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Learning Objectives
• To describe how excess amino acids are degraded.
The liver is the principal site of amino acid metabolism, but other tissues, such as the kidney, the small intestine, muscles, and adipose tissue, take part. Generally, the first step in the breakdown of amino acids is the separation of the amino group from the carbon skeleton, usually by a transamination reaction. The carbon skeletons resulting from the deaminated amino acids are used to form either glucose or fats, or they are converted to a metabolic intermediate that can be oxidized by the citric acid cycle. The latter alternative, amino acid catabolism, is more likely to occur when glucose levels are low—for example, when a person is fasting or starving.
Transamination
Transamination is an exchange of functional groups between any amino acid (except lysine, proline, and threonine) and an α-keto acid. The amino group is usually transferred to the keto carbon atom of pyruvate, oxaloacetate, or α-ketoglutarate, converting the α-keto acid to alanine, aspartate, or glutamate, respectively. Transamination reactions are catalyzed by specific transaminases (also called aminotransferases), which require pyridoxal phosphate as a coenzyme.
In an α-keto acid, the carbonyl or keto group is located on the carbon atom adjacent to the carboxyl group of the acid.
Oxidative Deamination
In the breakdown of amino acids for energy, the final acceptor of the α-amino group is α-ketoglutarate, forming glutamate. Glutamate can then undergooxidative deamination, in which it loses its amino group as an ammonium (NH4+) ion and is oxidized back to α-ketoglutarate (ready to accept another amino group):
This reaction occurs primarily in liver mitochondria. Most of the NH4+ ion formed by oxidative deamination of glutamate is converted to urea and excreted in the urine in a series of reactions known as the urea cycle.
The synthesis of glutamate occurs in animal cells by reversing the reaction catalyzed by glutamate dehydrogenase. For this reaction nicotinamide adenine dinucleotide phosphate (NADPH) acts as the reducing agent. The synthesis of glutamate is significant because it is one of the few reactions in animals that can incorporate inorganic nitrogen (NH4+) into an α-keto acid to form an amino acid. The amino group can then be passed on through transamination reactions, to produce other amino acids from the appropriate α-keto acids.
The Fate of the Carbon Skeleton
Any amino acid can be converted into an intermediate of the citric acid cycle. Once the amino group is removed, usually by transamination, the α-keto acid that remains is catabolized by a pathway unique to that acid and consisting of one or more reactions. For example, phenylalanine undergoes a series of six reactions before it splits into fumarate and acetoacetate. Fumarate is an intermediate in the citric acid cycle, while acetoacetate must be converted to acetoacetyl-coenzyme A (CoA) and then to acetyl-CoA before it enters the citric acid cycle.
Those amino acids that can form any of the intermediates of carbohydrate metabolism can subsequently be converted to glucose via a metabolic pathway known as gluconeogenesis. These amino acids are called glucogenic amino acids. Amino acids that are converted to acetoacetyl-CoA or acetyl-CoA, which can be used for the synthesis of ketone bodies but not glucose, are called ketogenic amino acids. Some amino acids fall into both categories. Leucine and lysine are the only amino acids that are exclusively ketogenic. Figure \(2\) summarizes the ultimate fates of the carbon skeletons of the 20 amino acids.
Career Focus: Exercise Physiologist
An exercise physiologist works with individuals who have or wish to prevent developing a wide variety of chronic diseases, such as diabetes, in which exercise has been shown to be beneficial. Each individual must be referred by a licensed physician. An exercise physiologist works in a variety of settings, such as a hospital or in a wellness program at a commercial business, to design and monitor individual exercise plans. A registered clinical exercise physiologist must have an undergraduate degree in exercise physiology or a related degree. Some job opportunities require a master’s degree in exercise physiology or a related degree.
Summary
Generally the first step in the breakdown of amino acids is the removal of the amino group, usually through a reaction known as transamination. The carbon skeletons of the amino acids undergo further reactions to form compounds that can either be used for the synthesis of glucose or the synthesis of ketone bodies. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.07%3A_Stage_II_of_Protein_Catabolism.txt |
Concept Review Exercise
1. Why is ATP referred to as the energy currency of the cell?
Answer
1. ATP is the principal molecule involved in energy exchange reactions in biological systems.
Exercises
1. How do ATP and ADP differ in structure?
2. Why does the hydrolysis of ATP to ADP involve the release of energy?
3. Identify whether each compound would be classified as a high-energy phosphate compound.
1. ATP
2. glucose 6-phosphate
3. creatine phosphate
4. Identify whether each compound would be classified as a high-energy phosphate compound.
1. ADP
2. AMP
3. glucose 1-phosphate
Answers
1. ATP has a triphosphate group attached, while ADP has only a diphosphate group attached.
1. yes
2. no
3. yes
Concept Review Exercises
1. Distinguish between each pair of compounds.
1. pepsin and pepsinogen
2. chymotrypsin and trypsin
3. aminopeptidase and carboxypeptidase
2. What are the primary end products of each form of digestion?
1. carbohydrate digestion
2. lipid digestion
3. protein digestion
3. In what section of the digestive tract does most of the carbohydrate, lipid, and protein digestion take place?
Answers
1. Pepsinogen is an inactive form of pepsin; pepsin is the active form of the enzyme.
2. Both enzymes catalyze the hydrolysis of peptide bonds. Chymotrypsin catalyzes the hydrolysis of peptide bonds following aromatic amino acids, while trypsin catalyzes the hydrolysis of peptide bonds following lysine and arginine.
3. Aminopeptidase catalyzes the hydrolysis of amino acids from the N-terminal end of a protein, while carboxypeptidase catalyzes the hydrolysis of amino acids from the C-terminal end of a protein.
1. glucose, fructose, and galactose
2. monoglycerides and fatty acids
3. amino acids
1. the small intestine
Exercises
1. What are the products of digestion (or stage I of catabolism)?
2. What is the general type of reaction used in digestion?
3. Give the site of action and the function of each enzyme.
1. chymotrypsin
2. lactase
3. pepsin
4. maltase
4. Give the site of action and the function of each enzyme.
1. α-amylase
2. trypsin
3. sucrase
4. aminopeptidase
1. What is the meaning of the following statement? “Bile salts act to emulsify lipids in the small intestine.”
2. Why is emulsification important?
5. Using chemical equations, describe the chemical changes that triglycerides undergo during digestion.
6. What are the expected products from the enzymatic action of chymotrypsin on each amino acid segment?
1. gly-ala-phe-thr-leu
2. ala-ile-tyr-ser-arg
3. val-trp-arg-leu-cys
7. What are the expected products from the enzymatic action of trypsin on each amino acid segment?
1. leu-thr-glu-lys-ala
2. phe-arg-ala-leu-val
3. ala-arg-glu-trp-lys
Answers
1. proteins: amino acids; carbohydrates: monosaccharides; fats: fatty acids and glycerol
1. Chymotrypsin is found in the small intestine and catalyzes the hydrolysis of peptide bonds following aromatic amino acids.
2. Lactase is found in the small intestine and catalyzes the hydrolysis of lactose.
3. Pepsin is found in the stomach and catalyzes the hydrolysis of peptide bonds, primarily those that occur after aromatic amino acids.
4. Maltase is found in the small intestine and catalyzes the hydrolysis of maltose.
1. Bile salts aid in digestion by dispersing lipids throughout the aqueous solution in the small intestine.
2. Emulsification is important because lipids are not soluble in water; it breaks lipids up into smaller particles that can be more readily hydrolyzed by lipases.
1. gly-ala-phe and thr-leu
2. ala-ile-tyr and ser-arg
3. val-trp and arg-leu-cys
Concept Review Exercises
1. What is a metabolic pathway?
2. What vitamin is required to make coenzyme A?
3. What is the net yield of Glycolysis as far as ATP?
4. Name the enzymes that are key regulatory sites in Glycolysis.
5. Why are the enzymes in the previous question targets for regulation?
6. Why is the priming phase necessary?
7. Draw the entire pathway for glycolysis including enzymes, reactants and products for each step.
8. Where does beta-oxidation occur?
9. What is the average net yield of ATP per carbon?
10. Where exactly is water formed during the process of fatty acid degradation? (Hint: H2O is formed when when the one of the products of beta-oxidation is passed through another of the metabolic pathways)
11. During the process of beta-oxidation, why is it that [FAD] is used to oxidize an alkane to an alkene while NAD+ is used to oxidize an alchol to a carbonyl
12. Draw out the entire process of the degradation of a triglyceride, include enzymes and products and reactants for each step.
Answers
1. A metabolic pathway is a series of biochemical reactions by which an organism converts a given reactant to a specific end product.
2. pantothenic acid
Concept Review Exercises
1. What is the main function of the citric acid cycle?
2. Two carbon atoms are fed into the citric acid cycle as acetyl-CoA. In what form are two carbon atoms removed from the cycle?
3. What are mitochondria and what is their function in the cell?
Answers
1. the complete oxidation of carbon atoms to carbon dioxide and the formation of a high-energy phosphate compound, energy rich reduced coenzymes (NADH and FADH2), and metabolic intermediates for the synthesis of other compounds
2. as carbon dioxide
3. Mitochondria are small organelles with a double membrane that contain the enzymes and other molecules needed for the production of most of the ATP needed by the body.
Exercises
1. Replace each question mark with the correct compound.
1. $\mathrm{?\xrightarrow{aconitase}isocitrate}$
2. $\mathrm{?\, +\, ? \xrightarrow{citrate\: synthase} citrate + coenzyme\: A}$
3. $\mathrm{fumarate \xrightarrow{fumarase}\, ?}$
4. $\mathrm{isocitrate + NAD^+ \xrightarrow{?} \alpha\textrm{-ketoglurate} + NADH + CO_2}$
2. Replace each question mark with the correct compound.
1. $\mathrm{malate + NAD^+ \xrightarrow{?} oxaloacetate + NADH}$
2. $\mathrm{?\, +\, ? \xrightarrow{nucleoside\: diphosphokinase} GDP + ATP}$
3. $\mathrm{\textrm{succinyl-CoA} \xrightarrow{\textrm{succinyl-CoA synthetase}} \,?\, +\, ?}$
4. $\mathrm{succinate + FAD \xrightarrow{succinate\: dehydrogenase}\, ? + FADH_2}$
3. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. isomerization
2. hydration
3. synthesis
4. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. oxidation
2. decarboxylation
3. phosphorylation
5. What similar role do coenzyme Q and cytochrome c serve in the electron transport chain?
6. What is the electron acceptor at the end of the electron transport chain? To what product is this compound reduced?
7. What is the function of the cytochromes in the electron transport chain?
1. What is meant by this statement? “Electron transport is tightly coupled to oxidative phosphorylation.”
2. How are electron transport and oxidative phosphorylation coupled or linked?
Answers
1. citrate
2. oxaloacetate + acetyl-CoA
3. malate
4. α-ketoglutarate hydrogenase complex
1. reaction in 1a
2. reaction in 1c
3. reaction in 1b
1. Both molecules serve as electron shuttles between the complexes of the electron transport chain.
1. Cytochromes are proteins in the electron transport chain and serve as one-electron carriers.
• Concept Review Exercises
1. In glycolysis, how many molecules of pyruvate are produced from one molecule of glucose?
2. In vertebrates, what happens to pyruvate when
1. plenty of oxygen is available?
2. oxygen supplies are limited?
3. In anaerobic glycolysis, how many molecules of ATP are produced from one molecule of glucose?
Answers
1. two
1. Pyruvate is completely oxidized to carbon dioxide.
2. Pyruvate is reduced to lactate, allowing for the reoxidation of NADH to NAD+.
2. There is a net production of two molecules of ATP.
Exercises
1. Replace each question mark with the correct compound.
1. $\mathrm{fructose\: 1,6\textrm{-bisphosphate} \xrightarrow{aldolase}\, ?\, +\, ?}$
2. $\mathrm{? + ADP \xrightarrow{pyruvate\: kinase} pyruvate + ATP}$
3. $\mathrm{dihydroxyacetone\: phosphate \xrightarrow{?} glyceraldehyde\: 3\textrm{-phosphate}}$
4. $\mathrm{glucose + ATP \xrightarrow{hexokinase} \, ? + ADP}$
2. Replace each question mark with the correct compound.
1. $\mathrm{fructose\: 6\textrm{-phosphate} + ATP \xrightarrow{?} fructose\: 1,6\textrm{-bisphosphate} + ADP}$
2. $\mathrm{? \xrightarrow{phosphoglucose\: isomerase} fructose\: 6\textrm{-phosphate}}$
3. $\mathrm{glyceraldehyde\: 3\textrm{-phosphate} + NAD^+ + P_i \xrightarrow{?} 1,3\textrm{-bisphosphoglycerate} + NADH}$
4. $\mathrm{3\textrm{-phosphoglycerate} \xrightarrow{phosphoglyceromutase} \, ?}$
3. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. hydrolysis of a high-energy phosphate compound
2. synthesis of ATP
4. From the reactions in Exercises 1 and 2, select the equation(s) by number and letter in which each type of reaction occurs.
1. isomerization
2. oxidation
5. What coenzyme is needed as an oxidizing agent in glycolysis?
6. Calculate
1. the total number of molecules of ATP produced for each molecule of glucose converted to pyruvate in glycolysis.
2. the number of molecules of ATP hydrolyzed in phase I of glycolysis.
3. the net ATP production from glycolysis alone.
7. How is the NADH produced in glycolysis reoxidized when oxygen supplies are limited in
1. muscle cells?
2. yeast?
1. Calculate the number of moles of ATP produced by the aerobic oxidation of 1 mol of glucose in a liver cell.
2. Of the total calculated in Exercise 9a, determine the number of moles of ATP produced in each process.
1. glycolysis alone
2. the citric acid cycle
3. the electron transport chain and oxidative phosphorylation
Answers
1. glyceraldehyde 3-phosphate + dihydroxyacetone phosphate
2. phosphoenolpyruvate
3. triose phosphate isomerase
4. glucose 6-phosphate
1. reactions 1b, 1d, and 2a
2. reaction 1b
1. NAD+
1. Pyruvate is reduced to lactate, and NADH is reoxidized to NAD+.
2. Pyruvate is converted to ethanol and carbon dioxide, and NADH is reoxidized to NAD+.
• Concept Review Exercises
1. How are fatty acids activated prior to being transported into the mitochondria and oxidized?
2. Draw the structure of hexanoic (caproic) acid and identify the α-carbon and the β-carbon.
Answers
1. They react with CoA to form fatty acyl-CoA molecules.
Key Takeaways
• Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of reactions known as β-oxidation.
• In each round of β-oxidation, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH2 are produced.
• The acetyl-CoA, NADH, and FADH2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP.
Exercises
1. For each reaction found in β-oxidation, identify the enzyme that catalyzes the reaction and classify the reaction as oxidation-reduction, hydration, or cleavage.
2. What are the products of β-oxidation?
3. How many rounds of β-oxidation are necessary to metabolize lauric acid (a saturated fatty acid with 12 carbon atoms)?
4. How many rounds of β-oxidation are necessary to metabolize arachidic acid (a saturated fatty acid with 20 carbon atoms)?
5. When myristic acid (a saturated fatty acid with 14 carbon atoms) is completely oxidized by β-oxidation, how many molecules of each are formed?
1. acetyl-CoA
2. FADH2
3. NADH
6. When stearic acid (a saturated fatty acid with 18 carbon atoms) is completely oxidized by β-oxidation, how many molecules of each are formed?
1. acetyl-CoA
2. FADH2
3. NADH
7. What is the net yield of ATP from the complete oxidation, in a liver cell, of one molecule of myristic acid?
8. What is the net yield of ATP from the complete oxidation, in a liver cell, of one molecule of stearic acid?
Answers
1. enoyl-CoA hydratase; hydration
2. thiolase; cleavage
3. acyl-CoA dehydrogenase; oxidation-reduction
1. five rounds
1. 7 molecules
2. 6 molecules
3. 6 molecules
1. 112 molecules
20.7: Stage II of Protein Catabolism
Concept Review Exercises
1. Write the equation for the transamination reaction between alanine and oxaloacetate.
2. Name the two products that are formed.
1. What is the purpose of oxidative deamination?
Answers
1. pyruvate and aspartate
1. Oxidative deamination provides a reaction in which the amino group [as the ammonium (NH4+) ion] is removed from a molecule, not simply transferred from one molecule to another. Most of the NH4+ ion is converted to urea and excreted from the body.
Exercises
1. Write the equation for the transamination reaction between valine and pyruvate.
2. Write the equation for the transamination reaction between phenylalanine and oxaloacetate.
3. What products are formed in the oxidative deamination of glutamate?
4. Determine if each amino acid is glucogenic, ketogenic, or both.
1. phenylalanine
2. leucine
3. serine
5. Determine if each amino acid is glucogenic, ketogenic, or both.
1. asparagine
2. tyrosine
3. valine
Answers
1. α-ketoglutarate, NADH, and NH4+
1. glucogenic
2. both
3. glucogenic
Additional Exercises
1. Hydrolysis of which compound—arginine phosphate or glucose 6-phosphate—would provide enough energy for the phosphorylation of ATP? Why?
2. If a cracker, which is rich in starch, is chewed for a long time, it begins to develop a sweet, sugary taste. Why?
3. Indicate where each enzymes would cleave the short peptide ala-ser-met-val-phe-gly-cys-lys-asp-leu.
1. aminopeptidase
2. chymotrypsin
4. Indicate where each enzymes would cleave the short peptide ala-ser-met-val-phe-gly-cys-lys-asp-leu.
1. trypsin
2. carboxypeptidase
5. If the methyl carbon atom of acetyl-CoA is labeled, where does the label appear after the acetyl-CoA goes through one round of the citric acid cycle?
6. If the carbonyl carbon atom of acetyl-CoA is labeled, where does the label appear after the acetyl-CoA goes through one round of the citric acid cycle?
7. The average adult consumes about 65 g of fructose daily (either as the free sugar or from the breakdown of sucrose). In the liver, fructose is first phosphorylated to fructose 1-phosphate, which is then split into dihydroxyacetone phosphate and glyceraldehyde. Glyceraldehyde is then phosphorylated to glyceraldehyde 3-phosphate, with ATP as the phosphate group donor. Write the equations (using structural formulas) for these three steps. Indicate the type of enzyme that catalyzes each step.
8. What critical role is played by both BPG and PEP in glycolysis?
9. How is the NADH produced in glycolysis reoxidized when oxygen supplies are abundant?
10. When a triglyceride is hydrolyzed to form three fatty acids and glycerol, the glycerol can be converted to glycerol 3-phosphate and then oxidized to form dihydroxyacetone phosphate, an intermediate of glycolysis. (In this reaction, NAD+ is reduced to NADH.) If you assume that there is sufficient oxygen to completely oxidize the pyruvate formed from dihydroxyacetone phosphate, what is the maximum amount of ATP formed from the complete oxidation of 1 mol of glycerol?
11. How is the FADH2 from β-oxidation converted back to FAD?
12. If 1 mol of alanine is converted to pyruvate in a muscle cell (through transamination) and the pyruvate is then metabolized via the citric acid cycle, the electron transport chain, and oxidative phosphorylation, how many moles of ATP are produced?
13. If the essential amino acid leucine (2-amino-4-methylpentanoic acid) is lacking in the diet, an α-keto acid can substitute for it. Give the structure of the α-keto acid and the probable reaction used to form leucine from this α-keto acid.
Answers
1. The hydrolysis of arginine phosphate releases more energy than is needed for the synthesis of ATP, while hydrolysis of glucose 6-phosphate does not.
1.
1. The enzyme will cleave off amino acids one at a time beginning with alanine (the N-terminal end).
2. following phenylalanine
1. Half of the label will be on the second carbon atom of oxaloacetate, while the other half will be on the third carbon atom.
1. When oxygen is abundant, NADH is reoxidized through the reactions of the electron transport chain.
1. FADH2 is reoxidized back to FAD via the electron transport chain. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.E%3A_Energy_Metabolism_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Metabolism is the general term for all chemical reactions in living organisms. The two types of metabolism are catabolism—those reactions in which complex molecules (carbohydrates, lipids, and proteins) are broken down to simpler ones with the concomitant release of energy—and anabolism—those reactions that consume energy to build complex molecules. Metabolism is studied by looking at individual metabolic pathways, which are a series of biochemical reactions in which a given reactant is converted to a desired end product.
The oxidation of fuel molecules (primarily carbohydrates and lipids), a process called respiration, is the source of energy used by cells. Catabolic reactions release energy from food molecules and use some of that energy for the synthesis of adenosine triphosphate (ATP); anabolic reactions use the energy in ATP to create new compounds. Catabolism can be divided into three stages. In stage I, carbohydrates, lipids, and proteins are broken down into their individual monomer units—simple sugars, fatty acids, and amino acids, respectively. In stage II, these monomer units are broken down by specific metabolic pathways to form a common end product acetyl-coenzyme A (CoA). In stage III, acetyl-CoA is completely oxidized to form carbon dioxide and water, and ATP is produced.
The digestion of carbohydrates begins in the mouth as α-amylase breaks glycosidic linkages in carbohydrate molecules. Essentially no carbohydrate digestion occurs in the stomach, and food particles pass through to the small intestine, where α-amylase and intestinal enzymes convert complex carbohydrate molecules (starches) to monosaccharides. The monosaccharides then pass through the lining of the small intestine and into the bloodstream for transport to all body cells.
Protein digestion begins in the stomach as pepsinogen in gastric juice is converted to pepsin, the enzyme that hydrolyzes peptide bonds. The partially digested protein then passes to the small intestine, where the remainder of protein digestion takes place through the action of several enzymes. The resulting amino acids cross the intestinal wall into the blood and are carried to the liver.
Lipid digestion begins in the small intestine. Bile salts emulsify the lipid molecules, and then lipases hydrolyze them to fatty acids and monoglycerides. The hydrolysis products pass through the intestine and are repackaged for transport in the bloodstream.
In cells that are operating aerobically, acetyl-CoA produced in stage II of catabolism is oxidized to carbon dioxide. The citric acid cycle describes this oxidation, which takes place with the formation of the coenzymes reduced nicotinamide adenine dinucleotide (NADH) and reduced flavin adenine dinucleotide (FADH2). The sequence of reactions needed to oxidize these coenzymes and transfer the resulting electrons to oxygen is called the electron transport chain, or the respiratory chain. The compounds responsible for this series of oxidation-reduction reactions include proteins known as cytochromes, Fe·S proteins, and other molecules that ultimately result in the reduction of molecular oxygen to water. Every time a compound with two carbon atoms is oxidized in the citric acid cycle, a respiratory chain compound accepts the electrons lost in the oxidation (and so is reduced) and then passes them on to the next metabolite in the chain. The energy released by the electron transport chain is used to transport hydrogen (H+) ions from the mitochondrial matrix to the intermembrane space. The flow of H+ back through ATP synthase leads to the synthesis and release of ATP from adenosine diphosphate (ADP) and inorganic phosphate ions (Pi) in a process known as oxidative phosphorylation. Electron transport and oxidative phosphorylation are tightly coupled to each other. The enzymes and intermediates of the citric acid cycle, the electron transport chain, and oxidative phosphorylation are located in organelles called mitochondria.
The oxidation of carbohydrates is the source of over 50% of the energy used by cells. Glucose is oxidized to two molecules of pyruvate through a series of reactions known as glycolysis. Some of the energy released in these reactions is conserved by the formation of ATP from ADP. Glycolysis can be divided into two phases: phase I consists of the first five reactions and requires energy to “prime” the glucose molecule for phase II, the last five reactions in which ATP is produced through substrate-level phosphorylation.
The pyruvate produced by glycolysis has several possible fates, depending on the organism and whether or not oxygen is present. In animal cells, pyruvate can be further oxidized to acetyl-CoA and then to carbon dioxide (through the citric acid cycle) if oxygen supplies are sufficient. When oxygen supplies are insufficient, pyruvate is reduced to lactate. In yeast and other microorganisms, pyruvate is not converted to lactate in the absence of oxygen but instead is converted to ethanol and carbon dioxide.
The amount of ATP formed by the oxidation of glucose depends on whether or not oxygen is present. If oxygen is present, glucose is oxidized to carbon dioxide, and 36–38 ATP molecules are produced for each glucose molecule oxidized, using the combined pathways of glycolysis, the citric acid cycle, the electron transport chain, and oxidative phosphorylation. Thus, approximately 42% of the energy released by the complete oxidation of glucose is conserved by the synthesis of ATP. In the absence of oxygen, only 2 molecules of ATP are formed for each molecule of glucose converted to lactate (2 molecules), and the amount of energy conserved is much less (2%).
Fatty acids, released by the degradation of triglycerides and other lipids, are converted to fatty acyl-CoA, transported into the mitochondria, and oxidized by repeated cycling through a sequence of four reactions known as β-oxidation. In each round of β-oxidation, the fatty acyl-CoA is shortened by two carbon atoms as one molecule of acetyl-CoA is formed. The final round of β-oxidation, once the chain has been shortened to four carbon atoms, forms two molecules of acetyl-CoA. β-oxidation also forms the reduced coenzymes FADH2 and NADH, whose reoxidation through the electron transport chain and oxidative phosphorylation leads to the synthesis of ATP. The efficiency of fatty acid oxidation in the human body is approximately 41%.
Amino acids from the breakdown of proteins can be catabolized to provide energy. Amino acids whose carbon skeletons are converted to intermediates that can be converted to glucose through gluconeogenesis are known as glucogenic amino acids. Amino acids whose carbon skeletons are broken down to compounds used to form ketone bodies are known as ketogenic amino acids.
The first step in amino acid catabolism is separation of the amino group from the carbon skeleton. In a transamination, the amino acid gives its NH2 to pyruvate, α-ketoglutarate, or oxaloacetate. The products of this reaction are a new amino acid and an α-keto acid containing the carbon skeleton of the original amino acid. Pyruvate is transaminated to alanine, α-ketoglutarate to glutamate, and oxaloacetate to aspartate. The amino groups used to form alanine and aspartate are ultimately transferred to α-ketoglutarate, forming glutamate. The glutamate then undergoes oxidative deamination to yield α-ketoglutarate and ammonia.
Template:HideOrg | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.S%3A_Energy_Metabolism_%28Summary%29.txt |
What is chemistry? Simply put, chemistry is the study of the interactions of matter with other matter and with energy. This seems straightforward enough. However, the definition of chemistry includes a wide range of topics that must be understood to gain a mastery of the topic or even take additional courses in chemistry. In this book, we will lay the foundations of chemistry in a topic-by-topic fashion to provide you with the background you need to successfully understand chemistry.
• 1.1: Prelude to Chemistry
Get ready for a fantastic journey through a world of wonder, delight, and knowledge. One of the themes of this book is "chemistry is everywhere," and indeed it is. You would not be alive if it were not for chemistry, because your body is a big chemical machine. If you do not believe it, do not worry. Every chapter in this book contains examples that will show you how chemistry is, in fact, everywhere. So enjoy the ride, and enjoy chemistry.
• 1.2: Basic Definitions
Chemistry is the study of matter and its interactions with other matter and energy. Matter is anything that has mass and takes up space. Matter can be described in terms of physical properties and chemical properties. Physical properties and chemical properties of matter can change. Matter is composed of elements and compounds. Combinations of different substances are called mixtures. Elements can be described as metals, nonmetals, and semimetals.
• 1.3: Chemistry as a Science
Science is a process of knowing about the natural universe through observation and experiment. Scientists go through a rigorous process to determine new knowledge about the universe; this process is generally referred to as the scientific method. Science is broken down into various fields, of which chemistry is one. Science, including chemistry, is both qualitative and quantitative.
• 1.E: What Is Chemistry? (Exercises)
These are exercises and select solutions to accompany the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
01: What Is Chemistry
If you are reading these words, you are likely starting a chemistry course. Get ready for a fantastic journey through a world of wonder, delight, and knowledge. One of the themes of this book is "chemistry is everywhere," and indeed it is; you would not be alive if it were not for chemistry, because your body is a big chemical machine.
If you do not believe it, do not worry. Every chapter in this book contains examples that will show you how chemistry is, in fact, everywhere. So enjoy the ride, and enjoy chemistry.
1.02: Basic Definitions
Learning Objective
• Learn the basic terms used to describe matter
The definition of chemistry—the study of the interactions of matter with other matter and with energy—uses some terms that should also be defined. We start the study of chemistry by defining basic terms.
Matter
Matter is anything that has mass and takes up space. A book is matter, a computer is matter, food is matter, and dirt in the ground is matter. Sometimes matter may be difficult to identify. For example, air is matter, but because it is so thin compared to other matter (e.g., a book, a computer, food, and dirt), we sometimes forget that air has mass and takes up space. Things that are not matter include thoughts, ideas, emotions, and hopes.
Example \(1\)
Which of the following is matter and not matter?
1. a hot dog
2. love
3. a tree
Solution
1. A hot dog has mass and takes up space, so it is matter.
2. Love is an emotion, and emotions are not matter.
3. A tree has mass and takes up space, so it is matter.
Exercise \(1\)
Which of the following is matter and not matter?
1. the moon
2. an idea for a new invention
Answer a
The moon is matter.
Answer b
The invention itself may be matter, but the idea for it is not.
To understand matter and how it changes, we need to be able to describe matter. There are two basic ways to describe matter: physical properties and chemical properties.
Physical properties
Physical properties are characteristics that describe matter as it exists. Some physical characteristics of matter are shape, color, size, and temperature. An important physical property is the phase (or state) of matter. The three fundamental phases of matter are solid, liquid, and gas (Figure \(1\)).
Chemical Properties
Chemical properties are characteristics of matter that describe how matter changes form in the presence of other matter. Does a sample of matter burn? Burning is a chemical property. Does it behave violently when put in water? This reaction is a chemical property as well (Figure \(2\)). In the following chapters, we will see how descriptions of physical and chemical properties are important aspects of chemistry.
Physical Change
A physical change occurs when a sample of matter changes one or more of its physical properties. For example, a solid may melt (Figure \(3\)), or alcohol in a thermometer may change volume as the temperature changes. A physical change does not affect the chemical composition of matter.
Chemical Change
Chemical change is the process of demonstrating a chemical property, such as the burning match in Figure \(2\) "Chemical Properties". As the matter in the match burns, its chemical composition changes, and new forms of matter with new physical properties are created. Note that chemical changes are frequently accompanied by physical changes, as the new matter will likely have different physical properties from the original matter.
Example \(2\)
Describe each process as a physical change or a chemical change.
1. Water in the air turns into snow.
2. A person's hair is cut.
3. Bread dough becomes fresh bread in an oven.
Solution
1. Because the water is going from a gas phase to a solid phase, this is a physical change.
2. Your long hair is being shortened. This is a physical change.
3. Because of the oven's temperature, chemical changes are occurring in the bread dough to make fresh bread. These are chemical changes. (In fact, a lot of cooking involves chemical changes.)
Exercise \(2\)
Identify each process as a physical change or a chemical change.
1. A fire is raging in a fireplace.
2. Water is warmed to make a cup of coffee.
Answer a
chemical change
Answer b
physical change
Substance
A sample of matter that has the same physical and chemical properties throughout is called a substance. Sometimes the phrase pure substance is used, but the word pure isn't needed. The definition of the term substance is an example of how chemistry has a specific definition for a word that is used in everyday language with a different, vaguer definition. Here, we will use the term substance with its strict chemical definition.
Chemistry recognizes two different types of substances: elements and compounds.
Element
An element is the simplest type of chemical substance; it cannot be broken down into simpler chemical substances by ordinary chemical means. There are 118 elements known to science, of which 80 are stable. (The other elements are radioactive, a condition we will consider in Chapter 15.) Each element has its own unique set of physical and chemical properties. Examples of elements include iron, carbon, and gold.
Compound
A compound is a combination of more than one element. The physical and chemical properties of a compound are different from the physical and chemical properties of its constituent elements; that is, it behaves as a completely different substance. There are over 50 million compounds known, and more are being discovered daily. Examples of compounds include water, penicillin, and sodium chloride (the chemical name for common table salt).
Mixtures
Physical combinations of more than one substance are called mixtures. Elements and compounds are not the only ways in which matter can be present. We frequently encounter objects that are physical combinations of more than one element or compound—mixtures. There are two types of mixtures.
Heterogeneous Mixture
A heterogeneous mixture is a mixture composed of two or more substances. It is easy to tell, sometimes by the naked eye, that more than one substance is present.
Homogeneous Mixture/ Solution
A homogeneous mixture is a combination of two or more substances that is so intimately mixed, that the mixture behaves as a single substance. Another word for a homogeneous mixture is a solution. Thus, a combination of salt and steel wool is a heterogeneous mixture because it is easy to see which particles of the matter are salt crystals and which are steel wool. On the other hand, if you take salt crystals and dissolve them in water, it is very difficult to tell that you have more than one substance present just by looking—even if you use a powerful microscope. The salt dissolved in water is a homogeneous mixture, or a solution (Figure \(3\)).
Example \(3\)
Identify the following combinations as heterogeneous mixtures or homogenous mixtures.
1. soda water (carbon dioxide is dissolved in water)
2. a mixture of iron metal filings and sulfur powder (both iron and sulfur are elements)
Solution
1. Because carbon dioxide is dissolved in water, we can infer from the behavior of salt crystals dissolved in water that carbon dioxide dissolved in water is (also) a homogeneous mixture.
2. Assuming that the iron and sulfur are simply mixed together, it should be easy to see what is iron and what is sulfur, so this is a heterogeneous mixture.
Exercise \(3\)
1. the human body
2. an amalgam, a combination of some other metals dissolved in a small amount of mercury
Answer a
heterogeneous mixture
Answer b
homogeneous mixture
There are other descriptors that we can use to describe matter, especially elements. We can usually divide elements into metals and nonmetals, and each set shares certain (but not always all) properties.
Metal
A metal is an element that conducts electricity and heat well and is shiny, silvery, solid, ductile, and malleable. At room temperature, metals are solid (although mercury is a well-known exception). A metal is ductile because it can be drawn into thin wires (a property called ductility); and malleable because it can be pounded into thin sheets (a property called malleability).
Nonmetal
A non-metal is an element that is brittle when solid, and does not conduct electricity or heat very well. Non-metals cannot be made into thin sheets or wires (Figure \(4\)). Nonmetals also exist in a variety of phases and colors at room temperature.
Semi-metals
Some elements have properties of both metals and nonmetals and are called semi-metals (or metalloids). We will see later how these descriptions can be assigned rather easily to various elements.
Describing Matter Flowchart
"Describing Matter" is a flowchart of the relationships among the different ways of describing matter.
Example \(1\): Chemistry is Everywhere: In the Morning
Most people have a morning ritual, a process that they go through every morning to get ready for the day. Chemistry appears in many of these activities.
• If you take a shower or bath in the morning, you probably use soap, shampoo, or both. These items contain chemicals that interact with the oil and dirt on your body and hair to remove them and wash them away. Many of these products also contain chemicals that make you smell good; they are called fragrances.
• When you brush your teeth in the morning, you usually use toothpaste, a form of soap, to clean your teeth. Toothpastes typically contain tiny, hard particles called abrasives that physically scrub your teeth. Many toothpastes also contain fluoride, a substance that chemically interacts with the surface of the teeth to help prevent cavities.
• Perhaps you take vitamins, supplements, or medicines every morning. Vitamins and other supplements contain chemicals your body needs in small amounts to function properly. Medicines are chemicals that help combat diseases and promote health.
• Perhaps you make some fried eggs for breakfast. Frying eggs involves heating them enough so that a chemical reaction occurs to cook the eggs.
• After you eat, the food in your stomach is chemically reacted so that the body (mostly the intestines) can absorb food, water, and other nutrients.
• If you drive or take the bus to school or work, you are using a vehicle that probably burns gasoline, a material that burns fairly easily and provides energy to power the vehicle. Recall that burning is a chemical change.
These are just a few examples of how chemistry impacts your everyday life. And we haven't even made it to lunch yet!
Key Takeaways
• Chemistry is the study of matter and its interactions with other matter and energy.
• Matter is anything that has mass and takes up space.
• Matter can be described in terms of physical properties and chemical properties.
• Physical properties and chemical properties of matter can change.
• Matter is composed of elements and compounds.
• Combinations of different substances are called mixtures.
• Elements can be described as metals, nonmetals, and semi-metals. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/01%3A_What_Is_Chemistry/1.01%3A_Prelude_to_Chemistry.txt |
Learning Objective
• Learn what science is and how it works.
Chemistry is a branch of science. Although science itself is difficult to define exactly, the following definition can serve as a starting point. Science is the process of knowing about the natural universe through observation and experiment. Science is not the only process of knowing (e.g., the ancient Greeks simply sat and thought), but it has evolved over more than 350 years into the best process that humanity has devised, to date, to learn about the universe around us.
