2.1: Introduction

Last updated
Save as PDF

Page ID: 46537

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

When you can measure what you are
speaking about, and express it in numbers,
you know something about it; but when
you cannot measure it, when you cannot
express it in numbers your knowledge is of
a meagre and unsatisfactory kind; it may
be the beginning of knowledge, but you
have scarcely, in your thoughts, advanced
to the stage of science.''

William Thomson, Lord Kelvin (1891)

Introduction

The concept of atoms goes back to the Greek philosophers. Democritus (470-360? b.c.) proposed that all matter is made up of separate, indestructible atoms, that different kinds of atoms have different structures and behavior, and that the observed properties of substances arise because of the way their individual atoms arrange themselves and combine with one another. His theories are essentially a primitive version of the material in Chapter 1. Why, then, did the ancient Greeks not use the theories of Democritus and go on to develop atomic energy? Why did 2000 years pass before modem science began to develop?

The answer, in large part, is that the Greeks did not think quantitatively about atoms, and they were not experimentalists. Their science was a philosophical explanation of the universe, rather than a pragmatic tool with which to manipulate the world around them. Cheap human labor kept them from having to worry about developing a scientific technology.

The Greek scientist Heron of Alexandria invented several steam-driven mechanisms that could have led directly to the steam engine, but he used them only as toys and novelties.

The atomic theory of Democritus was sterile because it did not lead to quantitative predictions that could be tested. It failed to develop beyond abstract concepts because it did not have the feedback from successful and unsuccessful experiments in the real world to challenge and improve it.

A scientific theory, to be useful, must be quantitative. It must predict: "If I do this, then that will happen, and to an extent that I can calculate in advance." Such a prediction is testable. It can be seen to be correct, increasing our confidence in the theory behind it; or what frequently is more important, it can be seen to be incorrect, causing us to revise and improve the theory.

Scientific theories grow by continual destruction and rebuilding. A theory that predicts nothing that could possibly be tested conveys no information, and is worthless.

The importance of precise measurement of mass in chemical reactions escaped the Greek philosophers. It also escaped the medieval European alchemists, metallurgists, and iatrochemists (medicinal chemists). The great French chemist Antoine Lavoisier (1743-1794) was the first to realize that mass was the fundamental quantity conserved during chemical reactions.

The total mass of all products formed must be precisely the same as the total mass of the starting materials. With this principle Lavoisier demolished the long-accepted phlogiston theory of heat (see Chapter 6) by showing that when a substance burns, it combines with another element, oxygen, rather than decomposing and giving off a mysterious universal substance called phlogiston. The principle of the conservation of mass is the cornerstone of all chemistry. More than just total mass is conserved; the same number of each type of atom must be present both before and after a chemical reaction, no matter how intricately these atoms may combine and rearrange into molecules.

Energy also must be conserved in chemical reactions. To the chemist, this means that the heat that is absorbed or given off in a particular chemical reaction (the heat of reaction) must be the same no matter how that reaction is carried out—in one step or in several. For example, the heat given off when hydrogen gas and graphite (a form of carbon) are burned must be the same as that given off when hydrogen and carbon are used to make synthetic gasoline, and this gasoline is then used as fuel for an automobile engine.

If the heat given off in the two variations of the reaction were not the same, then the more efficient reaction could be run in the forward direction and the less efficient one could be run in reverse. The result would be a cyclical no-fuel furnace that would pour out endless quantities of heat at no cost to the operator. Perpetual-motion schemes of all types vanish as soon as one becomes quantitative about heat, energy, and work. This is the basis of thermodynamics, which is covered in detail in Chapters 15-17.

In this chapter we shall look at the consequences, for chemistry, of two principles:

Atoms are neither created nor destroyed during chemical reactions (conservation of mass).
Heats of reaction are additive. If two reactions can be added to give a third, then the heat of the third reaction is equal to the sum of the heats of the first two reactions (conservation of energy).

Both these principles may seem obvious at first, but they are also quite powerful tools in explaining chemical behavior.

Atomic Weights, Molecular Weights, and Moles

Figure 2-1. Two molecules of hydrogen gas combine with one molecule of oxygen to yield two molecules of water. Avogadro's principle tells us that equal volumes of different gases contain equal numbers of molecules, at a specified temperature and pressure. Hence two volumes of H₂ gas will combine with one volume of 0₂, to produce two volumes of water vapor, and two moles of H₂ will combine with one mole of 0₂ to produce two moles of water vapor. From Dickerson and Geis, Chemistry, Matter, and the Universe

As soon as chemists realized that mass — not volume, density or some other measurable property — was the fundamental property that was conserved during chemical reactions, they began to try to establish a correct scale of atomic masses (atomic weights) for all the elements. How they did this is described in Chapter 6; the result of their years of work is the table of natural atomic weights on the inside back cover of this book. As we saw in Chapter 1, the molecular weights of molecular compounds and the formula weights of nonmolecular compounds (such as salts) are found by adding the atomic weights of all the constituent atoms.

Central to all chemical calculations is the concept of the mole. As defined in Chapter 1, a mole of any substance is the quantity that contains as many particles of the substance as there are atoms in exactly 12 g of carbon-12. Thus a mole of a substance is a quantity in grams that is numerically equal to its molecular weight expressed in atomic mass units. The number of particles in a mole is called Avogadro's number, and the experiments of Millikan and Faraday described at the end of Chapter 1 are one means of establishing its value:

N = 6.022 X 10²³ particles mole^-1

Moles are a way of manipulating atoms or molecules in bundles of 6.022 X 10²³. The molecular weights of H₂, 0₂, and H₂0 were worked out in Chapter 1. If we know that two molecules of hydrogen gas, H₂, react with one molecule of oxygen gas, 0₂, to produce two molecules of water, H₂0, then we can predict that 2 moles of H₂, or 4.032 g, will react with 1 mole of 0₂, or 31.999 g, to yield 2 moles of water, or 36.031 g (Figure 2-1). The check addition, 4.032 + 31.999 = 36.031, verifies the conservation of mass during the reaction. The chemist measures substances in grams, by weighing them.

Yet it is more meaningful to convert these quantities from grams to moles, because then one is working with relative molecular proportions, scaled up by a uniform factor of N.

Chemical Analyses. Percent Composition and Empirical Formulas

Chemical analysis involves breaking a substance down into its elements and then measuring the relative amount of each element present, either in grams per 100 g of original compound, or as a percent by weight. One way of doing this, if the compound is a hydrocarbon (made up only of carbon and hydrogen), is to burn a known amount of the substance in oxygen, and measure the quantities of C0₂ (carbon dioxide) and H₂0 that result.

**Example 1**
When 25.00 g of an unknown hydrocarbon is burned, 68.58 g of CO₂ and 56.15 g of H₂0 are produced. How many grams of carbon and hydrogen did the original sample contain?
Solution The atomic weight of carbon is 12.011 g mole^-1, and the molecular weight of CO₂ is 44.010 g mole^-1. First we find the percentage of carbon in carbon dioxide: $\textstyle{(\frac{12.011}{44.010})} \times 100% = 27.29%\mbox{ carbon}$ If 27.29% of CO₂ is carbon, then the quantity of carbon in 68.58 g of CO₂ will be $27.29% \times 68.58\mbox{ g} = 18.72\mbox{ g carbon.}$ A similar calculation for hydrogen in water gives: $\textstyle{(\frac{2 \times 1.008}{18.015})} \times 100% = 11.19%\mbox{ hydrogen}$ $11.19% \times 56.15\mbox{ g} = 6.283\mbox{ g hydrogen.}$ As a check: 18.72 g + 6.283 g = 25.00 g.

