unit 5 how do we predict chemical change?

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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


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


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)


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



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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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?



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?

unit 5 how do we predict chemical change?

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