The process of science is usually stated as the scientific method, which is rather naively described as follows:
1. state a hypothesis,
2. test the hypothesis, and
3. refine the hypothesis
In actuality, the process is not that simple. (For example, a scientist does not go into their lab every day and exclaim, "I am going to state a hypothesis today and spend the day testing it!") The process is not that simple because science and scientists have a body of knowledge that has already been identified as coming from the highest level of understanding, and most scientists build from that body of knowledge.
An educated guess about how the natural universe works is called a hypothesis. A scientist who is familiar with how part of the natural universe works—say, a chemist—is interested in furthering that knowledge. That person makes a reasonable guess—a hypothesis—that is designed to see if the universe works in a new way as well. Here's an example of a hypothesis: "if I mix one part of hydrogen with one part of oxygen, I can make a substance that contains both elements."
For a hypothesis to be termed a scientific hypothesis, it has to be something that can be supported or refuted through carefully crafted experimentation or observation.
Most good hypotheses are grounded in previously understood knowledge and represent a testable extension of that knowledge. The scientist then devises ways to test if that hypothesis is correct or not. That is, the scientist plans experiments. Experiments are tests of the natural universe to see if a guess (hypothesis) is correct. An experiment to test our previous hypothesis would be to actually mix hydrogen and oxygen and see what happens. Most experiments include observations of small, well-defined parts of the natural universe designed to see results of the experiments.
A Scientific Hypothesis
A hypothesis is often written in the form of an if/then statement that gives a possibility (if) and explains what may happen because of the possibility (then). For example, if eating elemental sulfur repels ticks, then someone that is eating sulfur every day will not get ticks.
Why do we have to do experiments? Why do we have to test? Because the natural universe is not always so obvious, experiments are necessary. For example, it is fairly obvious that if you drop an object from a height, it will fall. Several hundred years ago (coincidentally, near the inception of modern science), the concept of gravity explained that test. However, is it obvious that the entire natural universe is composed of only about 115 fundamental chemical building blocks called elements? This wouldn't seem true if you looked at the world around you and saw all the different forms matter can take. In fact, the concept of the element is only about 200 years old, and the last naturally occurring element was identified about 80 years ago. It took decades of tests and millions of experiments to establish what the elements actually are. These are just two examples; a myriad of such examples exists in chemistry and science in general.
When enough evidence has been collected to establish a general principle of how the natural universe works, the evidence is summarized in a theory. A theory is a general statement that explains a large number of observations. "All matter is composed of atoms" is a general statement, a theory, that explains many observations in chemistry. A theory is a very powerful statement in science. There are many statements referred to as "the theory of _______" or the "______ theory" in science (where the blanks represent a word or concept). When written in this way, theories indicate that science has an overwhelming amount of evidence of its correctness. We will see several theories in the course of this text.
A specific statement that is thought to never be violated by the entire natural universe is called a law. A scientific law is the highest understanding of the natural universe that science has and is thought to be inviolate. The fact that all matter attracts all other matter—the law of gravitation—is one such law. Note that the terms theory and law used in science have slightly different meanings from those in common usage; where theory is often used to mean hypothesis ("I have a theory…"), and a law is an arbitrary limitation that can be broken but with potential consequences (such as speed limits). Here again, science uses these terms differently, and it is important to apply their proper definitions when you use these words in science. (Figure \(1\))
There is an additional phrase in our definition of science: "the natural universe." Science is concerned only with the natural universe. What is the natural universe? It's anything that occurs around us, well, naturally. Stars, planets, the appearance of life on earth; as well as how animals, plants, and other matter function are all part of the natural universe. Science is concerned with that—and only that.
Of course, there are other things that concern us. For example, is the English language part of science? Most of us can easily answer no; English is not science. English is certainly worth knowing (at least for people in predominantly English-speaking countries), but why isn't it science? English, or any human language, is not science because ultimately it is contrived; it is made up. Think of it: the word spelled b-l-u-e represents a certain color, and we all agree what color that is. But what if we used the word h-a-r-d to describe that color? (Figure \(2\)) That would be fine—as long as everyone agreed. Anyone who has learned a second language must initially wonder why a certain word is used to describe a certain concept; ultimately, the speakers of that language agreed that a particular word would represent a particular concept. It was contrived.
That doesn't mean language isn't worth knowing. It is very important in society. But it's not science. Science deals only with what occurs naturally.
Example \(1\): Identifying Science
Which of the following fields would be considered science?
1. geology, the study of the earth
2. ethics, the study of morality
3. political science, the study of governance
4. biology, the study of living organisms
Solution
1. Because the earth is a natural object, the study of it is indeed considered part of science.
2. Ethics is a branch of philosophy that deals with right and wrong. Although these are useful concepts, they are not science.
3. There are many forms of government, but all are created by humans. Despite the fact that the word science appears in its name, political science is not true science.
4. Living organisms are part of the natural universe, so the study of them is part of science.
Exercise \(1\)
Which is part of science, and which is not?
1. dynamics, the study of systems that change over time
2. aesthetics, the concept of beauty
Answer A
science
Answer B
not science
The field of science has gotten so big that it is common to separate it into more specific fields. First, there is mathematics, the language of science. All scientific fields use mathematics to express themselves—some more than others. Physics and astronomy are scientific fields concerned with the fundamental interactions between matter and energy. Chemistry, as defined previously, is the study of the interactions of matter with other matter and with energy. Biology is the study of living organisms, while geology is the study of the earth. Other sciences can be named as well. Understand that these fields are not always completely separate; the boundaries between scientific fields are not always readily apparent. A scientist may be labeled a biochemist if he or she studies the chemistry of biological organisms.
Finally, understand that science can be either qualitative or quantitative. Qualitative implies a description of the quality of an object. For example, physical properties are generally qualitative descriptions: sulfur is yellow, your math book is heavy, or that statue is pretty. A quantitative description represents the specific amount of something; it means knowing how much of something is present, usually by counting or measuring it. Some quantitative descriptions include: 25 students in a class, 650 pages in a book, or a velocity of 66 miles per hour. Quantitative expressions are very important in science; they are also very important in chemistry.
Example \(2\): qualitative vs. quantitative Descriptions
Identify each statement as either a qualitative description or a quantitative description.
1. Gold metal is yellow.
2. A ream of paper has 500 sheets in it.
3. The weather outside is snowy.
4. The temperature outside is 24 degrees Fahrenheit.
Solution
1. Because we are describing a physical property of gold, this statement is qualitative.
2. This statement mentions a specific amount, so it is quantitative.
3. The word snowy is a description of how the day is; therefore, it is a qualitative statement.
4. In this case, the weather is described with a specific quantity—the temperature. Therefore, it is quantitative.
Exercise \(2\)
Are these qualitative or quantitative statements?
1. Roses are red, and violets are blue.
2. Four score and seven years ago….
Answer A
qualitative
Answer B
quantitative
Food and Drink Application: Carbonated Beverages
Some of the simple chemical principles discussed in this chapter can be illustrated with carbonated beverages: sodas, beer, and sparkling wines. Each product is produced in a different way, but they all have one thing in common: they are solutions of carbon dioxide dissolved in water.
Carbon dioxide is a compound composed of carbon and oxygen. Under normal conditions, it is a gas. If you cool it down enough, it becomes a solid known as dry ice. Carbon dioxide is an important compound in the cycle of life on earth.
Even though it is a gas, carbon dioxide can dissolve in water, just like sugar or salt can dissolve in water. When that occurs, we have a homogeneous mixture, or a solution, of carbon dioxide in water. However, very little carbon dioxide can dissolve in water. If the atmosphere were pure carbon dioxide, the solution would be only about 0.07% carbon dioxide. In reality, the air is only about 0.03% carbon dioxide, so the amount of carbon dioxide in water is reduced proportionally.
However, when soda and beer are made, manufacturers do two important things: they use pure carbon dioxide gas, and they use it at very high pressures. With higher pressures, more carbon dioxide can dissolve in the water. When the soda or beer container is sealed, the high pressure of carbon dioxide gas remains inside the package. (Of course, there are more ingredients in soda and beer besides carbon dioxide and water.)
When you open a container of soda or beer, you hear a distinctive hiss as the excess carbon dioxide gas escapes. But something else happens as well. The carbon dioxide in the solution comes out of solution as a bunch of tiny bubbles. These bubbles impart a pleasing sensation in the mouth, so much so that the soda industry sold over 225 billion servings of soda in the United States alone in 2009.
Some sparkling wines are made in the same way—by forcing carbon dioxide into regular wine. Some sparkling wines (including champagne) are made by sealing a bottle of wine with some yeast in it. The yeast ferments, a process by which the yeast converts sugars into energy and excess carbon dioxide. The carbon dioxide produced by the yeast dissolves in the wine. Then, when the champagne bottle is opened, the increased pressure of carbon dioxide is released, and the drink bubbles just like an expensive glass of soda.
Key Takeaways
• Science is a process of knowing about the natural universe through observation and experiment.
• Scientists go through a rigorous process to determine new knowledge about the universe; this process is generally referred to as the scientific method.
• Science is broken down into various fields, of which chemistry is one.
• Science, including chemistry, is both qualitative and quantitative. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/01%3A_What_Is_Chemistry/1.03%3A_Chemistry_as_a_Science.txt |
1.1: Basic Definitions
Q1.1.1
1. Identify each as either matter or not matter.
1. a book
2. hate
3. light
4. a car
5. a fried egg
Q1.1.2
1. Give an example of matter in each phase: solid, liquid, or gas.
1. Does each statement represent a physical property or a chemical property?
1. Sulfur is yellow.
2. Steel wool burns when ignited by a flame.
3. A gallon of milk weighs over eight pounds.
2. Does each statement represent a physical property or a chemical property?
1. A pile of leaves slowly rots in the backyard.
2. In the presence of oxygen, hydrogen can interact to make water.
3. Gold can be stretched into very thin wires.
3. Does each statement represent a physical change or a chemical change?
1. Water boils and becomes steam.
2. Food is converted into usable form by the digestive system.
3. The alcohol in many thermometers freezes at about −40 degrees Fahrenheit.
4. Does each statement represent a physical change or a chemical change?
1. Graphite, a form of elemental carbon, can be turned into diamond, another form of carbon, at very high temperatures and pressures.
2. The house across the street has been painted a new color.
3. The elements sodium and chlorine come together to make a new substance called sodium chloride.
5. Distinguish between an element and a compound. About how many of each are known?
6. What is the difference between a homogeneous mixture and a heterogeneous mixture?
7. Identify each as a heterogeneous mixture or a homogeneous mixture.
1. Salt is mixed with pepper.
2. Sugar is dissolved in water.
3. Pasta is cooked in boiling water.
8. Identify each as a heterogeneous mixture or a homogeneous mixture.
1. air
2. dirt
3. a television set
9. In Exercise 8, which choices are also solutions?
10. In Exercise 9, which choices are also solutions?
11. Why is iron considered a metal?
12. Why is oxygen considered a nonmetal?
13. Distinguish between a metal and a nonmetal.
14. What properties do semimetals have?
15. Elemental carbon is a black, dull-looking solid that conducts heat and electricity well. It is very brittle and cannot be made into thin sheets or long wires. Of these properties, how does carbon behave as a metal? How does carbon behave as a nonmetal?
16. Pure silicon is shiny and silvery but does not conduct electricity or heat well. Of these properties, how does silicon behave as a metal? How does silicon behave as a nonmetal?
Answers
1. matter
2. not matter
3. not matter
4. matter
5. matter
1. chemical change
2. chemical property
3. physical property
1. physical change
2. physical change
3. chemical change
1. An element is a fundamental chemical part of a substance; there are about 115 known elements. A compound is a combination of elements that acts as a different substance; there are over 50 million known substances.
1. homogeneous
2. heterogeneous
3. heterogeneous
1. Choice a) is a solution.
1. Iron is a metal because it is solid, is shiny, and conducts electricity and heat well.
1. Metals are typically shiny, conduct electricity and heat well, and are malleable and ductile; nonmetals are a variety of colors and phases, are brittle in the solid phase, and do not conduct heat or electricity well.
1. Carbon behaves as a metal because it conducts heat and electricity well. It is a nonmetal because it is black and brittle and cannot be made into sheets or wires.
1.2: Chemistry as a Science
1. Describe the scientific method.
2. "A hypothesis is just a guess"—is this an inadequate definition?
3. Why do scientists need to perform experiments?
4. What is the scientific definition of a theory? How is this word misused in general conversation?
5. What is the scientific definition of a law? How does it differ from the everyday definition of a law?
6. Name an example of a field that is not considered a science.
7. Which of the following fields are studies of the natural universe?
1. biophysics (a mix of biology and physics)
2. art
3. business
8. Which of the following fields are studies of the natural universe?
1. accounting
2. geochemistry (a mix of geology and chemistry)
3. astronomy (the study of stars and planets [but not the earth])
9. Which of these statements are qualitative descriptions?
1. Titanic was the largest passenger ship build at that time.
2. The population of the United States is about 306,000,000 people.
3. The peak of Mount Everest is 29,035 feet above sea level.
10. Which of these statements are qualitative descriptions?
1. A regular movie ticket in Cleveland costs \$6.00.
2. The weather in the Democratic Republic of the Congo is the wettest in all of Africa.
3. The deepest part of the Pacific Ocean is the Mariana Trench.
11. Of the statements in Exercise 9, which are quantitative?
12. Of the statements in Exercise 10, which are quantitative?
Answers
1. Simply stated, the scientific method includes three steps: (1) stating a hypothesis, (2) testing the hypothesis, and (3) refining the hypothesis.
1. Scientists perform experiments to test their hypotheses because sometimes the nature of natural universe is not obvious.
1. A scientific law is a specific statement that is thought to be never violated by the entire natural universe. Everyday laws are arbitrary limits that society puts on its members.
1. yes
2. no
3. no
1. qualitative
2. not qualitative
3. not qualitative
1. Statements b and c are quantitative. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/01%3A_What_Is_Chemistry/1.E%3A_What_Is_Chemistry_%28Exercises%29.txt |
In 1983, an Air Canada airplane had to make an emergency landing because it unexpectedly ran out of fuel; ground personnel had filled the fuel tanks with a certain number of pounds of fuel, not kilograms of fuel. In 1999, the Mars Climate Orbiter spacecraft was lost whilst attempting to orbit Mars because the thrusters were programmed in terms of English units, even though the engineers built the spacecraft using metric units. In 1993, a nurse mistakenly administered 23 units of morphine to a patient rather than the "2–3" units prescribed (the patient ultimately survived). These incidents occurred because people were not paying attention to quantities.
Chemistry, like all sciences, is quantitative. It deals with quantities, things that have amounts and units. Dealing with quantities is very important in chemistry, as is relating quantities to each other. In this chapter, we will discuss how we deal with numbers and units, including how they are combined and manipulated.
• 2.1: Prelude to Measurements
Chemistry, like all sciences, is quantitative. It deals with quantities, things that have amounts and units. Dealing with quantities is very important in chemistry, as is relating quantities to each other. In this chapter, we will discuss how we deal with numbers and units, including how they are combined and manipulated.
• 2.2: Expressing Numbers
Standard notation expresses a number normally. Scientific notation expresses a number as a coefficient times a power of 10. The power of 10 is positive for numbers greater than 1, and negative for numbers between 0 and 1.
• 2.3: Expressing Units
Numbers tell "how much," and units tell "of what." Chemistry uses a set of fundamental units and derived units from SI units. Chemistry uses a set of prefixes that represent multiples or fractions of units. Units can be multiplied and divided to generate new units for quantities.
• 2.4: Significant Figures
Significant figures in a quantity indicate the number of known values plus one place that is estimated. There are rules for which numbers in a quantity are significant and which are not significant. In calculations involving addition and subtraction, limit significant figures based on the rightmost place that all values have in common. In calculations involving multiplication and division, limit significant figures to the least number of significant figures in all the data values.
• 2.5: Converting Units
Units can be converted to other units using the proper conversion factors. Conversion factors are constructed from equalities that relate two different units. Conversions can be a single step or multi-step. Unit conversion is a powerful mathematical technique in chemistry that must be mastered. Exact numbers do not affect the determination of significant figures.
• 2.6: Other Units - Temperature and Density
Chemistry uses the Celsius and Kelvin scales to express temperature. A temperature on the Kelvin scale is the Celsius temperature plus 273.15. The minimum possible temperature is absolute zero and is assigned 0 K on the Kelvin scale. Density relates the mass and volume of a substance. Density can be used to calculate volume from a given mass, or mass from a given volume.
• 2.E: Measurements (Exercises)
These are exercises and select solutions to accompany Chapter 2 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
02: Measurements
Data suggest that a male child will weigh 50% of his adult weight at about 11 years of age. However, he will reach 50% of his adult height at only 2 years of age. It is obvious, then, that people eventually stop growing up but continue to grow out. Data also suggest that the average human height has been increasing over time. In industrialized countries, the average height of people increased 5.5 inches from 1810 to 1984. Most scientists attribute this simple, basic measurement of the human body to better health and nutrition. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.01%3A_Prelude_to_Measurements.txt |
Learning Objective
• Learn to express numbers properly
Quantities have two parts: the number and the unit. The number tells "how many." It is important to be able to express numbers properly so that the quantities can be communicated properly.
Standard Notation
Standard notation is the straightforward expression of a number. Numbers such as 17, 101.5, and 0.00446 are expressed in standard notation. For relatively small numbers, standard notation is fine. However, for very large numbers, such as 306,000,000, or for very small numbers, such as 0.000000419, standard notation can be cumbersome because of the number of zeros needed to place nonzero numbers in the proper position.
Scientific Notation
Scientific notation is an expression of a number using powers of 10. Powers of 10 are used to express numbers that have many zeros:
Table shows the powers of 10. Left side has 10 to a specific power and right side has the equation for the specified power.
100 = 1
101 = 10
102 = 100 = 10 × 10
103 = 1,000 = 10 × 10 × 10
104 = 10,000 = 10 × 10 × 10 × 10
and so forth.
The raised number to the right of the 10 indicates the number of factors of 10 in the original number. (Scientific notation is sometimes called exponential notation.) The exponent's value is equal to the number of zeros in the number expressed in standard notation.
Small numbers can also be expressed in scientific notation but with negative exponents:
Table shows the powers of 10 with negative exponents. Left side has 10 to a specific power and right side has the equation for the specified power.
10−1 = 0.1 = 1/10
10−2 = 0.01 = 1/100
10−3 = 0.001 = 1/1,000
10−4 = 0.0001 = 1/10,000
and so forth. Again, the value of the exponent is equal to the number of zeros in the denominator of the associated fraction. A negative exponent implies a decimal number less than one.
A number is expressed in scientific notation by writing the first nonzero digit, then a decimal point, and then the rest of the digits. The part of a number in scientific notation that is multiplied by a power of 10 is called the coefficient. We determine the power of 10 needed to make that number into the original number and multiply the written number by the proper power of 10. For example, to write 79,345 in scientific notation,
$79,345 = 7.9345 \times 10,000 = 7.9345 \times 10^4\nonumber$
Thus, the number in scientific notation is $7.9345 \times 10^4$. For small numbers, the same process is used, but the exponent for the power of 10 is negative:
$0.000411 = 4.11 \times \dfrac{1}{10,000} = 4.11 \times 10^{−4}\nonumber$
Typically, the extra zero digits at the end or the beginning of a number are not included (Figure $1$).
Example $1$: Expressing Numbers in Scientific Notation
Express these numbers in scientific notation.
1. 306,000
2. 0.00884
3. 2,760,000
4. 0.000000559
Solution
1. The number 306,000 is 3.06 times 100,000, or 3.06 times 105. In scientific notation, the number is 3.06 × 105.
2. The number 0.00884 is 8.84 times 1/1,000, which is 8.84 times 10−3. In scientific notation, the number is 8.84 × 10−3.
3. The number 2,760,000 is 2.76 times 1,000,000, which is the same as 2.76 times 106. In scientific notation, the number is written as 2.76 × 106. Note that we omit the zeros at the end of the original number.
4. The number 0.000000559 is 5.59 times 1/10,000,000, which is 5.59 times 10−7. In scientific notation, the number is written as 5.59 × 10−7.
Exercise $1$
Express these numbers in scientific notation.
1. 23,070
2. 0.0009706
Answer a
2.307 × 104
Answer b
9.706 × 10−4
Another way to determine the power of 10 in scientific notation is to count the number of places you need to move the decimal point to get a numerical value between 1 and 10. The number of places equals the power of 10. This number is positive if you move the decimal point to the right and negative if you move the decimal point to the left:
Many quantities in chemistry are expressed in scientific notation. When performing calculations, you may have to enter a number in scientific notation into a calculator. Be sure you know how to correctly enter a number in scientific notation into your calculator. Different models of calculators require different actions for properly entering scientific notation. If in doubt, consult your instructor immediately (Figure $2$).
Key Takeaways
• Standard notation expresses a number normally.
• Scientific notation expresses a number as a coefficient times a power of 10.
• The power of 10 is positive for numbers greater than 1 and negative for numbers between 0 and 1. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.02%3A_Expressing_Numbers.txt |
Learning Objective
• Learn the units that go with various quantities
• Express units using their abbreviations
• Make new units by combining numerical prefixes with units
A number indicates "how much," but the unit indicates "of what." The "of what" is important when communicating a quantity. For example, if you were to ask a friend how close you are to Lake Erie and your friend says "six," then your friend isn't giving you complete information. Six what? Six miles? Six inches? Six city blocks? The actual distance to the lake depends on what units you use.
Chemistry, like most sciences, uses the International System of Units, or SI for short. (The letters SI stand for the French "le Système International d'unités.") SI specifies certain units for various types of quantities, based on seven fundamental units. We will use most of the fundamental units in chemistry. Initially, we will deal with three fundamental units. The meter (m) is the SI unit of length. It is a little longer than a yard (Figure \(1\)). The SI unit of mass is the kilogram (kg), which is about 2.2 pounds (lb). The SI unit of time is the second (s).
To express a quantity, you need to combine a number with a unit. If you have a length that is 2.4 m, then you express that length as simply 2.4 m. A time of 15,000 s can be expressed as 1.5 × 104 s in scientific notation.
Sometimes, a given unit is not an appropriate size to easily express a quantity. For example, the width of a human hair is very small, and it doesn't make much sense to express it in meters. SI also defines a series of numerical prefixes, referring to multiples or fractions of a fundamental unit, to make a unit more conveniently sized for a specific quantity. Table \(1\) lists the prefixes, their abbreviations, and their multiplicative factors. Some of the prefixes, such as kilo-, mega-, and giga-, represent more than one of the fundamental unit, while other prefixes, such as centi-, milli-, and micro-, represent fractions of the original unit. Note, too, that once again we are using powers of 10. Each prefix is a multiple of or fraction of a power of 10.
Table \(1\): Multiplicative Prefixes for SI Units
Prefix Abbreviation Multiplicative Amount Power of Ten
giga- G 1,000,000,000 × 109 ×
mega- M 1,000,000 × 106 ×
kilo- k 1,000 × 103 ×
deci- d 1/10 × ×
centi- c 1/100 × ×
milli- m 1/1,000 × ×
micro- μ* 1/1,000,000 × ×
nano- n 1/1,000,000,000 × ×
pico- p 1/1,000,000,000,000 × x
* The letter μ is the Greek letter lowercase equivalent to an m and is called "mu" (pronounced "myoo").
To use the fractions to generate new units, simply combine the prefix with the unit itself; the abbreviation for the new unit is the combination of the abbreviation for the prefix and the abbreviation of the unit. For example, the kilometer (km) is 1,000 × meter, or 1,000 m. Thus, 5 kilometers (5 km) is equal to 5,000 m. Similarly, a millisecond (ms) is 1/1,000 × second, or one-thousandth of a second. Thus, 25 ms is 25 thousandths of a second. You will need to become proficient in combining prefixes and units. (You may recognize that one of our fundamental units, the kilogram, automatically has a prefix-unit combination. The word kilogram means 1,000 g.)
In addition to the fundamental units, SI also allows for derived units based on a fundamental unit or units. There are many derived units used in science. For example, the derived unit for area comes from the idea that area is defined as width times height. Because both width and height are lengths, they both have the fundamental unit of meter, so the unit of area is meter × meter, or meter2 (m2). This is sometimes spoken as "square meters." A unit with a prefix can also be used to derive a unit for area, so we can also have cm2, mm2, or km2 as acceptable units for area.
Volume is defined as length times width times height, so it has units of meter × meter × meter, or meter3 (m3)—sometimes spoken as "cubic meters." The cubic meter is a rather large unit, however, so another unit is defined that is somewhat more manageable: the liter (L). A liter is 1/1,000th of a cubic meter and is a little more than 1 quart in volume (Figure \(2\)). Prefixes can also be used with the liter unit, so we can speak of milliliters (1/1,000th of a liter; mL) and kiloliters (1,000 L; kL).
Another definition of a liter is one-tenth of a meter cubed. Because one-tenth of a meter is 10 cm, then a liter is equal to 1,000 cm3 (Figure \(3\)). Because 1 L equals 1,000 mL, we conclude that 1 mL equals 1 cm3; thus, these units are interchangeable.
Units are not only multiplied together—they can also be divided. For example, if you are traveling at one meter for every second of time elapsed, your velocity is 1 meter per second, or 1 m/s. The word per implies division, so velocity is determined by dividing a distance quantity by a time quantity. Other units for velocity include kilometers per hour (km/h) or even micrometers per nanosecond (μm/ns). Later, we will see other derived units that can be expressed as fractions.
Example \(1\)
1. A human hair has a diameter of about 6.0 × 10−5 m. Suggest an appropriate unit for this measurement and write the diameter of a human hair in terms of that unit.
2. What is the velocity of a car if it goes 25 m in 5.0 s?
Solution
1. The scientific notation 10−5 is close to 10−6, which defines the micro- prefix. Let us use micrometers as the unit for hair diameter. The number 6.0 × 10−5 can be written as 60 × 10−6, and a micrometer is 10−6 m, so the diameter of a human hair is about 60 μm.
2. If velocity is defined as a distance quantity divided by a time quantity, then velocity is 25 meters/5.0 seconds. Dividing the numbers gives us 25/5.0 = 5.0, and dividing the units gives us meters/second, or m/s. The velocity is 5.0 m/s.
Exercise \(1\)
1. Express the volume of an Olympic-sized swimming pool, 2,500,000 L, in more appropriate units.
2. A common garden snail moves about 6.1 m in 30 min. What is its velocity in meters per minute (m/min)?
Answer a
• 2.5 ML
Answer b
• 0.203 m/min
Key Takeaways
• Numbers tell "how much," and units tell "of what."
• Chemistry uses a set of fundamental units and derived units from SI units.
• Chemistry uses a set of prefixes that represent multiples or fractions of units.
• Units can be multiplied and divided to generate new units for quantities. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.03%3A_Expressing_Units.txt |
Learning Objective
• Apply the concept of significant figures to limit a measurement to the proper number of digits.
• Recognize the number of significant figures in a given quantity.
• Limit mathematical results to the proper number of significant figures.
If you use a calculator to evaluate the expression 337/217, you will get the following:
$337\div 217=1.5529953917 \nonumber \nonumber$
and so on for many more digits. Although this answer is correct, it is somewhat presumptuous. You start with two values that each have three digits, and the answer has twelve digits? That does not make much sense from a strict numerical point of view.
Consider using a ruler to measure the width of an object, as shown in Figure $1$. The object is definitely more than 1 cm long, so we know that the first digit in our measurement is 1. We see by counting the tick marks on the ruler that the object is at least three ticks after the 1. If each tick represents 0.1 cm, then we know the object is at least 1.3 cm wide. But our ruler does not have any more ticks between the 0.3 and the 0.4 marks, so we can't know exactly how much the next decimal place is. But with a practiced eye we can estimate it. Let us estimate it as about six-tenths of the way between the third and fourth tick marks, which estimates our hundredths place as 6, so we identify a measurement of 1.36 cm for the width of the object.
Does it make any sense to try to report a thousandths place for the measurement? No, it doesn't; we are not exactly sure of the hundredths place (after all, it was an estimate only), so it would be fruitless to estimate a thousandths place. Our best measurement, then, stops at the hundredths place, and we report 1.36 cm as proper measurement.
This concept of reporting the proper number of digits in a measurement or a calculation is called significant figures. Significant figures (sometimes called significant digits) represent the limits of what values of a measurement or a calculation we are sure of. The convention for a measurement is that the quantity reported should be all known values and the first estimated value. The conventions for calculations are discussed as follows.
Example $1$
Use each diagram to report a measurement to the proper number of significant figures.
Solution
1. The arrow is between 4.0 and 5.0, so the measurement is at least 4.0. The arrow is between the third and fourth small tick marks, so it's at least 0.3. We will have to estimate the last place. It looks like about one-third of the way across the space, so let us estimate the hundredths place as 3. Combining the digits, we have a measurement of 4.33 psi (psi stands for "pounds per square inch" and is a unit of pressure, like air in a tire). We say that the measurement is reported to three significant figures.
2. The rectangle is at least 1.0 cm wide but certainly not 2.0 cm wide, so the first significant digit is 1. The rectangle's width is past the second tick mark but not the third; if each tick mark represents 0.1, then the rectangle is at least 0.2 in the next significant digit. We have to estimate the next place because there are no markings to guide us. It appears to be about halfway between 0.2 and 0.3, so we will estimate the next place to be a 5. Thus, the measured width of the rectangle is 1.25 cm. Again, the measurement is reported to three significant figures.
Exercise $1$
What would be the reported width of this rectangle?
Answer
0.63 cm
In many cases, you will be given a measurement. How can you tell by looking what digits are significant? For example, the reported population of the United States is 306,000,000. Does that mean that it is exactly three hundred six million, or is some estimation occurring?
The following conventions dictate which numbers in a reported measurement are significant and which are not significant:
1. Any nonzero digit is significant.
2. Any zeros between nonzero digits (i.e., embedded zeros) are significant.
3. Zeros at the end of a number without a decimal point (i.e., trailing zeros) are not significant; they serve only to put the significant digits in the correct positions. However, zeros at the end of any number with a decimal point are significant.
4. Zeros at the beginning of a decimal number (i.e., leading zeros) are not significant; again, they serve only to put the significant digits in the correct positions.
So, by these rules, the population figure of the United States has only three significant figures: the 3, the 6, and the zero between them. The remaining six zeros simply put the 306 in the millions position. (See Figure $2$ for another example.)
Example $2$
Give the number of significant figures in each measurement.
1. 36.7 m
2. 0.006606 s
3. 2,002 kg
4. 306,490,000 people
Solution
1. By rule 1, all nonzero digits are significant, so this measurement has three significant figures.
2. By rule 4, the first three zeros are not significant, but by rule 2 the zero between the sixes is; therefore, this number has four significant figures.
3. By rule 2, the two zeros between the twos are significant, so this measurement has four significant figures.
4. The four trailing zeros in the number are not significant, but the other five numbers are, so this number has five significant figures.
Exercise $2$
Give the number of significant figures in each measurement.
1. 0.000601 m
2. 65.080 kg
Answer a
three significant figures
Answer b
five significant figures
How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.21, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column.
We drop the last digit—the 1—because it is not significant to the final answer.
The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater, and rounded down if the first dropped digit is less than 5.
Example $3$
Express the final answer to the proper number of significant figures.
1. 101.2 + 18.702 = ?
2. 202.88 − 1.013 = ?
Solution
1. If we use a calculator to add these two numbers, we would get 119.902. However, most calculators do not understand significant figures, and we need to limit the final answer to the tenths place. Thus, we drop the 02 and report a final answer of 119.9 (rounding down).
2. A calculator would answer 201.867. However, we have to limit our final answer to the hundredths place. Because the first number being dropped is 7, which is greater than 5, we round up and report a final answer of 201.87.
Exercise $3$
Express the answer for
$3.445 + 90.83 − 72.4 \nonumber \nonumber$
to the proper number of significant figures.
Answer
21.9
If the operations being performed are multiplication or division, the rule is as follows: limit the answer to the number of significant figures that the data value with the least number of significant figures has. So if we are dividing 23 by 448, which have two and three significant figures respectively, we should limit the final reported answer to two significant figures (the lesser of two and three significant figures):
$23\div 448= 0.051339286 \approx 0.051 \nonumber \nonumber$
The same rounding rules apply in multiplication and division as they do in addition and subtraction.
Example $4$: Significant Figures
Express the final answer to the proper number of significant figures.
1. 76.4 × 180.4 = ?
2. 934.9 ÷ 0.00455 = ?
Solution
1. The first number has three significant figures, while the second number has four significant figures. Therefore, we limit our final answer to three significant figures: 76.4 × 180.4 = 13,782.56 = 13,800.
2. The first number has four significant figures, while the second number has three significant figures. Therefore, we limit our final answer to three significant figures: 934.9 ÷ 0.00455 = 205,472.5275… = 205,000.
Exercise $4$
Express the final answer to the proper number of significant figures.
1. 22.4 × 8.314 = ?
2. 1.381 ÷ 6.02 = ?
Answer a
186
Answer b
0.229
As you have probably realized by now, the biggest issue in determining the number of significant figures in a value is the zero. Is the zero significant or not? One way to unambiguously determine whether a zero is significant or not is to write a number in scientific notation. Scientific notation will include zeros in the coefficient of the number only if they are significant. Thus, the number 8.666 × 106 has four significant figures. However, the number 8.6660 × 106 has five significant figures. That last zero is significant; if it were not, it would not be written in the coefficient. So, when in doubt about expressing the number of significant figures in a quantity, use scientific notation and include the number of zeros that are truly significant.
Summary
• Significant figures in a quantity indicate the number of known values plus one place that is estimated.
• There are rules for which numbers in a quantity are significant and which are not significant.
• In calculations involving addition and subtraction, limit significant figures based on the rightmost place that all values have in common.
• In calculations involving multiplication and division, limit significant figures to the least number of significant figures in all the data values. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.04%3A_Significant_Figures.txt |
Learning Objective
• Convert from one unit to another unit of the same type.
In Section 2.2, we showed some examples of how to replace initial units with other units of the same type to get a numerical value that is easier to comprehend. In this section, we will formalize the process.
Consider a simple example: how many feet are there in 4 yards? Most people will almost automatically answer that there are 12 feet in 4 yards. How did you make this determination? Well, if there are 3 feet in 1 yard and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards.
This is correct, of course, but it is informal. Let us formalize it in a way that can be applied more generally. We know that 1 yard (yd) equals 3 feet (ft):
$1\, yd = 3\, ft\nonumber$
In math, this expression is called an equality. The rules of algebra say that you can change (i.e., multiply or divide or add or subtract) the equality (as long as you do not divide by zero) and the new expression will still be an equality. For example, if we divide both sides by 2, we get:
$\dfrac{1}{2}\,yd= \dfrac{3}{2}\, ft\nonumber$
We see that one-half of a yard equals 3/2, or one and a half, feet—something we also know to be true—so the above equation is still an equality. Going back to the original equality, suppose we divide both sides of the equation by 1 yard (number and unit):
$\dfrac{1\,yd}{1\,yd}= \dfrac{3\,ft}{1\,yd}\nonumber$
The expression is still an equality, by the rules of algebra. The left fraction equals 1. It has the same quantity in the numerator and the denominator, so it must equal 1. The quantities in the numerator and denominator cancel, both the number and the unit:
$\dfrac{1\,yd}{1\,yd}= \dfrac{3\,ft}{1\,yd}\nonumber$
When everything cancels in a fraction, the fraction reduces to 1:
$1= \dfrac{3\,ft}{1\,yd}\nonumber$
Conversion Factors
We have an expression that equals 1.
$\dfrac{3\,ft}{1\,yd}=1\nonumber$
This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units.
The expression
$\dfrac{3\,ft}{1\,yd}\nonumber$
is called a conversion factor and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.)
To see how this happens, let us start with the original quantity: 4 yd.
Now let us multiply this quantity by 1. When you multiply anything by 1, you do not change the value of the quantity. Rather than multiplying by just 1, let us write 1 as:
$\dfrac{3\,ft}{1\,yd}\nonumber$
$4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber$
The 4 yd term can be thought of as 4yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard:
$4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber$
That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:
$\dfrac{4\times 3\, ft}{1}= \dfrac{12\,ft}{1}= 12\,ft\nonumber$
Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.
