Chapter 5 1

Chapter 5 – Chemical Reactions & EquationsOne of the fundamental realizations in the development chemistry was the “Law of Conservation of Matter” which simply states that “matter is neither created nor destroyed during the course of a chemical reaction”.

This law could also be called the “law of conservation of atoms” because during chemical reactions, atoms are just re-distributed into new chemical species. The total number of atoms of each element in the reactants is the same as the total number of atoms of each element in the products.

This is both one of the most important and one of the most misunderstood facts of science!

Chapter 5 2

In Chapter 2.6, we defined a “chemical reaction” or “chemical change” as a process in which one or more new substances are formed as a result of redistribution of atoms.

For most reactions, this would be sufficient. However, some reactions involve the redistribution of ions and, in some instances, these are polyatomic ions. In this case, we can almost think of the polyatomic ion as an “element” – that is, it is a chemical species that is not changed during a reaction.

i.e. 2NaOH + H2SO4 Na2SO4 + 2H2O

NaOH + CH3COOH NaCH3COO + H2O

Chapter 5 3

Another class of reactions, called “redox reactions” or “reduction-oxidation reactions”, do not necessarily result in a redistribution of atoms but of electrons:

i.e. Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s)

Two electrons move from the zinc metal to the copper ion, resulting in a zinc ion and metallic copper.

So, a better definition of a chemical reaction might be something like “a process in which new chemical species are formed from the redistribution of atoms, ions, or electrons.”

Chapter 5 4

One of the corollaries of the Law of Conservation of Matter is that mass is also conserved during the course of a chemical reaction. That is, if we start with 14.236 grams of reactants, we will end up with 14.236 grams of products. This is not always obvious, as some times there is a change in state and/or the system is “open”.

For example, consider a camp fire. You add wood to keep the fire going but, in the morning, all that is left is a pile of ashes which don’t weigh anywhere near as much as the wood you added. Where did the mass go? It was converted to carbon dioxide and water – both of which blew away as vapour (gas).

If you burnt a fire in an enclosed container, the total mass of the container wouldn’t change.

Chapter 5 5

A “chemical equation” is our model for a chemical reaction. It tells us certain things and doesn’t tell us others.

A balanced chemical equation must:- satisfy the “law of conservation of atoms” in that it

tells us exactly how many atoms of each element are present and where they are distributed in the reactants and products

- satisfy charge requirements so that all of the charges balance out (the number of electrons on both sides of the equation must be the same)

Balancing chemical equations is something that you should have learned in Grade 10…..

Chapter 5 6

A quick review of balancing equations:

1) Start with an element which is involved in only one species on each side of the equation

2) Assign appropriate stoichiometric coefficients to balance the species containing that element

3) Using the stoichiometry of the species that have just been balanced, balance the other elements present in that species

4) Repeat until all of the atoms have been accounted for

i.e. H2 + O2 H2O

C4H10 + O2 CO2 + H2O

Chapter 5 7

There are no “hard and fast” rules for balancing equations. Sometimes, it is just a matter of “trial and error”.

However:a) Quickly scan the equation to identify elements which

occur only once on each side – this is best starting point

b) Metal atoms (cations) often dictate the nature of a reaction, so they are good starting point

c) Assigning the first stoichiometric coefficient means that at least one other species has its coefficient defined

d) Try to balance entire groups if possible – such as polyatomic ions

e) Minimize the ratio of the coefficients where possible

Chapter 5 8

Examples:

Chapter 5 9

0

5

10

15

20

25

30

35

400 10 20 30 40 50 60 70 80 90 100

Mor

e

Freq

uenc

y

Bin

Histogram for Mid-term 1Section 1

Frequency

0

5

10

15

20

25

30

35

0 10 20 30 40 50 60 70 80 90 100

Mor

e

Freq

uenc

y

Bin

Histogram of 1st Mid-termSection 2

Frequency

Chapter 5 10

Chapter 5 12

What a balanced equation can tell us….

1) What the reactants are (their molecular or ionic formula along with their molar masses)

2) What the products are (their molecular or ionic formula along with their molar masses)

3) What their physical states are (although this is optional as it is not intrinsic knowledge in the equation)

Chapter 5 13

4) The stoichiometric relationship of all of the elements (which is also their molar ratios)

5) The atom economy (from which we can obtain the yield of the reaction and possible byproducts)

6) The relative ratio of the masses of all of the species present (from which we can work out limiting reagents)

Chapter 5 14

What a balanced equation can’t tell us….

1) It only tells us the “relative ratio” of reactants and of products – not absolutes. That is, it tell us that 2 moles of hydrogen will react with 1 mole of oxygen to give 2 moles of water but not that we have 1 mole or 2 moles.

2) It tells us what would happen if a reaction occurs but not that the reaction will occur.

3) It doesn’t tell us anything about whether the reaction will absorb or release energy.

4) It doesn’t tell us how fast a reaction will go.

5) It doesn’t tell us anything about the individual steps required for the reaction to occur – which is called the mechanism.

