevant features we need to consider:
Bond strength Chemical composition Charge distribution State of matter Intermolecular Forces
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Energetic FactorsLet’s analyze more carefully the effect of
bond strength on chemical stability.
Compounds with strong bonds tend to
be more energetically stable. Their decomposition
requires more energy input.
Bond Energy (kJ/mol)
Bond Energy (kJ/mol)
C-C 347 O-H 464
O-O 142 C-O 360
N-N 163 N-H 389
Cl-Cl 243 I-I 151
Let′s think!
What seems to be the effect of both the types of atoms (e.g. O-O vs. C-O) involved
and bond length on energetic stability?
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Bond Strength
Bond Energy (kJ/mol)
Bond Energy (kJ/mol)
C-C 347 O-H 464
O-O 142 C-O 360
N-N 163 N-H 389
Cl-Cl 243 I-I 151
In general, A-A bonds are weaker than A-B bonds.
In general, longer bonds are weaker than shorter bonds.
Thus, energy considerations favor shorter A-B bonds.
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Bond Energy (kJ/mol)
N-H 389
O-H 464
F-H 565
C-N 305
C-O 360
C-F 485
Let’s Think
Consider these bond energies.
How may you explain these values? What other atomic properties may be used to predict bond strength?
Bond polarity (~)
Electronegativity
Strength
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ILet’s Think
2 H-O-H 2 H-H + O=O Water maybe?
What is more energetically stable, the mixture of
reactants or the product?
Predict first, then calculate using bond energies
The amount of O2 in the primitive atmosphere was likely pretty small. However, in order for aerobic organism to develop, there must have been a source of O2 for them.
Bond Dissoc.Energy
(KJ/mol)
O-H 464
H-H 432
O=O 498
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ILet’s Think
2 H-O-H 2 H-H + O=O
Bond Dissoc.Energy
(KJ/mol)
O-H 464
H-H 432
O=O 498 2 H2O
-4 x 464-1856 kJ
2 H2 + O2
-2 x 432 - 498-1362 kJ
Hrxn = 494 kJEp
0
This process is not favored from the energetic point of view, but may occur if we supply the
energy (e.g., sun radiation).
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ILet’s Think
H-H + X-X 2 H-X
Consider the formation these three substances: HF, HCl, and HBr (HX)
Is the formation of these compounds
energetically favored?
What do you predict for the compound
HI? Why?
Bond Energy (kJ/mol)
Bond Energy (kJ/mol)
H-H 432 H-F 565
F-F 159 H-Cl 431
Cl-Cl 243 H-Br 364
Br-Br 193
Based on these results, what experimental data about a chemical process could be used to
evaluate how “energetically favored” a reaction is?
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Heat of Reaction
The more energy is released during a chemical reaction (exothermic processes), the more
energetically stable the products are relative to the reactants.
The more stable the products, the more likely the reaction to proceed to completion (extent).
Thus, the heat of reaction (or Enthalpy change Hrxn)
is an important piece of information in deciding
whether a chemical process will be favored or not.
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Standard Enthalpy ChangesMeasurements of the energy absorbed or released
in the form of heat during a chemical reaction are of central importance in making predictions about the
extent of a chemical process.
These measurements are commonly done using
substances in their standard state at 1 atm and 25 oC. This heat of reaction is identified as the standard enthalpy
change Horxn.
Calorimetry: Heat transfer is indirectly measured by quantifying changes in temperature.
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Standard Enthalpy of Formation
One particular useful quantity is the change in enthalpy when 1 mole of substance is formed from
its constituent elements in their standard state.
½ H2(g) + ½ F2(g) HF(g) Hof = -273.3 kJ/mol
½ H2(g) + ½ Cl2(g) HCl(g) Hof = -92.3 kJ/mol
Hof = -36.3 kJ/mol½ H2(g) + ½ Br2(l) HBr(g)
½ H2(g) + ½ I2(s) HI(g) Hof = +26.5 kJ/mol
The more negative the standard enthalpy of formation (Ho
f), the more energetically stable the compound is with respect to the simple elements.