**Example 2**
How many grams of carbon and hydrogen per 100.0 g of sample are there in the hydrocarbon of Example 1?
Solution $\textstyle{(\frac{100\mbox{ g}}{25\mbox{ g}})} \times 18.72\mbox{ g carbon} = 74.88\mbox{ g carbon per 100 g sample}$ $\textstyle{(\frac{100\mbox{ g}}{25\mbox{ g}})} \times 6.28\mbox{ g carbon} = 25.12\mbox{ g hydrogen per 100 g sample}$

Figure 2-2. Seven different molecules with the empirical formula CH. A simple elemental analysis could not distinguish between them. An approximate molecular weight could distinguish between C₂H₂, C₄H₄, and C₆H₆, but even more information would be required to identify the particular C₆H₆ molecule present.

**Example 3**
What is the percent composition by weight of the hydrocarbon in Example 1?
Solution $\left ( \frac{18.72\mbox{ g carbon}}{25\mbox{ g total}} \right ) \times 100% = 74.88%\mbox{ carbon}$ $\left ( \frac{6.28\mbox{ g hydrogen}}{25\mbox{ g total}} \right ) \times 100% = 25.12%\mbox{ hydrogen}$

**Example 4**
Calculate the relative number of carbon and hydrogen atoms in the hydrocarbon of Example 3.
Solution It is easiest to work with 100.0 g of the substance, so the percent elemental composition figures become grams of the respective elements. First we divide each amount of carbon and hydrogen by their atomic weights: $\frac{74.88\mbox{ g carbon}}{12.011\mbox{ g/mole}} = 6.234\mbox{ moles carbon}$ $\frac{25.12\mbox{ g hydrogen}}{1.008\mbox{ g/mole}} = 24.92\mbox{ moles hydrogen}$ These are the relative numbers of moles of carbon and hydrogen, and it is here that the concept of moles becomes useful. These numbers must also be the relative numbers oí atoms of carbon and hydrogen. For every 6.234 atoms of carbon in the unknown hydrocarbon, there are 24.92 atoms of hydrogen. If we look for a common factor for these two numbers, we see that they are in a 1:4 ratio. Dividing both numbers by the lower number, 6.234, we find that for every one atom of carbon there are $\frac{24.92}{6.234} = 3.997\mbox{ atoms of hydrogen.}$

**Example 5**
A common liquid is 11.19% hydrogen and 88.81% oxygen by weight. What are the relative numbers of hydrogen and oxygen atoms?
Solution Again working with 100.0 g of the substance, we calculate the number of moles of each element: $\frac{74.88\mbox{ g hydrogen}}{1.008\mbox{ g/mole}} = 11.10\mbox{ moles hydrogen}$ $\frac{25.12\mbox{ g oxygen}}{15.999\mbox{ g/mole}} = 5.551\mbox{ moles oxygen}$ Dividing both numbers by the smaller, to search for a common factor, we find that there are two atoms of hydrogen for every atom of oxygen.

**Example 6**
A common laboratory solvent, a hydrocarbon, is made up of 92.26% carbon and 7.74% hydrogen. What are the relative numbers of carbon and hydrogen atoms in the substance?
Solution The answer is that one carbon is found for each hydrogen.

An elemental analysis, by itself, is not enough to decide the correct molecular formula of a compound. The formula for methane is CH₄, which would fit the results of the calculation in Example 4. But the analytical results would also be compatible with the molecules C₂H₈, C₃H₁₂, or C₄H₁₆, if they could exist. The substance in Example 5 might be water, H₂0, but it could also be H₄0₂ or some higher multiple. If you recognize, correctly, that only CH₄ and H₂0 are chemically sensible, then you are bringing to bear new chemical information that is not present in the analytical data alone. Most chemists would assume that the molecule in Example 6 was benzene, C₆H₆. But it could also be acetylene, C₂H₂ (except for the fact that acetylene is a gas at room temperature, and the unknown hydrocarbon was said to be a common laboratory solvent, which would exclude acetylene) or any of the five other less common hydrocarbon molecules shown in Figure 2-2.

A chemical formula that gives the relative number of each type of atom, as integers with no common factor, is called the empirical formula of the substance. It is the empirical formula that results from an elemental analysis of a substance, not the molecular formula, which could be the same as the empirical formula or could be some integral multiple of it. The empirical formula is the same as the molecular formula for methane, CH₄, and for water, H₂O; the empirical formulas of acetylene and benzene are both CH, but the molecular formulas are C₂H₂ and C₆H₆, respectively. It frequently happens that some simple physical measurement can give a rough approximation of the molecular weight of a substance. Gas densities (Chapter 3), freezing-point depression, and osmotic pressure measurements (Chapter 18) are useful in this regard. If such an approximate molecular weight is available, then it can be used along with the empirical formula to decide the true molecular formula.

**Example 7**
Glucose is 40.00% carbon by weight, 6.71% hydrogen, and 53.29% oxygen. What is its empirical formula, and what is its molecular formula?
Solution Working with 100.0 g of glucose, we first find the number of moles of each element: $\frac{40.00\mbox{ g carbon}}{12.011\mbox{ g/mole}} = 3.330\mbox{ moles carbon}$ $\frac{6.71\mbox{ g hydrogen}}{1.008\mbox{ g/mole}} = 6.66\mbox{ moles hydrogen}$ $\frac{53.29\mbox{ g hydrogen}}{15.999\mbox{ g/mole}} = 3.331\mbox{ moles hydrogen}$ This is obviously a molar ratio of one carbon to two hydrogens to one oxygen, so the empirical formula is CH₂O. With the information provided we have no way of knowing whether this, or some multiple of this, is the true molecular formula.

**Example 8**
From other experiments we know that glucose has a molecular weight of approximately 175 g mole^-1. Use this information and the results of Example 7 to find the molecular formula and the exact molecular weight of glucose.
Solution The weight corresponding to the empirical formula is $12.011 + (2 \times 1.008) + 15.999 = 30.026\mbox{ g/mole}$ The approximate molecular weight is roughly six times this value, so the precise molecular weight is $6 \times 30.026\mbox{ g/mole} = 180.16\mbox{ g/mole,}$ and the molecular formula is C₆H₁₂0₆.

Chemical Equations. Balancing Equations and Conservation of Mass

When propane gas, C₃H₈, is burned in oxygen, the products are carbon dioxide and water. This can be written as a chemical equation:

      C₃H₈ + O₂ → CO₂ + H₂O                   (2-1)

If chemistry were not a quantitative science, then this description of the reaction, identifying both the reactants and the products, would be adequate. But we expect more from a chemical equation. How many molecules of oxygen are required per molecule of propane, and how many molecules of carbon dioxide and water result? Equation 2-1 is an unbalanced equation. When we add numerical coefficients (placed to the left of the formula) that tell how many of each kind of molecule are involved, then there will be the same number of each kind of atom on the left and right sides of the equation, since atoms are neither created nor destroyed in a chemical reaction. The result will be a balanced equation.

To balance equation 2-1, we note first that the 3 carbon atoms on the left will lead to 3 molecules of CO₂ as products, each requiring 2 oxygen atoms, or 6 oxygens in all. Similarly, the 8 hydrogen atoms in propane will produce 4 molecules of water, requiring 4 more oxygen atoms. This total of 10 oxygens on the right must come from 5 molecules of 0₂. The correct coefficients for the four substances in equation 2-1 are therefore 1,5,3, and 4:

      C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O             (2-2)

Each side of this balanced equation contains 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms.