How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter, which is:
$1\,mm= \dfrac{1}{1000\,m}\nonumber$
The 1/1000 is what the prefix milli- means. Most people are more comfortable working without fractions, so we will rewrite this equation by bringing the 1,000 into the numerator of the other side of the equation:
$1000\,mm=1\,m\nonumber$
Now we construct a conversion factor by dividing one quantity into both sides. But now a question arises: which quantity do we divide by? It turns out that we have two choices, and the two choices will give us different conversion factors, both of which equal 1:
$\dfrac{1000\,mm}{1000\,mm}= \dfrac{1\,m}{1000\,mm} \nonumber$
or
$\dfrac{1000\,mm}{1\,m}= \dfrac{1\,m}{1\,m}\nonumber$
$1=\dfrac{1\,m}{1000\,mm}\nonumber$
or
$\dfrac{1000\,mm}{1\,m}=1\nonumber$
Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Therefore, we will use the second conversion factor. Canceling units and performing the mathematics, we get:
$14.66m\times \dfrac{1000\,mm}{1\,m}= 14660\,mm\nonumber$
Note how $m$ cancels, leaving $mm$, which is the unit of interest.
The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses.
Example $1$
1. Convert 35.9 kL to liters.
2. Convert 555 nm to meters.
Solution
1. We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1000L/ 1kL. Applying this conversion factor, we get:
$35.9\, kL\times \dfrac{1000\,L}{1\,kL}= 35,900\, L \nonumber \nonumber$
1. We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator:
$\dfrac{1\,m}{10^{9}\,nm} \nonumber \nonumber$
Applying this conversion factor, we get:
$555\,nm\times \dfrac{1m}{10^{9}nm}= 0.000000555\,m= 5.55\times 10^{-7}\,m \nonumber \nonumber$
In the final step, we expressed the answer in scientific notation.
Exercise $1$
1. Convert 67.08 μL to liters.
2. Convert 56.8 m to kilometers.
Answer a
6.708 × 10−5 L
Answer b
5.68 × 10−2 km
What if we have a derived unit that is the product of more than one unit, such as m2? Suppose we want to convert square meters to square centimeters? The key is to remember that m2 means m × m, which means we have two meter units in our derived unit. That means we have to include two conversion factors, one for each unit. For example, to convert 17.6 m2 to square centimeters, we perform the conversion as follows:
\begin{align} 17.6m^{2} &= 17.6(m\times m)\times \dfrac{100cm}{1m}\times \dfrac{100cm}{1m} \nonumber \[4pt] &= 176000\,cm \times cm \nonumber \[4pt] &= 1.76\times 10^{5} \,cm^2\end{align}\nonumber
Example $2$
How many cubic centimeters are in 0.883 m3?
Solution
With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have:
$0.883m^{3}\times \dfrac{100\,cm}{1\,m}\times \dfrac{100\,cm}{1\,m} \times \dfrac{100\,cm}{1\,m}= 883000\,cm^{3} = 8.83\times 10^{5}\,cm^{3}\nonumber$
You should demonstrate to yourself that the three meter units do indeed cancel.
Exercise $2$
How many cubic millimeters are present in 0.0923 m3?
Answer
9.23 × 107 mm3
Suppose the unit you want to convert is in the denominator of a derived unit—what then? Then, in the conversion factor, the unit you want to remove must be in the numerator. This will cancel with the original unit in the denominator and introduce a new unit in the denominator. The following example illustrates this situation.
Example $3$
Convert 88.4 m/min to meters/second.
Solution
We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1min/60s. Apply and perform the math:
$\dfrac{88.4m}{min}\times \dfrac{1\,min}{60\,s}= 1.47\dfrac{m}{s}\nonumber$
Notice how the 88.4 automatically goes in the numerator. That's because any number can be thought of as being in the numerator of a fraction divided by 1.
Exercise $3$
Convert 0.203 m/min to meters/second.
Answer
0.00338 m/s
or
3.38 × 10−3 m/s
Sometimes there will be a need to convert from one unit with one numerical prefix to another unit with a different numerical prefix. How do we handle those conversions? Well, you could memorize the conversion factors that interrelate all numerical prefixes. Or you can go the easier route: first convert the quantity to the base unit—the unit with no numerical prefix—using the definition of the original prefix. Then, convert the quantity in the base unit to the desired unit using the definition of the second prefix. You can do the conversion in two separate steps or as one long algebraic step. For example, to convert 2.77 kg to milligrams:
$2.77\,kg\times \dfrac{1000\,g}{1\,kg}= 2770\,g\nonumber$ (convert to the base units of grams)
$2770\,g\times \dfrac{1000\,mg}{1\,g}= 2770000\,mg = 2.77\times 10^{6}\,mg\nonumber$ (convert to desired unit)
Alternatively, it can be done in a single multi-step process:
\begin{align} 2.77\, \cancel{kg}\times \dfrac{1000\,\cancel{g}}{1\,\cancel{kg}}\times \dfrac{1000\,mg}{1\,\cancel{g}} &= 2770000\, mg \nonumber \[4pt] &= 2.77\times 10^{6}\,mg \end{align}\nonumber
You get the same answer either way.
Example $4$
How many nanoseconds are in 368.09 μs?
Solution
You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-,
$368.0\,\mu s\times \dfrac{1\,s}{1000000\,\mu s}\times \dfrac{1000000000}{1\,s}= 3.6809\times 10^{5}\,ns\nonumber$
Exercise $4$
How many milliliters are in 607.8 kL?
Answer
6.078 × 108 mL
When considering the significant figures of a final numerical answer in a conversion, there is one important case where a number does not impact the number of significant figures in a final answer: the so-called exact number. An exact number is a number from a defined relationship, not a measured one. For example, the prefix kilo- means 1,000-exactly 1,000, no more or no less. Thus, in constructing the conversion factor:
$\dfrac{1000\,g}{1\,kg}\nonumber$
neither the 1,000 nor the 1 enter into our consideration of significant figures. The numbers in the numerator and denominator are defined exactly by what the prefix kilo- means. Another way of thinking about it is that these numbers can be thought of as having an infinite number of significant figures, such as:
$\dfrac{1000.0000000000 \dots \,g}{1.0000000000 \ldots \,kg}\nonumber$
The other numbers in the calculation will determine the number of significant figures in the final answer.
Example $5$
A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.
Solution
Area is defined as the product of the two dimensions, which we then have to convert to square meters, and express our final answer to the correct number of significant figures—which in this case will be three.
$36.7\,cm\times 128.8\,cm\times \dfrac{1\,m}{100\,cm}\times \dfrac{1\,m}{100\,cm}= 0.472696\,m^{2}= 0.473\,m^{2}\nonumber$
The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.
Exercise $5$
What is the volume of a block in cubic meters with the dimensions 2.1 cm × 34.0 cm × 118 cm?
Answer
0.0084 m3
Chemistry is Everywhere: The Gimli Glider
On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed "the Gimli Glider."
The Gimli Glider is the Boeing 767 that ran out of fuel and glided to safety at Gimli Airport. The aircraft ran out of fuel because of confusion over the units used to express the amount of fuel. Source: Photo courtesy of Will F., (CC BY-SA 2.5; Aero Icarus).
The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.
What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.
Key Takeaways
• Units can be converted to other units using the proper conversion factors.
• Conversion factors are constructed from equalities that relate two different units.
• Conversions can be a single step or multi-step.
• Unit conversion is a powerful mathematical technique in chemistry that must be mastered.
• Exact numbers do not affect the determination of significant figures. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.05%3A_Converting_Units.txt |
Learning Objective
• Learn about the various temperature scales that are commonly used in chemistry.
• Define density and use it as a conversion factor.
There are other units in chemistry that are important, and we will cover others over the course of the entire book. One of the fundamental quantities in science is temperature. Temperature is a measure of the average amount of energy of motion, or kinetic energy, a system contains. Temperatures are expressed using scales that use units called degrees, and there are several temperature scales in use. In the United States, the commonly used temperature scale is the Fahrenheit scale (symbolized by °F and spoken as "degrees Fahrenheit"). On this scale, the freezing point of liquid water (the temperature at which liquid water turns to solid ice) is 32°F, and the boiling point of water (the temperature at which liquid water turns to steam) is 212°F.
Science also uses other scales to express temperature. The Celsius scale (symbolized by °C and spoken as "degrees Celsius") is a temperature scale where 0°C is the freezing point of water and 100°C is the boiling point of water; the scale is divided into 100 divisions between these two landmarks and extended higher and lower. By comparing the Fahrenheit and Celsius scales, a conversion between the two scales can be determined:
\begin{align} \ce{^{\circ}C} &= \ce{(^{\circ}F-32)\times 5/9} \label{eq1} \[4pt] \ce{ ^{\circ}F} &= \left(\ce{^{\circ}C \times 9/5 } \right)+32 \label{eq2} \end{align}
Example $1$: Conversions
1. What is 98.6 °F in degrees Celsius?
2. What is 25.0 °C in degrees Fahrenheit?
Solution
1. Using Equation \ref{eq1}, we have
\begin{align*} ^{\circ}C &=(98.6-32)\times \dfrac{5}{9} \[4pt] &= 66.6\times \dfrac{5}{9} \[4pt] &= 37.0^{\circ}C \end{align*}\nonumber
1. Using Equation \ref{eq2}, we have
\begin{align*} ^{\circ}F &= \left(25.0\times \dfrac{9}{5}\right)+32 \[4pt] &= 45.0+32 \[4pt] &= 77.0^{\circ}F \end{align*}\nonumber
Exercise $1$
1. Convert 0 °F to degrees Celsius.
2. Convert 212 °C to degrees Fahrenheit.
Answer a
−17.8 °C
Answer b
414 °F
The fundamental unit of temperature (another fundamental unit of science, bringing us to four) in SI is the kelvin (K). The Kelvin temperature scale (note that the name of the scale capitalizes the word Kelvin, but the unit itself is lowercase) uses degrees that are the same size as the Celsius degree, but the numerical scale is shifted up by 273.15 units. That is, the conversion between the Kelvin and Celsius scales is as follows:
$K = {^{\circ}C + 273.15}\nonumber$
For most purposes, it is acceptable to use 273 instead of 273.15. Note that the Kelvin scale does not use the word degrees; a temperature of 295 K is spoken of as "two hundred ninety-five kelvins" and not "two hundred ninety-five degrees Kelvin."
The reason that the Kelvin scale is defined this way is because there exists a minimum possible temperature called absolute zero (zero kelvins). The Kelvin temperature scale is set so that 0 K is absolute zero, and temperature is counted upward from there. Normal room temperature is about 295 K, as seen in the following example.
Example $2$: Room Temperature
If normal room temperature is 72.0°F, what is room temperature in degrees Celsius and kelvin?
Solution
First, we use the formula to determine the temperature in degrees Celsius:
\begin{align*} ^{\circ}C &= (72.0-32)\times \dfrac{5}{9} \nonumber \[4pt] &= 40.0\times \dfrac{5}{9} \nonumber \[4pt] &= 22.2^{\circ}C \end{align*}\nonumber
Then we use the appropriate formula above to determine the temperature in the Kelvin scale:
\begin{align*} K &= 22.2^{\circ}C+273.15 \nonumber \[4pt] &= 295.4K \end{align*}\nonumber
So, room temperature is about 295 K.
Exercise $2$
What is 98.6°F on the Kelvin scale?
Answer
310.2 K
Figure $1$ compares the three temperature scales. Note that science uses the Celsius and Kelvin scales almost exclusively; virtually no practicing chemist expresses laboratory-measured temperatures with the Fahrenheit scale. In fact, the United States is one of the few countries in the world that still uses the Fahrenheit scale on a daily basis. The other two countries are Liberia and Myanmar (formerly Burma). People driving near the borders of Canada or Mexico may pick up local radio stations on the other side of the border that express the daily weather in degrees Celsius, so do not get confused by their weather reports.
Density
Density is a physical property that is defined as a substance's mass divided by its volume:
$density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{v}\nonumber$
Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm3, g/L, kg/L, and even kg/m3. Densities for some common substances are listed in Table $1$.
Table $1$: Densities of Some Common Substances
Substance Density (g/mL or g/cm3)
water 1.0
gold 19.3
mercury 13.6
air 0.0012
cork 0.22–0.26
aluminum 2.7
iron 7.87
Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm3. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table $1$), the volume units will cancel and leave you with mass units, telling you the mass of the sample:
$7.88\,\cancel{cm^{3}}\times \dfrac{2.7\,g}{\cancel{cm^{3}}}= 21\, g \text{ of aluminium} \nonumber \nonumber$
where we have limited our answer to two significant figures.
Example $3$: Mercury
What is the mass of 44.6 mL of mercury?
Solution
Use the density from Table $1$ "Densities of Some Common Substances" as a conversion factor to go from volume to mass:
$44.6\,\cancel{mL}\times \dfrac{13.6\,g}{\cancel{mL}}= 607\,g \nonumber \nonumber$
The mass of the mercury is 607 g.
Exercise $3$
What is the mass of 25.0 cm3 of iron?
Answer
197 g
Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:
$d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \nonumber$
Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator:
$d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \nonumber$
That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units, but also the numbers:
$\dfrac{1}{d}= \dfrac{1\,mL}{19.3\,g} \nonumber \nonumber$
This reciprocal density is still a useful conversion factor, but now the mass unit will cancel and the volume unit will be introduced. Thus, if we want to know the volume of 45.9 g of gold, we would set up the conversion as follows:
$45.9\,\cancel{g}\times \dfrac{1\,mL}{19.3\cancel{g}}= 2.38\,mL \nonumber \nonumber$
Note how the mass units cancel, leaving the volume unit, which is what we are looking for.
Example $4$: Calculating Volume from Density
A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/cm3, what is the volume of the cork?
Solution
To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find:
$\dfrac{1}{d}= \dfrac{1\,cm^{3}}{0.22\,g} \nonumber \nonumber$
We can use this expression as the conversion factor. So
$3.78\,\cancel{g}\times \dfrac{1\,cm^{3}}{0.22\,\cancel{g}}= 17.2\,cm^{3} \nonumber \nonumber$
Exercise $4$
What is the volume of 3.78 g of gold?
Answer
0.196 cm3
Care must be used with density as a conversion factor. Make sure the mass units are the same, or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.
Food and Drink Application: Cooking Temperatures
Because degrees Fahrenheit is the common temperature scale in the United States, kitchen appliances, such as ovens, are calibrated in that scale. A cool oven may be only 150°F, while a cake may be baked at 350°F and a chicken roasted at 400°F. The broil setting on many ovens is 500°F, which is typically the highest temperature setting on a household oven.
People who live at high altitudes, typically 2,000 ft above sea level or higher, are sometimes urged to use slightly different cooking instructions on some products, such as cakes and bread, because water boils at a lower temperature the higher in altitude you go, meaning that foods cook slower. For example, in Cleveland water typically boils at 212°F (100°C), but in Denver, the Mile-High City, water boils at about 200°F (93.3°C), which can significantly lengthen cooking times. Good cooks need to be aware of this.
A meat thermometer with a dial. Notice the markings for Fahrenheit (outer scale) and Celsius (inner scale) temperatures. Recipes for cooking food in an oven can use very different numbers, depending on the country you're in. (CC BY2.0 Bev Sykes)
At the other end is pressure cooking. A pressure cooker is a closed vessel that allows steam to build up additional pressure, which increases the temperature at which water boils. A good pressure cooker can get to temperatures as high as 252°F (122°C); at these temperatures, food cooks much faster than it normally would. Great care must be used with pressure cookers because of the high pressure and high temperature. (When a pressure cooker is used to sterilize medical instruments, it is called an autoclave.)
Other countries use the Celsius scale for everyday purposes. Therefore, oven dials in their kitchens are marked in degrees Celsius. It can be confusing for US cooks to use ovens abroad—a 425°F oven in the United States is equivalent to a 220°C oven in other countries. These days, many oven thermometers are marked with both temperature scales.
Key Takeaways
• Chemistry uses the Celsius and Kelvin scales to express temperatures.
• A temperature on the Kelvin scale is the Celsius temperature plus 273.15.
• The minimum possible temperature is absolute zero and is assigned 0 K on the Kelvin scale.
• Density relates the mass and volume of a substance.
• Density can be used to calculate volume from a given mass or mass from a given volume. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.06%3A_Other_Units_-_Temperature_and_Density.txt |
2.1: Expressing Numbers
1. Express these numbers in scientific notation.
1. 56.9
2. 563,100
3. 0.0804
4. 0.00000667
2. Express these numbers in scientific notation.
1. −890,000
2. 602,000,000,000
3. 0.0000004099
4. 0.000000000000011
3. Express these numbers in scientific notation.
1. 0.00656
2. 65,600
3. 4,567,000
4. 0.000005507
4. Express these numbers in scientific notation.
1. 65
2. −321.09
3. 0.000077099
4. 0.000000000218
5. Express these numbers in standard notation.
1. 1.381 × 105
2. 5.22 × 10−7
3. 9.998 × 104
6. Express these numbers in standard notation.
1. 7.11 × 10−2
2. 9.18 × 102
3. 3.09 × 10−10
7. Express these numbers in standard notation.
1. 8.09 × 100
2. 3.088 × 10−5
3. −4.239 × 102
8. Express these numbers in standard notation.
1. 2.87 × 10−8
2. 1.78 × 1011
3. 1.381 × 10−23
9. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.
1. 72.44 × 103
2. 9,943 × 10−5
3. 588,399 × 102
10. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.
1. 0.000077 × 10−7
2. 0.000111 × 108
3. 602,000 × 1018
11. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.
1. 345.1 × 102
2. 0.234 × 10−3
3. 1,800 × 10−2
12. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.
1. 8,099 × 10−8
2. 34.5 × 100
3. 0.000332 × 104
13. Write these numbers in scientific notation by counting the number of places the decimal point is moved.
1. 123,456.78
2. 98,490
3. 0.000000445
14. Write these numbers in scientific notation by counting the number of places the decimal point is moved.
1. 0.000552
2. 1,987
3. 0.00000000887
15. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.
1. 456 × (7.4 × 108) = ?
2. (3.02 × 105) ÷ (9.04 × 1015) = ?
3. 0.0044 × 0.000833 = ?
16. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.
1. 98,000 × 23,000 = ?
2. 98,000 ÷ 23,000 = ?
3. (4.6 × 10−5) × (2.09 × 103) = ?
17. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.
1. 45 × 132 ÷ 882 = ?
2. [(6.37 × 104) × (8.44 × 10−4)] ÷ (3.2209 × 1015) = ?
18. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.
1. (9.09 × 108) ÷ [(6.33 × 109) × (4.066 × 10−7)] = ?
2. 9,345 × 34.866 ÷ 0.00665 = ?
Answers
1. 5.69 × 101
2. 5.631 × 105
3. 8.04 × 10−2
4. 6.67 × 10−6
1. 6.56 × 10−3
2. 6.56 × 104
3. 4.567 × 106
4. 5.507 × 10−6
1. 138,100
2. 0.000000522
3. 99,980
1. 8.09
2. 0.00003088
3. −423.9
1. 7.244 × 104
2. 9.943 × 10−2
3. 5.88399 × 107
1. 3.451 × 104
2. 2.34 × 10−4
3. 1.8 × 101
1. 1.2345678 × 105
2. 9.849 × 104
3. 4.45 × 10−7
1. 3.3744 × 1011
2. 3.3407 × 10−11
3. 3.665 × 10−6
1. 6.7346 × 100
2. 1.6691 × 10−14
2.2: Expressing Units
1. Identify the unit in each quantity.
• 2 boxes of crayons
• 3.5 grams of gold
2. Identify the unit in each quantity.
• 32 oz of cheddar cheese
• 0.045 cm3 of water
3. Identify the unit in each quantity.
• 9.58 s (the current world record in the 100 m dash)
• 6.14 m (the current world record in the pole vault)
4. Identify the unit in each quantity.
• 2 dozen eggs
• 2.4 km/s (the escape velocity of the moon, which is the velocity you need at the surface to escape the moon's gravity)
5. Indicate what multiplier each prefix represents.
• k
• m
• M
6. Indicate what multiplier each prefix represents.
• c
• G
• μ
7. Give the prefix that represents each multiplier.
• 1/1,000th ×
• 1,000 ×
• 1,000,000,000 ×
8. Give the prefix that represents each multiplier.
• 1/1,000,000,000th ×
• 1/100th ×
• 1,000,000 ×
1. 9. Complete the following table with the missing information.
Table with Units on the right and Abbreviation on the left side. Missing units on the second and third rows on the right side of the table. Missing abbreviations on the first and fourth rows on the left side of the table.
Unit Abbreviation
kilosecond
mL
Mg
centimeter
2. 10. Complete the following table with the missing information.
Table with Units on the right and Abbreviation on the left side. Missing units on the third and fourth rows on the right side of the table. Missing abbreviations on the first, second, and fifth row on the left side of the table.
Unit Abbreviation
kilometer per second
second
cm3
μL
nanosecond
11. Express each quantity in a more appropriate unit. There may be more than one acceptable answer.
• 3.44 × 10−6 s
• 3,500 L
• 0.045 m
12. Express each quantity in a more appropriate unit. There may be more than one acceptable answer.
• 0.000066 m/s (Hint: you need consider only the unit in the numerator.)
• 4.66 × 106 s
• 7,654 L
13. Express each quantity in a more appropriate unit. There may be more than one acceptable answer.
• 43,600 mL
• 0.0000044 m
• 1,438 ms
14. Express each quantity in a more appropriate unit. There may be more than one acceptable answer.
• 0.000000345 m3
• 47,000,000 mm3
• 0.00665 L
15. Multiplicative prefixes are used for other units as well, such as computer memory. The basic unit of computer memory is the byte (b). What is the unit for one million bytes?
16. You may have heard the terms microscale or nanoscale to represent the sizes of small objects. What units of length do you think are useful at these scales? What fractions of the fundamental unit of length are these units?
17. Acceleration is defined as a change in velocity per time. Propose a unit for acceleration in terms of the fundamental SI units.
18. Density is defined as the mass of an object divided by its volume. Propose a unit of density in terms of the fundamental SI units.
Answers
1.
• boxes of crayons
• grams of gold
3.
• seconds
• meters
5.
• 1,000 x
• 1/1,000 x
• 1,000,000 x
7.
• milli-
• kilo-
• giga-
9.
Table with Units on the right and Abbreviation on the left side.
1. Unit
Abbreviation
1. kilosecond
1. ks
1. milliliter
1. mL
1. megagram
1. Mg
1. centimeter
1. cm
11.
• 3.44 μs
• 3.5 kL
• 4.5 cm
13.
• 43.6 mL
• 4.4 m
• 1.438 s
15. megabytes (Mb
17. meters/second2
2.3: Significant Figures
1. 1. Express each measurement to the correct number of significant figures.
1. 2. Express each measurement to the correct number of significant figures.
1. How many significant figures do these numbers have?
• 23
• 23.0
• 0.00023
• 0.0002302
1. How many significant figures do these numbers have?
• 5.44 × 108
• 1.008 × 10−5
• 43.09
• 0.0000001381
5. How many significant figures do these numbers have?
• 765,890
• 765,890.0
• 1.2000 × 105
• 0.0005060
6. How many significant figures do these numbers have?
• 0.009
• 0.0000009
• 65,444
• 65,040
7. Compute and express each answer with the proper number of significant figures, rounding as necessary.
• 56.0 + 3.44 = ?
• 0.00665 + 1.004 = ?
• 45.99 − 32.8 = ?
• 45.99 − 32.8 + 75.02 = ?
8. Compute and express each answer with the proper number of significant figures, rounding as necessary.
• 1.005 + 17.88 = ?
• 56,700 − 324 = ?
• 405,007 − 123.3 = ?
• 55.5 + 66.66 − 77.777 = ?
9. Compute and express each answer with the proper number of significant figures, rounding as necessary.
• 56.7 × 66.99 = ?
• ÷ 77 = ?
• ÷ 77.0 = ?
• 6.022 × 1.89 = ?
10. Compute and express each answer with the proper number of significant figures, rounding as necessary.
• 0.000440 × 17.22 = ?
• 203,000 ÷ 0.044 = ?
• 67 × 85.0 × 0.0028 = ?
• 999,999 ÷ 3,310 = ?
11. Write the number 87,449 in scientific notation with four significant figures.
12. Write the number 0.000066600 in scientific notation with five significant figures.
13. Write the number 306,000,000 in scientific notation to the proper number of significant figures.
14. Write the number 0.0000558 in scientific notation with two significant figures.
15. Perform each calculation and limit each answer to three significant figures.
• 67,883 × 0.004321 = ?
• (9.67 × 103) × 0.0055087 = ?
16. Perform each calculation and limit each answer to four significant figures.
• 18,900 × 76.33 ÷ 0.00336 = ?
• 0.77604 ÷ 76,003 × 8.888 = ?
Answers
1.
• 375 psi
• 1.30 cm
3.
• two
• three
• two
• four
5.
• five
• seven
• five
• four
7.
• 59.4
• 1.011
• 13.2
• 88.2
9.
• 3.80 × 103
• 0.013
• 0.0130
• 11.4
11.
• 8.745 × 104
• 6.6600 × 10−5
15.
• 293
• 53.3
2.4: Converting Units
1. Write the two conversion factors that exist between the two given units.
2. milliliters and liters
3. microseconds and seconds
4. kilometers and meters
1. Write the two conversion factors that exist between the two given units.
• kilograms and grams
• milliseconds and seconds
• centimeters and meters
1. Perform the following conversions.
• 5.4 km to meters
• 0.665 m to millimeters
• 0.665 m to kilometers
1. Perform the following conversions.
• 90.6 mL to liters
• 0.00066 ML to liters
• 750 L to kiloliters
1. Perform the following conversions.
• 17.8 μg to grams
• 7.22 × 102 kg to grams
• 0.00118 g to nanograms
1. Perform the following conversions.
• 833 ns to seconds
• 5.809 s to milliseconds
• 2.77 × 106 s to megaseconds
1. Perform the following conversions.
• 9.44 m2 to square centimeters
• 3.44 × 108 mm3 to cubic meters
1. Perform the following conversions.
• 0.00444 cm3 to cubic meters
• 8.11 × 102 m2 to square nanometers
1. Why would it be inappropriate to convert square centimeters to cubic meters?
2. Why would it be inappropriate to convert from cubic meters to cubic seconds?
1. Perform the following conversions.
1. 45.0 m/min to meters/second
2. 0.000444 m/s to micrometers/second
3. 60.0 km/h to kilometers/second
1. Perform the following conversions.
• 3.4 × 102 cm/s to centimeters/minute
• 26.6 mm/s to millimeters/hour
• 13.7 kg/L to kilograms/milliliters
1. Perform the following conversions.
• 0.674 kL to milliliters
• 2.81 × 1012 mm to kilometers
• 94.5 kg to milligrams
1. Perform the following conversions.
• 6.79 × 10−6 kg to micrograms
• 1.22 mL to kiloliters
• 9.508 × 10−9 ks to milliseconds
1. Perform the following conversions.
• 6.77 × 1014 ms to kiloseconds
• 34,550,000 cm to kilometers
1. Perform the following conversions.
• 4.701 × 1015 mL to kiloliters
• 8.022 × 10−11 ks to microseconds
1. Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.
• 88 ft/s to miles/hour (Hint: use 5,280 ft = 1 mi.)
• 0.00667 km/h to meters/second
1. Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.
• 3.88 × 102 mm/s to kilometers/hour
• 1.004 kg/L to grams/milliliter
1. What is the area in square millimeters of a rectangle whose sides are 2.44 cm × 6.077 cm? Express the answer to the proper number of significant figures.
2. What is the volume in cubic centimeters of a cube with sides of 0.774 m? Express the answer to the proper number of significant figures.
3. The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square centimeters if its base is 1.007 m and its height is 0.665 m? Express the answer to the proper number of significant figures.
The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square meters if its base is 166 mm and its height is 930.0 mm? Express the answer to the proper number of significant figures.
Answers
1. $\frac{1000mL}{1L} and \frac{1L}{1000mL}\nonumber$
2. $\frac{1000000\mu s}{1s}and \frac{1s}{1000000\mu s}\nonumber$
3. $\frac{1000m}{1km}and \frac{1km}{1000m}\nonumber$
1. 5,400 m
2. 665 mm
3. 6.65 × 10−4 km
•
1. 1.78 × 10−5 g
2. 7.22 × 105 g
3. 1.18 × 106 ng
•
1. 94,400 cm2
2. 0.344 m3
•
1. One is a unit of area, and the other is a unit of volume.
2.
1.
1. 0.75 m/s
2. 444 µm/s
3. 1.666 × 10−2 km/s
•
1. 674,000 mL
2. 2.81 × 106 km
3. 9.45 × 107 mg
•
1. 6.77 × 108 ks
2. 345.5 km
•
1. 6.0 × 101 mi/h
2. 0.00185 m/s
•
1. 1.48 × 103 mm2
2.
3. 3.35 × 103 cm2
2.5: Other Units - Temperature and Density
1. Perform the following conversions.
2. 255°F to degrees Celsius
3. −255°F to degrees Celsius
4. 50.0°C to degrees Fahrenheit
5. −50.0°C to degrees Fahrenheit
1. Perform the following conversions.
• 1,065°C to degrees Fahrenheit
• −222°C to degrees Fahrenheit
• 400.0°F to degrees Celsius
• 200.0°F to degrees Celsius
1. Perform the following conversions.
• 100.0°C to kelvins
• −100.0°C to kelvins
• 100 K to degrees Celsius
• 300 K to degrees Celsius
1. Perform the following conversions.
• 1,000.0 K to degrees Celsius
• 50.0 K to degrees Celsius
• 37.0°C to kelvins
• −37.0°C to kelvins
1. Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius?
2. Convert 0 K to degrees Fahrenheit. What is the significance of the temperature in degrees Fahrenheit?
3. The hottest temperature ever recorded on the surface of the earth was 136°F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins?
4. The coldest temperature ever recorded on the surface of the earth was −128.6°F in Vostok, Antarctica, in 1983. What is the temperature in degrees Celsius and in kelvins?
5. Give at least three possible units for density.
6. What are the units when density is inverted? Give three examples.
7. A sample of iron has a volume of 48.2 cm3. What is its mass?
8. A sample of air has a volume of 1,015 mL. What is its mass?
9. The volume of hydrogen used by the Hindenburg, the German airship that exploded in New Jersey in 1937, was 2.000 × 108 L. If hydrogen gas has a density of 0.0899 g/L, what mass of hydrogen was used by the airship?
10. The volume of an Olympic-sized swimming pool is 2.50 × 109 cm3. If the pool is filled with alcohol (d = 0.789 g/cm3), what mass of alcohol is in the pool?
11. A typical engagement ring has 0.77 cm3 of gold. What mass of gold is present?
12. A typical mercury thermometer has 0.039 mL of mercury in it. What mass of mercury is in the thermometer?
13. What is the volume of 100.0 g of lead if lead has a density of 11.34 g/cm3?
14. What is the volume of 255.0 g of uranium if uranium has a density of 19.05 g/cm3?
15. What is the volume in liters of 222 g of neon if neon has a density of 0.900 g/L?
16. What is the volume in liters of 20.5 g of sulfur hexafluoride if sulfur hexafluoride has a density of 6.164 g/L?
1. Which has the greater volume, 100.0 g of iron (d = 7.87 g/cm3) or 75.0 g of gold (d = 19.3 g/cm3)?
2. Which has the greater volume, 100.0 g of hydrogen gas (d = 0.0000899 g/cm3) or 25.0 g of argon gas (d = 0.00178 g/cm3)?
Answers
1. 124°C
2. −159°C
3. 122°F
4. −58°F
•
1. 373 K
2. 173 K
3. −173°C
4. 27°C
•
1. −273°C. This is the lowest possible temperature in degrees Celsius.
2.
3. 57.8°C; 331 K
4.
5. g/mL, g/L, and kg/L (answers will vary)
6.
7. 379 g
8.
9. 1.80 × 107 g
10.
11. 15 g
12.
13. 8.818 cm3
14.
15. 247 L
16.
17. The 100.0 g of iron has the greater volume
Additional Exercises
1. Evaluate 0.00000000552 × 0.0000000006188 and express the answer in scientific notation. You may have to rewrite the original numbers in scientific notation first.
2. Evaluate 333,999,500,000 ÷ 0.00000000003396 and express the answer in scientific notation. You may need to rewrite the original numbers in scientific notation first.
3. Express the number 6.022 × 1023 in standard notation.
4. Express the number 6.626 × 10−34 in standard notation.
5. When powers of 10 are multiplied together, the powers are added together. For example, 102 × 103 = 102+3 = 105. With this in mind, can you evaluate (4.506 × 104) × (1.003 × 102) without entering scientific notation into your calculator?
6. When powers of 10 are divided into each other, the bottom exponent is subtracted from the top exponent. For example, 105/103 = 105−3 = 102. With this in mind, can you evaluate (8.552 × 106) ÷ (3.129 × 103) without entering scientific notation into your calculator?
7. Consider the quantity two dozen eggs. Is the number in this quantity "two" or "two dozen"? Justify your choice.
8. Consider the quantity two dozen eggs. Is the unit in this quantity "eggs" or "dozen eggs"? Justify your choice.
9. Fill in the blank: 1 km = ______________ μm.
10. Fill in the blank: 1 Ms = ______________ ns.
11. Fill in the blank: 1 cL = ______________ ML.
12. Fill in the blank: 1 mg = ______________ kg.
13. Express 67.3 km/h in meters/second.
14. Express 0.00444 m/s in kilometers/hour.
15. Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 mi/h into kilometers/hour.
16. Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 km/h into miles/hour.
17. Convert 52.09 km/h into meters/second.
18. Convert 2.155 m/s into kilometers/hour.
19. Use the formulas for converting degrees Fahrenheit into degrees Celsius to determine the relative size of the Fahrenheit degree over the Celsius degree.
20. Use the formulas for converting degrees Celsius into kelvins to determine the relative size of the Celsius degree over kelvins.
21. What is the mass of 12.67 L of mercury?
22. What is the mass of 0.663 m3 of air?
23. What is the volume of 2.884 kg of gold?
24. What is the volume of 40.99 kg of cork? Assume a density of 0.22 g/cm3.
Answers
1. 3.42 × 10−18
2.
3. 602,200,000,000,000,000,000,000
4.
5. 4.520 × 106
6.
7. The quantity is two; dozen is the unit.
8.
9. 1,000,000,000
10.
11. 1/100,000,000
12.
13. 18.7 m/s
14.
15. 96.1 km/h
16.
17. 14.47 m/s
18.
19. One Fahrenheit degree is nine-fifths the size of a Celsius degree.
20.
21. 1.72 × 105 g
22.
23. 149 mL | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.E%3A_Measurements_%28Exercises%29.txt |
The basic building block of all matter is the atom. Curiously, the idea of atoms was first proposed in the fifth century BCE, when the Greek philosophers Leucippus and Democritus proposed their existence in a surprisingly modern fashion. However, their ideas never took hold among their contemporaries, and it wasn't until the early 1800s that evidence amassed to make scientists reconsider the idea. Today, the concept of the atom is central to the study of matter.
• 3.1: Prelude to Atoms, Molecules, and Ions
The angstrom unit is named after Anders Jonas Ångström, a nineteenth-century Swedish physicist. Ångström's research dealt with light being emitted by glowing objects, including the sun. Ångström studied the brightness of the different colors of light that the sun emitted and was able to deduce that the sun is composed of the same kinds of matter that are present on the earth. By extension, we now know that all matter throughout the universe is similar to the matter that exists on our own planet.
• 3.2: Atomic Theory
Chemistry is based on the modern atomic theory, which states that all matter is composed of atoms. Atoms themselves are composed of protons, neutrons, and electrons. Each element has its own atomic number, which is equal to the number of protons in its nucleus. Isotopes of an element contain different numbers of neutrons. Elements are represented by an atomic symbol. The periodic table is a chart that organizes all the elements.
• 3.3: Molecules and Chemical Nomenclature
Molecules are groups of atoms that behave as a single unit. Some elements exist as molecules: hydrogen, oxygen, sulfur, and so forth. There are rules that can express a unique name for any given molecule, and a unique formula for any given name.
• 3.4: Masses of Atoms and Molecules
The atomic mass unit (u) is a unit that describes the masses of individual atoms and molecules. The atomic mass is the weighted average of the masses of all isotopes of an element. The molecular mass is the sum of the masses of the atoms in a molecule.
• 3.5: Ions and Ionic Compounds
Ions form when atoms lose or gain electrons. Ionic compounds have positive ions and negative ions. Ionic formulas balance the total positive and negative charges. Ionic compounds have a simple system of naming. Groups of atoms can have an overall charge and make ionic compounds.