?
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ILet’s Think
H2O(l)H2O(g)
-241.8 kJ/mol -286.0 kJ/mol
C2H6O(l)C2H6O(l)
-271 kJ/mol -277 kJ/mol
How would you justify the differences in Hof in
these cases?
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Intermolecular Forces
The strength of the intermolecular forces between particles, determined by proximity between
molecules and types of interactions, affects the potential energy of the bulk material.
Stronger IMFS More negative Hfo
The effect of interparticle interactions is particularly
important in ionic compounds, where the substance is
comprised of a network of interacting ions.
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Ionic CompoundsSome scientist have proposed that some ionic
compounds, either in crystal form or dissolved in water, played a central role in the origin of life.
How may we make predictions about the energetic stability of these types of compounds?
They may have served as catalysts or as
replicators of “information.”
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IIonic Compounds
Similar ideas about interparticle strength and enthalpy of formation can be used to make judgments of
stability in the case of ionic compounds.
However, in this case we need to remember that we are not dealing
with molecules but ionic networks.
The properties of an ionic compound are determined by the electrostatic forces among
its ions, and between these ions any surrounding particle (atom, ion, or molecule).
Coulomb’s Law
221
r
qqF
+
r
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Based on Coulomb’s Law, one may expect forces between ions to be stronger:
the larger the charge of the ions (larger q1, q2);
the smaller the size of the ions (smaller r);
Forces and Energy
The stronger the forces, the lower the potential energy of the lattice.
The lower the potential energy, the more
energetically stable the ionic compound. Ep
+
+
2+
0
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ILattice Energy
The energy released during the formation of an ionic compound starting from its ions in the gas
phase is a good measure of the potential energy of the compound.
A+(g) + B-(g) AB(s) Hrxn= Lattice Energy
Rank these sets of compounds from lower (less
negative) to higher (more negative) lattice energy.
Discuss which compounds are more energetically stable.
NaF, NaBr, NaCl
KF, LiF, NaF
NaF, MgO, SrO
Let′s think!
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Lat
tice
En
erg
y (k
J/m
ol)
-600
-900
-1000
-800
-700
-1100
Li+
76 pm
Na+
102 pm
K+
138 pm
Rb+
152 pm
F-
133 pmI-
220 pmBr-
196 pmCl-
181 pm
Lattice Energy
MgO (-3795 kJ/mol)SrO (-3217 kJ/mol)
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ILet’s Think
Cs(s) + 1/2 Cl2(g) CsCl(s)
Based on our discussion:
Rank the following synthesis reactions in order of more to less “energetically favored”
(from more to less negative Hof).
Ba(s) + 1/2 O2(g) BaO(s)
2 Al(s) + 3/2 O2(g) Al2O3(s)
Hof = -438 kJ/mol
Hof = -1676 kJ/mol
Hof = -548 kJ/mol
Ca(s) + 1/2 O2(g) CaO(s) Hof = -635 kJ/mol
4
1
3
2
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Summary
In making judgments about relative stability, we need to consider :
ENERGETIC FACTORS
ENTROPIC FACTORS
Bond: strength, length,
heterogeneity, polarityIMFs:
StrengthIon:
Size, charge
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Entropic FactorsIn general, the stability of a system increases the
larger the number of ways the system has to distribute both its matter and energy.
Consider these systems with the same total kinetic energy:
Let′s think!
Arrange these systems from fewer to larger possible distinguishable “configurations”
(different ways to distribute matter and energy).
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Entropic Factors
The larger the number of different types of particles
which can move and interact in different ways, the more
distinguishable “configurations.”
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IEntropy
The Entropy (S) of a system is an indirect measure of the number of different configurations that its
matter and energy can take.
The more possible configurations a system has (distributions of matter and energy),
the larger its entropy (larger entropic stability).