**Example 9**
Trinitrotoluene (TNT), C₇H₅N₃O₆, combines violently with oxygen to produce CO₂, water, and N₂. Write a balanced chemical equation for the explosion.
Solution The unbalanced equation is C₇H₅N₃0₆ + O₂ → C0₂ + H₂O + N₂ Since there are odd numbers of hydrogen and nitrogen atoms on the left, and even numbers on the right, it will be easier to balance the equation on the basis of two molecules of TNT: 2 C₇H₅N₃O₆ + O₂ → CO₂ + H₂O + N₂ The 14 carbons, 10 hydrogens, and 6 nitrogens then will result in 14 carbon dioxide, 5 water, and 3 nitrogen molecules: 2 C₇H₅N₃0₆ + O₂ → 14 CO₂ + 5 H₂O + 3 N₂ Now all atoms are balanced on left and right sides of the equation except for oxygen. Of the 33 oxygens on the right, 12 are provided on the left by the 2 starting molecules of TNT, and 21 must be supplied by $\textstyle{10\tfrac{1}{2}}$ 0₂. The final, balanced equation is 2 C₇H₅N₃0₆ + 10 $\tfrac{1}{2}$ O₂ → 14 CO₂ + 5 H₂O + 3 N₂

Example 9 led to an equation with a fractional coefficient for oxygen. This can be removed by multiplying all coefficients on both sides by 2:

      4 C₇H₅N₃O₆ + 21 O₂ -> 28 CO₂ + 10 H₂O + 6 N₂           (2-3)

but this is not necessary, since there is no reason why all coefficients must be integers. It would even be correct to base the equation on a single molecule of TNT:

      C₇H₅N₃O₆ + 21/4 O₂ → 7 CO₂ + 5/2 H₂O + 3/2 N₂         (2-4)

A balanced chemical equation such as equation 2-3 has several levels of meaning. Most simply, it describes the starting materials and the products. It also tells us that the number of each kind of atom entering the reaction is the same as the number leaving. Each type of atom individually is conserved during the reaction. Equation 2-3 is also a statement that for every 4 molecules of TNT, 21 molecules of oxygen are required, and the products are 28 molecules of CO₂, 10 molecules of water, and 6 molecules of N₂. Scaling the reaction up by a factor of 6.022 X 10²³ to go from molecules to moles, 4 moles of TNT react with 21 moles of O₂ to produce 28 moles of CO₂, 10 moles of H₂O, and 6 moles of N₂. The individual molecular weights are

C₇H₅N₃O₆	227.13 g mole^-1
O₂	31.999 g mole^-1
CO₂	44.010 g mole^-1
H₂O	18.015 g mole^-1
N₂	28.013 g mole^-1

Hence equation 2-3 also tells us that 4 X 227.13 g = 908.52 g of TNT requires 21 X 31.999 g = 671.98 g of oxygen for complete reaction. It also tells us that the products will be

28 X 44.010 g = 1232.3 g of CO₂ 10 X 18.015 g = 180.15 g of H₂O 6 x 28.013 g = 168.08 g of N₂

We can verify that mass is indeed conserved:

Reactants: 908.52 g + 671.98 g = 1580.5 g Products: 1232.3 g + 180.15 g + 16.08 g = 1580.5 g

What a balanced chemical equation does not tell us is the molecular mechanism or course of events by which the reaction takes place. Equation 2-3 should not be construed as suggesting that 4 TNT molecules must collide simultaneously with 21 oxygen molecules. Even three-body collisions are so much rarer than two-body collisions that they can be dismissed from consideration in most chemical reactions. An elaborate series of individual steps could take place, as long as the overall net reaction was described correctly by equation 2-3 or equation 2-4.

The reactants and products need not be molecules:

      CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O                      (2-5)

Equation 2-5 describes the reaction of CaCO₃, calcium carbonate (limestone), and HCl, hydrochloric acid, to produce an aqueous solution of calcium chloride, CaCl₂, and carbon dioxide. The equation is balanced, because the number of each type of atom is the same on both sides. The molar meaning is clear: 1 mole or 100.09 g of calcium carbonate requires 2 moles or 72.92 g of hydrochloric acid for complete reaction, and the products will be 1 mole each of calcium chloride (110.99 g mole^-1), carbon dioxide (44.01 g mole^-1), and water (18.02 g mole^-1). You can verify from these figures that mass is conserved during the reaction. The molecular interpretation is less straightforward, since calcium carbonate is a salt and not a molecular compound, Equation 2-5 should not be taken as meaning that one molecule of calcium carbonate reacts with two molecules of HCl. Although HCl exists as discrete molecules in the gas phase, in solution the molecules dissociate into H+ and Cl- ions. A better approximation of what actually happens at the molecular level is

      CaCO₃(s) + 2 H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l)        (2-6)

The letters in parentheses describe the physical state of each species (s, solid; aq, a hydrated ion in aqueous solution; g, gas; l, liquid). This equation says that solid calcium carbonate reacts with two hydrated protons (hydrogen ions) in aqueous solution to produce hydrated calcium ions, gaseous carbon dioxide, and liquid water. The chloride ions remain as hydrated chloride ions in solution before and after the reaction; hence they are omitted from the equation. Equation 2-5, like other balanced chemical equations, is most useful in describing the amounts of materials involved, rather than the molecular mechanism of reaction. Equation 2-6, although a better description of what is going on at the level of atoms and ions, is less useful in keeping track of the quantities of matter involved.

**Example 10**
Metallic sodium reacts with water to produce hydrogen gas and sodium hydroxide solution (a mixture of Na⁺ and OH^- ions). Write (a) a balanced equation for the overall reaction and (b) an equation that more accurately describes the actual atomic or ionic species present.
Solution The balanced equation is Na + H₂O → 1/2 H₂ + NaOH or 2 Na + 2 H₂O → H₂ + 2 NaOH A better description of what is actually present would be 2 Na(s) + 2 H₂O(l) → H₂(g) + 2 Na⁺(aq) + 2 OH^-(aq)

Calculations of Reaction Yields

Balanced chemical equations are primarily used to calculate the expected yield (quantity of product) from a reaction, and to determine whether any of the reactants will remain unused when the other reactants are depleted.

**Example 11**
How many grams of hydrogen are needed to combine with 100.0 g of carbon to make benzene, C₆H₆? How many moles and how many grams of benzene will be produced?
Solution The balanced equation for the reaction is 6 C + 3 H₂ → C₂H₆ The number of moles of carbon present is $\frac{100.0\mbox{ g carbon}}{12.011\mbox{ g/mole}}=8.326\mbox{ moles carbon}$ The balanced equation tells us that half as many moles of H₂ are needed as moles of C, so we need 4.163 moles of H₂ $4.163\mbox{ moles hydrogen}\times2.016\mbox{ g/mole}=8.393\mbox{ g hydrogen}$ The molecular weight of benzene is (6 X 12.011 g) + (6 X 1.008 g) = 78.11 g mole^-1 One-sixth as many moles of benzene are produced as moles of carbon used up, or 8.326/6 = 1.388 moles of benzene. Hence the amount of benzene produced is $1.388\mbox{ moles} \times 78.11\mbox{ g/mole} = 108.4\mbox{ g benzene}$

As a check on arithmetic, note that 100.0 g of carbon and 8.4 g of hydrogen combine to produce 108.4 g of benzene. Again, mass is conserved during a chemical reaction.

Example 12

How many grams of silver sulfide (Ag₂S) can be formed by the reaction

2Ag + S → Ag₂S

if we start with 10.00 g of silver (Ag) and 1.00 g of sulfur (S)? Which starting material, if any, will be left over, and how much?