• 3.6: Acids
An acid is a compound of the H+ ion dissolved in water. Acids have their own naming system. Acids have certain chemical properties that distinguish them from other compounds.
• 3.E: Atoms, Molecules, and Ions (Exercises)
These are exercises and select solutions to accompany Chapter 3 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
03: Atoms Molecules and Ions
Although not an SI unit, the angstrom (Å) is a useful unit of length. It is one ten-billionth of a meter, or 10−10 m. Why is it a useful unit? The ultimate particles that compose all matter are about 10−10 m in size, or about 1 Å. This makes the angstrom a natural—though not approved—unit for describing these particles.
The angstrom unit is named after Anders Jonas Ångström, a nineteenth-century Swedish physicist. Ångström's research dealt with light being emitted by glowing objects, including the sun. Ångström studied the brightness of the different colors of light that the sun emitted and was able to deduce that the sun is composed of the same kinds of matter that are present on the earth. By extension, we now know that all matter throughout the universe is similar to the matter that exists on our own planet. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.01%3A_Prelude_to_Atoms_Molecules_and_Ions.txt |
Learning Objectives
• State the modern atomic theory.
• Learn how atoms are constructed.
The smallest piece of an element that maintains the identity of that element is called an atom. Individual atoms are extremely small. It would take about fifty million atoms in a row to make a line that is 1 cm long. The period at the end of a printed sentence has several million atoms in it. Atoms are so small that it is difficult to believe that all matter is made from atoms—but it is.
The concept that atoms play a fundamental role in chemistry is formalized by the modern atomic theory, first stated by John Dalton, an English scientist, in 1808. It consists of three parts:
1. All matter is composed of atoms.
2. Atoms of the same element are the same; atoms of different elements are different.
3. Atoms combine in whole-number ratios to form compounds.
These concepts form the basis of chemistry. Although the word atom comes from a Greek word that means "indivisible," we understand now that atoms themselves are composed of smaller parts called subatomic particles. The first part to be discovered was the electron, a tiny subatomic particle with a negative charge. It is often represented as e, with the right superscript showing the negative charge. Later, two larger particles were discovered. The proton is a more massive (but still tiny) subatomic particle with a positive charge, represented as p+. The neutron is a subatomic particle with about the same mass as a proton, but no charge. It is represented as either n or n0. We now know that all atoms of all elements are composed of electrons, protons, and (with one exception) neutrons. Table $1$ summarizes the properties of these three subatomic particles.
Table $1$: Properties of the Three Subatomic Particles
Name Symbol Mass (approx.; kg) Charge
Proton p+ 1.6 × 10−27 1+
Neutron n, n0 1.6 × 10−27 none
Electron e 9.1 × 10−31 1−
How are these particles arranged in atoms? They are not arranged at random. Experiments by Ernest Rutherford in England in the 1910s pointed to a nuclear model with atoms that has the protons and neutrons in a central nucleus with the electrons in orbit about the nucleus. The relatively massive protons and neutrons are collected in the center of an atom, in a region called the nucleus of the atom (plural nuclei). The electrons are outside the nucleus and spend their time orbiting in space about the nucleus. (Figure $1$).
The modern atomic theory states that atoms of one element are the same, while atoms of different elements are different. What makes atoms of different elements different? The fundamental characteristic that all atoms of the same element share is the number of protons. All atoms of hydrogen have one and only one proton in the nucleus; all atoms of iron have 26 protons in the nucleus. This number of protons is so important to the identity of an atom that it is called the atomic number. The number of protons in an atom is the atomic number of the element. Thus, hydrogen has an atomic number of 1, while iron has an atomic number of 26. Each element has its own characteristic atomic number.
Atoms of the same element can have different numbers of neutrons, however. Atoms of the same element (i.e., atoms with the same number of protons) with different numbers of neutrons are called isotopes. Most naturally occurring elements exist as isotopes. For example, most hydrogen atoms have a single proton in their nucleus. However, a small number (about one in a million) of hydrogen atoms have a proton and a neutron in their nuclei. This particular isotope of hydrogen is called deuterium. A very rare form of hydrogen has one proton and two neutrons in the nucleus; this isotope of hydrogen is called tritium. The sum of the number of protons and neutrons in the nucleus is called the mass number of the isotope.
Neutral atoms have the same number of electrons as they have protons, so their overall charge is zero. However, as we shall see later, this will not always be the case.
Example $1$
1. The most common carbon atoms have six protons and six neutrons in their nuclei. What are the atomic number and the mass number of these carbon atoms?
2. An isotope of uranium has an atomic number of 92 and a mass number of 235. What are the number of protons and neutrons in the nucleus of this atom?
Solution
1. If a carbon atom has six protons in its nucleus, its atomic number is 6. If it also has six neutrons in the nucleus, then the mass number is 6 + 6, or 12.
2. If the atomic number of uranium is 92, then that is the number of protons in the nucleus. Because the mass number is 235, then the number of neutrons in the nucleus is 235 − 92, or 143.
Exercise $1$
The number of protons in the nucleus of a tin atom is 50, while the number of neutrons in the nucleus is 68. What are the atomic number and the mass number of this isotope?
Answer
Atomic number = 50, mass number = 118
When referring to an atom, we simply use the element's name: the term sodium refers to the element as well as an atom of sodium. But it can be unwieldy to use the name of elements all the time. Instead, chemistry defines a symbol for each element. The atomic symbol is a one- or two-letter representation of the name of an element. By convention, the first letter of an element's symbol is always capitalized, while the second letter (if present) is lowercase. Thus, the symbol for hydrogen is H, the symbol for sodium is Na, and the symbol for nickel is Ni. Most symbols come from the English name of the element, although some symbols come from an element's Latin name. (The symbol for sodium, Na, comes from its Latin name, natrium.) Table $2$ lists some common elements and their symbols. You should memorize the symbols in Table $2$, as this is how we will be representing elements throughout chemistry.
Table $2$: Names and Symbols of Common Elements
Element Name Symbol Element Name Symbol
Aluminum Al Mercury Hg
Argon Ar Molybdenum Mo
Arsenic As Neon Ne
Barium Ba Nickel Ni
Beryllium Be Nitrogen N
Bismuth Bi Oxygen O
Boron B Palladium Pd
Bromine Br Phosphorus P
Calcium Ca Platinum Pt
Carbon C Potassium K
Chlorine Cl Radium Ra
Chromium Cr Radon Rn
Cobalt Co Rubidium Rb
Copper Cu Scandium Sc
Fluorine F Selenium Se
Gallium Ga Silicon Si
Germanium Ge Silver Ag
Gold Au Sodium Na
Helium He Strontium Sr
Hydrogen H Sulfur S
Iodine I Tantalum Ta
Iridium Ir Tin Sn
Iron Fe Titanium Ti
Krypton Kr Tungsten W
Lead Pb Uranium U
Lithium Li Xenon Xe
Magnesium Mg Zinc Zn
Manganese Mn Zirconium Zr
The elements are grouped together in a special chart called the periodic table of all the elements. A simple periodic table is shown in Figure $2$, while one may view a more extensive periodic table from another source. The elements on the periodic table are listed in order of ascending atomic number. The periodic table has a special shape that will become important to us when we consider the organization of electrons in atoms (Chapter 8). One immediate use of the periodic table helps us identify metals and nonmetals. Nonmetals are in the upper right-hand corner of the periodic table, on one side of the heavy line splitting the right-side part of the chart. All other elements are metals.
There is an easy way to represent isotopes using the atomic symbols. We use the construction:
$\ce{_{Z}^{A}X}\nonumber$
where $X$ is the symbol of the element, $A$ is the mass number, and $Z$ is the atomic number. Thus, for the isotope of carbon that has 6 protons and 6 neutrons, the symbol is:
$\ce{_{6}^{12}C}\nonumber$
where $C$ is the symbol for the element, 6 represents the atomic number, and 12 represents the mass number.
Example $2$
1. What is the symbol for an isotope of uranium that has an atomic number of 92 and a mass number of 235?
2. How many protons and neutrons are in $\ce{_{26}^{56}Fe}$
Solution
1. The symbol for this isotope is $\ce{_{92}^{235}U}$
2. This iron atom has 26 protons and 56 − 26 = 30 neutrons.
Exercise $2$
How many protons are in $\ce{_{11}^{23} Na}$
Answer
11 protons
It is also common to state the mass number after the name of an element to indicate a particular isotope. Carbon-12 represents an isotope of carbon with 6 protons and 6 neutrons, while uranium-238 is an isotope of uranium that has 146 neutrons.
Key Takeaways
• Chemistry is based on the modern atomic theory, which states that all matter is composed of atoms.
• Atoms themselves are composed of protons, neutrons, and electrons.
• Each element has its own atomic number, which is equal to the number of protons in its nucleus.
• Isotopes of an element contain different numbers of neutrons.
• Elements are represented by an atomic symbol.
• The periodic table is a chart that organizes all the elements. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.02%3A_Atomic_Theory.txt |
Learning Objectives
• Define molecule.
• Name simple molecules based on their formulas.
• Determine a formula of a molecule based on its name.
There are many substances that exist as two or more atoms connected together so strongly that they behave as a single particle. These multiatom combinations are called molecules. A molecule is the smallest part of a substance that has the physical and chemical properties of that substance. In some respects, a molecule is similar to an atom. A molecule, however, is composed of more than one atom.
Table \(1\): Elements That Exist as Diatomic Molecules
Hydrogen (\(\ce{H2}\)) Oxygen (\(\ce{O2}\)) Nitrogen (\(\ce{N2}\)) Fluorine (\(\ce{F2}\))
Chlorine (\(\ce{Cl2}\)) Bromine (\(\ce{Br2}\)) Iodine (\(\ce{I2}\))
Some elements exist naturally as molecules. For example, hydrogen and oxygen exist as two-atom molecules. Other elements also exist naturally as diatomic molecules (Table \(1\)). As with any molecule, these elements are labeled with a molecular formula, a formal listing of what and how many atoms are in a molecule. (Sometimes only the word formula is used, and its meaning is inferred from the context.) For example, the molecular formula for elemental hydrogen is H2, with H being the symbol for hydrogen and the subscript 2 implying that there are two atoms of this element in the molecule. Other diatomic elements have similar formulas: O2, N2, and so forth. Other elements exist as molecules—for example, sulfur normally exists as an eight-atom molecule, S8, while phosphorus exists as a four-atom molecule, P4 (Figure \(1\)). Otherwise, we will assume that elements exist as individual atoms, rather than molecules. It is assumed that there is only one atom in a formula if there is no numerical subscript on the right side of an element's symbol.
Figure \(1\) shows two examples of how we will be representing molecules in this text. An atom is represented by a small ball or sphere, which generally indicates where the nucleus is in the molecule. A cylindrical line connecting the balls represents the connection between the atoms that make this collection of atoms a molecule. This connection is called a chemical bond and is the connection between two atoms in a molecule.
Many compounds exist as molecules. In particular, when nonmetals connect with other nonmetals, the compounds typically exist as molecules. (Compounds between a metal and a nonmetal are different and will be considered in Section 3.4.) In some cases, there are many different kinds of molecules that can be formed between any given elements, with all the different molecules having different chemical and physical properties. How do we tell them apart?
The answer is a very specific system of naming compounds, called chemical nomenclature. By following the rules of nomenclature, each and every compound has its own unique name, and each name refers to one and only one compound. Here, we will start with relatively simple molecules that have only two elements in them, the so-called binary compounds:
1. Identify the elements in the molecule from its formula.
2. Begin the name with the element name of the first element. If there is more than one atom of this element in the molecular formula, use a numerical prefix to indicate the number of atoms, as listed in Table \(2\). Do not use the prefix mono- if there is only one atom of the first element.
3. Name the second element by using three pieces:
• a numerical prefix indicating the number of atoms of the second element, plus
• the stem of the element name (e.g., ox for oxygen, chlor for chlorine, etc.), plus
• the suffix -ide
4. Combine the two words, leaving a space between them.
Table \(2\): Numerical Prefixes Used in Naming Molecular Compounds
The Number of Atoms of an Element Prefix
1 mono-
2 di-
3 tri-
4 tetra-
5 penta-
6 hexa-
7 hepta-
8 octa-
9 nona-
10 deca-
Let us see how these steps work for a molecule whose molecular formula is SO2, which has one sulfur atom and two oxygen atoms—this completes step 1. According to step 2, we start with the name of the first element—sulfur. Remember, we do not use the mono- prefix for the first element. Now for step 3, we combine the numerical prefix di- (see Table \(2\)) with the stem ox- and the suffix -ide, to make dioxide. Bringing these two words together, we have the unique name for this compound—sulfur dioxide.
Why all this trouble? There is another common compound consisting of sulfur and oxygen whose molecular formula is SO3, so the compounds need to be distinguished. SO3 has three oxygen atoms in it, so it is a different compound with different chemical and physical properties. The system of chemical nomenclature is designed to give this compound its own unique name. Its name, if you go through all the steps, is sulfur trioxide. Different compounds have different names.
In some cases, when a prefix ends in a or o; and the element name begins with o, we drop the a or o on the prefix. So we see monoxide or pentoxide, rather than monooxide or pentaoxide in molecule names.
Example \(1\)
Name each molecule.
1. PF3
2. CO
3. Se2Br2
Solution
1. A molecule with a single phosphorus atom and three fluorine atoms is called phosphorus trifluoride.
2. A compound with one carbon atom and one oxygen atom is properly called carbon monoxide, not carbon monooxide.
3. There are two atoms of each element, selenium and bromine. According to the rules, the proper name here is diselenium dibromide.
Exercise \(1\)
Name each molecule.
1. SF4
2. P2S5
Answer a
sulfur tetrafluoride
Answer b
diphosphorus pentasulfide
One great thing about this system is that it works both ways. From the name of a compound, you should be able to determine its molecular formula. Simply list the element symbols, with a numerical subscript if there is more than one atom of that element, in the order of the name (we do not use the subscript 1 if there is only one atom of the element present; 1 is implied). From the name nitrogen trichloride, you should be able to get NCl3 as the formula for this molecule. From the name diphosphorus pentoxide, you should be able to get the formula P2O5 (note the numerical prefix on the first element, indicating there is more than one atom of phosphorus in the formula).
Example \(2\)
Give the formula for each molecule.
1. carbon tetrachloride
2. silicon dioxide
3. trisilicon tetranitride
Solution
1. The name carbon tetrachloride implies one carbon atom and four chlorine atoms, so the formula is CCl4.
2. The name silicon dioxide implies one silicon atom and two oxygen atoms, so the formula is SiO2.
3. We have a name that has numerical prefixes on both elements. Tri- means three, and tetra- means four, so the formula of this compound is Si3N4.
Exercise \(2\)
Give the formula for each molecule.
1. disulfur difluoride
2. iodine pentabromide
Answer a
\(\ce{S2F2 }\)
Answer b
\(\ce{IBr5}\)
Some simple molecules have common names that we use as part of the formal system of chemical nomenclature. For example, H2O is given the name water, not dihydrogen monoxide. NH3 is called ammonia, while CH4 is called methane. We will occasionally see other molecules that have common names; we will point them out as they occur.
Key Takeaways
• Molecules are groups of atoms that behave as a single unit.
• Some elements exist as molecules: hydrogen, oxygen, sulfur, and so forth.
• There are rules that can express a unique name for any given molecule, and a unique formula for any given name. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.03%3A_Molecules_and_Chemical_Nomenclature.txt |
Learning Objective
• Express the masses of atoms and molecules.
Because matter is defined as anything that has mass and takes up space, it should not be surprising to learn that atoms and molecules have mass.
Individual atoms and molecules, however, are very small, and the masses of individual atoms and molecules are also very small. For macroscopic objects, we use units such as grams and kilograms to state their masses, but these units are much too big to comfortably describe the masses of individual atoms and molecules. Another scale is needed.
Atomic Mass Unit
The atomic mass unit (u; some texts use amu, but this older style is no longer accepted) is defined as one-twelfth of the mass of a carbon-12 atom, an isotope of carbon that has six protons and six neutrons in its nucleus. By this scale, the mass of a proton is 1.00728 u, the mass of a neutron is 1.00866 u, and the mass of an electron is 0.000549 u. There will not be much error if you estimate the mass of an atom by simply counting the total number of protons and neutrons in the nucleus (i.e., identify its mass number) and ignore the electrons. Thus, the mass of carbon-12 is about 12 u, the mass of oxygen-16 is about 16 u, and the mass of uranium-238 is about 238 u. More exact masses are found in scientific references—for example, the exact mass of uranium-238 is 238.050788 u, so you can see that we are not far off by using the whole-number value as the mass of the atom.
What is the mass of an element? This is somewhat more complicated because most elements exist as a mixture of isotopes, each of which has its own mass. Thus, although it is easy to speak of the mass of an atom, when talking about the mass of an element, we must take the isotopic mixture into account.
Atomic Mass
The atomic mass of an element is a weighted average of the masses of the isotopes that compose an element. What do we mean by a weighted average? Well, consider an element that consists of two isotopes, 50% with mass 10 u and 50% with mass 11 u. A weighted average is found by multiplying each mass by its fractional occurrence (in decimal form) and then adding all the products. The sum is the weighted average and serves as the formal atomic mass of the element. In this example, we have the following:
0.50 × 10 u = 5.0 u
0.50 × 11 u = 5.5 u
Sum = 10.5 u = the atomic mass of our element
Note that no atom in our hypothetical element has a mass of 10.5 u; rather, that is the average mass of the atoms, weighted by their percent occurrence.
This example is similar to a real element. Boron exists as about 20% boron-10 (five protons and five neutrons in the nuclei) and about 80% boron-11 (five protons and six neutrons in the nuclei). The atomic mass of boron is calculated similarly to what we did for our hypothetical example, but the percentages are different:
0.20 × 10 u = 2.0 u
0.80 × 11 u = 8.8 u
Sum = 10.8 u = the atomic mass of boron
Thus, we use 10.8 u for the atomic mass of boron.
Virtually all elements exist as mixtures of isotopes, so atomic masses may vary significantly from whole numbers. Table \(1\) lists the atomic masses of some elements. The atomic masses in Table \(1\) are listed to three decimal places where possible, but in most cases, only one or two decimal places are needed. Note that many of the atomic masses, especially the larger ones, are not very close to whole numbers. This is, in part, the effect of an increasing number of isotopes as the atoms increase in size. (The record number is 10 isotopes for tin.)
Table \(1\): Selected Atomic Masses of Some Elements
Element Name Atomic Mass (u) Element Name Atomic Mass (u)
Aluminum 26.981 Molybdenum 95.94
Argon 39.948 Neon 20.180
Arsenic 74.922 Nickel 58.693
Barium 137.327 Nitrogen 14.007
Beryllium 9.012 Oxygen 15.999
Bismuth 208.980 Palladium 106.42
Boron 10.811 Phosphorus 30.974
Bromine 79.904 Platinum 195.084
Calcium 40.078 Potassium 39.098
Carbon 12.011 Radium n/a
Chlorine 35.453 Radon n/a
Cobalt 58.933 Rubidium 85.468
Copper 63.546 Scandium 44.956
Fluorine 18.998 Selenium 78.96
Gallium 69.723 Silicon 28.086
Germanium 72.64 Silver 107.868
Gold 196.967 Sodium 22.990
Helium 4.003 Strontium 87.62
Hydrogen 1.008 Sulfur 32.065
Iodine 126.904 Tantalum 180.948
Iridium 192.217 Tin 118.710
Iron 55.845 Titanium 47.867
Krypton 83.798 Tungsten 183.84
Lead 207.2 Uranium 238.029
Lithium 6.941 Xenon 131.293
Magnesium 24.305 Zinc 65.409
Manganese 54.938 Zirconium 91.224
Mercury 200.59 Molybdenum 95.94
Note: Atomic mass is given to three decimal places, if known.
Molecular Mass
Now that we understand that atoms have mass, it is easy to extend the concept to the mass of molecules. The molecular mass is the sum of the masses of the atoms in a molecule. This may seem like a trivial extension of the concept, but it is important to count the number of each type of atom in the molecular formula. Also, although each atom in a molecule is a particular isotope, we use the weighted average, or atomic mass, for each atom in the molecule.
For example, if we were to determine the molecular mass of dinitrogen trioxide, N2O3, we would need to add the atomic mass of nitrogen two times with the atomic mass of oxygen three times:
2 N masses = 2 × 14.007 u = 28.014 u
3 O masses = 3 × 15.999 u = 47.997 u
Total = 76.011 u = the molecular mass of N2O3
We would not be far off if we limited our numbers to one or even two decimal places.
Example \(1\)
What is the molecular mass of each substance?
1. \(\ce{NBr3}\)
2. \(\ce{C2H6}\)
Solution
Add one atomic mass of nitrogen and three atomic masses of bromine:
Solutions to Example 3.4.1
1 N mass = 14.007 u
3 Br masses = 3 × 79.904 u = 239.712 u
Total = 253.719 u = the molecular mass of NBr3
Add two atomic masses of carbon and six atomic masses of hydrogen:
Solutions to Example 3.4.1
2 C masses = 2 × 12.011 u = 24.022 u
6 H masses = 6 × 1.008 u = 6.048 u
Total = 30.070 u = the molecular mass of C2H6
The compound \(\ce{C2H6}\) also has a common name—ethane.
Exercise \(1\)
What is the molecular mass of each substance?
1. \(\ce{SO2}\)
2. \(\ce(PF3}\)
Answer a
64.063 u
Answer b
87.968 u
Chemistry is Everywhere: Sulfur Hexafluoride
On March 20, 1995, the Japanese terrorist group Aum Shinrikyo (Sanskrit for "Supreme Truth") released some sarin gas in the Tokyo subway system; twelve people were killed, and thousands were injured (Figure \(\PageIndex{2a}\)). Sarin (molecular formula \(\ce{C4H10FPO2}\)) is a nerve toxin that was first synthesized in 1938. It is regarded as one of the most deadly toxins known, estimated to be about 500 times more potent than cyanide. Scientists and engineers who study the spread of chemical weapons such as sarin (yes, there are such scientists) would like to have a similar less dangerous chemical to study, indeed one that is nontoxic, so they are not at risk themselves.
Sulfur hexafluoride is used as a model compound for sarin. \(\ce{SF6}\) (a molecular model of which is shown Figure \(\PageIndex{2b}\)) has a similar molecular mass (about 146 u) as sarin (about 140 u), so it has similar physical properties in the vapor phase. Sulfur hexafluoride is also very easy to accurately detect, even at low levels, and it is not a normal part of the atmosphere, so there is little potential for contamination from natural sources. Consequently, \(\ce{SF6}\) is also used as an aerial tracer for ventilation systems in buildings. It is nontoxic and very chemically inert, so workers do not have to take special precautions other than watching for asphyxiation.
Sulfur hexafluoride also has another interesting use: a spark suppressant in high-voltage electrical equipment. High-pressure \(\ce{SF6}\) gas is used in place of older oils that may have contaminants that are not environmentally friendly (Figure \(\PageIndex{2c}\)).
Key Takeaways
• The atomic mass unit (u) is a unit that describes the masses of individual atoms and molecules.
• The atomic mass is the weighted average of the masses of all isotopes of an element.
• The molecular mass is the sum of the masses of the atoms in a molecule. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.04%3A_Masses_of_Atoms_and_Molecules.txt |
Learning Objectives
• Know how ions form.
• Learn the characteristic charges that ions have.
• Construct a proper formula for an ionic compound.
• Generate a proper name for an ionic compound.
So far, we have discussed elements and compounds that are electrically neutral. They have the same number of electrons as protons, so the negative charges of the electrons are balanced by the positive charges of the protons. However, this is not always the case. Electrons can move from one atom to another; when they do, species with overall electric charges are formed. Such species are called ions. Species with overall positive charges are termed cations, while species with overall negative charges are called anions. Remember that ions are formed only when electrons move from one atom to another; a proton never moves from one atom to another. Compounds formed from positive and negative ions are ionic compounds.
Individual atoms can gain or lose electrons. When they do, they become monatomic ions. When atoms gain or lose electrons, they usually gain or lose a characteristic number of electrons and so take on a characteristic overall charge. Table \(1\) lists some common ions in terms of how many electrons they lose (making cations) or gain (making anions). There are several things to notice about the ions in Table \(1\). First, each element that forms a cation is a metal, except for one (hydrogen), while each element that forms an anion is a nonmetal. This is actually one of the chemical properties of metals and nonmetals: metals tend to form cations, while nonmetals tend to form anions. Second, most atoms form ions of a single characteristic charge. When sodium atoms form ions, they always form a 1+ charge, never a 2+ or 3+ or even 1− charge. Thus, if you commit the information in Table \(1\) to memory, you will always know what charges most atoms form.
Table \(1\): Monatomic Ions of Various Charges
Ions formed by losing a single electron H+
Na+
K+
Rb+
Ag+
Au+
Ions formed by losing two electrons Mg2+
Ca2+
Sr2+
Fe2+
Co2+
Ni2+
Cu2+
Zn2+
Sn2+
Hg2+
Pb2+
Ions formed by losing three electrons Sc3+
Fe3+
Co3+
Ni3+
Au3+
Al3+
Cr3+
Ions formed by losing four electrons Ti4+
Sn4+
Pb4+
Ions formed by gaining a single electron F
Cl
Br
I
Ions formed by gaining two electrons O2−
S2−
Se2
Ions formed by gaining three electrons N3−
P3−
Third, there are some exceptions to the previous point. A few elements, all of which are metals, can form more than one possible charge. For example, iron atoms can form 2+ cations or 3+ cations. Cobalt is another element that can form more than one possible charged ion (2+ and 3+), while lead can form 2+ or 4+ cations. Unfortunately, there is little understanding which two charges a metal atom may take, so it is best to just memorize the possible charges a particular element can have.
Note the convention for indicating an ion. The magnitude of the charge is listed as a right superscript next to the symbol of the element. If the charge is a single positive or negative one, the number 1 is not written; if the magnitude of the charge is greater than 1, then the number is written before the + or − sign. An element symbol without a charge written next to it is assumed to be the uncharged atom.
Naming an ion is straightforward. For a cation, simply use the name of the element and add the word ion (or if you want to be more specific, add cation) after the element's name. So Na+ is the sodium ion; Ca2+ is the calcium ion. If the element has more than one possible charge, the value of the charge comes after the element name and before the word ion. Thus, Fe2+ is the iron two ion, while Fe3+ is the iron three ion. In print, we use roman numerals in parentheses to represent the charge on the ion; so these two iron ions would be represented as the iron(II) cation and the iron(III) cation, respectively.
For a monatomic anion, use the stem of the element name and append the suffix -ide to it, and then add ion. This is similar to how we named molecular compounds. Thus, Cl is the chloride ion, and N3− is the nitride ion.
Example \(1\)
Name each species.
1. O2−
2. Co
3. Co2+
Solution
1. This species has a 2− charge on it, so it is an anion. Anions are named using the stem of the element name with the suffix -ide added. This is the oxide anion.
2. Because this species has no charge, it is an atom in its elemental form. This is cobalt.
3. In this case, there is a 2+ charge on the atom, so it is a cation. We note from Table \(1\) that cobalt cations can have two possible charges, so the name of the ion must specify which charge the ion has. This is the cobalt(II) cation.
Exercise \(1\)
Name each species.
1. P3−
2. Sr2+
Answers
1. the phosphide anion
2. the strontium cation
Ionic Formulas
Chemical formulas for ionic compounds are called ionic formulas. A proper ionic formula has a cation and an anion in it; an ionic compound is never formed between two cations only or two anions only. The key to writing proper ionic formulas is simple: the total positive charge must balance the total negative charge. Because the charges on the ions are characteristic, sometimes we have to have more than one of a cation or an anion to balance the overall positive and negative charges. It is conventional to use the lowest ratio of ions that are needed to balance the charges.
For example, consider the ionic compound between Na+ and Cl. Each ion has a single charge, one positive and one negative, so we need only one ion of each to balance the overall charge. When writing the ionic formula, we follow two additional conventions: (1) write the formula for the cation first and the formula for the anion second, but (2) do not write the charges on the ions. Thus, for the compound between Na+ and Cl, we have the ionic formula NaCl (Figure \(1\)). The formula Na2Cl2 also has balanced charges, but the convention is to use the lowest ratio of ions, which would be one of each. (Remember from our conventions for writing formulas that we do not write a 1 subscript if there is only one atom of a particular element present.) For the ionic compound between magnesium cations (Mg2+) and oxide anions (O2−), again we need only one of each ion to balance the charges. By convention, the formula is MgO.
For the ionic compound between Mg2+ ions and Cl ions, we now consider the fact that the charges have different magnitudes: 2+ on the magnesium ion and 1− on the chloride ion. To balance the charges with the lowest number of ions possible, we need to have two chloride ions to balance the charge on the one magnesium ion. Rather than write the formula MgClCl, we combine the two chloride ions and write it with a 2 subscript: MgCl2.
What is the formula MgCl2 telling us? There are two chloride ions in the formula. Although chlorine as an element is a diatomic molecule, Cl2, elemental chlorine is not part of this ionic compound. The chlorine is in the form of a negatively charged ion, not the neutral element. The 2 subscript is in the ionic formula because we need two Cl ions to balance the charge on one Mg2+ ion.
Example \(2\)
Write the proper ionic formula for each of the two given ions.
1. Ca2+ and Cl
2. Al3+ and F
3. Al3+ and O2−
Solution
1. We need two Cl ions to balance the charge on one Ca2+ ion, so the proper ionic formula is CaCl2.
2. We need three F ions to balance the charge on the Al3+ ion, so the proper ionic formula is AlF3.
3. With Al3+ and O2−, note that neither charge is a perfect multiple of the other. This means we have to go to a least common multiple, which in this case will be six. To get a total of 6+, we need two Al3+ ions; to get 6−, we need three O2− ions. Hence the proper ionic formula is Al2O3.
Exercise \(2\)
Write the proper ionic formulas for each of the two given ions.
1. Fe2+ and S2−
2. Fe3+ and S2−
Answers
1. FeS
2. Fe2S3
Naming ionic compounds is simple: combine the name of the cation and the name of the anion, in both cases omitting the word ion. Do not use numerical prefixes if there is more than one ion necessary to balance the charges. NaCl is sodium chloride, a combination of the name of the cation (sodium) and the anion (chloride). MgO is magnesium oxide. MgCl2 is magnesium chloride—not magnesium dichloride.
When naming ionic compounds whose cations can have more than one possible charge, we must also include the charge, in parentheses and in roman numerals, as part of the name. Hence FeS is iron(II) sulfide, while Fe2S3 is iron(III) sulfide. Again, no numerical prefixes appear in the name. The number of ions in the formula is dictated by the need to balance the positive and negative charges.
Example \(3\)
Name each ionic compound.
1. CaCl2
2. AlF3
3. Co2O3
Solution
1. Using the names of the ions, this ionic compound is named calcium chloride. It is not calcium(II) chloride, because calcium forms only one cation when it forms an ion, and it has a characteristic charge of 2+.
2. The name of this ionic compound is aluminum fluoride.
3. We know that cobalt can have more than one possible charge; we just need to determine what it is. Oxide always has a 2− charge, so with three oxide ions, we have a total negative charge of 6−. This means that the two cobalt ions have to contribute 6+, which for two cobalt ions means that each one is 3+. Therefore, the proper name for this ionic compound is cobalt(III) oxide.
Exercise \(3\)
Name each ionic compound.
1. Sc2O3
2. AgCl
Answers
1. scandium oxide
2. silver chloride
How do you know whether a formula—and by extension, a name—is for a molecular compound or for an ionic compound? Molecular compounds form between nonmetals and nonmetals, while ionic compounds form between metals and nonmetals. The periodic table can be used to determine which elements are metals and nonmetals.
There also exists a group of ions that contain more than one atom. These are called polyatomic ions. Table \(2\) lists the formulas, charges, and names of some common polyatomic ions. Only one of them, the ammonium ion, is a cation; the rest are anions. Most of them also contain oxygen atoms, so sometimes they are referred to as oxyanions. Some of them, such as nitrate and nitrite, and sulfate and sulfite, have very similar formulas and names, so care must be taken to get the formulas and names correct. Note that the -ite polyatomic ion has one less oxygen atom in its formula than the -ate ion but with the same ionic charge.
Table \(2\): Common Polyatomic Ions
Name Formula and Charge Name Formula and Charge
ammonium NH4+ hydroxide OH
acetate C2H3O2, or CH3COO nitrate NO3
bicarbonate (hydrogen carbonate) HCO3 nitrite NO2
bisulfate (hydrogen sulfate) HSO4 peroxide O22
carbonate CO32 perchlorate ClO4
chlorate ClO3 phosphate PO43
chromate CrO42 sulfate SO42
cyanide CN sulfite SO32
dichromate Cr2O72 triiodide I3
The naming of ionic compounds that contain polyatomic ions follows the same rules as the naming for other ionic compounds: simply combine the name of the cation and the name of the anion. Do not use numerical prefixes in the name if there is more than one polyatomic ion; the only exception to this is if the name of the ion itself contains a numerical prefix, such as dichromate or triiodide.
Writing the formulas of ionic compounds has one important difference. If more than one polyatomic ion is needed to balance the overall charge in the formula, enclose the formula of the polyatomic ion in parentheses and write the proper numerical subscript to the right and outside of the parentheses. Thus, the formula between calcium ions, Ca2+, and nitrate ions, NO3, is properly written Ca(NO3)2, not CaNO32 or CaN2O6. Use parentheses where required. The name of this ionic compound is simply calcium nitrate.
Example \(4\)
Write the proper formula and give the proper name for each ionic compound formed between the two listed ions.
1. NH4+ and S2−
2. Al3+ and PO43
3. Fe2+ and PO43
Solution
1. Because the ammonium ion has a 1+ charge and the sulfide ion has a 2− charge, we need two ammonium ions to balance the charge on a single sulfide ion. Enclosing the formula for the ammonium ion in parentheses, we have (NH4)2S. The compound's name is ammonium sulfide.
2. Because the ions have the same magnitude of charge, we need only one of each to balance the charges. The formula is AlPO4, and the name of the compound is aluminum phosphate.
3. Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6. To get 6+, we need three iron(II) ions, and to get 6−, we need two phosphate ions. The proper formula is Fe3(PO4)2, and the compound's name is iron(II) phosphate.
Exercise \(4\)
Write the proper formula and give the proper name for each ionic compound formed between the two listed ions.
1. NH4+ and PO43
2. Co3+ and NO2
Answers
1. (NH4)3PO4, ammonium phosphate
2. Co(NO2)3, cobalt(III) nitrite
Food and Drink Application: Sodium in Your Food
The element sodium, at least in its ionic form as Na+, is a necessary nutrient for humans to live. In fact, the human body is approximately 0.15% sodium, with the average person having one-twentieth to one-tenth of a kilogram in their body at any given time, mostly in fluids outside cells and in other bodily fluids.
Sodium is also present in our diet. The common table salt we use on our foods is an ionic sodium compound. Many processed foods also contain significant amounts of sodium added to them as a variety of ionic compounds. Why are sodium compounds used so much? Usually sodium compounds are inexpensive, but, more importantly, most ionic sodium compounds dissolve easily. This allows processed food manufacturers to add sodium-containing substances to food mixtures and know that the compound will dissolve and distribute evenly throughout the food. Simple ionic compounds such as sodium nitrite (NaNO2) are added to cured meats, such as bacon and deli-style meats, while a compound called sodium benzoate is added to many packaged foods as a preservative. Table \(3\) is a partial list of some sodium additives used in food. Some of them you may recognize after reading this chapter. Others you may not recognize, but they are all ionic sodium compounds with some negatively charged ion also present.