Let′s think!
Which features of a substance, or its
structural units (ions, molecules),
may affect its entropy?
Molecular size
Complexity
State of Matter
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Fortunately, it is possible to measure the standard entropy of formation So
f of chemical substances.
Basic Assumption:
So (perfect crystal) = 0 at 0 K.
Standard Entropy of Formation
Sof for any substance is a measure of the
different configurations that matter and energy can take in 1 mole of the substance at 25 oC and
1 atm (measure of entropic stability).
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Consider this data:
Substance Sof (J/(mol k)
H2O(g) 188
H2O(l) 70.0
C(s,diamond) 2.4
C(s,graphite) 5.7
C(g) 158.1
Let′s think!
What patterns do you observe? How do you explain them?
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Entropic Trends
S increases slightly with T
S increases a large amount
with phase changes
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Entropic Trends
Let′s think!
What trends do you observe? How would you explain these results?
Substance Sof
(J/(mol k)Substance So
f
(J/(mol k)
NH3(g) 192.8 NH3(aq) 111.3
CO2(g) 213.8 CO2(aq) 117.6
NaBr(s) 86.8 NaBr(aq) 141.8
NaCl(s) 72.1 NaCl(aq) 115.5
MgO(s) 26.9
In general, the less constrained the atoms, molecules, or ions in a system, the larger the entropy.
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Entropic Trends
Substance Sof
(J/(mol k)Substance So
f
(J/(mol k)
He(g) 126.2 H2(g) 130.7
Ne(g) 146.1 N2(g) 191.6
Ar(g) 154.8 O2(g) 205.2
Kr(g) 163.8 Cl2(g) 223.1
Xe(g) 169.4 Br2(g) 245.5
Consider this data:
Let′s think!
What two major patterns do you observe? How do you explain them?
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Entropic Trends
For a given state of matter, entropy
generally increases with
Molar mass
and
Molecular complexity.
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Sof So
f Sof
Ar(g)
40 g/mol
154.8 O2(g)
32 g/mol
205.2 NO(g)
30 g/mol
210.8
Let′s think!
How would you explain these results?
Sof So
f Sof
N2(g)
28 g/mol
191.6 CO(g)
28 g/mol
197.7 C2H4(g)
28 g/mol
219.3
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Entropy and Reaction Extent
Chemical reactions in which the total entropy of the products is higher than the total entropy of the
reactants are “entropically” favored.
Let’s consider the reaction: A + B C + D
Sorxn = So
products – Soreactants
= (SoC + So
D) – (SoA+So
B) > 0 Entropically favored
Sorxn > 0 1) Fewer constraints (g, aq);
2) Larger molar mass;
3) More complexity (Number, Types).
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I Assess what you know
Let′s apply!
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Fuel Synthesis
C4H10
Butane
C8H18
Octane
C6H12O6
Glucose
H2
Hydrogen
The following molecules could have been used as fuels to generate energy by primitive organisms
in our planet.
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Compare the energetic stability of reactants and products and predict the signs of Ho
rxn and Sorxn
Let′s apply! Predict
CH4(g) + H2O(g) CO(g) + 3 H2(g)
6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g)
8 C(s) + 9 H2(g) C8H18(l)
4 C(s) + 5 H2(g) C4H10(g)
SorxnHo
rxnChemical Reaction
- -
--
+ -
++
Which of these reaction can we expect to be favored?
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Summarize the main energetic and entropic factors that can help
predict relative stability.
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Analyzing Structure
Relative stability depends on:
ENERGETIC FACTORS ENTROPIC FACTORS
Summary
Constraints: structural, dynamic
Molar MassComplexity:
structural, dynamic(number and types of
atoms)
Bond: strength, length,
heterogeneity, polarityIMFs:
StrengthIon:
Size, charge
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For next class,
Investigate what the Second Law of Thermodynamics is about.
How can this law be used to decide on the directionality of a chemical process?