Solution

Reaction:	2 Ag + S → Ag₂S
Masses:	215.7 g + 32.06 g = 247.8 g

The quantity of sulfur required to react with 10.00 g of silver is
$\frac{32.06\mbox{ g S}}{215.7\mbox{ g Ag}}\times10.00\mbox{ g Ag}=1.486\mbox{ g S}$
But we only have 1.00 g sulfur, so not all of the silver can react. Turning the problem around, the amount of silver needed to react with 1.00 g sulfur is
$\frac{215.7\mbox{ g Ag}}{32.06\mbox{ g S}}\times1.00\mbox{ g S}=6.73\mbox{ g Ag}$
Hence, there will be $10.00 - 6.73 = 3.72\mbox{ g Ag}$ left over. The quantity of silver sulfide produced will be
$\frac{247.8\mbox{ g silver sulfide}}{32.06\mbox{ g S}}\times1.00\mbox{ g S}=7.73\mbox{ g silver sulfide}$

Notice that Example 12 was worked a different way: First the total masses of all reactants and products were written under the balanced chemical equation, and then the ratios of these masses were worked out to find the desired answers. The problem can also be solved using moles, and the choice is one of convenience.

Alternative Solution

First we find the number of moles of silver and sulfur:
$\frac{10.00\mbox{ g Ag}}{107.9\mbox{ g/mole}}=0.0927\mbox{ mole Ag}$
$\frac{1.00\mbox{ g S}}{32.06\mbox{ g/mole}}=0.0312\mbox{ mole S}$
Since 2 moles of silver are required for every mole of sulfur, and there are more than twice as many moles of silver as moles of sulfur, some of the silver must be left behind when all of the sulfur has been used. The 0.0312 mole of sulfur will combine with 0.0624 mole of silver, adn form 0.0312 mole of Ag₂S. Left behind are 0.0927 - 0.0624 = 0.0303 mole of silver. Translating these quantities back to grams, we have
$0.0303\mbox{ mole Ag}\times107.9\mbox{ g/mole}=3.27\mbox{ g Ag left over}$
$0.0312\mbox{ mole Ag}_2\mbox{S}\times247.8\mbox{ g/mole}=7.73\mbox{ g Ag}_2\mbox{S produced}$
We arrived at the same answers in the first solution to the example. The mole method is surer, but slower. The ratio method is faster, but you can go astray more easily if you are not absolutely sure of what you are doing. Use the mole method until you are proficient in chemical calculations.

Solutions as Chemical Reagents

Liquid solutions are convenient media for chemical reactions. Rapid mixing of the liquid means that potential reactants are brought close to one another frequently, so collisions and chemical reactions can take place much faster than they would in a crystalline solid. Moreover, a given number of molecules in a liquid is confined to a smaller space than the same number of molecules in a gas, so reactant molecules in a liquid have more of a chance to come in contact. Water is an especially good solvent for chemical reactions because its molecules are polar. The H₂O molecules, and the H⁺ and OH^- ions into which water dissociates to a small extent, can help to polarize bonds in other molecules, weaken bonds, and encourage chemical attack. It is no accident that life evolved in the oceans rather than in the upper atmosphere or on dry land. If life had been forced to evolve using solid-state crystal reactions, the 4.5 billion years of earth's history to date might barely have been time enough for the process to begin.

Concentration Units: Molarity and Molality

In solutions involving a liquid and a gas or solid, the liquid component is called the solvent, and the other component is called the solute. If the solution is made up of two liquids, the distinction is less clear, but the substance present in the greater amount is usually considered as the solvent. The most common way of expressing concentration in solution is molarity, or the number of moles of solute per liter of solution. ^[1] The symbol M is read as "moles per liter of solution," as in 1.5M NaCl. The symbol c is used to denote concentration in moles per liter, as is the chemical symbol in brackets [H], although such brackets are sometimes used to represent concentration in any units. Hence the expression c_NaC1 would be read as "the concentration of sodium chloride in moles per liter of solution." This is not the solution that would result from adding 1 mole of NaCl to a liter of water, since the total volume after mixing would be a little more than 1 liter. Sodium and chloride ions take up room, even when dissolved in water. The proper procedure in making a 1.0M solution would be to dissolve the salt in less than a liter of water, and then slowly add more water, with mixing, until the total volume reached 1.00 liter.

For many salts, we can use the approximation that volumes are additive, or that the volume of a solution will be equal to the original volume of the solvent plus that of the crystals that were dissolved.

**Example 13**
If 264 g of ammonium sulfate, (NH₄)₂SO₄, are dissolved in 1.000 liter of water, what will the approximate final volume and the approximate molarity of the solution be, assuming additivity of volumes? The density of crystalline (NH₄)₂SO₄ is 1.76 g ml^-1.
Solution The volume of solid ammonium sulfate added is $\frac{264\mbox{ g}}{1.76\mbox{ g/mL}}=150\mbox{ mL or 0.150 L}$ The final solution volume then will be 1.000 + 0.150 = 1.150 liters. The number of moles of solute is $\frac{264\mbox{ g}}{132\mbox{ g/mole}}=2.00\mbox{ moles ammonium sulfate}$ The molarity of ammonium sulfate then is $\frac{2.00\mbox{ moles}}{1.15\mbox{ liters}}=1.74\mbox{ moles/liter or 1.74}M$

The approximation of additivity of volumes must be used with care. In this example, the true molarity of such a solution is 1.80M, so the approximation is only 3.3% in error. But for liquids whose molecules interact strongly, such as ethyl alcohol and water, the total volume may shrink after mixing because of molecular attractions. Additivity of volumes should be used only as a rough guide to molarity.

An alternative expression of concentration, molality, is based on the amount of solvent used rather than the solution that results. The molality of a solute is the number of moles of solute in 1 kg of solvent (not of solution). The density of water is 1.00 g mL^-1, so 1 kg of water occupies a volume of 1 liter. Hence the ammonium sulfate solution of Example 13 is a 2.00 molal solution, since it was made up from 2.00 moles of solute in a kilogram (1 liter) of water. For solvents other than water, we must use the density of the liquid to convert from kilograms to liters.

**Example 14**
Suppose 5.00 g of acetic acid, C₂H₄O₂, are dissolved in 1 liter of ethanol. Calculate the molality of the resulting solution. The density of ethanol is 0.789 g mL^-1. Can you calculate the molarity from the information given?
Solution The molecular weight of acetic acid is 60.05 g mole^-1, so the number of moles is $\frac{5.00\mbox{ g}}{60.05\mbox{ g/mole}}=0.0833\mbox{ mole acetic acid}$ The number of kilograms of solvent used is $1.00 \mbox{ liter}\times0.789\mbox{ kg/liter}=0.789\mbox{ kg ethanol}$ Notice that 1 g mL^-1 is the same as 1 kg liter^-1, since there are 1000 g in a kilogram and 1000 mL in a liter. The molality then is $\frac{0.0833\mbox{ mole solute}}{0.789\mbox{ kg solvent}}=0.106\mbox{ mole/kg}$ The solution is therefore 0.106 molal. The molarity of the solution cannot be calculated because we know neither the volume of the acetic acid nor whether volumes are additive when acetic acid is dissolved in ethanol.