Table \(3\): Some Sodium Compounds Added to Food
Sodium Compound Use in Food
Sodium acetate preservative, acidity regulator
Sodium adipate food acid
Sodium alginate thickener, vegetable gum, stabilizer, gelling agent, emulsifier
Sodium aluminum phosphate acidity regulator, emulsifier
Sodium aluminosilicate anti-caking agent
Sodium ascorbate antioxidant
Sodium benzoate preservative
Sodium bicarbonate mineral salt
Sodium bisulfite preservative, antioxidant
Sodium carbonate mineral salt
Sodium carboxymethylcellulose emulsifier
Sodium citrates food acid
Sodium dehydroacetate preservative
Sodium erythorbate antioxidant
Sodium erythorbin antioxidant
Sodium ethyl para-hydroxybenzoate preservative
Sodium ferrocyanide anti-caking agent
Sodium formate preservative
Sodium fumarate food acid
Sodium gluconate stabilizer
Sodium hydrogen acetate preservative, acidity regulator
Sodium hydroxide mineral salt
Sodium lactate food acid
Sodium malate food acid
Sodium metabisulfite preservative, antioxidant, bleaching agent
Sodium methyl para-hydroxybenzoate preservative
Sodium nitrate preservative, color fixative
Sodium nitrite preservative, color fixative
Sodium orthophenyl phenol preservative
Sodium propionate preservative
Sodium propyl para-hydroxybenzoate preservative
Sodium sorbate preservative
Sodium stearoyl lactylate emulsifier
Sodium succinates acidity regulator, flavor enhancer
Sodium salts of fatty acids emulsifier, stabilizer, anti-caking agent
Sodium sulfite mineral salt, preservative, antioxidant
Sodium sulfite preservative, antioxidant
Sodium tartrate food acid
Sodium tetraborate preservative
The use of so many sodium compounds in prepared and processed foods has alarmed some physicians and nutritionists. They argue that the average person consumes too much sodium from his or her diet. The average person needs only about 500 mg of sodium every day; most people consume more than this—up to 10 times as much. Some studies have implicated increased sodium intake with high blood pressure; newer studies suggest that the link is questionable. However, there has been a push to reduce the amount of sodium most people ingest every day: avoid processed and manufactured foods, read labels on packaged foods (which include an indication of the sodium content), avoid oversalting foods, and use other herbs and spices besides salt in cooking.
Key Takeaways
• Ions form when atoms lose or gain electrons.
• Ionic compounds have positive ions and negative ions.
• Ionic formulas balance the total positive and negative charges.
• Ionic compounds have a simple system of naming.
• Groups of atoms can have an overall charge and make ionic compounds. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.05%3A_Ions_and_Ionic_Compounds.txt |
Learning Objectives
• Define acid.
• Name a simple acid.
There is one other group of compounds that is important to us—acids—and these compounds have interesting chemical properties. Initially, we will define an acid as an ionic compound of the H+ cation dissolved in water. To indicate that something is dissolved in water, we will use the phase label (aq) next to a chemical formula (where aq stands for "aqueous," a word that describes something dissolved in water). If the formula does not have this label, then the compound is treated as a molecular compound rather than an acid.
Acids have their own nomenclature system. If an acid is composed of only hydrogen and one other element, the name is hydro- + the stem of the other element + -ic acid. For example, the compound HCl(aq) is hydrochloric acid, while H2S(aq) is hydrosulfuric acid. If these acids were not dissolved in water, the compounds would be called hydrogen chloride and hydrogen sulfide, respectively. Both of these substances are well known as molecular compounds; when dissolved in water, however, they are treated as acids.
If a compound is composed of hydrogen ions and a polyatomic anion, then the name of the acid is derived from the stem of the polyatomic ion's name. Typically, if the anion name ends in -ate, the name of the acid is the stem of the anion name plus -ic acid; if the related anion's name ends in -ite, the name of the corresponding acid is the stem of the anion name plus -ous acid. Table \(1\) lists the formulas and names of a variety of acids that you should be familiar with. You should recognize most of the anions in the formulas of the acids.
Table \(1\) Names and Formulas of Acids
Formula Name
HC2H3O2 acetic acid
HClO3 chloric acid
HCl hydrochloric acid
HBr hydrobromic acid
HI hydriodic acid
HF hydrofluoric acid
HNO3 nitric acid
H2C2O4 oxalic acid
HClO4 perchloric acid
H3PO4 phosphoric acid
H2SO4 sulfuric acid
H2SO3 sulfurous acid
Note: The "aq" label is omitted for clarity.
Example \(1\)
Name each acid without consulting Table 3.9.
1. HBr
2. H2SO4
Solution
1. As a binary acid, the acid's name is hydro- + stem name + -ic acid. Because this acid contains a bromine atom, the name is hydrobromic acid.
2. Because this acid is derived from the sulfate ion, the name of the acid is the stem of the anion name + -ic acid. The name of this acid is sulfuric acid.
Exercise \(1\)
Name each acid.
1. HF
2. HNO2
Answer a
hydrofluoric acid
Answer b
nitrous acid
All acids have some similar properties. For example, acids have a sour taste; in fact, the sour taste of some of our foods, such as citrus fruits and vinegar, is caused by the presence of acids in food. Many acids react with some metallic elements to form metal ions and elemental hydrogen. Acids make certain plant pigments change colors; indeed, the ripening of some fruits and vegetables is caused by the formation or destruction of excess acid in the plant. In Chapter 12, we will explore the chemical behavior of acids.
Acids are very prevalent in the world around us. We have already mentioned that citrus fruits contain acid; among other compounds, they contain citric acid, H3C6H5O7(aq). Oxalic acid, H2C2O4(aq), is found in spinach and other green leafy vegetables. Hydrochloric acid not only is found in the stomach (stomach acid), but also can be bought in hardware stores as a cleaner for concrete and masonry. Phosphoric acid is an ingredient in some soft drinks.
Key Takeaways
• An acid is a compound of the H+ ion dissolved in water.
• Acids have their own naming system.
• Acids have certain chemical properties that distinguish them from other compounds. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.06%3A_Acids.txt |
3.1: Atomic Theory
1. List the three statements that make up the modern atomic theory.
2. Explain how atoms are composed.
3. Which is larger, a proton or an electron?
4. Which is larger, a neutron or an electron?
5. What are the charges for each of the three subatomic particles?
6. Where is most of the mass of an atom located?
7. Sketch a diagram of a boron atom, which has five protons and six neutrons in its nucleus.
8. Sketch a diagram of a helium atom, which has two protons and two neutrons in its nucleus.
9. Define atomic number. What is the atomic number for a boron atom?
10. What is the atomic number of helium?
11. Define isotope and give an example.
12. What is the difference between deuterium and tritium?
13. Which pair represents isotopes?
1. \[_{2}^{4}\textrm{He} \, and\: \, _{2}^{3}\textrm{He}\]
2. \[_{26}^{56}\textrm{Fe} \, and\: \, _{25}^{56}\textrm{Mn}\]
3. \[_{14}^{28}\textrm{Si} \, and\: \, _{15}^{31}\textrm{P}\]
14. Which pair represents isotopes?
1. \[_{20}^{40}\textrm{Ca} \, and\: \, _{19}^{40}\textrm{K}\]
2. \[_{26}^{56}\textrm{Fe} \, and\: \, _{28}^{56}\textrm{Fe}\]
3. \[_{92}^{238}\textrm{U} \, and\: \, _{92}^{235}\textrm{U}\]
1. Give complete symbols of each atom, including the atomic number and the mass number.
a. an oxygen atom with 8 protons and 8 neutrons
b. a potassium atom with 19 protons and 20 neutrons
c. a lithium atom with 3 protons and 4 neutron
2. Give complete symbols of each atom, including the atomic number and the mass number.
a. a magnesium atom with 12 protons and 12 neutrons
b. a magnesium atom with 12 protons and 13 neutrons
c. a xenon atom with 54 protons and 77 neutron
3. Americium-241 is an isotope used in smoke detectors. What is the complete symbol for this isotope?
18. Carbon-14 is an isotope used to perform radioactive dating tests on previously living material. What is the complete symbol for this isotope?
• Give atomic symbols for each element.
a. sodium
b. argon
c. nitrogen
d. radon
• Give atomic symbols for each element.
a. silver
b. gold
c. mercury
d. iodine
• Give the name of the element.
a.Si
b. Mn
c. Fe
d. Cr
• Give the name of the element.
a. F
b. Cl
c. Br
d. I
Answers
1. All matter is composed of atoms; atoms of the same element are the same, and atoms of different elements are different; atoms combine in whole-number ratios to form compounds.
2.
3. A proton is larger than an electron.
4. proton: 1+; electron: 1−; neutron: 0
5. The atomic number is the number of protons in a nucleus. Boron has an atomic number of five.
6.
7. Isotopes are atoms of the same element but with different numbers of neutrons. \[_{1}^{1}\textrm{H} \, and\: \, _{1}^{2}\textrm{H}\]
1. isotopes
2. not isotopes
3. not isotopes
1.
8.
1. \[_{8}^{16}\textrm{O}\]
2. \[_{19}^{39}\textrm{K}\]
3. \[_{3}^{7}\textrm{Li}\]
9. \[_{95}^{241}\textrm{Am}\]
10.
1. Na
2. Ar
3. N
4. Rn
11.
1. silicon
2. manganese
3. iron
4. chromium
3.2: Molecules and Chemical Nomenclature
1. Which of these formulas represent molecules? State how many atoms are in each molecule.
1. Fe
2. PCl3
3. P4
4. Ar
2. Which of these formulas represent molecules? State how many atoms are in each molecule.
1. I2
2. He
3. H2O
4. Al
3. What is the difference between CO and Co?
4. What is the difference between H2O and H2O2 (hydrogen peroxide)?
5. Give the proper formula for each diatomic element.
6. In 1986, when Halley’s comet last passed the earth, astronomers detected the presence of S2 in their telescopes. Why is sulfur not considered a diatomic element?
7. What is the stem of fluorine used in molecule names? CF4 is one example.
1. What is the stem of selenium used in molecule names? SiSe2 is an example.
2. Give the proper name for each molecule.
1. PF3
2. TeCl2
3. N2O3
3. Give the proper name for each molecule.
1. NO
2. CS2
3. As2O3
4. Give the proper name for each molecule.
1. XeF2
2. O2F2
3. SF6
5. Give the proper name for each molecule.
1. P4O10
2. B2O3
3. P2S3
6. Give the proper name for each molecule.
1. N2O
2. N2O4
3. N2O5
7. Give the proper name for each molecule.
1. SeO2
2. Cl2O
3. XeF6
8. Give the proper formula for each name.
1. dinitrogen pentoxide
2. tetraboron tricarbide
3. phosphorus pentachloride
9. Give the proper formula for each name.
1. nitrogen triiodide
2. diarsenic trisulfide
3. iodine trichloride
10. Give the proper formula for each name.
1. dioxygen dichloride
2. dinitrogen trisulfide
3. xenon tetrafluoride
11. Give the proper formula for each name.
1. chlorine dioxide
2. selenium dibromide
3. dinitrogen trioxide
12. Give the proper formula for each name.
1. iodine trifluoride
2. xenon trioxide
3. disulfur decafluoride
13. Give the proper formula for each name.
1. germanium dioxide
2. carbon disulfide
3. diselenium dibromide
Answers
1. not a molecule
2. a molecule; four atoms total
3. a molecule; four atoms total
1.
2. CO is a compound of carbon and oxygen; Co is the element cobalt.
3.
4. H2, O2, N2, F2, Cl2, Br2, I2
5.
6. fluor-
7.
1. phosphorus trifluoride
2. tellurium dichloride
3. dinitrogen trioxide
8.
1. xenon difluoride
2. dioxygen difluoride
3. sulfur hexafluoride
9.
1. dinitrogen monoxide
2. dinitrogen tetroxide
3. dinitrogen pentoxide
10.
1. N2O5
2. B4C3
3. PCl5
11.
1. O2Cl2
2. N2S3
3. XeF4
12.
1. IF3
2. XeO3
3. S2F10
3.3: Masses of Atoms and Molecules
1. Define atomic mass unit. What is its abbreviation?
2. Define atomic mass. What is its unit?
3. Estimate the mass, in whole numbers, of each isotope.
1. hydrogen-1
2. hydrogen-3
3. iron-56
4. Estimate the mass, in whole numbers, of each isotope.
1. phosphorus-31
2. carbon-14
3. americium-241
5. Determine the atomic mass of each element, given the isotopic composition.
1. lithium, which is 92.4% lithium-7 (mass 7.016 u) and 7.60% lithium-6 (mass 6.015 u)
2. oxygen, which is 99.76% oxygen-16 (mass 15.995 u), 0.038% oxygen-17 (mass 16.999 u), and 0.205% oxygen-18 (mass 17.999 u)
6. Determine the atomic mass of each element, given the isotopic composition.
1. neon, which is 90.48% neon-20 (mass 19.992 u), 0.27% neon-21 (mass 20.994 u), and 9.25% neon-22 (mass 21.991 u)
2. uranium, which is 99.27% uranium-238 (mass 238.051 u) and 0.720% uranium-235 (mass 235.044 u)
7. How far off would your answer be from Exercise 5a if you used whole-number masses for individual isotopes of lithium?
8. How far off would your answer be from Exercise 6b if you used whole-number masses for individual isotopes of uranium?
1. What is the atomic mass of an oxygen atom?
2. What is the molecular mass of oxygen in its elemental form?
1. What is the atomic mass of bromine?
2. What is the molecular mass of bromine in its elemental form?
9. Determine the mass of each substance.
1. F2
2. CO
3. CO2
10. Determine the mass of each substance.
1. Kr
2. KrF4
3. PF5
11. Determine the mass of each substance.
1. Na
2. B2O3
3. S2Cl2
12. Determine the mass of each substance.
1. IBr3
2. N2O5
3. CCl4
13. Determine the mass of each substance.
1. GeO2
2. IF3
3. XeF6
14. Determine the mass of each substance.
1. NO
2. N2O4
3. Ca
Answers
1. The atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom. Its abbreviation is u.
2.
1. 1
2. 3
3. 56
3.
1. 6.940 u
2. 16.000 u
4.
5. We would get 6.924 u.
6.
1. 15.999 u
2. 31.998 u
7.
1. 37.996 u
2. 28.010 u
3. 44.009 u
8.
1. 22.990 u
2. 69.619 u
3. 135.036 u
9.
1. 104.64 u
2. 183.898 u
3. 245.281 u
3.4: Ions and Ionic Compounds
1. Explain how cations form.
2. Explain how anions form.
3. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.
1. K
2. O
3. Co
4. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.
1. Ca
2. I
3. Fe
5. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.
1. Ag
2. Au
3. Br
6. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.
1. S
2. Na
3. H
7. Name the ions from Exercise 3.
8. Name the ions from Exercise 4.
9. Name the ions from Exercise 5.
10. Name the ions from Exercise 6.
11. Give the formula and name for each ionic compound formed between the two listed ions.
1. Mg2+ and Cl
2. Fe2+ and O2−
3. Fe3+ and O2−
12. Give the formula and name for each ionic compound formed between the two listed ions.
1. K+ and S2−
2. Ag+ and Br
3. Sr2+ and N3−
13. Give the formula and name for each ionic compound formed between the two listed ions.
1. Cu2+ and F
2. Ca2+ and O2−
3. K+ and P3−
14. Give the formula and name for each ionic compound formed between the two listed ions.
1. Na+ and N3−
2. Co2+ and I
3. Au3+ and S2−
15. Give the formula and name for each ionic compound formed between the two listed ions.
1. K+ and SO42
2. NH4+ and S2−
3. NH4+ and PO43
16. Give the formula and name for each ionic compound formed between the two listed ions.
1. Ca2+ and NO3
2. Ca2+ and NO2
3. Sc3+ and C2H3O2
17. Give the formula and name for each ionic compound formed between the two listed ions.
1. Pb4+ and SO42
2. Na+ and I3
3. Li+ and Cr2O72
18. Give the formula and name for each ionic compound formed between the two listed ions.
1. NH4+ and N3−
2. Mg2+ and CO32
3. Al3+ and OH
19. Give the formula and name for each ionic compound formed between the two listed ions.
1. Ag+ and SO32
2. Na+ and HCO3
3. Fe3+ and ClO3
20. Give the formula and name for each ionic compound formed between the two listed ions.
1. Rb+ and O22
2. Au3+ and HSO4
3. Sr2+ and NO2
21. What is the difference between SO3 and SO32?
22. What is the difference between NO2 and NO2?
Answers
1. Cations form by losing electrons.
2.
1. 1+
2. 2−
3. 2+, 3+
3.
1. 1+
2. 1+, 3+
3. 1−
4.
1. the potassium ion
2. the oxide ion
3. the cobalt(II) and cobalt(III) ions, respectively
5.
1. the silver ion
2. the gold(I) and gold(III) ions, respectively
3. the bromide ion
6.
1. magnesium chloride, MgCl2
2. iron(II) oxide, FeO
3. iron(III) oxide, Fe2O3
7.
1. copper(II) fluoride, CuF2
2. calcium oxide, CaO
3. potassium phosphide, K3P
8.
1. potassium sulfate, K2SO4
2. ammonium sulfide, (NH4)2S
3. ammonium phosphate, (NH4)3PO4
9.
1. lead(IV) sulfate, Pb(SO4)2
2. sodium triiodide, NaI3
3. lithium dichromate, Li2Cr2O7
10.
1. silver sulfite, Ag2SO3
2. sodium hydrogen carbonate, NaHCO3
3. iron(III) chlorate, Fe(ClO3)3
11.
12. SO3 is sulfur trioxide, while SO32 is the sulfite ion.
3.5: Acids
1. Give the formula for each acid.
1. perchloric acid
2. hydriodic acid
2. Give the formula for each acid.
1. hydrosulfuric acid
2. phosphorous acid
3. Name each acid.
1. HF(aq)
2. HNO3(aq)
3. H2C2O4(aq)
4. Name each acid.
1. H2SO4(aq)
2. H3PO4(aq)
3. HCl(aq)
5. Name an acid found in food.
6. Name some properties that acids have in common.
Answers
1. HClO4(aq)
2. HI(aq)
1.
1. hydrofluoric acid
2. nitric acid
3. oxalic acid
2.
3. oxalic acid (answers will vary)
Additional Exercises
1. How many electrons does it take to make the mass of one proton?
2. How many protons does it take to make the mass of a neutron?
3. Dalton’s initial version of the modern atomic theory says that all atoms of the same element are the same. Is this actually correct? Why or why not?
4. How are atoms of the same element the same? How are atoms of the same element different?
5. Give complete atomic symbols for the three known isotopes of hydrogen.
6. A rare isotope of helium has a single neutron in its nucleus. Write the complete atomic symbol of this isotope.
7. Use its place on the periodic table to determine if indium, In, atomic number 49, is a metal or a nonmetal.
8. Only a few atoms of astatine, At, atomic number 85, have been detected. On the basis of its position on the periodic table, would you expect it to be a metal or a nonmetal?
9. Americium-241 is a crucial part of many smoke detectors. How many neutrons are present in its nucleus?
1. Potassium-40 is a radioactive isotope of potassium that is present in the human body. How many neutrons are present in its nucleus?
2. Determine the atomic mass of ruthenium from the given abundance and mass data.
Ruthenium-96 5.54% 95.907 u
Ruthenium-98 1.87% 97.905 u
Ruthenium-99 12.76% 98.906 u
Ruthenium-100 12.60% 99.904 u
Ruthenium-101 17.06% 100.906 u
Ruthenium-102 31.55% 101.904 u
Ruthenium-104 18.62% 103.905 u
1. Determine the atomic mass of tellurium from the given abundance and mass data.
Tellurium-120 0.09% 119.904 u
Tellurium-122 2.55% 121.903 u
Tellurium-123 0.89% 122.904 u
Tellurium-124 4.74% 123.903 u
Tellurium-125 7.07% 124.904 u
Tellurium-126 18.84% 125.903 u
Tellurium-128 31.74% 127.904 u
Tellurium-130 34.08% 129.906 u
1. One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one atom of sodium?
2. One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one atom of uranium?
• One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one molecule of H2O?
• One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one molecule of PF5?
• From their positions on the periodic table, will Cu and I form a molecular compound or an ionic compound?
1. From their positions on the periodic table, will N and S form a molecular compound or an ionic compound?
2. Mercury is an unusual element in that when it takes a 1+ charge as a cation, it always exists as the diatomic ion.
1. Propose a formula for the mercury(I) ion.
2. What is the formula of mercury(I) chloride?
3. Propose a formula for hydrogen peroxide, a substance used as a bleaching agent. (Curiously, this compound does not behave as an acid, despite its formula. It behaves more like a classic nonmetal-nonmetal, molecular compound.)
4. The uranyl cation has the formula UO22+. Propose formulas and names for the ionic compounds between the uranyl cation and F, SO42, and PO43.
5. The permanganate anion has the formula MnO4. Propose formulas and names for the ionic compounds between the permanganate ion and K+, Ca2+, and Fe3+.
Answers
1. about 1,800 electrons
2.
3. It is not strictly correct because of the existence of isotopes.
4.
5. \[_{1}^{1}\textrm{H},\; _{1}^{2}\textrm{H},\, and\; _{1}^{3}\textrm{H}\]
6.
7. It is a metal.
8.
9. 146 neutrons
10.
11. 101.065 u
12.
13. 3.817 × 10−23 g
14.
15. 2.991 × 10−23 g
16.
17. ionic
18.
19.
1. Hg22+
2. Hg2Cl2
20.
21. uranyl fluoride, UO2F2; uranyl sulfate, UO2SO4; uranyl phosphate, (UO2)3(PO4)2 | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.E%3A_Atoms_Molecules_and_Ions_%28Exercises%29.txt |
Chemistry is largely about chemical changes. Indeed, if there were no chemical changes, chemistry as such would not exist! Chemical changes are a fundamental part of chemistry. Because chemical changes are so central, it may be no surprise that chemistry has developed some special ways of presenting them.
• 4.1: Prelude to Chemical Reactions
The space shuttle—and any other rocket-based system—uses chemical reactions to propel itself into space and maneuver itself when it goes into orbit. The rockets that lift the orbiter are of two different types. Although the solid rocket boosters each have a significantly lower mass than the liquid oxygen and liquid hydrogen tanks, they provide over 80% of the lift needed to put the shuttle into orbit—all because of chemical reactions.
• 4.2: The Chemical Equation
A chemical equation is a concise description of a chemical reaction. Proper chemical equations are balanced.
• 4.3: Types of Chemical Reactions - Single and Double Replacement Reactions
A single-replacement reaction replaces one element for another in a compound. The periodic table or an activity series can help predict whether single-replacement reactions occur. A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate. Solubility rules are used to predict whether some double-replacement reactions will occur.
• 4.4: Ionic Equations - A Closer Look
Ionic compounds that dissolve separate into individual ions. Complete ionic equations show dissolved ionic solids as separated ions. Net ionic equations show only the ions and other substances that change in a chemical reaction.
• 4.5: Composition, Decomposition, and Combustion Reactions
A composition reaction produces a single substance from multiple reactants. A decomposition reaction produces multiple products from a single reactant. Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make \(N_2\)).
• 4.6: Neutralization Reactions
The Arrhenius definition of an acid is a substance that increases the amount of H+ in an aqueous solution. The Arrhenius definition of a base is a substance that increases the amount of OH- in an aqueous solution. Neutralization is the reaction of an acid and a base, which forms water and a salt. Net ionic equations for neutralization reactions may include solid acids, solid bases, solid salts, and water.
• 4.7: Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another. Oxidation numbers are used to keep track of electrons in atoms. There are rules for assigning oxidation numbers to atoms. Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number (gain of electrons).
• 4.E: Chemical Reactions and Equations (Exercises)
These are exercises and select solutions to accompany Chapter 4 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
Thumbnail: Reaction of sodium and water breaks the glass vessel. (CC SA-BY-3.0; Tony Mach).
04: Chemical Reactions and Equations
The space shuttle—and any other rocket-based system—uses chemical reactions to propel itself into space and maneuver itself when it gets into orbit. The rockets that lift the orbiter are of two different types. The three main engines are powered by reacting liquid hydrogen with liquid oxygen to generate water. Then there are the two solid rocket boosters, which use a solid fuel mixture that contains mainly ammonium perchlorate and powdered aluminum. The chemical reaction between these substances produces aluminum oxide, water, nitrogen gas, and hydrogen chloride. Although the solid rocket boosters each have a significantly lower mass than the liquid oxygen and liquid hydrogen tanks, they provide over 80% of the lift needed to put the shuttle into orbit—all because of chemical reactions. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.01%3A_Prelude_to_Chemical_Reactions.txt |
Learning Objectives
• Define chemical equation.
• Identify the parts of a chemical equation.
A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows:
$\text{hydrogen reacts with oxygen to make water}\nonumber$
We can represent this chemical change more succinctly as:
$\text{hydrogen} + \text{oxygen} → \text{water}\nonumber$
where the + sign means that the two substances interact chemically with each other and the → symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as:
$\ce{H_2 + O_2 → H_2O}\nonumber$
This is an example of a chemical equation, which is a concise way of representing a chemical reaction. The initial substances are called reactants and the final substances are called products.
Unfortunately, it is also an incomplete chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants, but only one oxygen atom in the products.
What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No—you cannot change the formulas of individual substances, because the chemical formula for a given substance is characteristic of that substance. What you can do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side.
To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products:
$\ce{H_2 + O_2 → 2H_2O}\nonumber$
The 2 in front of the formula for water is called a coefficient. Now there are the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem, simply go back to the reactant side of the equation, and add a coefficient in front of the H2. The coefficient that works is 2:
$\ce{2H_2 + O_2 → 2H_2O}\nonumber$
There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is balanced. All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more "back and forth" work when balancing a chemical equation.
Example $1$
Write and balance the chemical equation for each given chemical reaction.
1. Hydrogen and chlorine react to make $\ce{HCl}$
2. Ethane, $\ce{C2H6}$, reacts with oxygen to make carbon dioxide and water.
Solution
1. Let us start by simply writing a chemical equation in terms of the formulas of the substances, remembering that both elemental hydrogen and chlorine are diatomic: $\ce{H2 + Cl2 → HCl} \nonumber \nonumber$
There are two hydrogen atoms and two chlorine atoms in the reactants and one of each atom in the product. We can fix this by including the coefficient 2 on the product side:
$\ce{H2 + Cl2 → 2HCl} \nonumber \nonumber$
Now there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced.
2. Start by writing the chemical equation in terms of the substances involved: $\ce{C2H6 + O2 → CO2 + H2O} \nonumber \nonumber$
We have two carbon atoms on the left, so we need two carbon dioxide molecules on the product side, so that each side has two carbon atoms; that element is balanced. We have six hydrogen atoms in the reactants, so we need six hydrogen atoms in the products. We can get this by having three water molecules:
$\ce{C2H6 + O2 → 2CO2 + 3H2O} \nonumber \nonumber$
Now we have seven oxygen atoms in the products (four from the CO2 and three from the H2O). This means we need seven oxygen atoms in the reactants. However, because oxygen is a diatomic molecule, we can only get an even number of oxygen atoms at a time. We can achieve this by multiplying the other coefficients by 2:
$\ce{2C2H6 + O2 → 4CO2 + 6H2O} \nonumber \nonumber$
By multiplying everything else by 2, we do not unbalance the other elements, and we now get an even number of oxygen atoms in the product—14. We can get 14 oxygen atoms on the reactant side by having 7 oxygen molecules:
$\ce{2C2H6 + 7O2 → 4CO2 + 6H2O} \nonumber \nonumber$
As a check, recount everything to determine that each side has the same number of atoms of each element. This chemical equation is now balanced.
Exercise $1$
Write and balance the chemical equation that represents nitrogen and hydrogen reacting to produce ammonia, NH3.
Answer
$\ce{N2 + 3H2 → 2NH3 \nonumber \nonumber$
Many chemical equations also include phase labels for the substances: (s) for solid, (ℓ) for liquid, (g) for gas, and (aq) for aqueous (i.e., dissolved in water). Special conditions, such as temperature, may also be listed above the arrow. For example,
$\ce{ 2NaHCO3 (s)} \overset{200^{\circ}C}{\rightarrow} \ce{Na2CO3 (s) + CO2(g) + H2O(l) }\nonumber$
Key Takeaways
• A chemical equation is a concise description of a chemical reaction.
• Proper chemical equations are balanced. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.02%3A_The_Chemical_Equation.txt |
Learning Objectives
• Recognize chemical reactions as single-replacement reactions and double-replacement reactions.
• Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.
Up until now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be.
A single-replacement reaction is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. Presented below:
$\ce{2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)}\nonumber$
is an example of a single-replacement reaction. The hydrogen atoms in $\ce{HCl}$ are replaced by $\ce{Zn}$ atoms, and in the process a new element—hydrogen—is formed. Another example of a single-replacement reaction is
$\ce{2NaCl(aq) + F2(g) → 2NaF(s) + Cl2(g)}\nonumber$
Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.
Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the halogens and are in the next-to-last column on the periodic table (Figure $1$). The elements on top of the column will replace the elements below them on the periodic table, but not the other way around. Thus, the reaction represented by
$\ce{CaI2(s) + Cl2(g) → CaCl2(s) + I2(s)}\nonumber$
will occur; but the reaction
$\ce{CaF2(s) + Br2(ℓ) → CaBr2(s) + F2(g)}\nonumber$
will not, because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us to understand chemistry.
Example $1$
Will a single-replacement reaction occur? If so, identify the products.
1. MgCl2 + I2 → ?
2. CaBr2 + F2 → ?
Solution
1. Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.
2. Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF2 and Br2.
Exercise $1$
Will a single-replacement reaction occur? If so, identify the products.
$\ce{FeI2 + Cl2 → }\nonumber$
Answer
Yes; FeCl2 and I2
Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds—simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the activity series does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.
Activity Series for Cation Replacement in Single-Replacement Reactions
• Li
• K
• Ba
• Sr
• Ca
• Na
• Mg
• Al
• Mn
• Zn
• Cr
• Fe
• Ni
• Sn
• Pb
• H2
• Cu
• Hg
• Ag
• Pd
• Pt
• Au
Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.
Example $2$
Use the activity series to predict the products, if any, of each equation.
1. FeCl2 + Zn → ?
2. HNO3 + Au → ?
Solution
1. Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl2 and Fe.
2. Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.
Exercise $2$
Use the activity series to predict the products, if any, of this equation.
$\ce{AlPO4 + Mg → }\nonumber$
Answer
Mg3(PO4)2 and Al
A double-replacement reaction occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is
$\ce{CuCl2(aq) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2AgCl(s)}\nonumber$
There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion, and not a cation with a cation or an anion with an anion.
Example $3$
Predict the products of this double-replacement equation:
$\ce{BaCl2 + Na2SO4 → }\nonumber$
Solution
Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO4 and NaCl.
Exercise $3$
Predict the products of this double-replacement equation:
$\ce{KBr + AgNO3 → }\nonumber$
Answer
KNO3 and AgBr
Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate. The formation of a solid precipitate is the driving force that makes the reaction proceed.
To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use solubility rules, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble, or insoluble). Table $1$ lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules. If a compound is soluble, we use the (aq) label with it, indicating that it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.
Table $1$: Some Useful Solubility Rules (soluble)
These compounds generally dissolve in water (are soluble): Exceptions:
All compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ None
All compounds of NO3 and C2H3O2 None
Compounds of Cl, Br, I Ag+, Hg22+, Pb2+
Compounds of SO42 Hg22+, Pb2+, Sr2+, Ba2+
Table $2$: Some Useful Solubility Rules (insoluble)
These compounds generally do not dissolve in water (are insoluble): Exceptions:
Compounds of CO32 and PO43 Compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+
Compounds of OH Compounds of Li+, Na+, K+, Rb+, Cs+, NH4+, Sr2+, and Ba2+
For example, consider the possible double-replacement reaction between Na2SO4 and SrCl2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble, except for Ag+, Hg22+, and Pb2+, which are not being considered here. Therefore, Na2SO4 and SrCl2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO4? Compounds of the sulfate ion are generally soluble, but Sr2+ is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be:
$\ce{Na2SO4(aq) + SrCl2(aq) → 2NaCl(aq) + SrSO4(s)}\nonumber$
You would expect to see a visual change corresponding to SrSO4 precipitating out of solution (Figure $2$).
Example $4$
Will a double-replacement reaction occur? If so, identify the products.
1. Ca(NO3)2 + KBr → ?
2. NaOH + FeCl2 → ?
Solution
1. According to the solubility rules, both Ca(NO3)2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr2 and KNO3. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.
2. According to the solubility rules, both NaOH and FeCl2 are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)2. NaCl is soluble, but, according to the solubility rules, Fe(OH)2 is not. Therefore, a reaction would occur, and Fe(OH)2(s) would precipitate out of solution. The balanced chemical equation is $\ce{2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)}\nonumber$
Exercise $4$
$\ce{Sr(NO3)2 + KCl → }\nonumber$
Answer
No reaction; all possible products are soluble.
Key Takeaways
• A single-replacement reaction replaces one element for another in a compound.
• The periodic table or an activity series can help predict whether single-replacement reactions occur.
• A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.
• A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.
• Solubility rules are used to predict whether some double-replacement reactions will occur. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.03%3A_Types_of_Chemical_Reactions_-_Single_and_Double_Replacement_Reactions.txt |
Learning Objectives
• Write ionic equations for chemical reactions between ionic compounds.
• Write net ionic equations for chemical reactions between ionic compounds.
For single-replacement and double-replacement reactions, many of the reactions included ionic compounds—compounds between metals and nonmetals, or compounds that contained recognizable polyatomic ions. Now, we take a closer look at reactions that include ionic compounds.
One important aspect about ionic compounds that differs from molecular compounds has to do with dissolution in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, the ions physically separate from each other. We can use a chemical equation to represent this process—for example, with NaCl:
$\ce{ NaCl(s) ->[\ce{H2O}] Na^{+}(aq) + Cl^{-}(aq)}\nonumber$
When NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (Figure $1$). This process is called dissociation; we say that the ions dissociate.
All ionic compounds that dissolve behave this way. This behavior was first suggested by the Swedish chemist Svante August Arrhenius [1859–1927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry. Keep in mind that when the ions separate, all the ions separate. Thus, when CaCl2 dissolves, the one Ca2+ ion and the two Cl ions separate from one another:
$CaCl_{2}(s)\overset{H_{2}O}{\rightarrow}Ca^{2+}(aq)+Cl^{-}(aq)+Cl^{-}(aq)\nonumber$
$CaCl_{2}(s)\overset{H_{2}O}{\rightarrow}Ca^{2+}(aq)+2Cl^{-}(aq)\nonumber$
That is, the two chloride ions go off on their own. They do not remain as Cl2 (that would be elemental chlorine; these are chloride ions), and they do not stick together to make Cl2 or Cl22. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved.
Example $1$
Write the chemical equation that represents the dissociation of each ionic compound.
1. KBr
2. Na2SO4
Solution
1. KBr(s) → K+(aq) + Br(aq)
2. Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is Na2SO4(s) → 2Na+(aq) + SO42(aq)
Exercise $1$
Write the chemical equation that represents the dissociation of (NH4)2S.
Answer
(NH4)2S(s) → 2NH4+(aq) + S2−(aq)
When chemicals in solution react, the proper way of writing the chemical formulas of the dissolved ionic compounds is in terms of the dissociated ions, not the complete ionic formula. A complete ionic equation is a chemical equation in which the dissolved ionic compounds are written as separated ions. Solubility rules are very useful in determining which ionic compounds are dissolved and which are not. For example, when NaCl(aq) reacts with AgNO3(aq) in a double-replacement reaction to precipitate AgCl(s) and form NaNO3(aq), the complete ionic equation includes NaCl, AgNO3, and NaNO3 written as separate ions:
$\ce{Na^{+}(aq) + Cl^{−}(aq) + Ag^{+}(aq) + NO3^{−}(aq) → AgCl(s) + Na^{+}(aq) + NO3^{−}(aq)}\nonumber$
This is more representative of what is occurring in the solution.
Example $1$
Write the complete ionic equation for each chemical reaction.
1. KBr(aq) + AgC2H3O2(aq) → KC2H3O2(aq) + AgBr(s)
2. MgSO4(aq) + Ba(NO3)2(aq) → Mg(NO3)2(aq) + BaSO4(s)
Solution
For any ionic compound that is aqueous, we will write the compound as separated ions.
1. The complete ionic equation is K+(aq) + Br(aq) + Ag+(aq) + C2H3O2(aq) → K+(aq) + C2H3O2(aq) + AgBr(s)
2. The complete ionic equation is Mg2+(aq) + SO42(aq) + Ba2+(aq) + 2NO3(aq) → Mg2+(aq) + 2NO3(aq) + BaSO4(s)
Exercise $1$
Write the complete ionic equation for
$\ce{CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)}\nonumber$
Answer
Ca2+(aq) + 2Cl(aq) + Pb2+(aq) + 2NO3(aq) → Ca2+(aq) + 2NO3(aq) + PbCl2(s)
You may notice that in a complete ionic equation, some ions do not change their chemical form; they stay exactly the same on the reactant and product sides of the equation. For example, in
Na+(aq) + Cl(aq) + Ag+(aq) + NO3(aq) → AgCl(s) + Na+(aq) + NO3(aq)
the Ag+(aq) and Cl(aq) ions become AgCl(s), but the Na+(aq) ions and the NO3(aq) ions stay as Na+(aq) ions and NO3(aq) ions. These two ions are examples of spectator ions—ions that do nothing in the overall course of a chemical reaction. They are present, but they do not participate in the overall chemistry. It is common to cancel spectator ions (something also done with algebraic quantities) on the opposite sides of a chemical equation:
$\cancel{Na^{+}(aq)}+Cl^{-}(aq)+Ag^{+}(aq)+\cancel{NO_{3}^{-}}(aq)\rightarrow AgCl(s)+\cancel{Na}^{+}(aq)+\cancel{NO}_{3}^{-}(aq)\nonumber$
What remains when the spectator ions are removed is called the net ionic equation, which represents the actual chemical change occurring between the ionic compounds:
Cl(aq) + Ag+(aq) → AgCl(s)
It is important to reiterate that the spectator ions are still present in solution, but they do not experience any net chemical change, so they are not written in a net ionic equation.