Chapter 5 15

Chapter 5 16

We can use other information to determine the direction of a chemical reaction – the direction that it will be spontaneous. Note that if a reaction is “spontaneous” in one direction, it will be “non-spontaneous” in the opposite direction.

There are a number of ways to illustrate this:

But the general idea is that the reactants have more chemical potential than the products – more energy.

Chapter 5 17

Note that these potentials are “relative”.

Saying that “sodium chloride” (NaCl(s)) is more stable than “sodium” (Na(s)) and “chlorine” (Cl2 (g)) is not the same thing as saying that sodium and chlorine are unstable. Indeed, sodium metal can be stored in oil for years and years – without reacting.

Chapter 5 18

This view of chemical reactions leads to the idea that reactions only travel in one direction. In fact, this is not the case. All reactions are “reversible” – and are deemed to be “dynamic chemical equilibrium”.

This is a bit at odds with the view expressed in the textbook (5.5) but, at a microscale, it makes sense that at any given point in a reaction, you can turn around and go back. This is called “the principle of microreversibility”.

For some reactions, it is fairly easy to see how they could be reversible. For example:

NaCl(s) Na+(aq) + Cl-(aq)

A saturated solution of sodium chloride has both solid and ionic forms.

Chapter 5 19

It is a little harder to see something, such as combustion, as “reversible”. Burning a hydrocarbon, such as methane, results in the production of both carbon dioxide and water in gaseous form. It is not easy to imagine that these molecules would come back together again and generate methane:

CO2 (g) + 2H2O(g) CH4 (g) + 2O2 (g)

However, this reaction can be observed when catalyzed. That is, it is not that it is “impossible” but that it is “improbable”.

In principle, we can regard all reactions as ones that achieve the condition of chemical equilibrium, even if there is only an infinitesimal amounts of reactants or products.

Chapter 5 20

Chapter 5

Stoichiometry:“The relation between the quantities of substances that take part in a reaction or form a compound (typically a ratio of whole integers).”

The calculation of the amounts and masses of reactants and products involved in a chemical reaction depends upon the stoichiometry of all of the chemical species.

Under ideal conditions:2H2 + O2 2H2O

2 moles 1 mole 2 moles1 mole 0.5 mole 1 mole

4.0318 g 31.9988 g 36.0306 g2.0159 g 15.9994 g 18.0153 g0.003 mole 0.0015 mole 0.003 mole

21

Chapter 5 22

The book uses the reaction:

P4 (s) + 6Cl2 (g) 4PCl3 (g)

Giving a table:

Note that this is structurally similar to the “ICE” tables with which you may already be familiar.

Chapter 5 23

“Everything that irritates us about others can lead us to an understanding of ourselves.”

Carl Jung

Chapter 5 24

The point is that a reaction equation tells us exactly how much of each of the reactants will react with each other to give how much of the products.

It tells us what the ratios should be to get exactly “zero” reactants and the “maximum” amount of products. (Realizing that reactions rarely go to completion…..)

But what if the ratios aren’t perfect? What if they don’t match the stoichiometry?

2CO(g) + O2 (g) 2CO2 (g)

Chapter 5 25

In a practical chemical reaction (laboratory experiment), one of the species will invariably represent a “limiting reagent”. This is the species that runs out first during a chemical reaction. Or, put another way, it is the reactant that limits the extent of the reaction because it goes to zero before the other chemical reactants.

Chapter 5 26

Calculating Limiting Reagents

When there is a limiting reagent, the amount of products produced and the amount of other reactants is determined by the moles of limiting reagent.

Strategy

1) Balance the chemical equation.2) From the masses of the reactants and their respective molar masses, calculate the number of moles of each reactant.3) Use the stoichiometric factors to calculate which of the reactants is in short supply.4) Use the molar quantity of the limiting reagent to work out the exact mass of all of the remaining species.

Chapter 5 27

Example:

A balloon container holds 4.62 grams of hydrogen and 48.7 grams of oxygen. If the mixture is ignited, it will generate water. How much water will be produced?

1) Balance equation:2H2 (g) + O2 (g) 2H2O(g)

2) Calculate moles:moles H2: 4.62 grams/2.0159 g/mol = 2.292 mol H2moles O2: 48.7 grams/31.9988 g/mol = 1.522 mol O2

3) Determine limiting regent:for 1.522 mol O2 to react, 2 x 1.522 mol or 3.044 mol

of H2 – but since we only have 2.292 mol H2, we don’t have enough to react all of the O2 – H2 is limiting reagent.

Chapter 5 28

Put a different way:Since hydrogen and oxygen react in a 2:1 ratio,

2.292 mol of H2 will only react with 1.146 mol of O2.