The symbol m is used for concentration expressed as molality. We would write the results of Example 14 as

m_{acetic acid} = 0.106 mole kg^-1

Dilution Problems

If we dilute a solution (add more solvent), the number of moles of solute does not change. If c is the molarity (not molality) of the solution and V is the volume in liters, then the number of moles of solute is

c (moles liter^-1) X V(liters) = cV(moles)

If we use the subscript 1 to represent a solution before it is diluted with more solvent, and the subscript 2 for the diluted solution, then

Moles of solute = c_l V_l = c₂ V₂

**Example 15**
To what volume must 5.00 mL of 6.00M HCl be diluted to make the concentration 0.100M?
Solution $V_2=\frac{c_1}{c_2}\times V_1=\frac{6.00M}{0.100M}\times 5.00\mbox{ mL}=300\mbox{ mL}$ This does not mean that 300 mL of water must be added, but that the total volume of solution must be brought up to 300 mL

**Example 16**
If 175 ML of a 2.00M solution are diluted to 1.00 liter, what will the molarity be?
Solution $c_2=\frac{V_1}{V_2}\times c_1=\frac{175\mbox{ mL}}{1000\mbox{ mL}}\times 2.00M=0.350M$

Acid-Base Neutralization

Probably the most familiar definition of acids and bases is that by the Swedish physicist and chemist Svante Arrhenius (1859-1927): An acid is a substance that increases the hydrogen ion concentration, [H⁺], when added to water, and a base is a substance that increases the hydroxide ion concentration, [OH^-], when added to water. Some of the more common acids and bases are listed in Tables 2-1 and 2-2. The first 11 acids in Table 2-1, from HF to HNO₃, dissociate in aqueous solution to release one proton or hydrogen ion:

HNO₃	→ H⁺(aq) +	NO₃^-(aq)
nitric acid		nitrate ion

**Table 2-1. Common Acids**
HF	Hydrofluoric
HCl	Hydrochloric
HClO	Hypochlorous
HClO₂	Chlorous
HClO₃	Chloric
HClO₄	Perchloric
HBr	Hydrobromic
HBrO₃	Bromic
HI	Hydroiodic
HNO₂	Nitrous
HNO₃	Nitric
H₂CO₃	Carbonic
H₂SO₃	Sulfurous
H₂SO₄	Sulfuric
H₃PO₂	Hypophosphorous
H₃PO₃	Phosphorous
H₃PO₄	Phosphoric
H₃BO₃	Boric
HCOOH	Formic
CH₃COOH	Acetic

**Table 2-2. Common Bases**
LiOH	Lithium hydroxide
NaOH	Sodium hydroxide
Mg(OH)₂	Magnesium hydroxide
Ca(OH)₂	Calcium hydroxide
Ba(OH)₂	Barium hydroxide
NH₃	Ammonia

The abbreviation (aq) is a reminder that the ions are hydrated, but it is really not necessary since every ion in aqueous solution is hydrated, and we shall omit it in the rest of this discussion. Remember that the water molecules are always present, surrounding each ion and helping to stabilize it in solution. Carbonic, sulfurous, and sulfuric acids release two protons in two stages, and the three phosphorus-containing acids produce three protons:

H₂CO₃ → H⁺ + HCO₃^- → 2 H⁺ + CO₃^2-

carbonic acid

H₂SO₄ → H⁺ + HSO₄^- → 2 H⁺ + SO₄^2-

sulfuric acid

H₃PO₄ → H⁺ + H₂PO₄^- → 2 H⁺ + HPO₄^2- → 3 H⁺ + PO₄^3-

phosphoric acid

Carbonic acid is classed as a weak acid because its loss of protons is only partial; the species present in aqueous solution are a mixture of carbonate and bicarbonate ions and a small amount of undissociated carbonic acid. In contrast, sulfuric acid is a strong acid because the loss of the first of the two H+ is complete in aqueous solution. (Acid-dissociation equilibria are considered in detail in Chapter 5.) Nitric and hydrochloric acids are common strong acids, and phosphoric acid is weak. Organic acids such as formic and acetic release a proton from their —COOH carboxyl groups:

CH₃COOH	→	CH₃COO^- + H⁺
acetic acid		acetate ion

It is common to use the abbreviation HOAc for acetic acid and OAC^- for the acetate ion.

Hydroxide bases such as sodium hydroxide and magnesium hydroxide dissolve in water to release hydroxide ions:

NaOH →	Na⁺ + OH^-
Mg(OH)₂ →	Mg²⁺ + 2 OH^-

The hydroxide ions are already present in solid NaOH, just as chloride ions are present in NaCl. Ammonia, NH₃, is also a base, but it has no hydroxide ions of its own. Instead, it produces them by reacting with water molecules:

NH₃	+ H₂O →	NH₄⁺	+	OH^-
ammonia		ammonium ion		hydroxide ion

Ammonia is sometimes written as ammonium hydroxide, NH₄OH, to make it resemble the metal hydroxide bases (such as sodium hydroxide, NaOH). But this is incorrect; there is no such substance as ammonium hydroxide; there is only ammonia.

Acids and bases are useful because the H⁺ and OH^- ions that they produce can attack molecules in solution and bring about chemical changes that would be difficult or slow in their absence. When acids and bases react with one another, the H⁺ and OH^- ions combine to form water molecules. This is called neutralization:

H⁺ + OH^- → H₂O

The easiest way to determine how much of an acid or base is present is to find out how much of a base or acid of known concentration is required to neutralize it completely. This is the process of acid-base titration. One equivalent (equiv) of an acid is the quantity of acid that will release 1 mole of protons or H⁺ in neutralizing a base, and 1 equiv of a base is the quantity that will produce 1 mole of OH^- ions. Complete neutralization occurs when the same number of equivalents of acid and base react with one another. For acids that release one proton per molecule, such as HCl and HNO₃, the equivalent is the same as the mole, and 1 equivalent weight is the same as the molecular weight. But since H₂SO₄ is capable of releasing two H⁺ ions, 1 mole of H₂SO₄ corresponds to 2 equiv, and the equivalent weight of sulfuric acid in acid-base neutralizations is half the molecular weight. The equivalent weight of phosphoric acid, H₃PO₄, or the quantity that will produce 1 mole of H⁺ ions, is one-third the molecular weight. Similarly, the mole and the equivalent are identical for NaOH, KOH, and NH₃, but the equivalent weight of Ca(OH)₂ is half its molecular weight. We can appreciate the usefulness of the concept of equivalents by looking at the neutralization of phosphoric acid by magnesium hydroxide:

	2 H₃PO₄	+	3 Mg(OH)₂	→	Mg₃(PO₄)₂	+	6 H₂O
Molecular weight:	98.0 g		58.3 g		262.9 g		18.0 g
Equivalent weight:	32.7 g		29.2 g

One mole or 98.0 g of phosphoric acid will not neutralize 1 mole or 58.3 g of magnesium hydroxide, but 1 equiv or 32.7 g of phosphoric acid will neutralize 1 equiv or 29.2 g of magnesium hydroxide. This is the same answer that would be obtained by using the balanced equation shown. Since 2 moles of acid react with 3 moles of base as shown, 2 X 98.0 = 196 g of phosphoric acid will neutralize 3 X 58.3 = 175 g of magnesium hydroxide. These numbers are just the numbers obtained by using equivalents, but scaled up by a factor of 6.