Example $1$
Write the net ionic equation for each chemical reaction.
1. K+(aq) + Br(aq) + Ag+(aq) + C2H3O2(aq) → K+(aq) + C2H3O2(aq) + AgBr(s)
2. Mg2+(aq) + SO42(aq) + Ba2+(aq) + 2NO3(aq) → Mg2+(aq) + 2NO3(aq) + BaSO4(s)
Solution
1. In the first equation, the K+(aq) and C2H3O2(aq) ions are spectator ions, so they are canceled:
$\cancel{K^{+}(aq)}+Br^{-}(aq)+Ag^{+}(aq)+\cancel{C_{2}H_{3}O_{2}^{-}(aq)}\rightarrow K^{+}(aq)+\cancel{C_{2}H_{3}O_{2}^{-}(aq)}+AgBr(s)\nonumber$
The net ionic equation is
Br(aq) + Ag+(aq) → AgBr(s)
1. In the second equation, the Mg2+(aq) and NO3(aq) ions are spectator ions, so they are canceled:
$\cancel{Mg^{2+}(aq)}+SO_{4}^{2-}(aq)+Ba^{2+}(aq)+\cancel{2NO_{3}^{-}(aq)}\rightarrow Mg^{2+}(aq)+\cancel{2NO_{3}^{-}(aq)}+BaSo_{4}(s)\nonumber$
The net ionic equation is
SO42(aq) + Ba2+(aq) → BaSO4(s)
Exercise $1$
Write the net ionic equation for
CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)
Answer
Pb2+(aq) + 2Cl(aq) → PbCl2(s)
Chemistry is Everywhere: Soluble and Insoluble Ionic Compounds
The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound that can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider $\ce{NaCl}$ soluble but $\ce{AgCl}$ insoluble.
One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO3). However, CaCO3 has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO3 can precipitate if there is enough of it in the water. This precipitate, called limescale, can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction.
Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility!
Key Takeaways
• Ionic compounds that dissolve separate into individual ions.
• Complete ionic equations show dissolved ionic solids as separated ions.
• Net ionic equations show only the ions and other substances that change in a chemical reaction. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.04%3A_Ionic_Equations_-_A_Closer_Look.txt |
Learning Objectives
• Recognize composition, decomposition, and combustion reactions.
• Predict the products of a combustion reaction.
Three classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.
A composition reaction (sometimes also called a combination reaction or a synthesis reaction) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction
$\ce{2H_2(g) + O_2(g) → 2H_2O(ℓ)}\nonumber$
water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance—water—as a product. So this is a composition reaction.
A decomposition reaction starts from a single substance and produces more than one substance; that is, it decomposes. The key characteristics of a decomposition reaction are: one substance as a reactant and more than one substance as the products. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate):
$\ce{2NaHCO_3(s) → Na_2CO_3(s) + CO_2(g) + H_2O(ℓ) }\nonumber$
sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.
Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.
Example $1$: Identifying Reactions
Identify each equation as a composition reaction, a decomposition reaction, or neither.
1. $\ce{Fe2O3 + 3SO3 → Fe2(SO4)3}$
2. $\ce{NaCl + AgNO3 → AgCl + NaNO3}$
3. $\ce{(NH4)2Cr2O7 → Cr2O3 + 4H2O + N2}$
Solution
1. In this equation, two substances combine to make a single substance. This is a composition reaction.
2. Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.
3. A single substance reacts to make multiple substances. This is a decomposition reaction.
Exercise $1$
Identify the equation as a composition reaction, a decomposition reaction, or neither.
$\ce{C3H8 → C3H4 + 2H2} \nonumber$
Answer
decomposition
A combustion reaction occurs when a reactant combines with oxygen, many times from the atmosphere, to produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N2. Many reactants, called fuels, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO2 and H2O. For example, the balanced chemical equation for the combustion of methane, CH4, is as follows:
$\ce{CH_4 + 2O_2 → CO_2 + 2H_2O}\nonumber$
Kerosene can be approximated with the formula $\ce{C_{12}H_{26}}$, and its combustion equation is:
$\ce{2C_{12}H_{26} + 37O_2 → 24CO-2 + 26H_2O}\nonumber$
Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, $\ce{C_2H_5OH}$, whose combustion equation is:
$\ce{C_2H_5OH + 3O_2 → 2CO_2 + 3H_2O}\nonumber$
If nitrogen is present in the original fuel, it is converted to $\ce{N_2}$, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is $\ce{C_2H_2N_2O_4}$, we have:
$\ce{2C_2H_2N_2O_4 + O_2 → 4CO_2 + 2H_2O + 2N_2}\nonumber$
Example $2$: Combustion Reactions
Complete and balance each combustion equation.
1. the combustion of propane ($\ce{C3H8}$)
2. the combustion of ammonia ($\ce{NH3}$)
Solution
1. The products of the reaction are CO2 and H2O, so our unbalanced equation is $\ce{C3H8 + O2 → CO2 + H2O} \nonumber$
Balancing (and you may have to go back and forth a few times to balance this), we get $\ce{C3H8 + 5O2 → 3CO2 + 4H2O} \nonumber$
2. The nitrogen atoms in ammonia will react to make N2, while the hydrogen atoms will react with O2 to make H2O: $\ce{NH3 + O2 → N2 + H2O} \nonumber$
To balance this equation without fractions (which is the convention), we get $\ce{4NH3 + 3O2 → 2N2 + 6H2O} \nonumber$
Exercise $2$
Complete and balance the combustion equation for cyclopropanol (\ce{C3H6O}\)).
Answer
$\ce{C_3H_6O + 4O_2 → 3CO_2 + 3H_2O}\nonumber$
Key Takeaways
• A composition reaction produces a single substance from multiple reactants.
• A decomposition reaction produces multiple products from a single reactant.
• Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make N2). | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.05%3A_Composition_Decomposition_and_Combustion_Reactions.txt |
Learning Objectives
• Identify an acid and a base.
• Identify a neutralization reaction and predict its products.
In Chapter 3, we defined an acid as an ionic compound that contains H+ as the cation. This is slightly incorrect, but until additional concepts were developed, a better definition needed to wait. Now we can redefine an acid: an acid is any compound that increases the amount of hydrogen ion (H+) in an aqueous solution. The chemical opposite of an acid is a base. The equivalent definition of a base is that a base is a compound that increases the amount of hydroxide ion (OH) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the Arrhenius definition of an acid and a base, respectively.
You may recognize that, based on the description of a hydrogen atom, an H+ ion is a hydrogen atom that has lost its lone electron; that is, H+ is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H+ ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the hydronium ion as H3O+, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way that a hydrogen ion appears in an aqueous solution, although in many chemical reactions H+ and H3O+ are treated equivalently.
The reaction of an acid and a base is called a neutralization reaction. Although acids and bases have their own unique chemistries, the acid and base cancel each other's chemistry to produce a rather innocuous substance—water. In fact, the general reaction between an acid and a base is:
$\ce{acid + base → water + salt}\nonumber$
where the term salt is generally used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. (In chemistry, the word salt refers to more than just table salt.) For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is
$\ce{HCl(aq) + KOH(aq) → H2O(ℓ) + KCl(aq)}\nonumber$
where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)2(aq), additional molecules of HCl and H2O are required to balance the chemical equation:
$\ce{2HCl(aq) + Mg(OH)2(aq) → 2H2O(ℓ) + MgCl2(aq)}\nonumber$
Here, the salt is MgCl2. (This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid.)
Example $1$
Write the neutralization reactions between each acid and base.
1. HNO3(aq) and Ba(OH)2(aq)
2. H3PO4(aq) and Ca(OH)2(aq)
Solution
First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.
1. The expected products are water and barium nitrate, so the initial chemical reaction is $\ce{HNO3(aq) + Ba(OH)2(aq) → H2O(ℓ) + Ba(NO3)2(aq)}\nonumber$
To balance the equation, we need to realize that there will be two H2O molecules, so two HNO3 molecules are required: $\ce{2HNO3(aq) + Ba(OH)2(aq) → 2H2O(ℓ) + Ba(NO3)2(aq)}\nonumber$
This chemical equation is now balanced.
2. The expected products are water and calcium phosphate, so the initial chemical equation is $\ce{H3PO4(aq) + Ca(OH)2(aq) → H2O(ℓ) + Ca3(PO4)2(s)}\nonumber$
According to the solubility rules, Ca3(PO4)2 is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions. We end up with six water molecules to balance the equation: $\ce{2H3PO4(aq) + 3Ca(OH)2(aq) → 6H2O(ℓ) + Ca3(PO4)2(s)}\nonumber$
This chemical equation is now balanced.
Exercise $1$
Write the neutralization reaction between H2SO4(aq) and Sr(OH)2(aq).
Answer
$\ce{H2SO4(aq) + Sr(OH)2(aq) → 2H2O(ℓ) + SrSO4(aq)}\nonumber$
Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)3(s) still proceeds according to the equation:
$\ce{3HCl(aq) + Fe(OH)3(s) → 3H2O(ℓ) + FeCl3(aq)}\nonumber$
even though Fe(OH)3 is not soluble. When one realizes that Fe(OH)3(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids—the neutralization reaction produces products that are soluble and wash away. (Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!)
Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq),
$\ce{HCl(aq) + NaOH(aq) → H2O(ℓ) + NaCl(aq)}\nonumber$
the complete ionic reaction is
$\ce{H^{+}(aq) + Cl^{−}(aq) + Na^{+}(aq) + OH^{−}(aq) → H2O(ℓ) + Na^{+}(aq) + Cl^{−}(aq)}\nonumber$
The Na+(aq) and Cl(aq) ions are spectator ions, so we can remove them to have
$\ce{H^{+}(aq) + OH^{−}(aq) → H2O(ℓ)}\nonumber$
as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H3O+(aq), we would write it as
$\ce{H3O^{+}(aq) + OH^{−}(aq) → 2H2O(ℓ)}\nonumber$
With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.
However, for the reaction between HCl(aq) and Cr(OH)2(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:
$\ce{2H^{+}(aq) + 2Cl^{−}(aq) + Cr(OH)2(s) → 2H2O(ℓ) + Cr^{2+}(aq) + 2Cl^{−}(aq)}\nonumber$
The chloride ions are the only spectator ions here, so the net ionic equation is
$\ce{2H^{+}(aq) + Cr(OH)2(s) → 2H2O(ℓ) + Cr^{2+}(aq)}\nonumber$
Example $2$
Oxalic acid, H2C2O4(s), and Ca(OH)2(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? (The anion in oxalic acid is the oxalate ion, C2O42.)
Solution
The products of the neutralization reaction will be water and calcium oxalate:
H2C2O4(s) + Ca(OH)2(s) → 2H2O(ℓ) + CaC2O4(s)
Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.
Exercise $2$
What is the net ionic equation between HNO3(aq) and Ti(OH)4(s)?
Answer
4H+(aq) + Ti(OH)4(s) → 4H2O(ℓ) + Ti4+(aq)
Key Takeaways
• The Arrhenius definition of an acid is a substance that increases the amount of H+ in an aqueous solution.
• The Arrhenius definition of a base is a substance that increases the amount of OH in an aqueous solution.
• Neutralization is the reaction of an acid and a base, which forms water and a salt.
• Net ionic equations for neutralization reactions may include solid acids, solid bases, solid salts, and water. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.06%3A_Neutralization_Reactions.txt |
Learning Objectives
• Define oxidation and reduction.
• Assign oxidation numbers to atoms in simple compounds.
• Recognize a reaction as an oxidation-reduction reaction.
Consider this chemical reaction:
$\ce{Mg(s) + Cl2(g) → MgCl2}\nonumber$
The reactants are two electrically neutral elements; they have the same number of electrons as protons. The product, however, is ionic; it is composed of Mg2+ and Cl ions. Somehow, the individual Mg atoms lose two electrons to make the Mg2+ ion, while the Cl atoms gain an electron to become Cl ions. This reaction involves the transfer of electrons between atoms.
The process of losing and gaining electrons occurs simultaneously. However, mentally we can separate the two processes. Oxidation is defined as the loss of one or more electrons by an atom. Reduction is defined as the gain of one or more electrons by an atom. So oxidation and reduction always occur together; it is only mentally that we can separate them. Chemical reactions that involve the transfer of electrons are called oxidation-reduction (or redox) reactions.
Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate.
Assigning Oxidation Numbers
The rules for assigning oxidation numbers to atoms are as follows:
1. Atoms in their elemental state are assigned an oxidation number of 0.
2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges.
3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion [H], in which case rule 2 prevails).
4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral).
Here are some examples for practice. In H2, both H atoms have an oxidation number of 0 by rule 1. In MgCl2, magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1 by rule 2. In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H2O2) has an oxidation number of +1, while each O atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO2. Each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.
Example $1$
Assign oxidation numbers to the atoms in each substance.
1. Cl2
2. GeO2
3. Ca(NO3)2
Solution
1. Cl2 is the elemental form of chlorine. Rule 1 states each atom has an oxidation number of 0.
2. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4.
3. Ca(NO3)2 can be separated into two parts: the Ca2+ ion and the NO3 ion. Considering these separately, the Ca2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3 ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation x + 3(−2) = −1
where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for x,
x + (−6) = −1x = +5
Thus the oxidation number on the N atom in the NO3 ion is +5.
Exercise $1$: Phosphoric Acid
Assign oxidation numbers to the atoms in H3PO4.
Answer
H: +1; O: −2; P: +5
All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively.
Example $2$
Identify what is being oxidized and reduced in this redox reaction.
$\ce{2Na + Br2 → 2NaBr} \nonumber \nonumber$
Solution
Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br ions have an oxidation number of −1.
$2\underset{0}{Na}+\underset{0}{Br_{2}}\rightarrow 2\underset{+1 -1}{NaBr} \nonumber \nonumber$
Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced:
Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom).
Exercise $2$
Identify what is being oxidized and reduced in this redox reaction.
$\ce{C + O2 → CO2}\nonumber \nonumber$
Answer
C is being oxidized from 0 to +4; O is being reduced from 0 to −2.
Oxidation reactions can become quite complex, as attested by the following redox reaction:
$6H^{+}(aq)+2\underset{+7}{MnO_{4}^{-}}(aq)+5\underset{-1}{H_{2}O_{2}}(l)\rightarrow 2\underset{+2}{Mn^{2+}}(aq)+5\underset{0}{O_{2}}(g)+8H_{2}O(l)\nonumber$
To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation. Eventually, we will need to learn techniques for writing correct (i.e., balanced) redox reactions.
Food and Drink Application: Fortifying Food with Iron
Iron is an essential mineral in our diet; iron-containing compounds like the heme protein in hemoglobin could not function without it. Most biological iron has the form of the Fe2+ ion; iron with other oxidation numbers is almost inconsequential in human biology (although the body does contain an enzyme to reduce Fe3+ to Fe2+, so Fe3+ must have some biological significance, albeit minor). To ensure that we ingest enough iron, many foods are enriched with iron. Although Fe2+ compounds are the most logical substances to use, some foods use "reduced iron" as an ingredient (bread and breakfast cereals are the most well-known examples). Reduced iron is simply iron metal; iron is added as a fine metallic powder. The metallic iron is oxidized to Fe2+ in the digestive system and then absorbed by the body, but the question remains: Why are we ingesting metallic iron? Why not just use Fe2+ salts as an additive?
Although it is difficult to establish conclusive reasons, a search of scientific and medical literature suggests a few reasons. One reason is that fine iron filings do not affect the taste of the product. The size of the iron powder (several dozen micrometers) is not noticeable when chewing iron-supplemented foods, and the tongue does not detect any changes in flavor that can be detected when using Fe2+ salts. Fe2+ compounds can affect other properties of foodstuffs during preparation and cooking, like dough pliability, yeast growth, and color. Finally, of the common iron substances that might be used, metallic iron is the least expensive. These factors appear to be among the reasons why metallic iron is the supplement of choice in some foods.
Key Takeaways
• Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another.
• Oxidation numbers are used to keep track of electrons in atoms.
• There are rules for assigning oxidation numbers to atoms.
• Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number (gain of electrons). | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.07%3A_Oxidation-Reduction_Reactions.txt |
4.2: The Chemical Equation
1. From the statement “nitrogen and hydrogen react to produce ammonia,” identify the reactants and the products.
2. From the statement “sodium metal reacts with water to produce sodium hydroxide and hydrogen,” identify the reactants and the products.
3. From the statement “magnesium hydroxide reacts with nitric acid to produce magnesium nitrate and water,” identify the reactants and the products.
4. From the statement “propane reacts with oxygen to produce carbon dioxide and water,” identify the reactants and the products.
5. Write and balance the chemical equation described by Exercise 1.
6. Write and balance the chemical equation described by Exercise 2.
7. Write and balance the chemical equation described by Exercise 3.
8. Write and balance the chemical equation described by Exercise 4. The formula for propane is C3H8.
9. Balance: ___NaClO3 → ___NaCl + ___O2
10. Balance: ___N2 + ___H2 → ___N2H4
11. Balance: ___Al + ___O2 → ___Al2O3
12. Balance: ___C2H4 + ___O2 → ___CO2 + ___H2O
13. How would you write the balanced chemical equation in Exercise 10 if all substances were gases?
1. How would you write the balanced chemical equation in Exercise 12 if all the substances except water were gases and water itself were a liquid?
Answers
1. reactants: nitrogen and hydrogen; product: ammonia
2.
3. reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water
4.
5. N2 + 3H2 → 2NH3
6.
7. Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H2O
8.
9. 2NaClO3 → 2NaCl + 3O2
10.
11. 4Al + 3O2 → 2Al2O3
12.
13. N2(g) + 3H2(g) → 2NH3(g)
4.3: Types of Chemical Reactions - Single and Double Displacement Reactions
1. What are the general characteristics that help you recognize single-replacement reactions?
2. What are the general characteristics that help you recognize double-replacement reactions?
3. Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Zn + Fe(NO3)2 → ?
2. F2 + FeI3 → ?
4. Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Li + MgSO4 → ?
2. NaBr + Cl2 → ?
5. Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Sn + H2SO4 → ?
2. Al + NiBr2 → ?
6. Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Mg + HCl → ?
2. HI + Br2 → ?
7. Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.
1. FeCl2 + Br2 → ?
2. Fe(NO3)3 + Al → ?
8. Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.
1. Zn + Fe3(PO4)2 → ?
2. Ag + HNO3 → ?
9. Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.
1. NaI + Cl2 → ?
2. AgCl + Au → ?
10. Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.
1. Pt + H3PO4 → ?
2. Li + H2O → ? (Hint: treat H2O as if it were composed of H+ and OH ions.)
11. Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Zn(NO3)2 + NaOH → ?
2. HCl + Na2S → ?
12. Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Ca(C2H3O2)2 + HNO3 → ?
2. Na2CO3 + Sr(NO2)2 → ?
13. Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Pb(NO3)2 + KBr → ?
2. K2O + MgCO3 → ?
14. Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.
1. Sn(OH)2 + FeBr3 → ?
2. CsNO3 + KCl → ?
15. Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.
1. Pb(NO3)2 + KBr → ?
2. K2O + Na2CO3 → ?
16. Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.
1. Na2CO3 + Sr(NO2)2 → ?
2. (NH4)2SO4 + Ba(NO3)2 → ?
17. Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.
1. K3PO4 + SrCl2 → ?
2. NaOH + MgCl2 → ?
18. Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.
1. KC2H3O2 + Li2CO3 → ?
2. KOH + AgNO3 → ?
Answers
1. One element replaces another element in a compound.
2.
1. Zn + Fe(NO3)2 → Zn(NO3)2 + Fe
2. 3F2 + 2FeI3 → 3I2 + 2FeF3
3.
1. Sn + H2SO4 → SnSO4 + H2
2. 2Al + 3NiBr2 → 2AlBr3 + 3Ni
4.
1. No reaction occurs.
2. Fe(NO3)3 + Al → Al(NO3)3 + Fe
5.
1. 2NaI + Cl2 → 2NaCl + I2
2. No reaction occurs.
6.
1. Zn(NO3)2 + 2NaOH → Zn(OH)2 + 2NaNO3
2. 2HCl + Na2S → 2NaCl + H2S
7.
1. Pb(NO3)2 + 2KBr → PbBr2 + 2KNO3
2. K2O + MgCO3 → K2CO3 + MgO
8.
1. Pb(NO3)2 + 2KBr → PbBr2(s) + 2KNO3
2. No reaction occurs.
9.
1. 2K3PO4 + 3SrCl2 → Sr3(PO4)2(s) + 6KCl
2. 2NaOH + MgCl2 → 2NaCl + Mg(OH)2(s)
4.4: Ionic Equations - A Closer Look
1. Write a chemical equation that represents NaBr(s) dissociating in water.
2. Write a chemical equation that represents SrCl2(s) dissociating in water.
3. Write a chemical equation that represents (NH4)3PO4(s) dissociating in water.
4. Write a chemical equation that represents Fe(C2H3O2)3(s) dissociating in water.
5. Write the complete ionic equation for the reaction of FeCl2(aq) and AgNO3(aq). You may have to consult the solubility rules.
6. Write the complete ionic equation for the reaction of BaCl2(aq) and Na2SO4(aq). You may have to consult the solubility rules.
7. Write the complete ionic equation for the reaction of KCl(aq) and NaC2H3O2(aq). You may have to consult the solubility rules.
8. Write the complete ionic equation for the reaction of Fe2(SO4)3(aq) and Sr(NO3)2(aq). You may have to consult the solubility rules.
9. Write the net ionic equation for the reaction of FeCl2(aq) and AgNO3(aq). You may have to consult the solubility rules.
10. Write the net ionic equation for the reaction of BaCl2(aq) and Na2SO4(aq). You may have to consult the solubility rules.
11. Write the net ionic equation for the reaction of KCl(aq) and NaC2H3O2(aq). You may have to consult the solubility rules.
12. Write the net ionic equation for the reaction of Fe2(SO4)3(aq) and Sr(NO3)2(aq). You may have to consult the solubility rules.
13. Identify the spectator ions in Exercises 9 and 10.
1. Identify the spectator ions in Exercises 11 and 12.
Answers
1. NaBr(s) Na+(aq) + Br(aq)
2.
3. (NH4)3PO4(s) 3NH4+(aq) + PO43(aq)
4.
5. Fe2+(aq) + 2Cl(aq) + 2Ag+(aq) + 2NO3(aq) → Fe2+(aq) + 2NO3(aq) + 2AgCl(s)
6.
7. K+(aq) + Cl(aq) + Na+(aq) + C2H3O2(aq) → Na+(aq) + Cl(aq) + K+(aq) + C2H3O2(aq)
8.
9. 2Cl(aq) + 2Ag+(aq) → 2AgCl(s)
10.
11. There is no overall reaction.
12.
13. In Exercise 9, Fe2+(aq) and NO3(aq) are spectator ions; in Exercise 10, Na+(aq) and Cl(aq) are spectator ions.
4.5: Composition, Decomposition, and Combustion Reactions
1. Which is a composition reaction and which is not?
1. NaCl + AgNO3 → AgCl + NaNO3
2. CaO + CO2 → CaCO3
2. Which is a composition reaction and which is not?
1. H2 + Cl2 → 2HCl
2. 2HBr + Cl2 → 2HCl + Br2
3. Which is a composition reaction and which is not?
1. 2SO2 + O2 → 2SO3
2. 6C + 3H2 → C6H6
4. Which is a composition reaction and which is not?
1. 4Na + 2C + 3O2 → 2Na2CO3
2. Na2CO3 → Na2O + CO2
5. Which is a decomposition reaction and which is not?
1. HCl + NaOH → NaCl + H2O
2. CaCO3 → CaO + CO2
6. Which is a decomposition reaction and which is not?
1. 3O2 → 2O3
2. 2KClO3 → 2KCl + 3O2
7. Which is a decomposition reaction and which is not?
1. Na2O + CO2 → Na2CO3
2. H2SO3 → H2O + SO2
8. Which is a decomposition reaction and which is not?
1. 2C7H5N3O6 → 3N2 + 5H2O + 7CO + 7C
2. C6H12O6 + 6O2 → 6CO2 + 6H2O
9. Which is a combustion reaction and which is not?
1. C6H12O6 + 6O2 → 6CO2 + 6H2O
2. 2Fe2S3 + 9O2 → 2Fe2O3 + 6SO2
10. Which is a combustion reaction and which is not?
1. CH4 + 2F2 → CF4 + 2H2
2. 2H2 + O2 → 2H2O
11. Which is a combustion reaction and which is not?
1. P4 + 5O2 → 2P2O5
2. 2Al2S3 + 9O2 → 2Al2O3 + 6SO2
12. Which is a combustion reaction and which is not?
1. C2H4 + O2 → C2H4O2
2. C2H4 + Cl2 → C2H4Cl2
13. Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.
14. Is it possible for a decomposition reaction to also be a combustion reaction? Give an example to support your case.
15. Complete and balance each combustion equation.
1. C4H9OH + O2 → ?
2. CH3NO2 + O2 → ?
16. Complete and balance each combustion equation.
1. B2H6 + O2 → ? (The oxide of boron formed is B2O3.)
2. Al2S3 + O2 → ? (The oxide of sulfur formed is SO2.)
3. Al2S3 + O2 → ? (The oxide of sulfur formed is SO3.)
Answers
1. not composition
2. composition
1.
1. composition
2. composition
2.
1. not decomposition
2. decomposition
3.
1. not decomposition
2. decomposition
4.
1. combustion
2. combustion
5.
1. combustion
2. combustion
6.
7. Yes; 2H2 + O2 → 2H2O (answers will vary)
8.
1. C4H9OH + 6O2 → 4CO2 + 5H2O
2. 4CH3NO2 + 3O2 → 4CO2 + 6H2O + 2N2
4.6: Neutralization Reactions
1. What is the Arrhenius definition of an acid?
2. What is the Arrhenius definition of a base?
3. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.
1. HCl and KOH
2. H2SO4 and KOH
3. H3PO4 and Ni(OH)2
4. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.
1. HBr and Fe(OH)3
2. HNO2 and Al(OH)3
3. HClO3 and Mg(OH)2
5. Write a balanced chemical equation for each neutralization reaction in Exercise 3.
6. Write a balanced chemical equation for each neutralization reaction in Exercise 4.
7. Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.
1. HI(aq) + KOH(aq) → ?
2. H2SO4(aq) + Ba(OH)2(aq) → ?
8. Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.
1. HNO3(aq) + Fe(OH)3(s) → ?
2. H3PO4(aq) + CsOH(aq) → ?
9. Write the net ionic equation for each neutralization reaction in Exercise 7.
10. Write the net ionic equation for each neutralization reaction in Exercise 8.
11. Write the complete and net ionic equations for the neutralization reaction between HClO3(aq) and Zn(OH)2(s). Assume the salt is soluble.
12. Write the complete and net ionic equations for the neutralization reaction between H2C2O4(s) and Sr(OH)2(aq). Assume the salt is insoluble.
13. Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is the same as the net ionic equation for the neutralization reaction between HNO3(aq) and RbOH.
14. Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is different from the net ionic equation for the neutralization reaction between HCl(aq) and AgOH.
15. Write the complete and net ionic equations for the neutralization reaction between HCl(aq) and KOH(aq) using the hydronium ion in place of H+. What difference does it make when using the hydronium ion?
16. Write the complete and net ionic equations for the neutralization reaction between HClO3(aq) and Zn(OH)2(s) using the hydronium ion in place of H+. Assume the salt is soluble. What difference does it make when using the hydronium ion?
Answers
1. An Arrhenius acid increases the amount of H+ ions in an aqueous solution.
2.
1. KCl and H2O
2. K2SO4 and H2O
3. Ni3(PO4)2 and H2O
3.
1. HCl + KOH → KCl + H2O
2. H2SO4 + 2KOH → K2SO4 + 2H2O
3. 2H3PO4 + 3Ni(OH)2 → Ni3(PO4)2 + 6H2O
4.
1. HI(aq) + KOH(aq) → KI(aq) + H2O(ℓ)
2. H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O(ℓ)
5.
1. H+(aq) + OH(aq) → H2O(ℓ)
2. 2H+(aq) + SO42(aq) + Ba2+(aq) + 2OH(aq) → BaSO4(s) + 2H2O(ℓ)
6.
7. Complete ionic equation: 2H+(aq) + 2ClO3(aq) + Zn2+(aq) + 2OH(aq) → Zn2+(aq) + 2ClO3(aq) + 2H2O(ℓ)
Net ionic equation: 2H+(aq) + 2OH(aq) → 2H2O(ℓ)
1.
2. Because the salts are soluble in both cases, the net ionic reaction is just H+(aq) + OH(aq) → H2O(ℓ).
3.
4. Complete ionic equation: H3O+(aq) + Cl(aq) + K+(aq) + OH(aq) → 2H2O(ℓ) + K+(aq) + Cl(aq)
Net ionic equation: H3O+(aq) + OH(aq) → 2H2O(ℓ)
The difference is simply the presence of an extra water molecule as a product.
4.7: Oxidation-Reduction Reactions
1. Is the reaction
2K(s) + Br2(ℓ) → 2KBr(s)
an oxidation-reduction reaction? Explain your answer.
1. Is the reaction
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
an oxidation-reduction reaction? Explain your answer.
2. In the reaction
2Ca(s) + O2(g) → 2CaO
indicate what has lost electrons and what has gained electrons.
3. In the reaction
16Fe(s) + 3S8(s) → 8Fe2S3(s)
indicate what has lost electrons and what has gained electrons.
4. In the reaction
2Li(s) + O2(g) → Li2O2(s)
indicate what has been oxidized and what has been reduced.
5. In the reaction
2Ni(s) + 3I2(s) → 2NiI3(s)
indicate what has been oxidized and what has been reduced.
6. What are two different definitions of oxidation?
1. What are two different definitions of reduction?
2. Assign oxidation numbers to each atom in each substance.
1. P4
2. SO2
3. SO22−
4. Ca(NO3)2
3. Assign oxidation numbers to each atom in each substance.
1. PF5
2. (NH4)2S
3. Hg
4. Li2O2 (lithium peroxide)
4. Assign oxidation numbers to each atom in each substance.
1. CO
2. CO2
3. NiCl2
4. NiCl3
5. Assign oxidation numbers to each atom in each substance.
1. NaH (sodium hydride)
2. NO2
3. NO2
4. AgNO3
6. Assign oxidation numbers to each atom in each substance.
1. CH2O
2. NH3
3. Rb2SO4
4. Zn(C2H3O2)2
7. Assign oxidation numbers to each atom in each substance.
1. C6H6
2. B(OH)3
3. Li2S
4. Au
8. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
2NO + Cl2 → 2NOCl
1. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
Fe + SO3 → FeSO3
1. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
2KrF2 + 2H2O → 2Kr + 4HF + O2
2. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
SO3 + SCl2 → SOCl2 + SO2
3. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
2K + MgCl2 → 2KCl + Mg
1. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
C7H16 + 11O2 → 7CO2 + 8H2O
Answers
1. Yes; both K and Br are changing oxidation numbers.
2.
3. Ca has lost electrons, and O has gained electrons.
4.
5. Li has been oxidized, and O has been reduced.
6.
7. loss of electrons; increase in oxidation number
8.
1. P: 0
2. S: +4; O: −2
3. S: +2; O: −2
4. Ca: 2+; N: +5; O: −2
9.
1. C: +2; O: −2
2. C: +4; O: −2
3. Ni: +2; Cl: −1
4. Ni: +3; Cl: −1
10.
1. C: 0; H: +1; O: −2
2. N: −3; H: +1
3. Rb: +1; S: +6; O: −2
4. Zn: +2; C: 0; H: +1; O: −2
11.
12. N is being oxidized, and Cl is being reduced.
13.
14. O is being oxidized, and Kr is being reduced.
15.
16. K is being oxidized, and Mg is being reduced.
Additional Exercises
1. Chemical equations can also be used to represent physical processes. Write a chemical reaction for the boiling of water, including the proper phase labels.
2. Chemical equations can also be used to represent physical processes. Write a chemical reaction for the freezing of water, including the proper phase labels.
3. Explain why
4Na(s) + 2Cl2(g) → 4NaCl(s)
should not be considered a proper chemical equation.
4. Explain why
H2(g) + 1/2O2(g) → H2O(ℓ)
should not be considered a proper chemical equation.
5. Does the chemical reaction represented by
3Zn(s) + 2Al(NO3)3(aq) → 3Zn(NO3)2(aq) + 2Al(s)
proceed as written? Why or why not?
6. Does the chemical reaction represented by
2Au(s) + 2HNO3(aq) → 2AuNO3(aq) + H2(g)
proceed as written? Gold is a relatively useful metal for certain applications, such as jewelry and electronics. Does your answer suggest why this is so?
7. Explain what is wrong with this double-replacement reaction.
NaCl(aq) + KBr(aq) → NaK(aq) + ClBr(aq)
8. Predict the products of and balance this double-replacement reaction.
Ag2SO4(aq) + SrCl2(aq) → ?
9. Write the complete and net ionic equations for this double-replacement reaction.
BaCl2(aq) + Ag2SO4(aq) → ?
10. Write the complete and net ionic equations for this double-replacement reaction.
Ag2SO4(aq) + SrCl2(aq) → ?
11. Identify the spectator ions in this reaction. What is the net ionic equation?
NaCl(aq) + KBr(aq) → NaBr(aq) + KCl(aq)
12. Complete this reaction and identify the spectator ions. What is the net ionic equation?
3H2SO4(aq) + 2Al(OH)3(s) → ?
13. Can a reaction be a composition reaction and a redox reaction at the same time? Give an example to support your answer.
14. Can a reaction be a combustion reaction and a redox reaction at the same time? Give an example to support your answer.
15. Can a reaction be a decomposition reaction and a redox reaction at the same time? Give an example to support your answer.
16. Can a reaction be a combustion reaction and a double-replacement reaction at the same time? Give an example to support your answer.
17. Why is CH4 not normally considered an acid?
1. Methyl alcohol has the formula CH3OH. Why would methyl alcohol not normally be considered a base?
2. What are the oxidation numbers of the nitrogen atoms in these substances?
1. N2
2. NH3
3. NO
4. N2O
5. NO2
6. N2O4
7. N2O5
8. NaNO3
3. What are the oxidation numbers of the sulfur atoms in these substances?
1. SF6
2. Na2SO4
3. K2SO3
4. SO3
5. SO2
6. S8
7. Na2S
4. Disproportion is a type of redox reaction in which the same substance is both oxidized and reduced. Identify the element that is disproportionating and indicate the initial and final oxidation numbers of that element.
2CuCl(aq) → CuCl2(aq) + Cu(s)
1. Disproportion is a type of redox reaction in which the same substance is both oxidized and reduced. Identify the element that is disproportionating and indicate the initial and final oxidation numbers of that element.
3Cl2(g) + 6OH(aq) → 5Cl(aq) + ClO3(aq) + 3H2O(ℓ)
Answers
1. H2O(ℓ) → H2O(g)
2.
3. The coefficients are not in their lowest whole-number ratio.
4.
5. No; zinc is lower in the activity series than aluminum.
6.
7. In the products, the cation is pairing with the cation, and the anion is pairing with the anion.
8.
9. Complete ionic equation: Ba2+(aq) + 2Cl(aq) + 2Ag+(aq) + SO42(aq) → BaSO4(s) + 2AgCl(s)
Net ionic equation: The net ionic equation is the same as the complete ionic equation.
10.
11. Each ion is a spectator ion; there is no overall net ionic equation.
12.
13. Yes; H2 + Cl2 → 2HCl (answers will vary)
14.
15. Yes; 2HCl → H2 + Cl2 (answers will vary)
16.
17. It does not increase the H+ ion concentration; it is not a compound of H+.
18.
19.
1. 0
2. −3
3. +2
4. +1
5. +4
6. +4
7. +5
8. +5
20.