4) Determine the mass of the products (and of the oxygen) that is generated (consumed) during the reaction:

since H2 and H2O are in a 1:1 ratio, the number of moles of H2O produced is 2.292. Water has a molar mass of 18.0153 g/mol, so the amount of water produced is:

2.292 mol x 18.0153 g/mol = 41.291 grams

amount of oxygen consumed:

1.146 mol x 31.9988 g/mol = 36.671 grams

Chapter 5 29

Note that the amount of hydrogen consumed (4.62 grams) plus the amount of oxygen consumed (36.671 grams) adds up to the amount of water produced:

4.62 grams + 36.671 grams = 41.291 grams

And that the amount of oxygen left over is:

48.7 grams – 36.671 grams = 12.029 grams

which makes sense – since oxygen is in excess (hydrogen is the limiting reagent!).

Chapter 5 30

The book uses a “tabular approach”. This might be something that you are more familiar with and you may find more convenient.

i.e. 750 grams of NH3 is reacted with 750 grams of O2 – how much NO is produced? (pg. 128)

Chemical equation: 4NH3 + 5O2 4NO + 6H2O

Chapter 5 31

Yield:How much of a substance is produced.

Theoretical Yield:The amount of substance that we can theoretically

produce based on the mass of the reagents and the stoichiometric factors in the balanced equation.

Percent Yield: This is the usual way for reporting yields as it tells us

how much we actually produced compared to how much we could have obtained.

Chapter 5 32

We have looked at spectroscopy as a way to identify the presence of chemical compounds. But sometimes it is important to know “how much” of a chemical species is present – not just whether or not it is there.

“Quantitative Analysis” requires an analytical method that is exactly proportional to the amount of substance present.

That is, infrared spectroscopy can tell us that a sample contains dioxins but not how much (this is called “Qualitative Analysis”). To determine the quantity, we need to employ a technique that counts each and every molecule (mole) of the species that we are interested in.

Chapter 5 33

One of the most common and useful techniques for determining the quantity of a chemical species is “chromatography”. (Literally, “colour writing”.) Over the past century, techniques such as gas chromatography, ion selective chromatography, and high performance liquid chromatography have been refined to such an extent that analytical chemists can routinely measure substances in a “parts per billion” range and some substances, such as dioxins, can be measured at the “part per trillion” level. That is the equivalent of measuring1 ml in a trillion millilitres – which is a billion litres or a million cubic metres or a lake on 1 km by 1 km by 1 metre deep. (That’s a lot of water!)

More on chromatography later….

Chapter 5 34

Originally, chemists utilized chemical reactions and stoichiometry to work out quantitative analysis. For example, the amount of acid in a solution could be determined by the slow addition of a basic substance. This is called a “titration” and depends upon the stoichiometry of both the reactants and products.

For example:Sulphuric acid has a formula of “H2SO4” and, as a

consequence, reacts with two molecules of NaOH:

H2SO4 (aq) + 2NaOH(aq) Na2SO4 (aq) + 2H2O

This means that if we titrate a quantity of the acid with 1 mol of NaOH, then there must be 0.5 mol of H2SO4.

More on titrations later….

Chapter 5 35

A third form of quantitative analysis is called “gravimetric analysis”. Here, the species to be analyzed is mixed with a chemical species that will result in the formation of a precipitate which can be filtered and weighed. (Hence, “gravity” is key to the measurement.)

For example:

DMG – dimethylglyoxime(C4H8N2O2) reacts quantitative in a 2:1 ratio with Ni2+

(aq) to give Ni(DMG)2. If 0.163 g of DMG reacts with a solution containing nickel, how many grams of nickel were present?

Chapter 5 36

Balanced Equation:Ni2+

(aq) + 2DMG(aq) Ni(DMG)2 (s)

Convert “mass to moles”:

Molar mass of DMG = 99.977 g/mol

Moles of DMG = 0.163 g / 99.977 g/mol = 0.00163 mol

Convert “moles to moles”:

1 mol of DMG = 0.5 mol Ni2+(aq)

0.00163 mol DMG = 0.000815 mol Ni2+(aq)

Convert “moles to mass”:

Molar mass of Ni (or Ni2+(aq)): 58.6934 g/mol

Mass of Ni = 0.000815 mol x 58.6934 g/mol = 0.0478 g

Chapter 5 37

“Atom efficiency” and “atom economy” are relatively new concepts in chemistry and, perhaps more importantly, in industrial processes.

The principle of atom economy says that synthetic methods should be designed to maximize the incorporation of all materials used in production into the final product (as long as it makes sense….) and to minimize the amount of waste that is generated. Further, that where it is possible to do so, solvents and other chemical species should be recycled.

Atom efficiency measures this, by calculating for each element separately, the percentage of atoms that end up in the desired product.

Chapter 5 38

i.e. for a reaction such as producing aspirin:

Carbon atom efficiency = #of C atoms in product x 100%# of C atoms in reactant

Overall atom efficiency = mass of product x 100%total mass of all products

E-factor = total mass of waste materialsmass of product

Green Chemistry recommends that industrial processes should be designed to maximize the atom efficiency and the overall atom efficiency while minimizing the E-factor.

Chapter 5 39

Ancient Proverb: "Treat the earth well: it was not given to you by your parents, it was loaned to you by your children. We do not inherit the Earth from our Ancestors, we borrow it from our Children."