**Example 17**
Use equivalents to find the number of grams of nitric acid, HNO₃, needed to neutralize 100.0 g of barium hydroxide, Ba(OH)₂.
Solution The molecular weight of HNO₃ is 63.01 g mole^-1; of Ba(OH)₂, 171.34 g mole^-1. The corresponding equivalent weights are 63.01/1 = 63.01 g equiv^-1 for HNO₃, and 171.34/2 = 85.67 g equiv^-1 for Ba(OH)₂. The number of equivalents of barium hydroxide is $\frac{100.0\mbox{ g}}{85.67\mbox{g/equiv}} = 1.167\mbox{ equiv of Ba(OH)}_2$ The same number of equivalents of nitric acid is needed: $1.167\mbox{ equiv}\times63.01\mbox{g/equiv}=73.53\mbox{ g nitric acid}$
Alternative Solution This example could also be solved by using the balanced chemical equation: 2 HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2 H₂O The number of moles of barium hydroxide at the start is $\frac{100.0\mbox{ g}}{171.3\mbox{g/mole}}=0.5838\mbox{ mole Ba(OH)}_2$ The balanced equation tells us that twice as many moles of nitric acid are required: 1.167 moles HNO₃. In grams, this is $1.167\mbox{ moles}\times63.01\mbox{ g/mole}=73.53\mbox{ g nitric acid}$ The use of equivalents eliminates the need to work out a balanced equation for the reaction.

The normality of a solution, represented by N is the number of equivalents of solute per liter of solution. A 1.00 M solution of phosphoric acid is 3.00N, and a 0.010M solution of Ca(OH)₂ is 0.020N.

**Example 18**
If 4.00 g of sodium hydroxide are dissolved in water and the volume is brought up to 500 mL, find the molarity and the normality of the solution.
Solution Since the molecular weight of NaOH is 40.0 g mole^-1, $\frac{4.00\mbox{ g}}{40.0\mbox{ g/mole}}=0.010\mbox{ mole NaOH}$ $\frac{0.100\mbox{ mole NaOH}}{0.500\mbox{ liter solution}}=0.200\mbox{ mole/liter, or }0.200M\mbox{ NaOH}$ Because 1 mole of NaOH releases 1 mole of OH^- ions, the molarity and normality are the same. The solution is 0.200N.

**Example 19**
If 10.0 g of sulfuric acid (H₂SO₄) are mixed slowly with enough water to make a final volume of 750 mL, what are the molarity and normality of the resulting solution?
Solution Since the molecular weight of NaOH is 40.0 g mole^-1, $\frac{10.0\mbox{ g}}{98.1\mbox{ g/mole}}=0.102\mbox{ mole sulfuric acid}$ $\frac{0.102\mbox{ mole}}{0.750\mbox{ liter}}=0.136M\mbox{ sulfuric acid}$ Since each mole of sulfuric acid contributes 2 equiv, the solution is $2\times0.136M=0.272N\mbox{ H}_2\mbox{SO}_4$

Acid-Base Titration

Figure 2-3. An acid—base titration. The solution in the flask contains an unknown number of equivalents of base (or acid). The burette is calibrated to show volume to the nearest 0.001 cm³. It is filled with a solution of strong acid (or base) of known concentration. Small increments are added from the burette until, at the end point, one drop or less changes the indicator color permanently. (An indication that the equivalence point is being approached is the appearance — and disappearance on stirring — of the color that the indicator assumes beyond neutralization.) At the equivalence point, the total amount of acid (or base) is recorded from the burette readings. The number of equivalents of acid and base must be equal at the equivalence point.

Chemists frequently use titrations to compare relative concentrations of chemical equivalents in acid—base solutions (Figure 2-3). When enough acid solution from a burette (shown in the figure) has been added to neutralize the base in the sample being analyzed, the number of equivalents of acid and base involved must be the same. The point of neutralization is called the equivalence point. An acid—base indicator such as litmus or Phenolphthalein can be used to determine the equivalence point. From the volume of acid solution used and its normality, we can calculate the number of equivalents of base in the unknown sample. If N_A and N_B are the normalities of acid and base solutions, and V_A and V_B are the volumes of each at neutrality, then

      Number of equivalents = N_AV_A = N_BV_B             (2-7)

**Example 20**
If 25.00 ml of phosphoric acid (H₃PO₄) are just enough to neutralize 30.25 mL of a sodium hydroxide solution, what is the ratio of the normalities of the two solutions? What is the ratio of molarities?
Solution $\frac{N_A}{N_B}=\frac{V_B}{V_A}=0.102\mbox{ mole sulfuric acid}$ $\frac{0.102\mbox{ mole}}{0.750\mbox{ liter}}=0.136M\mbox{ sulfuric acid}$ Since each mole of sulfuric acid contributes 2 equiv, the solution is $2\times0.136M=0.272N\mbox{ H}_2\mbox{SO}_4$ Since the normality of the acid is three times its molarity, and the normality of the base is the same as its molarity, the molarity ratio is $\frac{c_A}{c_B}=\frac{N_A/3}{N_B}=0.403$

**Example 21**
In a titration, 25.00 mL of a solution of calcium hydroxide, Ca(OH)₂, require 10.81 mL of 0.100N HCl for neutralization. Calculate (a) the normality of the Ca(OH)₂ solution, (b) the molarity, and (c) the number of grams of Ca(OH)₂ present in the sample.
Solution The normality of the solution of Ca(OH)₂ is $N_B=\frac{V_A}{V_B}\times N_A=\frac{10.81\mbox{ mL}}{25.00\mbox{ mL}}\times0.100N=0.0432N$ Since 1 mole of calcium hydroxide yields 2 equiv OH^-, the molarity is half the normality, or 0.0216M Ca(OH)₂. The number of moles of Ca(OH)₂ is $0.0216\mbox{ mole/liter}\times0.02500\mbox{ liter}=0.000541\mbox{ mole}$ Since the molecular weight of Ca(OH)₂ is 74.1 g mole^-1, the mass present is $0.000541\mbox{ mole}\times74.1\mbox{ g/mole}=0.0401\mbox{g Ca(OH)}_2$

**Example 22**
An organic chemist synthesizes a new acid. She dissolves 0.500 g in a convenient volume of water and finds that it requires 15.73 mL of 0.437N NaOH for neutralization. What is the equivalent weight of the new compound as an acid? If it is known that the acid contains three ionizing -COOH groups, what is the molecular weight?
Solution The number of equivalents of the base are $0.01573\mbox{ liter}\times0.437\mbox{ eqiv/liter}=0.00687\mbox{ equiv}$ The equivalent weight is found from $\frac{0.500\mbox{ g}}{0.00687\mbox{ equiv}}=72.8\mbox{ g/equiv}$ If the equivalent weight of the acid is 72.8 g, and each mole yields 3 equiv, then the molecular weight is $3\times72.8=218\mbox{ g}$

Heats of Reaction

So far this chapter has been devoted to the consequences of the conservation of mass, and little has been said about energy. But the principle that heats of reaction are additive, that energy is conserved in a process whether the process is carried out in one step or in several, is an important one. Heat and work are both forms of energy, and are measured in the same units. If you do work on an object or collection of objects, you can increase the energy or make the system heat up, depending on how the work is done. Lifting a heavy object is a conversion of work to potential energy, and friction is a conversion of work to heat. Conversely, energy can be reconverted to work when a heavy object falls, and heat is converted to work in an automobile engine. Of these three — heat, work, and energy — the chemist usually is more concerned with heat: the heat that may be absorbed or given off when a chemical reaction takes place.

By Newton's laws of motion, the force on an object is the product of its mass and acceleration:

Force	=	mass	×	acceleration
F	=	m	×	a

The force that must be applied to a 1.00 kg mass to give it an acceleration of 1 meter per second per second (1 m s^-2) is defined as a force of 1 newton (N). Hence 1 N = 1 kg m s^-2. (SI units are based on length in meters and mass in kilograms.)