21. Copper is disproportionating. Initially, its oxidation number is +1; in the products, its oxidation numbers are +2 and 0, respectively. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.E%3A_Chemical_Reactions_and_Equations_%28Exercises%29.txt |
We have already established that quantities are important in science, especially in chemistry. It is important to make accurate measurements of a variety of quantities when performing experiments. However, it is also important to be able to relate one measured quantity to another, unmeasured quantity. In this chapter, we will consider how we manipulate quantities to relate them to each other.
• 5.1: Introduction
• 5.2: Stoichiometry
Quantities of substances can be related to each other using balanced chemical equations.
• 5.3: The Mole
The mole is a key unit in chemistry. The molar mass of a substance, in grams, is numerically equal to one atom's or molecule's mass in atomic mass units.
• 5.4: The Mole in Chemical Reactions
Balanced chemical reactions are balanced in terms of moles. A balanced chemical reaction gives equivalences in moles that allow stoichiometry calculations to be performed.
• 5.5: Mole-Mass and Mass-Mass Calculations
Mole quantities of one substance can be related to mass quantities using a balanced chemical equation. Mass quantities of one substance can be related to mass quantities using a balanced chemical equation. In all cases, quantities of a substance must be converted to moles before the balanced chemical equation can be used to convert to moles of another substance.
• 5.6: Yields
Theoretical yield is the calculated yield using the balanced chemical reaction. Actual yield is what is actually obtained in a chemical reaction. Percent yield is a comparison of the actual yield with the theoretical yield.
• 5.7: Limiting Reagents
The limiting reagent is the reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced, and how much of the other reactants remain.
• 5.E: Stoichiometry and the Mole (Exercises)
These are exercises and select solutions to accompany Chapter 5 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
05: Stoichiometry and the Mole
At Contrived State University in Anytown, Ohio, a new building was dedicated in March 2010 to house the College of Education. The 100,000-square-foot building has enough office space to accommodate 86 full-time faculty members and 167 full-time staff. In a fit of monetary excess, the university administration offered to buy new furniture (desks and chairs) and computer workstations for all faculty and staff members moving into the new building. However, to save on long-term energy and materials costs, the university offered to buy only 1 laser printer per 10 employees, with the plan to network the printers together.
How many laser printers did the administration have to buy? It is rather simple to show that 26 laser printers are needed for all the employees. However, what if a chemist was calculating quantities for a chemical reaction? Interestingly enough, similar calculations can be performed for chemicals as well as laser printers. In filling a new office building with furniture and equipment, managers do calculations similar to those performed by scientists doing chemical reactions.
5.02: Stoichiometry
Learning Objectives
• Define stoichiometry.
• Relate quantities in a balanced chemical reaction on a molecular basis.
Consider a classic recipe for pound cake: 1 pound of eggs, 1 pound of butter, 1 pound of flour, and 1 pound of sugar. (That's why it's called "pound cake.") If you have 4 pounds of butter, how many pounds of sugar, flour, and eggs do you need? You would need 4 pounds each of sugar, flour, and eggs.
Now suppose you have 1.00 g H2. If the chemical reaction follows the balanced chemical equation
$\ce{2H2(g) + O2(g) → 2H2O(ℓ)}\nonumber$
then what mass of oxygen do you need to make water?
Curiously, this chemical reaction question is very similar to the pound cake question. Both of them involve relating a quantity of one substance to a quantity of another substance or substances. The relating of one chemical substance to another using a balanced chemical reaction is called stoichiometry. Using stoichiometry is a fundamental skill in chemistry; it greatly broadens your ability to predict what will occur and, more importantly, how much is produced.
Let us consider a more complicated example. A recipe for pancakes calls for 2 cups (c) of pancake mix, 1 egg, and 1/2 c of milk. We can write this in the form of a chemical equation:
2 c mix + 1 egg + 1/2 c milk → 1 batch of pancakes
If you have 9 c of pancake mix, how many eggs and how much milk do you need? It might take a little bit of work, but eventually you will find you need 4½ eggs and 2¼ c milk.
How can we formalize this? We can make a conversion factor using our original recipe and use that conversion factor to convert from a quantity of one substance to a quantity of another substance. This is similar to the way we constructed a conversion factor between feet and yards in Chapter 2. Because one recipe's worth of pancakes requires 2 c of pancake mix, 1 egg, and 1/2 c of milk, we actually have the following mathematical relationships that relate these quantities:
2 c pancake mix ⇔ 1 egg ⇔ 1/2 c milk
where ⇔ is the mathematical symbol for "is equivalent to." This does not mean that 2 c of pancake mix equal 1 egg. However, as far as this recipe is concerned, these are the equivalent quantities needed for a single recipe of pancakes. So, any possible quantities of two or more ingredients must have the same numerical ratio as the ratios in the equivalence.
We can deal with these equivalences in the same way we deal with equalities in unit conversions: we can make conversion factors that essentially equal 1. For example, to determine how many eggs we need for 9 c of pancake mix, we construct the conversion factor:
$\frac{1 \, egg}{2\, c \: pancake \: mix}\nonumber$
This conversion factor is, in a strange way, equivalent to 1 because the recipe relates the two quantities. Starting with our initial quantity and multiplying by our conversion factor,
$\cancel{9\, c\: pancake\: mix}\times \frac{1\, egg}{\cancel{2\, c\: pancake\: mix}}=4.5\, eggs\nonumber$
Note how the units cups pancake mix canceled, leaving us with units of eggs. This is the formal, mathematical way of getting our amounts to mix with 9 c of pancake mix. We can use a similar conversion factor for the amount of milk:
$\cancel{9\, c\: pancake\: mix}\times \frac{\frac{1}{2}c\, milk}{\cancel{2\, c\: pancake\: mix}}=2.25\, c\, milk\nonumber$
Again, units cancel, and new units are introduced.
A balanced chemical equation is nothing more than a recipe for a chemical reaction. The difference is that a balanced chemical equation is written in terms of atoms and molecules, not cups, pounds, and eggs.
For example, consider the following chemical equation:
$\ce{2H2(g) + O2(g) → 2H2O(ℓ)} \nonumber \nonumber$
We can interpret this as, literally, "two hydrogen molecules react with one oxygen molecule to make two water molecules." That interpretation leads us directly to some equivalencies, just as our pancake recipe did:
2H2 molecules ⇔ 1O2 molecule ⇔ 2H2O molecules
These equivalences allow us to construct conversion factors:
$\frac{2\, molecules\, H_{2}}{1\, molecule\, O_{2}}\; \frac{2\, molecules\, H_{2}}{2\, molecules\, H_{2}O}\; \frac{1\, molecule\, H_{2}}{2\, molecules\, H_{2}O}\nonumber$
and so forth. These conversions can be used to relate quantities of one substance to quantities of another. For example, suppose we need to know how many molecules of oxygen are needed to react with 16 molecules of H2. As we did with converting units, we start with our given quantity and use the appropriate conversion factor:
$\cancel{16\, molecules\, H_{2}}\times \frac{1\, molecules\, O_{2}}{\cancel{2\, molecules\, H_{2}}}=8\, molecules\, O_{2}\nonumber$
Note how the unit molecules H2 cancels algebraically, just as any unit does in a conversion like this. The conversion factor came directly from the coefficients in the balanced chemical equation. This is another reason why a properly balanced chemical equation is important.
Example $1$
How many molecules of SO3 are needed to react with 144 molecules of Fe2O3 given this balanced chemical equation?
$\ce{Fe2O3 + 3 SO3 → Fe2(SO4)3}\nonumber$
Solution
We use the balanced chemical equation to construct a conversion factor between Fe2O3 and SO3. The number of molecules of Fe2O3 goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO3 go on the top. Thus, the appropriate conversion factor is
$\frac{3\, molecules\, SO_{3}}{1\, molecule\, Fe_{2}O_{3}}\nonumber$
Starting with our given amount and applying the conversion factor, the result is
$144\, \cancel{molecules}\, Fe_{2}O_{3}\times \frac{3\, molecules\, SO_{3}}{1\, \cancel{molecule}\, Fe_{2}O_{3}}=432\, molecules\, SO_{3}\nonumber$
We need 432 molecules of SO3 to react with 144 molecules of Fe2O3.
Exercise $1$
How many molecules of H2 are needed to react with 29 molecules of N2 to make ammonia if the balanced chemical equation is:
$\ce{N2 + 3H2 → 2NH3}?\nonumber$
Answer
87 molecules
Chemical equations also allow us to make conversions regarding the number of atoms in a chemical reaction, because a chemical formula lists the number of atoms of each element in a compound. The formula H2O indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors:
$\frac{2\, atoms\, H}{1\, molecule\, H_{2}O}\; \frac{1\, molecule\, H_{2}O}{1\, atom\, O}\nonumber$
Conversion factors like this can also be used in stoichiometry calculations.
Example $2$
How many molecules of NH3 can you make if you have 228 atoms of H2?
Solution
From the formula, we know that one molecule of NH3 has three H atoms. Use that fact as a conversion factor:
$228\, \cancel{atoms\, H}\times \frac{1\, molecule\, NH_{3}}{3\cancel{atoms\, H}}=76\, NH_{3}\nonumber$
Exercise $2$
How many molecules of $\ce{Fe2(SO4)3}$ can you make from 777 atoms of S?
Answer
259 molecules
Summary
Quantities of substances can be related to each other using balanced chemical equations. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.01%3A_Introduction.txt |
Learning Objectives
• Describe the unit mole.
• Relate the mole quantity of substance to its mass.
So far, we have been talking about chemical substances in terms of individual atoms and molecules. Yet we do not typically deal with substances an atom or a molecule at a time; we work with millions, billions, and trillions of atoms and molecules at a time. We need a way to deal with macroscopic, rather than microscopic, amounts of matter. We need a unit of amount that relates quantities of substances on a scale that we can interact with.
Chemistry uses a unit called mole. A mole (mol) is a number of things equal to the number of atoms in exactly 12 g of carbon-12. Experimental measurements have determined that this number is very large:
1 mol = 6.02214179 × 1023 things
Understand that a mole means a specific number of things, just like a dozen means a certain number of things—twelve, in the case of a dozen. But a mole is a much larger number of things. These things can be atoms, or molecules, or eggs; however, in chemistry, we usually use the mole to refer to the amounts of atoms or molecules. Although the number of things in a mole is known to eight decimal places, it is usually fine to use only two or three decimal places in calculations. The numerical value of things in a mole is often called Avogadro's number (NA). Avogadro's number is also known as the Avogadro constant, after Amadeo Avogadro, an Italian chemist who first proposed its importance.
Example $1$
How many molecules are present in 2.76 mol of H2O? How many atoms is this?
Solution
The definition of a mole is an equality that can be used to construct a conversion factor. Also, because we know that there are three atoms in each molecule of H2O, we can also determine the number of atoms in the sample.
$2.76\, \cancel{mol\, H_{2}O}\times \frac{6.022\times 10^{23}molecules\, H_{2}O}{\cancel{mol\, H_{2}O}}=1.66\times 10^{24}molecules\, H_{2}O \nonumber \nonumber$
To determine the total number of atoms, we have
$1.66\times 10^{24}\cancel{molecules\, H_{2}O}\times \frac{3\, atoms}{1\, molecule}=4.99\times 10^{24}\, atoms \nonumber \nonumber$
Exercise $1$
How many molecules are present in 4.61 × 10−2 mol of $\ce{O2}$?
Answer
2.78 × 1022 molecules
How big is a mole? It is very large. Suppose you had a mole of dollar bills that need to be counted. If everyone on earth (about 6 billion people) counted one bill per second, it would take about 3.2 million years to count all the bills. A mole of sand would fill a cube about 32 km on a side. A mole of pennies stacked on top of each other would have about the same diameter as our galaxy, the Milky Way. A mole is a lot of things—but atoms and molecules are very tiny. One mole of carbon atoms would make a cube that is 1.74 cm on a side, small enough to carry in your pocket.
Why is the mole unit so important? It represents the link between the microscopic and the macroscopic, especially in terms of mass. A mole of a substance has the same mass in grams as one unit (atom or molecules) has in atomic mass units. The mole unit allows us to express amounts of atoms and molecules in visible amounts that we can understand.
For example, we already know that, by definition, a mole of carbon has a mass of exactly 12 g. This means that exactly 12 g of C has 6.022 × 1023 atoms:
12 g C = 6.022 × 1023 atoms C
We can use this equality as a conversion factor between the number of atoms of carbon and the number of grams of carbon. How many grams are there, say, in 1.50 × 1025 atoms of carbon? This is a one-step conversion:
$1.50\times 10^{25}\cancel{atoms\, C}\times \frac{12.0000\, g\, C}{6.022\times 10^{23}\cancel{atoms\, C}}=299\, g\, C\nonumber$
But it also goes beyond carbon. Previously we defined atomic and molecular masses as the number of atomic mass units per atom or molecule. Now we can do so in terms of grams. The atomic mass of an element is the number of grams in 1 mol of atoms of that element, while the molecular mass of a compound is the number of grams in 1 mol of molecules of that compound. Sometimes these masses are called molar masses to emphasize the fact that they are the mass for 1 mol of things. (The term molar is the adjective form of mole and has nothing to do with teeth.)
Here are some examples. The mass of a hydrogen atom is 1.0079 u; the mass of 1 mol of hydrogen atoms is 1.0079 g. Elemental hydrogen exists as a diatomic molecule, H2. One molecule has a mass of 1.0079 + 1.0079 = 2.0158 u, while 1 mol H2 has a mass of 2.0158 g. A molecule of H2O has a mass of about 18.01 u; 1 mol H2O has a mass of 18.01 g. A single unit of NaCl has a mass of 58.45 u; NaCl has a molar mass of 58.45 g. In each of these moles of substances, there are 6.022 × 1023 units: 6.022 × 1023 atoms of H, 6.022 × 1023 molecules of H2 and H2O, 6.022 × 1023 units of NaCl ions. These relationships give us plenty of opportunities to construct conversion factors for simple calculations.
Example $2$: Sugar
What is the molar mass of sugar ($\ce{C6H12O6}$)?
Solution
To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula; but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table.
Solutions to Example 5.3.2
6 C = 6 × 12.011 = 72.066
12 H = 12 × 1.0079 = 12.0948
6 O = 6 × 15.999 = 95.994
TOTAL = 180.155 g/mol
Per convention, the unit grams per mole is written as a fraction.
Exercise $2$
What is the molar mass of $\ce{AgNO3}$?
Answer
169.87 g/mol
Knowing the molar mass of a substance, we can calculate the number of moles in a certain mass of a substance and vice versa, as these examples illustrate. The molar mass is used as the conversion factor.
Example $3$
What is the mass of 3.56 mol of HgCl2? The molar mass of HgCl2 is 271.49 g/mol.
Solution
Use the molar mass as a conversion factor between moles and grams. Because we want to cancel the mole unit and introduce the gram unit, we can use the molar mass as given:
$3.56\, \cancel{mol\, HgCl_{2}}\times \frac{271.49\, g\, HgCl_{2}}{\cancel{mol\, HgCl_{2}}}=967\, g\, HgCl_{2} \nonumber \nonumber$
Exercise $3$
What is the mass of 33.7 mol of $\ce{H2O}$?
Answer
607 g
Example $4$
How many moles of H2O are present in 240.0 g of water (about the mass of a cup of water)?
Solution
Use the molar mass of H2O as a conversion factor from mass to moles. The molar mass of water is
(1.0079 + 1.0079 + 15.999) = 18.015 g/mol.
However, because we want to cancel the gram unit and introduce moles, we need to take the reciprocal of this quantity, or 1 mol/18.015 g:
$240.0\, \cancel{g\, H_{2}O}\times \frac{1\, mol\, H_{2}O}{18.015\cancel{g\, H_{2}O}}=13.32\, mol\, H_{2}O \nonumber \nonumber$
Exercise $4$
How many moles are present in 35.6 g of H2SO4 (molar mass = 98.08 g/mol)?
Answer
0.363 mol
Other conversion factors can be combined with the definition of mole—density, for example.
Example $5$
The density of ethanol is 0.789 g/mL. How many moles are in 100.0 mL of ethanol? The molar mass of ethanol is 46.08 g/mol.
Solution
Here, we use density to convert from volume to mass and then use the molar mass to determine the number of moles.
$100\cancel{ml}\: ethanol\times \frac{0.789\, g}{\cancel{ml}}\times \frac{1\, mol}{46.08\, \cancel{g}}=1.71\, mol\, ethanol\nonumber$
Exercise $5$
If the density of benzene, C6H6, is 0.879 g/mL, how many moles are present in 17.9 mL of benzene?
Answer
0.201 mol
Summary
The mole is a key unit in chemistry. The molar mass of a substance, in grams, is numerically equal to one atom's or molecule's mass in atomic mass units. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.03%3A_The_Mole.txt |
Learning Objectives
• Balance a chemical equation in terms of moles.
• Use the balanced equation to construct conversion factors in terms of moles.
• Calculate moles of one substance from moles of another substance using a balanced chemical equation.
Consider this balanced chemical equation:
$\ce{2H2 + O2 → 2H2O}\nonumber$
We interpret this as "two molecules of hydrogen react with one molecule of oxygen to make two molecules of water." The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:
$\ce{100H2 + 50O2 → 100H2O}\nonumber$
This equation is not conventional—because convention says that we use the lowest ratio of coefficients—but it is balanced. So is this chemical equation:
$\ce{5,000 H2 + 2,500 O2 → 5,000H2O}\nonumber$
Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:
$12.044 \times 10^{23} \ce{H2} + 6.022 \times 10^{23} \ce{O2} → 12.044 \times 10^{23} \ce{H2O}\nonumber$
These coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro's number, while the second number is Avogadro's number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?
$\ce{2H2 + O2 → 2H2O}\nonumber$
is the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles. We can just as easily read this chemical equation as "two moles of hydrogen react with one mole of oxygen to make two moles of water." All balanced chemical reactions are balanced in terms of moles.
Example $1$
Interpret this balanced chemical equation in terms of moles.
$\ce{P4 + 5O2 → P4O10}\nonumber$
Solution
The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as "one mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide."
Exercise $1$
Interpret this balanced chemical equation in terms of moles.
$\ce{N2 + 3H2 → 2NH3}\nonumber$
Answer
One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.
In Section 4.1, we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalents—equivalents between the reactants and the products. However, now we understand that these equivalents are expressed in terms of moles. Consider the chemical equation
$\ce{2H2 + O2 → 2H2O} \nonumber \nonumber$
This chemical reaction gives us the following equivalents:
2 mol H2 ⇔ 1 mol O2 ⇔ 2 mol H2O
Any two of these quantities can be used to construct a conversion factor that lets us relate the number of moles of one substance to an equivalent number of moles of another substance. If, for example, we want to know how many moles of oxygen will react with 17.6 mol of hydrogen, we construct a conversion factor between 2 mol of H2 and 1 mol of O2 and use it to convert from moles of one substance to moles of another:
$17.6\cancel{mol\, H_{2}}\times \frac{1\, mol\, O_{2}}{2\cancel{mol\, H_{2}}}=8.80\, mol\, O_{2}\nonumber$
Note how the mol H2 unit cancels, and mol O2 is the new unit introduced. This is an example of a mole-mole calculation, when you start with moles of one substance and convert to moles of another substance by using the balanced chemical equation. The example may seem simple because the numbers are small, but numbers won't always be so simple!
Example $2$
For the balanced chemical equation
$\ce{2C4H10(g) + 13O2 → 8CO2(g) + 10H2O(ℓ)}\nonumber$
if 154 mol of O2 are reacted, how many moles of CO2 are produced?
Solution
We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is
13 mol O2 ⇔ 8 mol CO2
We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:
$154\cancel{mol\, O_{2}}\times \frac{8\, mol\, CO_{2}}{13\cancel{mol\, O_{2}}}=94.8\, mol\, CO_{2}\nonumber$
The mol O2 unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they do not contribute to the number of significant figures in the final answer.
Exercise $2$
Using the above equation, how many moles of H2O are produced when 154 mol of O2 react?
Answer
118 mol
It is important to reiterate that balanced chemical equations are balanced in terms of moles. Not grams, kilograms, or liters—but moles. Any stoichiometry problem will likely need to work through the mole unit at some point, especially if you are working with a balanced chemical reaction.
Summary
Balanced chemical reactions are balanced in terms of moles. A balanced chemical reaction gives equivalents in moles that allow stoichiometry calculations to be performed. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.04%3A_The_Mole_in_Chemical_Reactions.txt |
Learning Objectives
• From a given number of moles of a substance, calculate the mass of another substance involved using the balanced chemical equation.
• From a given mass of a substance, calculate the moles of another substance involved using the balanced chemical equation.
• From a given mass of a substance, calculate the mass of another substance involved using the balanced chemical equation.
Mole-mole calculations are not the only type of calculations that can be performed using balanced chemical equations. Recall that the molar mass can be determined from a chemical formula and used as a conversion factor. We can add that conversion factor as another step in a calculation to make a mole-mass calculation, where we start with a given number of moles of a substance and calculate the mass of another substance involved in the chemical equation, or vice versa.
For example, suppose we have the balanced chemical equation
$2Al + 3Cl_2 → 2AlCl_3\nonumber$
Suppose we know we have 123.2 g of Cl2. How can we determine how many moles of AlCl3 we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl2 to an amount of AlCl3, we need to convert the given amount of Cl2 into moles. We know how to do this by simply using the molar mass of Cl2 as a conversion factor. The molar mass of Cl2 (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mol. We must invert this fraction so that the units cancel properly:
$123.2\, \cancel{g\, Cl_{2}}\times \frac{1\, mol\, Cl_{2}}{70.90\cancel{g\, Cl_{2}}}=1.738\, \, mol\, Cl_{2}\nonumber$
Now that we have the quantity in moles, we can use the balanced chemical equation to construct a conversion factor that relates the number of moles of Cl2 to the number of moles of AlCl3. The numbers in the conversion factor come from the coefficients in the balanced chemical equation:
$\frac{2\, mol\, AlCl_{3}}{3\, mol\, Cl_{2}}\nonumber$
Using this conversion factor with the molar quantity we calculated above, we get
$1.738\, \cancel{mol\, Cl_{2}}\times \frac{2\, mol\, AlCl_{3}}{3\, \cancel{mol\, Cl_{2}}}=1.159\, mol\, AlCl_{3}\nonumber$
So, we will get 1.159 mol of AlCl3 if we react 123.2 g of Cl2.
In this last example, we did the calculation in two steps. However, it is mathematically equivalent to perform the two calculations sequentially on one line:
$123.2\, \cancel{g\, Cl_{2}}\times \frac{1\, mol\, Cl_{2}}{70.90\cancel{g\, Cl_{2}}}\times \frac{2\, mol\, AlCl_{3}}{3\, \cancel{mol\, Cl_{2}}}=1.159\, mol\, AlCl_{3}\nonumber$
The units still cancel appropriately, and we get the same numerical answer in the end. Sometimes the answer may be slightly different from doing it one step at a time because of rounding of the intermediate answers, but the final answers should be effectively the same.
Example $1$
How many moles of HCl will be produced when 249 g of AlCl3 are reacted according to this chemical equation?
$2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g)\nonumber$
Solution
We will do this in two steps: convert the mass of AlCl3 to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl3 is 133.33 g/mol, which we have to invert to get the appropriate conversion factor:
$249\, \cancel{g\, AlCl_{3}}\times \frac{1\, mol\, AlCl_{3}}{133.33\, \cancel{g\, AlCl_{3}}}=1.87\, mol\, AlCl_{3}\nonumber$
Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl3 and the number of moles of HCl:
$frac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}\nonumber$
Applying this conversion factor to the quantity of AlCl3, we get
$1.87\, \cancel{mol\, AlCl_{3}}\times \frac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}=5.61\, mol\, HCl\nonumber$
Alternatively, we could have done this in one line:
$249\, \cancel{g\, AlCl_{3}}\times \frac{1\, \cancel{mol\, AlCl_{3}}}{133.33\, \cancel{g\, AlCl_{3}}}\times \frac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}=5.60\, mol\, HCl\nonumber$
The last digit in our final answer is slightly different because of rounding differences, but the answer is essentially the same.
Exercise $1$
How many moles of Al2O3 will be produced when 23.9 g of H2O are reacted according to this chemical equation?
$2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g)\nonumber$
Answer
0.442 mol
A variation of the mole-mass calculation is to start with an amount in moles and then determine an amount of another substance in grams. The steps are the same but are performed in reverse order.
Example $2$
How many grams of NH3 will be produced when 33.9 mol of H2 are reacted according to this chemical equation?
$N_2(g) + 3H_2(g) → 2NH_3(g)\nonumber$
Solution
The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have
$33.9\cancel{mol\, H_{2}}\times \frac{2\, mol\, NH_{3}}{3\cancel{mol\, H_{2}}}=22.6\, mol\, NH_{3}\nonumber$
Now, using the molar mass of NH3, which is 17.03 g/mol, we get
$22.6\cancel{mol\, NH_{3}}\times \frac{17.03\, g\, NH_{3}}{1\cancel{mol\, NH_{3}}}=385\, g\, NH_{3}\nonumber$
Exercise $2$
How many grams of N2 are needed to produce 2.17 mol of NH3 when reacted according to this chemical equation?
$\ce{N_2(g) + 3H_2(g) → 2NH_3(g)} \nonumber$
Answer
30.4 g (Note: here we go from a product to a reactant, showing that mole-mass problems can begin and end with any substance in the chemical equation.)
It should be a trivial task now to extend the calculations to mass-mass calculations, in which we start with a mass of some substance and end with the mass of another substance in the chemical reaction. For this type of calculation, the molar masses of two different substances must be used—be sure to keep track of which is which. Again, however, it is important to emphasize that before the balanced chemical reaction is used, the mass quantity must first be converted to moles. Then the coefficients of the balanced chemical reaction can be used to convert to moles of another substance, which can then be converted to a mass.
For example, let us determine the number of grams of SO3 that can be produced by the reaction of 45.3 g of SO2 and O2:
$2SO_2(g) + O_2(g) → 2SO_3(g)\nonumber$
First, we convert the given amount, 45.3 g of SO2, to moles of SO2 using its molar mass (64.06 g/mol):
$45.3\cancel{g\, SO_{2}}\times \frac{1\, mol\, SO_{2}}{64.06\cancel{g\, SO_{2}}}=0.707\, mol\, SO_{2}\nonumber$
Second, we use the balanced chemical reaction to convert from moles of SO2 to moles of SO3:
$0.707\cancel{mol\, SO_{2}}\times \frac{2\, mol\, SO_{3}}{2\cancel{mol\, SO_{2}}}=0.707\, mol\, SO_{3}\nonumber$
Finally, we use the molar mass of SO3 (80.06 g/mol) to convert to the mass of SO3:
$0.707\cancel{mol\, SO_{3}}\times \frac{80.06\, g\, SO_{3}}{1\cancel{mol\, SO_{3}}}=56.6\, g\, SO_{3}\nonumber$
We can also perform all three steps sequentially, writing them on one line as
$45.3\cancel{g\, SO_{2}}\times \frac{1\, mol\, SO_{2}}{64.06\cancel{g\, SO_{2}}}\times \frac{2\, mol\, SO_{3}}{2\cancel{mol\, SO_{2}}}\times \frac{80.06\, g\, SO_{3}}{1\cancel{mol\, SO_{3}}}=56.6\, g\, SO_{3}\nonumber$
We get the same answer. Note how the initial and all the intermediate units cancel, leaving grams of SO3, which is what we are looking for, as our final answer.
Example $3$
What mass of Mg will be produced when 86.4 g of K are reacted?
$\ce{MgCl2(s) + 2K(s) → Mg(s) + 2KCl(s)}\nonumber$
Solution
We will simply follow the steps
mass K → mol K → mol Mg → mass Mg
In addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line,
$86.4\cancel{g\, K}\times \frac{1\, mol\, K}{39.09\cancel{g\, K}}\times \frac{1\, \cancel{mol\, Mg}}{2\cancel{mol\, K}}\times \frac{24.31\, g\, Mg}{1\cancel{mol\, Mg}}=26.87\, g\, Mg\nonumber$
Exercise $3$
What mass of H2 will be produced when 122 g of Zn are reacted?
$\ce{Zn(s) + 2HCl(aq) → ZnCl_2(aq) + H_2(g)} \nonumber$
Answer
3.77 g
Summary
• Mole quantities of one substance can be related to mass quantities using a balanced chemical equation.
• Mass quantities of one substance can be related to mass quantities using a balanced chemical equation.
• In all cases, quantities of a substance must be converted to moles before the balanced chemical equation can be used to convert to moles of another substance. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.05%3A_Mole-Mass_and_Mass-Mass_Calculations.txt |
Learning Objective
• Define and determine theoretical yields, actual yields, and percent yields.
In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions:
1. The reaction goes exactly as written.
2. The reaction proceeds completely.
In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO2. However, we will continue to ignore side reactions, unless otherwise noted. The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons (some of which we will consider in Chapter 13). When we calculate an amount of product assuming that all the reactant reacts, we calculate the theoretical yield, an amount that is theoretically produced as calculated using the balanced chemical reaction.
In many cases, however, this is not what really happens. In many cases, less—sometimes, much less—of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the actual yield. By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made.
Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The percent yield is a comparison between the actual yield and the theoretical yield and is defined as
$\text{percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \label{yield}$
It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%. Again, if percent yield is greater than 100%, an error has been made.
Example $1$
A worker reacts 30.5 g of Zn with nitric acid and evaporates the remaining water to obtain 65.2 g of Zn(NO3)2. What are the theoretical yield, the actual yield, and the percent yield?
$\ce{Zn(s) + 2HNO_3(aq) → Zn(NO_3)_2(aq) + H_2(g)} \nonumber$
Solution
A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g/mol) and Zn(NO3)2 (189.41 g/mol). In three steps, the mass-mass calculation is:
$30.5\cancel{g\, Zn}\times \frac{1\, \cancel{mol\, Zn}}{65.39\cancel{g\, Zn}}\times \frac{1\, \cancel{mol\, Zn(NO_{3})_{2}}}{1\cancel{mol\, Zn}}\times \frac{189.41\, g\, Zn(NO_{3})_{2}}{1\cancel{mol\,Zn(NO_{3})_{2}}}=88.3\, g\, Zn(NO_{3})_{2}\nonumber$
Thus, the theoretical yield is 88.3 g of Zn(NO3)2. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO3)2. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100 (Equation \ref{yield}):
$\frac{65.2\, g\, Zn(NO_{3})_{2}}{88.3\, g\,Zn(NO_{3})_{2}}\times 100\%=73.8\%\nonumber$
The worker achieved almost three-fourths of the possible yield.
Exercise $1$
A synthesis produced 2.05 g of NH3 from 16.5 g of N2. What is the theoretical yield and the percent yield?
$N_2(g) + 3H_2(g) → 2NH_3(g)\nonumber$
*Technically, this is a reversible reaction (with double arrows), but for this exercise consider it irreversible (single arrow).
Answer
theoretical yield = 20.1 g; percent yield = 10.2%
Chemistry is Everywhere: Actual Yields in Drug Synthesis and Purification
Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.
Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C13H21NO2; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:
A child using an albuterol inhaler, a container of albuterol medication, and a molecular model of albuterol are shown in three combined images.
impure albuterol → intermediate A percent yield = 70%
intermediate A → intermediate B percent yield = 100%
intermediate B → intermediate C percent yield = 40%
intermediate C → intermediate D percent yield = 72%
intermediate D → purified albuterol percent yield = 35%
overall percent yield = 70% × 100% × 40% × 72% × 35% = 7.5%
That is, only about one-fourteenth of the original material was turned into the purified drug. This demonstrates one reason why some drugs are so expensive—a lot of material is lost in making a high-purity pharmaceutical.
Summary
Theoretical yield is the calculated yield using the balanced chemical reaction. Actual yield is what is actually obtained in a chemical reaction. Percent yield is a comparison of the actual yield with the theoretical yield. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.06%3A_Yields.txt |
Learning Objectives
• Identify a limiting reagent from a set of reactants.
• Calculate how much product will be produced from the limiting reagent.
• Calculate how much reactant(s) remains when the reaction is complete.
In addition to the assumption that reactions proceed all the way to completion, one additional assumption we have made about chemical reactions is that all the reactants are present in the proper quantities to react to products; this is not always the case. In Figure $2$ we are taking hydrogen atoms and oxygen atoms (left) to make water molecules (right). However, there are not enough oxygen atoms to use up all the hydrogen atoms. We run out of oxygen atoms and cannot make any more water molecules, so the process stops when we run out of oxygen atoms.
A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reagent; the other reactant or reactants are considered to be in excess. A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reagent and which is in excess.
The key to recognizing which reactant is the limiting reagent is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reagent. What we need to do is determine an amount of one product (either moles or mass), assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reagent. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful.
For example, consider this reaction:
$4As(s) + 3O_2(g) → 2As_2O_3(s)\nonumber$
Suppose we start a reaction with 50.0 g of As and 50.0 g of O2. Which one is the limiting reagent? We need to perform two mole-mass calculations, each assuming that each reactant reacts completely. Then we compare the amount of the product produced by each and determine which is less.
The calculations are as follows:
$50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{2\, mol\, As_{2}O_{3}}{4\cancel{mol\, As}}=0.334\, mol\, As_{2}O_{3}\nonumber$
$50.0\cancel{g\, O_{2}}\times \frac{1\cancel{mol\, O_{2}}}{32.00\cancel{g\, O_{2}}}\times \frac{2\, mol\, As_{2}O_{3}}{3\cancel{mol\, O_{2}}}=1.04\, mol\, As_{2}O_{3}\nonumber$
Comparing these two answers, it is clear that 0.334 mol of As2O3 is less than 1.04 mol of As2O3, so arsenic is the limiting reagent. If this reaction is performed under these initial conditions, the arsenic will run out before the oxygen runs out. We say that the oxygen is "in excess."
Identifying the limiting reagent, then, is straightforward. However, there are usually two associated questions: (1) what mass of product (or products) is then actually formed? and (2) what mass of what reactant is left over? The first question is straightforward to answer: simply perform a conversion from the number of moles of product formed to its mass, using its molar mass. For As2O3, the molar mass is 197.84 g/mol; knowing that we will form 0.334 mol of As2O3 under the given conditions, we will get
$0.334\cancel{mol\, As_{2}O_{3}}\times \frac{197.84\, g\, As_{2}}{\cancel{1\, mol\, As_{2}O_{3}}}=66.1\, g\, As_{2}O_{3}\nonumber$
The second question is somewhat more convoluted to answer. First, we must do a mass-mass calculation relating the limiting reagent (here, As) to the other reagent (O2). Once we determine the mass of O2 that reacted, we subtract that from the original amount to determine the amount left over. According to the mass-mass calculation,
$50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{3\cancel{mol\, O_{2}}}{4\cancel{mol\, As}}\times \frac{32.00\, g\, O_{2}}{\cancel{1\, mol\, O_{2}}}=16.0\, g\, O_{2}\; reacted\nonumber$
Because we reacted 16.0 g of our original O2, we subtract that from the original amount, 50.0 g, to get the mass of O2 remaining:
50.0 g O2 − 16.0 g O2 reacted = 34.0 g O2 left over
You must remember to perform this final subtraction to determine the amount remaining; a common error is to report the 16.0 g as the amount remaining.
Example $1$
A 5.00 g quantity of Rb is combined with 3.44 g of MgCl2 according to this chemical reaction:
$2R b(s) + MgCl_2(s) → Mg(s) + 2RbCl(s) \nonumber \nonumber$
What mass of Mg is formed, and what mass of what reactant is left over?
Solution
Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less.
$5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, Mg}}{2\cancel{mol\, Rb}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.711\, g\, Mg \nonumber$
$3.44\cancel{g\, MgCl_{2}}\times \frac{1\cancel{mol\, MgCl_{2}}}{95.21\cancel{g\, MgCl_{2}}}\times \frac{1\cancel{mol\, Mg}}{1\cancel{mol\, MgCl_{2}}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.878\, g\, Mg \nonumber$
The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb, and then subtract the amount reacted from the original amount.
$5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, MgCl_{2}}}{2\cancel{mol\, Rb}}\times \frac{95.21\, g\, Mg}{\cancel{1\, mol\, MgCl_{2}}}=2.78\, g\, MgCl_{2}\: \: reacted \nonumber$
Because we started with 3.44 g of MgCl2, we have
3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left
Exercise $1$
Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent?
$\underbrace{22.7\, g}_{MgO(s)}+\underbrace{17.9\, g}_{H_2S}\rightarrow MgS(s)+H_{2}O(l) \nonumber$
Answer
H2S is the limiting reagent; 1.5 g of MgO are left over.