**Example 23**
When a pitcher whips a 5.00 ounce (oz) baseball around an arc 5.00 m in circumference in order to accelerate it from zero to 90 miles hr^-1, what average acceleration does he give to the ball during the pitch, and what average force does he exert on it during his windup?
Solution Assume uniform acceleration on the ball from the time the windup begins until the ball leaves the pitcher's hand. For uniform acceleration from rest, v = at and s = at², where v is velocity, a is acceleration, t is time, and s is distance. Eliminating time from the two expressions yields a = v²/2s. If v = 40.2 m s^-1 (90 mph) and s = 5.00 m, then $a=\frac{40.2^2}{2\times5.00}=162\mbox{ m s}^{-2}$ Since the mass (m) is 0.142 kg (5.00 oz), the average force applied to the ball during the swing is $F=m\times a=0.142\times162\mbox{ kg m s}^{-2}=23.0N$

The work done when 1 N of force is exerted on an object for a distance of 1 m is defined as 1 joule (J). Hence 1 J = 1 N m = 1 kg m² s^-2. For a thrown object, all the work is converted into kinetic energy (energy of motion); in other circumstances, part or all the work can end as heat.

**Example 24**
How much work is done on the baseball in the pitch described in Example 23? How much kinetic energy does the ball have as it leaves the pitcher's hand?
Solution The work done on the baseball is $W=F\times s=23.0\mbox{ N} \times 5.00\mbox{ m}=115\mbox{ N m} = 115\mbox{ J}$ The ball ends with a kinetic energy of 115 J.

As a check on these results, we can calculate the kinetic energy directly:

E = \tfrac{1}{2} mv^2 = \tfrac{1}{2}\times0.142\times(40.2)^2\mbox{ kg m}^2\mbox{s}^{-2}=115\mbox{ J}

where E is energy. The advantage of the joule as a unit of heat is that it makes immediately apparent the connection between heat, work, and energy. An older unit of energy that arose from heat measurements is the calorie. One calorie (cal) is defined as the quantity of heat required to raise the temperature of 1 g of pure water by 1°C (from 14.5°C to 15.5°C, to be exact). This definition had no obvious connection with work, and in fact the calorie was defined in the nineteenth century, before anyone realized that heat and work were alternative forms of the same thing: energy. We will use only joules in this book, but you should be aware of calories since most of the preexisting literature uses that unit. The calorie is approximately four times as big as a joule: 1 cal = 4.184 J. Heats of reaction of mole quantities of substances are typically in the range of kilojoules (kJ) or kilocalories (kcal), where 1 kJ = 1000 J and 1 kcal = 1000 cal.

As an illustration of heats of reaction and the principle of additivity of heat, let us look at the decomposition of hydrogen peroxide, H₂O₂. When an aqueous solution of hydrogen peroxide reacts to form oxygen gas and liquid water, heat is given off. The amount of heat will vary somewhat with the temperature at which the reaction occurs, but at 25°C, the commonly accepted standard "room temperature" for measuring and tabulating heats of reaction, each mole of H₂O₂ that decomposes produces 94.7 kJ of heat. (If this energy could be used with perfect efficiency, it would be enough to accelerate 823 baseballs as described in Example 24.)

A schematic representation of a bomb calorimeter used for the measurement of heats of combustion. The weighed sample is placed in a crucible, which in turn is placed in the bomb. The sample is burned completely in oxygen under pressure. The sample is ignited by an iron wire ignition coll that glows when heated. The calorimeter is filled with fluid, usually water, and insulated by means of a jacket. The temperature of the water is measured with the thermometer. From the change in temperature, the heat of reaction can be calculated.

The heat involved in a chemical reaction carried out at constant pressure (or at least with the final pressure brought back to the starting value) is known as the change in the enthalpy of the reacting system, ΔH (read as "delta H"). As we shall see in Chapter 15, the energy change, ΔE, corresponds to the heat of the reaction if the reaction is carried out at constant volume, as in the bomb calorimeter shown in Figure 2-4. Enthalpy can be regarded as a "corrected" energy, the correction being for any work that the chemicals might do in pushing against the atmosphere if they expand. The difference between ΔE and ΔH is small but significant, but it is not important to us now. If heat is given off during the reaction, then the enthalpy of the reacting system of chemical falls; ΔH, the change in enthalpy, is negative. Such a reaction is called exothermic. In an endothermic reaction heat is absorbed and the enthalpy of the reaction mixture rises. For the hydrogen peroxide reaction, we can write

      H₂O₂(aq) → H₂O(l) + 1/2 O₂    ΔH = -94.7 kJ    (2-8)

This heat is released when 1 mole of hydrogen peroxide decomposes to 1 mole of water and 1/2 mole of oxygen gas, or for 1 mole of the reaction as just written. If all the coefficients of the reaction are doubled, then the heat of reaction must be doubled also, since it then refers to twice as much reaction:

      2 H₂O₂(aq) → 2 H₂O(l) + O₂    ΔH = -189.4 kJ   (2-9)

"One mole of the reaction as written" now means 2 moles of hydrogen peroxide decomposing to 2 moles of water and 1 mole of oxygen, because the coefficients in the equation have all been doubled. The heat of a reaction also depends on the physical state of the reactants and products. If hydrogen peroxide were to decompose to give oxygen gas and water vapor instead of liquid, part of the 94.7 kJ would be diverted to evaporating H₂O:

      H₂O(l) → H₂O(g) ΔH= +44.0 kJ           (2-10)

and less heat would be given off by decomposition of peroxide:

      H₂O₂(aq) → H₂O(g) + 1/2 O₂(g)     ΔH= -50.7 kJ       (2-11)

An important assumption is hidden here: that heats of reaction are additive (Figure 2-5). Equation 2-9 plus equation 2-10 gives equation 2-11, and so we have assumed that the heat of the third reaction will be the sum of the first two:

ΔH= -94.7 + 44.0 kJ = -50.7 kJ

The change in enthalpy for the reaction
H₂O₂(aq) → H₂O(g) + 1/2 O₂(g) can be obtained without actually measuring the enthalpy change for the reaction by adding ΔH for the two reactions
H₂O₂(aq) → H₂O(l) + 1/2 O₂(g) ΔH = -94.7 kJ/mol
H₂O(l) → H₂O(g) ΔH = +44.0 kJ/mol

The additivity of reaction heats follows directly from the first law of thermodynamics (Chapter 15): The change in energy or enthalpy between two

states depends only on the nature of those states, and not on how the change is carried out. A collection of chemicals in a given state has a certain energy and a certain enthalpy, neither of which depends in any way on how the chemicals were brought to that state. (That is, the past history of the chemicals can affect their present energy and enthalpy, but we do not need to know that history to measure the values of E and H.) Hence the difference between enthalpies of reactants and products, or the heat of reaction, can depend only on the nature of the starting and ending states, and not on the particular way that the reaction is carried out. This is sometimes called Hess' law of heat summation, which is a rather dignified name for a natural consequence of the first law of thermodynamics.

The additivity of heats of reaction makes a great amount of experimentation in thermochemistry (the chemistry of heat and energy) unnecessary. We need not measure and tabulate the enthalpy change of every conceivable chemical reaction. For example, if we know the heat of vaporization of liquid water (equation 2-10) and the heat of decomposition of hydrogen peroxide to liquid water (equation 2-9), then we never need to measure the heat of decomposition of hydrogen peroxide to water vapor; the answer can be calculated ahead of time. If a certain reaction is inconvenient to carry out, there may be a set of easier reactions whose sum is the reaction in question. After we have carried out the individual experiments, we can add the enthalpy changes in the same way as the chemical equations, to find the heat of the difficult-to-measure reaction.

Suppose that someone proposed a scheme for making diamonds by oxidizing methane:

CH₄(g) + O₂(g) → C_(di) + 2 H₂O(l)

[The notation (s) is not sufficient for carbon, since diamond (di) must be differentiated from graphite (gr).] You would like to find whether the reaction will liberate heat that must be allowed for in the design of the reaction vessel. This particular synthesis has never been carried out (and probably never will be), yet you can give your misguided friend his answer from a knowledge of the heats of easier reactions. The heat of combustion of a substance containing C, N, O, and H is the heat, per mole of substance, of the reaction with enough oxygen to produce CO₂, N₂, and liquid H₂O. Heats of combustion are easy to measure, and were among the first reaction heats to be measured and tabulated systematically. Extensive tables of heats of combustion can be found in books such as the CRC Handbook of Chemistry and Physics or Lange's Handbook of Chemistry. The heats of combustion of methane and diamond are

      CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)   ΔH = -890 kJ/mol    (2-12)
        C(di) + O₂(g) → CO₂(g)               ΔH = -395 kJ/mol    (2-13)

The desired diamond-synthesizing reaction is produced by subtracting the second reaction from the first, or by adding the first reaction to the reverse of the second, and the heat of reaction is found in the same way:

      CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)   ΔH = -890 kJ/mol    (2-12)
                CO₂(g) → C(di) + O₂(g)       ΔH = +395 kJ/mol    (2-14)
                                                             
        CH₄(g) + O₂(g) → C(di) + 2 H₂O(l)    ΔH = -495 kJ/mol    (2-15)

Notice that when a reaction is turned around and run in reverse, the heat of reaction changes sign, since a process that gave off 395 kJ in one direction must absorb 395 kJ in reverse.

Heats of Formation

Because of the additivity of heats of reaction, not all heats have to be tabulated — only those for the minimum set of reactions from which all others can be obtained. The set that has been agreed upon by scientists and engineers is made up of the heats of formation of compounds from their pure elements in standard states. For solids and liquids this standard state is the most common form of the element at 25 °C or 298 K and 1 atmosphere (atm)^[2] external pressure; gases are defined similarly but at 1 atm partial pressure.^[3] The standard state for thermodynamic measurements involving carbon is graphite (gr), not diamond (di). The heats of formation for all the compounds involved in the diamond synthesis are

      C(gr) + 2 H₂(g) → CH₄(g)     ΔH =  -74.8 kJ/mol          (2-16)
                C(gr) → C(di)     ΔH =   +1.9 kJ/mol          (2-17)
    H₂(g) + 1/2 O₂(g) → H₂O(l)     ΔH = -285.8 kJ/mol          (2-18)

A table of standard heats of formation of compounds from their pure elements is given in Appendix 3. In that table, the subscript 298 refers to the temperature (298 K), and the zero superscript signifies that reactants and products are all in their standard states. To illustrate how heats of general reactions are found from heats of formation, let us look again at the diamond synthesis, equation 2-15. That reaction can be obtained by adding equation 2-17 to twice equation 2-18 and the reverse of equation 2-16:

               C(gr) → C(di)                ΔH =   +1.9 kJ        (2-17)
      2 H₂(g) + O₂(g) → 2 H₂O(l)             ΔH = -571.6 kJ      2x(2-18)
               CH₄(g) → C(gr) + 2 H₂(g)      ΔH =  +74.8 kJ       -(2-16)
                                                            
      CH₄(g) + O₂(g) → C(di) + 2 H₂O(l)      ΔH = -494.9 kJ        (2-15)

The heat of reaction is found in exactly the same manner:

ΔH = (+1.9) + 2(-285.8) - (-74.8) = -494.9 kJ

As you can see, the consequences of not keeping track of signs and coefficients can be disastrous. The surest method is to write out each equation, with its heat of reaction, in such a way that the sum of the individual equations will be the desired reaction. If all the coefficients of an equation are multiplied by an arbitrary number n, then the heat of formation must also be multiplied by n, and if a formation equation is turned around and run in reverse, then the sign ΔH must be changed. If the individual equations add to give the desired reaction, then the individual heats add to give the corresponding overall heat of reaction.

A convenient shortcut is to think of the heat of formation of a compound as if it were, in a sense, the enthalpy of the compound itself. (Warning: This is possible only because the heats of formation of the elements are zero by definition.) Then the heat of a reaction becomes the sum of heats of formation of all the products, minus the heats of formation of all the reactants, each being multiplied by the coefficient of that substance in the balanced equation.

Example 25

What is the standard heat of the reaction by which ferric oxide is reduced by carbon to iron and carbon monoxide in a blast furnace?

Solution

The reaction is as follows, with the standard heat of formation per mole written below each compound:

	Fe₂O₃(s)	+	3 C(gr)	→	2 Fe(s)	+	3 CO(g)
ΔH (kJ/mole)	-822.1		0.0		0.0		-110.5

The standard heat of formation of the elements from themselves, of course is zero, by definition. For the reaction as written,

ΔH = 2(0.0) + 3(-110.5) - (-822.1) - 3(0.0) = +490.6 kJ

These results are consistent with the fact that much heat must be supplied to reduce iron ore to iron. Note, however, that 490.6 kJ is the net heat that would be absorbed if the reaction were run at 298 K, not the 1800 K of a blast furnace. Yet this calculated figure is also the heat absorbed if ferric oxide and carbon are heated from 298 K to 1800 K, allowed to react, and the products are cooled again to room temperature. The enthalpy change or heat of a reaction depends only on the initial and final states of the participants, and not on whether the temperature remained constant or went to blast-furnace levels in between. All that matters is that the temperature is brought back down to 298 K at the end.

As another example of the principle, the net heat evolved when water is made from hydrogen and oxygen will be the same whether a mixture of H2 and 02 at 298 K explodes violently and the resulting water is cooled back to 298 K, or whether the same mixture reacts slowly in the presence of finely divided platinum as a catalyst, never increasing its temperature. So, in referring to heats of reaction, when we say that the values are correct for the process carried out "at 1 atm pressure and 298 K," we require only that the reactants begin at these conditions, and that the products end there. This is why tables of heats of formation under standard conditions (Appendix 3) are useful.

Conservation Principles

References and Notes

The SI unit of length is the meter (m), divided into 10 decimeters (dm) or 100 centimeters (cm). The unit of volume is the cubic meter (m³). For laboratory work the cubic meter is too large to be convenient, so it is customary to use the liter, which in the SI is defined as 1 dm³, and the milliliter (mL), which equals 1 cm³ (or sometimes cc). By strict logic the liter is an extraneous unit in SI, but it is too convenient, and its use too deeply ingrained, to be eliminated. Scientists in the past tended to use milliliters for liquid volumes, and cubic centimeters for volumes of solids. Hence the volume of a sodium chloride solution would be measured in milliliters, but the density of rock salt (sodium chloride crystals) would be reported in grams per cubic centimeter, or g cm^-3. We shall use only milliters in this chapter, but thereafter shall feel free to use cubic centimeters wherever that unit seems more natural. Remember that 1 m³ = 1000 liters, 1 liter = 1000 mL, and 1 mL = 1 cm³. For more information on SI, see Appendix 1.
The conversion to the absolute or Kelvin temperature scale is considered in Chapter 3, as is the atmosphere as a unit of pressure. (By SI convention, no degree sign is used for the Kelvin scale.)
The partial pressure of a gas in a mixture is the pressure that the gas would show if all the other gases were removed and it were the only gas present.

Search

Text Color

Text Size

Margin Size

Font Type