Summary
The limiting reagent is the reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced and how much of the other reactants remain. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.07%3A_Limiting_Reagents.txt |
5.1: Stoichiometry
1. Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe?
2. Think back to the pancake recipe. What possible conversion factors can you construct relating the components of the recipe?
3. What are all the conversion factors that can be constructed from the balanced chemical reaction: $\ce{2H2(g) + O2(g) → 2H2O(ℓ)}?$
4. What are all the conversion factors that can be constructed from the balanced chemical reaction N2(g) + 3H2(g) → 2NH3(g)?
5. Given the chemical equation : Na(s) + H2O(ℓ) → NaOH(aq) + H2(g)
1. Balance the equation.
2. How many molecules of H2 are produced when 332 atoms of Na react?
6. Given the chemical equation: S(s) + O2(g) → SO3(g)
1. Balance the equation.
2. How many molecules of O2 are needed when 38 atoms of S react?
7. For the balanced chemical equation:
6H+(aq) + 2MnO4(aq) + 5H2O2(ℓ) → 2Mn2+(aq) + 5O2(g) + 8H2O(ℓ)
how many molecules of H2O are produced when 75 molecules of H2O2 react?
1. For the balanced chemical reaction
2C6H6(ℓ) + 15O2(g) → 12CO2(g) + 6H2O(ℓ)
how many molecules of CO2 are produced when 56 molecules of C6H6 react?
2. Given the balanced chemical equation
Fe2O3(s) + 3SO3(g) → Fe2(SO4)3
how many molecules of Fe2(SO4)3 are produced if 321 atoms of S are reacted?
3. For the balanced chemical equation
CuO(s) + H2S(g) → CuS + H2O(ℓ)
how many molecules of CuS are formed if 9,044 atoms of H react?
4. For the balanced chemical equation
Fe2O3(s) + 3SO3(g) → Fe2(SO4)3
suppose we need to make 145,000 molecules of Fe2(SO4)3. How many molecules of SO3 do we need?
1. One way to make sulfur hexafluoride is to react thioformaldehyde, CH2S, with elemental fluorine:
CH2S + 6F2 → CF4 + 2HF + SF6
If 45,750 molecules of SF6 are needed, how many molecules of F2 are required?
1. Construct the three independent conversion factors possible for these two reactions:
1. 2H2 + O2 → 2H2O
2. H2 + O2 → H2O2
Why are the ratios between H2 and O2 different?
The conversion factors are different because the stoichiometries of the balanced chemical reactions are different.
1. Construct the three independent conversion factors possible for these two reactions:
1. 2Na + Cl2 → 2NaCl
2. 4Na + 2Cl2 → 4NaCl
What similarities, if any, exist in the conversion factors from these two reactions?
Answers
1. $\frac{1\, pound\, butter}{1\, pound\, flour}$ or $\frac{1\, pound\, sugar}{1\, pound\, eggs}$ are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible.1 pound butter1 pound flour
2.
3. $\frac{2\, molecules\, H_{2}}{1\, molecule\, O_{2}}$ , $\frac{1\, molecule\, O_{2}}{2\, molecules\, H_{2}O}$ , $\frac{2\, molecules\, H_{2}}{2\, molecules\, H_{2}O}$ and their reciprocals are the conversion factors that can be constructed.
4.
5.
1. 2Na(s) + 2H2O(ℓ) → 2NaOH(aq) + H2(g)
2. 166 molecules
6.
7. 120 molecules
8.
9. 107 molecules
10.
11. 435,000 molecules
12.
13.
1. $\frac{2\, molecules\, H_{2}}{1\, molecule\, O_{2}}\ , \frac{1\, molecule\, O_{2}}{2\, molecules\, H_{2}O}\ , \frac{2\, molecules\, H_{2}}{2\, molecules\, H_{2}O}$
2. $\frac{1\, molecules\, H_{2}}{1\, molecule\, O_{2}}\ , \frac{1\, molecule\, O_{2}}{2\, molecules\, H_{2}O_{2}}\ , \frac{1\, molecule\, H_{2}}{1\, molecule\, H_{2}O_{2}}$
5.2: The Mole
1. How many atoms are present in 4.55 mol of Fe?
2. How many atoms are present in 0.0665 mol of K?
3. How many molecules are present in 2.509 mol of H2S?
4. How many molecules are present in 0.336 mol of acetylene (C2H2)?
5. How many moles are present in 3.55 × 1024 Pb atoms?
6. How many moles are present in 2.09 × 1022 Ti atoms?
7. How many moles are present in 1.00 × 1023 PF3 molecules?
8. How many moles are present in 5.52 × 1025 penicillin molecules?
9. Determine the molar mass of each substance.
1. Si
2. SiH4
3. K2O
10. Determine the molar mass of each substance.
1. Cl2
2. SeCl2
3. Ca(C2H3O2)2
11. Determine the molar mass of each substance.
1. Al
2. Al2O3
3. CoCl3
12. Determine the molar mass of each substance.
1. O3
2. NaI
3. C12H22O11
13. What is the mass of 4.44 mol of Rb?
14. What is the mass of 0.311 mol of Xe?
15. What is the mass of 12.34 mol of Al2(SO4)3?
16. What is the mass of 0.0656 mol of PbCl2?
17. How many moles are present in 45.6 g of CO?
18. How many moles are present in 0.00339 g of LiF?
19. How many moles are present in 1.223 g of SF6?
20. How many moles are present in 48.8 g of BaCO3?
21. How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?
22. How many moles are present in 56.83 mL of O2 if the density of O2 is 0.00133 g/mL?
Answers
1. 2.74 × 1024 atoms
2.
3. 1.511 × 1024 molecules
4.
5. 5.90 mol
6.
7. 0.166 mol
8.
9.
1. 28.086 g
2. 32.118 g
3. 94.195 g
10.
11.
1. 26.981 g
2. 101.959 g
3. 165.292 g
12.
13. 379 g
14.
15. 4,222 g
16.
17. 1.63 mol
18.
19. 0.008374 mol
20.
21. 3.72 mol
5.3: The Mole in Chemical Reactions
1. Express in mole terms what this chemical equation means: CH4 + 2O2 → CO2 + 2H2O
2. Express in mole terms what this chemical equation means.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
3. How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles?
4. How many molecules of each substance are involved in the equation in Exercise 2 if it is interpreted in terms of moles?
5. For the chemical equation
2C2H6 + 7O2 → 4CO2 + 6H2O
what equivalents can you write in terms of moles? Use the ⇔ sign.
6. For the chemical equation
2Al + 3Cl2 → 2AlCl3
what equivalents can you write in terms of moles? Use the ⇔ sign.
7. Write the balanced chemical reaction for the combustion of C5H12 (the products are CO2 and H2O) and determine how many moles of H2O are formed when 5.8 mol of O2 are reacted.
8. Write the balanced chemical reaction for the formation of Fe2(SO4)3 from Fe2O3 and SO3 and determine how many moles of Fe2(SO4)3 are formed when 12.7 mol of SO3 are reacted.
9. For the balanced chemical equation
3Cu(s) + 2NO3(aq) + 8H+(aq) → 3Cu2+(aq) + 4H2O(ℓ) + 2NO(g)
how many moles of Cu2+ are formed when 55.7 mol of H+ are reacted?
10. For the balanced chemical equation
Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s)
how many moles of Ag are produced when 0.661 mol of Al are reacted?
11. For the balanced chemical reaction
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(ℓ)
how many moles of H2O are produced when 0.669 mol of NH3 react?
12. For the balanced chemical reaction
4NaOH(aq) + 2S(s) + 3O2(g) → 2Na2SO4(aq) + 2H2O(ℓ)
how many moles of Na2SO4 are formed when 1.22 mol of O2 react?
13. For the balanced chemical reaction
4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g)
determine the number of moles of both products formed when 6.88 mol of KO2 react.
1. For the balanced chemical reaction
2AlCl3 + 3H2O(ℓ) → Al2O3 + 6HCl(g)
determine the number of moles of both products formed when 0.0552 mol of AlCl3 react.
Answers
1. One mole of CH4 reacts with 2 mol of O2 to make 1 mol of CO2 and 2 mol of H2O.
2.
3. 6.022 × 1023 molecules of CH4, 1.2044 × 1024 molecules of O2, 6.022 × 1023 molecules of CO2, and 1.2044 × 1024 molecules of H2O
4.
5. 2 mol of C2H6 ⇔ 7 mol of O2 ⇔ 4 mol of CO2 ⇔ 6 mol of H2O
6.
7. C5H12 + 8O2 → 5CO2 + 6H2O; 4.4 mol
8.
9. 20.9 mol
10.
11. 1.00 mol
12.
13. 3.44 mol of K2CO3; 5.16 mol of O2
5.4: Mole-Mass and Mass-Mass Calculations
1. What mass of CO2 is produced by the combustion of 1.00 mol of CH4?CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
2. What mass of H2O is produced by the combustion of 1.00 mol of CH4?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
3. What mass of HgO is required to produce 0.692 mol of O2?
2HgO(s) → 2Hg(ℓ) + O2(g)
4. What mass of NaHCO3 is needed to produce 2.659 mol of CO2?
2NaHCO3(s) → Na2CO3(s) + H2O(ℓ) + CO2(g)
5. How many moles of Al can be produced from 10.87 g of Ag?
Al(NO3) 3(s) + 3Ag → Al + 3AgNO3
6. How many moles of HCl can be produced from 0.226 g of SOCl2?
SOCl2(ℓ) + H2O(ℓ) → SO2(g) + 2HCl(g)
7. How many moles of O2 are needed to prepare 1.00 g of Ca(NO3)2?
Ca(s) + N2(g) + 3O2(g) → Ca(NO3) 2(s)
8. How many moles of C2H5OH are needed to generate 106.7 g of H2O?
C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O(ℓ)
9. What mass of O2 can be generated by the decomposition of 100.0 g of NaClO3?
2NaClO3 → 2NaCl(s) + 3O2(g)
10. What mass of Li2O is needed to react with 1,060 g of CO2?
Li2O(aq) + CO2(g) → Li2CO3(aq)
11. What mass of Fe2O3 must be reacted to generate 324 g of Al2O3?
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
12. What mass of Fe is generated when 100.0 g of Al are reacted?
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
13. What mass of MnO2 is produced when 445 g of H2O are reacted?
H2O(ℓ) + 2MnO4(aq) + Br(aq) → BrO3(aq) + 2MnO2(s) + 2OH(aq)
14. What mass of PbSO4 is produced when 29.6 g of H2SO4 are reacted?
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(ℓ)
15. If 83.9 g of ZnO are formed, what mass of Mn2O3 is formed with it?
Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s)
16. If 14.7 g of NO2 are reacted, what mass of H2O is reacted with it?
3NO2(g) + H2O(ℓ) → 2HNO3(aq) + NO(g)
17. If 88.4 g of CH2S are reacted, what mass of HF is produced?
CH2S + 6F2 → CF4 + 2HF + SF6
1. If 100.0 g of Cl2 are needed, what mass of NaOCl must be reacted?
NaOCl + HCl → NaOH + Cl2
Answers
1. 44.0 g
2.
3. 3.00 × 102 g
4.
5. 0.0336 mol
6.
7. 0.0183 mol
8.
9. 45.1 g
10.
11. 507 g
12.
13. 4.30 × 103 g
14.
15. 163 g
16.
17. 76.7 g
5.5: Yields
1. What is the difference between the theoretical yield and the actual yield?
2. What is the difference between the actual yield and the percent yield?
3. A worker isolates 2.675 g of SiF4 after reacting 2.339 g of SiO2 with HF. What are the theoretical yield and the actual yield?
SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(ℓ)
4. A worker synthesizes aspirin, C9H8O4, according to this chemical equation. If 12.66 g of C7H6O3 are reacted and 12.03 g of aspirin are isolated, what are the theoretical yield and the actual yield?
C7H6O3 + C4H6O3 → C9H8O4 + HC2H3O2
5. A chemist decomposes 1.006 g of NaHCO3 and obtains 0.0334 g of Na2CO3. What are the theoretical yield and the actual yield?
2NaHCO3(s) → Na2CO3(s) + H2O(ℓ) + CO2(g)
6. A chemist combusts a 3.009 g sample of C5H12 and obtains 3.774 g of H2O. What are the theoretical yield and the actual yield?
C5H12(ℓ) + 8O2(g) → 5CO2 + 6H2O(ℓ)
7. What is the percent yield in Exercise 3?
8. What is the percent yield in Exercise 4?
9. What is the percent yield in Exercise 5?
1. What is the percent yield in Exercise 6?
Answers
1. Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.
2.
3. theoretical yield = 4.052 g; actual yield = 2.675 g
4.
5. theoretical yield = 0.635 g; actual yield = 0.0334 g
6.
7. 66.02%
8.
9. 5.26%
5.6: Limiting Reagents
1. The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH3. What is the limiting reagent?
1. The box below shows a group of hydrogen and oxygen molecules that will react to produce water, H2O. What is the limiting reagent?
1. Given the statement “20.0 g of methane is burned in excess oxygen,” is it obvious which reactant is the limiting reagent?
2. Given the statement “the metal is heated in the presence of excess hydrogen,” is it obvious which substance is the limiting reagent despite not specifying any quantity of reactant?
3. Acetylene (C2H2) is formed by reacting 7.08 g of C and 4.92 g of H2.
2C(s) + H2(g) → C2H2(g)
What is the limiting reagent? How much of the other reactant is in excess?
4. Ethane (C2H6) is formed by reacting 7.08 g of C and 4.92 g of H2.
2C(s) + 3H2(g) → C2H6(g)
What is the limiting reagent? How much of the other reactant is in excess?
1. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?
$\underset{35.6\, g}{P_{4}O_{6}(s)}+6\underset{4.77\, g}{H_{2}O(l)}\rightarrow 4H_{3}PO_{4}$
1. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?
$\underset{377\, g}{3NO_{2}(g)}+\underset{244\, g}{H_{2}O(l)}\rightarrow 2HNO_{3}(aq)+NO(g)$
1. To form the precipitate PbCl2, 2.88 g of NaCl and 7.21 g of Pb(NO3)2 are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?
1. In a neutralization reaction, 18.06 g of KOH are reacted with 13.43 g of HNO3. What mass of H2O is produced, and what mass of which reactant is in excess?
Answers
1. Nitrogen is the limiting reagent.
2.
3. Yes; methane is the limiting reagent.
4.
5. C is the limiting reagent; 4.33 g of H2 are left over.
6.
7. H2O is the limiting reagent; 25.9 g of P4O6 are left over.
8.
9. 6.06 g of PbCl2 are formed; 0.33 g of NaCl is left over.
5.7: Additional Exercises
1. How many molecules of O2 will react with 6.022 × 1023 molecules of H2 to make water? The reaction is 2H2(g) + O2(g) → 2H2O(ℓ).
2. How many molecules of H2 will react with 6.022 × 1023 molecules of N2 to make ammonia? The reaction is N2(g) + 3H2(g) → 2NH3(g).
3. How many moles are present in 6.411 kg of CO2? How many molecules is this?
4. How many moles are present in 2.998 mg of SCl4? How many molecules is this?
5. What is the mass in milligrams of 7.22 × 1020 molecules of CO2?
6. What is the mass in kilograms of 3.408 × 1025 molecules of SiS2?
7. What is the mass in grams of 1 molecule of H2O?
8. What is the mass in grams of 1 atom of Al?
9. What is the volume of 3.44 mol of Ga if the density of Ga is 6.08 g/mL?
10. What is the volume of 0.662 mol of He if the density of He is 0.1785 g/L?
11. For the chemical reaction
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(ℓ)
assume that 13.4 g of C4H10 reacts completely to products. The density of CO2 is 1.96 g/L. What volume in liters of CO2 is produced?
12. For the chemical reaction
2GaCl3(s) + 3H2(g) → 2Ga(ℓ) + 6HCl(g)
if 223 g of GaCl3 reacts completely to products and the density of Ga is 6.08 g/mL, what volume in milliliters of Ga is produced?
13. Calculate the mass of each product when 100.0 g of CuCl react according to the reaction
2CuCl(aq) → CuCl2(aq) + Cu(s)
What do you notice about the sum of the masses of the products? What concept is being illustrated here?
14. Calculate the mass of each product when 500.0 g of SnCl2 react according to the reaction
2SnCl2(aq) → SnCl4(aq) + Sn(s)
What do you notice about the sum of the masses of the products? What concept is being illustrated here?
15. What mass of CO2 is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C8H18. Assume that there are 2,801 g of gasoline per gallon.
16. What mass of H2O is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C8H18. Assume that there are 2,801 g of gasoline per gallon.
17. A chemical reaction has a theoretical yield of 19.98 g and a percent yield of 88.40%. What is the actual yield?
18. A chemical reaction has an actual yield of 19.98 g and a percent yield of 88.40%. What is the theoretical yield?
1. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?
$\underset{35.0\, g}{P_{4}}+\underset{12.7\, g}{3NaOH}+\underset{9.33\, g}{3H_{2}O}\rightarrow 2Na_{2}HPO_{4}+PH_{3}$
1. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?
$\underset{46.3\, g}{2NaCrO_{2}}+\underset{88.2\, g}{3NaBrO_{4}}+\underset{32.5\, g}{2NaOH}\rightarrow 3NaBrO_{3}+2Na_{2}CrO_{4}+H_{2}O$
1. Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.
$\underset{26.3\, g}{C_{3}H_{8}}+\underset{21.8\, g}{5O_{2}}\rightarrow 3CO_{2}(g)+4H_{2}O(l)$
1. Just in case you suspect Exercise 21 is rigged, do it for another chemical reaction and verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.
$\underset{35.0\, g}{2P_{4}}+\underset{12.7\, g}{6NaOH}+\underset{9.33\, g}{6H_{2}O}\rightarrow 3Na_{2}HPO_{4}+5PH_{3}$
Answers
1. 1.2044 × 1024 molecules
2.
3. 145.7 mol; 8.77 × 1025 molecules
4.
5. 52.8 mg
6.
7. 2.99 × 10−23 g
8.
9. 39.4 mL
10.
11. 20.7 L
12.
13. 67.91 g of CuCl2; 32.09 g of Cu. The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.
14.
15. 8,632 g
16.
17. 17.66 g
18.
19. The limiting reagent is NaOH; 21.9 g of P4 and 3.61 g of H2O are left over.
20.
21. Both products predict that O2 is the limiting reagent; 20.3 g of C3H8 are left over. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.E%3A_Stoichiometry_and_the_Mole_%28Exercises%29.txt |
Of the three basic phases of matter—solids, liquids, and gases—only one of them has predictable physical properties: gases. In fact, the study of the properties of gases was the beginning of the development of modern chemistry from its alchemical roots. The interesting thing about some of these properties is that they are independent of the identity of the gas. That is, it doesn't matter if the gas is helium gas, oxygen gas, or sulfur vapors; some of their behavior is predictable and, as we will find, very similar. In this chapter, we will review some of the common behaviors of gases.
Let us start by reviewing some properties of gases. Gases have no definite shape or volume; they tend to fill whatever container they are in. They can compress and expand, sometimes to a great extent. Gases have extremely low densities, one-thousandth or less the density of a liquid or solid. Combinations of gases tend to mix together spontaneously; that is, they form solutions. Air, for example, is a solution of mostly nitrogen and oxygen. Any understanding of the properties of gases must be able to explain these characteristics.
• 6.1: Prelude to Gases
• 6.2: Kinetic Theory of Gases
The physical behavior of gases is explained by the kinetic theory of gases. An ideal gas adheres exactly to the kinetic theory of gases.
• 6.3: Pressure
Pressure is a force exerted over an area. Pressure has several common units that can be converted.
• 6.4: Gas Laws
The behavior of gases can be modeled with gas laws. Boyle's law relates a gas's pressure and volume at constant temperature and amount. Charles's law relates a gas's volume and temperature at constant pressure and amount. In gas laws, temperatures must always be expressed in kelvins.
• 6.5: Other Gas Laws
There are gas laws that relate any two physical properties of a gas. The combined gas law relates pressure, volume, and temperature of a gas.
• 6.6: The Ideal Gas Law and Some Applications
The ideal gas law relates the four independent physical properties of a gas at any time. The ideal gas law can be used in stoichiometry problems with chemical reactions that involve gases. Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases. At STP, gases have a volume of 22.4 L per mole. The ideal gas law can be used to determine densities of gases.
• 6.7: Gas Mixtures
The pressure of a gas in a gas mixture is termed the partial pressure. Dalton's law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures. Collecting gases over water requires that we take the vapor pressure of water into account. Mole fraction is another way to express the amounts of components in a mixture.
06: Gases
Perhaps one of the most spectacular chemical reactions involving a gas occurred on May 6, 1937, when the German airship Hindenburg exploded on approach to the Naval Air Station in Lakehurst, New Jersey. The actual cause of the explosion is still unknown, but the entire volume of hydrogen gas used to float the airship, about 200,000 m3, burned in less than one minute. Thirty-six people, including one on the ground, were killed. Hydrogen is the lightest known gas. Any balloon filled with hydrogen gas will float in air if its mass is not too great. This makes hydrogen an obvious choice for flying machines based on balloons—airships, dirigibles, and blimps.
The German airship Hindenburg (left) was one of the largest airships ever built. However, it was filled with hydrogen gas and exploded in Lakehurst, New Jersey, at the end of a transatlantic voyage in May 1937 (right).
However, hydrogen also has one obvious drawback—it burns in air according to the well-known chemical equation:
$\ce{2H_2(g) + O_2(g) → 2H_2O(ℓ)}\nonumber$
So although hydrogen is an obvious choice, it is also a dangerous choice. Helium gas is also lighter than air and has 92% of the lifting power of hydrogen. Why, then, was helium not used in the Hindenburg? In the 1930s, helium was much more expensive. In addition, the best source of helium at the time was the United States, which banned helium exports to pre–World War II Germany. Today all airships use helium, a legacy of the Hindenburg disaster. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/06%3A_Gases/6.01%3A_Prelude_to_Gases.txt |
Learning Objectives
• State the major concepts behind the kinetic theory of gases.
• Relate the general properties of gases to the kinetic theory.
Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all-encompassing theory. Today, that theory is the kinetic theory of gases. The kinetic theory of gases is based on the following statements:
1. Gases consist of tiny particles of matter that are in constant motion.
2. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic—that is, there is no net loss of energy from the collisions.
3. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them.
4. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.
5. The average speed of gas particles is dependent on the temperature of the gas.
Figure \(1\) shows a representation of how we mentally picture the gas phase.
This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other.
An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements, and the kinetic theory of gases is well accepted by the scientific community.
• The physical behavior of gases is explained by the kinetic theory of gases.
• An ideal gas adheres exactly to the kinetic theory of gases.
6.03: Pressure
Learning Objectives
• Define pressure.
• Learn the units of pressure and how to convert between them.
The kinetic theory of gases indicates that gas particles are always in motion and are colliding with other particles and the walls of the container holding them. Although collisions with container walls are elastic (i.e., there is no net energy gain or loss because of the collision), a gas particle does exert a force on the wall during the collision. The accumulation of all these forces, distributed over the area of the walls of the container, causes pressure. Pressure ($P$) is defined as the force of all the gas particle/wall collisions divided by the area of the wall:
$\text{pressure}=\frac{\text{force}}{\text{area}}\nonumber$
All gases exert pressure; it is one of the fundamental measurable quantities of this phase of matter. Even our atmosphere exerts pressure—in this case, the gas is being "held in" by the earth's gravity, rather than the gas being in a container. The pressure of the atmosphere is about 14.7 pounds of force for every square inch of surface area: 14.7 lb/in2.
Pressure has a variety of units. The formal, SI-approved unit of pressure is the pascal (Pa), which is defined as 1 N/m2 (one newton of force over an area of one square meter). However, this is usually too small in magnitude to be useful. A common unit of pressure is the atmosphere (atm), which was originally defined as the average atmospheric pressure at sea level.
However, "average atmospheric pressure at sea level" is difficult to pinpoint because of atmospheric pressure variations. A more reliable and common unit is millimeters of mercury (mmHg), which is the amount of pressure exerted by a column of mercury exactly 1 mm high. An equivalent unit is the torr, which equals 1 mmHg. (The torr is named after Evangelista Torricelli, a seventeenth-century Italian scientist who invented the mercury barometer.) With these definitions of pressure, the atmosphere unit is redefined: 1 atm is defined as exactly 760 mmHg, or 760 torr. We thus have the following equivalencies:
1 atm=760 mmHg=760 torr
We can use these equivalencies as with any equivalencies—to perform conversions from one unit to another. Relating these to the formal SI unit of pressure, 1 atm = 101,325 Pa.
Example $1$: Pressure Conversion
How many atmospheres are there in 595 torr?
Solution
Using the pressure equivalencies, we construct a conversion factor between torr and atmospheres:
$\frac{1\, atm}{760\, torr}\nonumber$
$595\, \cancel{torr}\times \frac{1\, atm}{760\, \cancel{torr}}=0.783\, atm\nonumber$
Because the numbers in the conversion factor are exact, the number of significant figures in the final answer is determined by the initial value of pressure.
Exercise $1$
How many atmospheres are there in 1,022 torr?
Answer
1.345 atm
Example $2$: Mars
The atmosphere on Mars is largely CO2 at a pressure of 6.01 mmHg. What is this pressure in atmospheres?
Solution
Use the pressure equivalencies to construct the proper conversion factor between millimeters of mercury and atmospheres.
$6.01\, \cancel{mmHg}\times \frac{1\, atm}{760\, \cancel{mmHg}}=0.00791\, atm=7.91\times 10^{-3}atm\nonumber$
At the end, we expressed the answer in scientific notation.
Exercise $2$
Atmospheric pressure is low in the eye of a hurricane. In a 1979 hurricane in the Pacific Ocean, a pressure of 0.859 atm was reported inside the eye. What is this pressure in torr?
Answer
652 torr
Summary
• Pressure is a force exerted over an area.
• Pressure has several common units that can be converted. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/06%3A_Gases/6.02%3A_Kinetic_Theory_of_Gases.txt |
Learning Objectives
• Learn what is meant by the term gas laws.
• Learn and apply Boyle's law.
• Learn and apply Charles's law.
When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas was kept constant, pressure and volume were related: As one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related.
There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:
P × V = constant at constant n and T
If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled P1 and V1 and the new conditions are labeled P2 and V2, we have
P1V1 = constant = P2V2
where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply
P1V1 = P2V2 at constant n and T
This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's law, after the English scientist Robert Boyle, who first announced it in 1662. Figure $1$ shows two representations of how Boyle's law works.
Boyle's law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle's law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won't matter what the unit is, but the unit must be the same on both sides of the equation.
Example $1$
A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?
Solution
First, determine what quantities we are given. We are given an initial pressure and an initial volume, so let these values be P1 and V1:
P1 = 2.44 atm and V1 = 4.01 L
We are given another quantity, final pressure of 1.93 atm, but not a final volume. This final volume is the variable we will solve for.
P2 = 1.93 atm and V2 = ? L
Substituting these values into Boyle's law, we get
(2.44 atm)(4.01 L) = (1.93 atm)V2
To solve for the unknown variable, we isolate it by dividing both sides of the equation by 1.93 atm—both the number and the unit:
$\frac{(2.44\, atm)(4.01\, L)}{1.93\, atm}=\frac{(1.93\, atm)\, V_{2}}{1.93\, atm}\nonumber$
Note that, on the left side of the equation, the unit atm is in the numerator and the denominator of the fraction. They cancel algebraically, just as a number would. On the right side, the unit atm and the number 1.93 are in the numerator and the denominator, so the entire quantity cancels:
$\frac{(2.44\, \cancel{atm})(4.01\, L)}{1.93\, \cancel{atm}}=\frac{(1.93\, \cancel{atm})\, V_{2}}{1.93\, \cancel{atm}}\nonumber$
What we have left is
$\frac{(2.44)(4.01\, L)}{1.93}=V_{2}\nonumber$
Now we simply multiply and divide the numbers together and combine the answer with the $L$ unit, which is a unit of volume. Doing so, we get $V_2 = 5.07\, L$
Does this answer make sense? We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle's law.
Exercise $1$
If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2?
Answer
124 mL
As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.
Example $2$
A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?
Solution
We can still use Boyle's law to answer this, but now the two volume quantities have different units. It does not matter which unit we change, as long as we perform the conversion correctly. Let us change the 0.663 L to milliliters:
$0.663\, L\times \frac{1000\, ml}{1\, L}=663\, ml\nonumber$
Now that both volume quantities have the same units, we can substitute into Boyle's law:
$(722\, torr)(88.8\, ml)=P_{2}(663\, ml)\nonumber$
$\frac{(722\, torr)(88.8)\, ml}{(663\, ml)}=P_{2}\nonumber$
The mL units cancel, and we multiply and divide the numbers to get P2 = 96.7 torr
The volume is increasing, and the pressure is decreasing, which is as expected for Boyle's law.
Exercise $2$
If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2?
Answer
119 mL
There are other measurable characteristics of a gas. One of them is temperature (T). Perhaps one can vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume (V). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related.
A mathematical relationship between V and T should be possible except for one thought: what temperature scale should we use? We know from Chapter 2 that science uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin, not its temperature in degrees Celsius. If the temperature of a gas is expressed in kelvins, then experiments show that the ratio of volume to temperature is a constant:
$\frac{V}{T}=constant\nonumber$
We can modify this equation as we modified Boyle's law: the initial conditions V1 and T1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V2 and T2, as long as pressure and the amount of the gas remain constant. Thus, we have another gas law:
$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\; at\; constant\; P\; and\; n\nonumber$
This gas law is commonly referred to as Charles's law, after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle's law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale:
K = °C + 273
where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius.
Example $3$
A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas.
Solution
First, we assign the given values to their variables. The initial volume is V1, so V1 = 34.8 mL, and the initial temperature is T1, so T1 = 315 K. The temperature is increased to 559 K, so the final temperature T2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles's law yields
$\frac{34.8\, ml}{315\, K}=\frac{V_{2}}{559\, K}\nonumber$
We solve for V2 by algebraically isolating the V2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side:
$\frac{(559\cancel{K})(34.8\, ml)}{315\, \cancel{K}}=\frac{V_{2}(\cancel{559\, K})}{\cancel{559\, K}}\nonumber$
The expression simplifies to
$\frac{(559)(34.8\, ml)}{315}=V_{2}\nonumber$
By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is
V2 = 61.8 mL
Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does.
Exercise $3$
If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K?
Answer
1.82 L
It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles's law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles's law. That is, rather than write it as
$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\nonumber$
write the equation as
$\frac{T_{1}}{V_{1}}=\frac{T_{2}}{V_{2}}\nonumber$
It is still an equality and a correct form of Charles's law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler.
Example $4$
A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must the temperature of the gas be for its volume to be 25.0 L?
Solution
Here, we are looking for a final temperature, so we will use the reciprocal form of Charles's law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins:
−67°C + 273 = 206 K
In using the gas law, we must use T1 = 206 K as the temperature. Substituting into the reciprocal form of Charles's law, we get
$\frac{206\, K}{34.8\, L}=\frac{T_{2}}{25.0\, L}\nonumber$
Bringing the 25.0 L quantity over to the other side of the equation, we get
$\frac{(25.0\cancel{L})(206\, K)}{34.8\cancel{L}}=T_{2}\nonumber$
The L units cancel, so our final answer is T2 = 148 K
This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example.
Exercise $4$
If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2?
Answer
235 K, or −38°C
Summary
• The behavior of gases can be modeled with gas laws.
• Boyle's law relates a gas's pressure and volume at constant temperature and amount.
• Charles's law relates a gas's volume and temperature at constant pressure and amount.
• In gas laws, temperatures must always be expressed in kelvins. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/06%3A_Gases/6.04%3A_Gas_Laws.txt |
Learning Objectives
• Review other simple gas laws.
• Learn and apply the combined gas law.
You may notice in Boyle's law and Charles's law that we actually refer to four physical properties of a gas: pressure (P), volume (V), temperature (T), and amount (in moles—n). We do this because these are the only four independent physical properties of a gas. There are other physical properties, but they are all related to one (or more) of these four properties.
Boyle's law is written in terms of two of these properties, with the other two being held constant. Charles's law is written in terms of two different properties, with the other two being held constant. It may not be surprising to learn that there are other gas laws that relate other pairs of properties—as long as the other two are held constant. In this section, we will mention a few.
Gay-Lussac's law relates pressure with absolute temperature. In terms of two sets of data, Gay-Lussac's law is
$\dfrac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}} \nonumber$
at constant $V$ and $n$.
Note that it has a structure very similar to that of Charles's law, only with different variables—pressure instead of volume. Avogadro's law introduces the last variable for amount. The original statement of Avogadro's law states that equal volumes of different gases at the same temperature and pressure contain the same number of particles of gas. Because the number of particles is related to the number of moles (1 mol = 6.022 × 1023 particles), Avogadro's law essentially states that equal volumes of different gases, at the same temperature and pressure, contain the same amount (moles, particles) of gas. Put mathematically into a gas law, Avogadro's law is
$\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\nonumber$
at constant $V$ and $T$.
(First announced in 1811, it was Avogadro's proposal that volume is related to the number of particles that eventually led to naming the number of things in a mole as Avogadro's number.) Avogadro's law is useful because for the first time we are seeing amount, in terms of the number of moles, as a variable in a gas law.
Example $1$
A 2.45 L volume of gas contains 4.5 × 1021 gas particles. How many gas particles are there in 3.87 L if the gas is at constant pressure and temperature?
Solution
We can set up Avogadro's law as follows:
$\frac{2.45\, L}{4.5\times 10^{21}\, \text{particles}}=\frac{3.87\, L}{n_{2}}\nonumber$
We algebraically rearrange to solve for $n_2$:
$n_{2}=\frac{(3.87\, \cancel{L})(4.5\times 10^{21} \, \text{particles})}{2.45\, \cancel{L}}\nonumber$
The L units cancel, so we solve for $n_2$:
$n_2 = 7.1 \times 10^{21}\, \text{particles} \nonumber$
Exercise $1$
A 12.8 L volume of gas contains 3.00 × 1020 gas particles. At constant temperature and pressure, what volume does 8.22 × 1018 gas particles fill?
Answer
0.351 L
The variable n in Avogadro's law can also stand for the number of moles of gas, in addition to number of particles.
One thing we notice about all gas laws, collectively, is that volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the combined gas law, and its mathematical form is
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\; at\; constant\; n\nonumber$
This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in kelvins.
Example $2$
A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?
Solution
We can use the combined gas law directly; all the units are consistent with each other, and the temperatures are given in Kelvin. Substituting,
$\frac{(1.82\, atm)(8.33\, L)}{286\, K}=\frac{P_{2}(5.72\, L)}{355\, K}\nonumber$
We rearrange this to isolate the P2 variable all by itself. When we do so, certain units cancel:
$\frac{(1.82\, atm)(8.33\, \cancel{L})(355\, \cancel{K})}{(286\, \cancel{K})(5.72\, \cancel{L})}=P_{2}\nonumber$
Multiplying and dividing all the numbers, we get
$P_2 = 3.29\, atm \nonumber$
Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing.
Exercise $2$
If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2?
Answer
105 mL
As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms, or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.
Summary
• There are gas laws that relate any two physical properties of a gas.
• The combined gas law relates pressure, volume, and temperature of a gas. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/06%3A_Gases/6.05%3A_Other_Gas_Laws.txt |
Learning Objectives
• Learn the ideal gas law.
• Apply the ideal gas law to any set of conditions of a gas.
• Apply the ideal gas law to molar volumes, density, and stoichiometry problems.
So far, the gas laws we have considered have all required that the gas change its conditions; then we predict a resulting change in one of its properties. Are there any gas laws that relate the physical properties of a gas at any given time? Consider a further extension of the combined gas law to include n. By analogy to Avogadro's law, n is positioned in the denominator of the fraction, opposite the volume. So,
$\frac{PV}{nT}=constant\nonumber$
Because pressure, volume, temperature, and amount are the only four independent physical properties of a gas, the constant in the above equation is truly a constant. Indeed, because we do not need to specify the identity of a gas to apply the gas laws, this constant is the same for all gases. We define this constant with the symbol R, so the previous equation is written as
$\dfrac{PV}{nT}=R\nonumber$
which is usually rearranged as
$PV = nRT\nonumber$
This equation is called the ideal gas law. It relates the four independent properties of a gas at any time. The constant $R$ is called the ideal gas law constant. Its value depends on the units used to express pressure and volume.
Table $1$: Values of the Ideal Gas Law Constant lists the numerical values of $R$.
Numerical Value Units
0.08205 $\dfrac{L.atm}{mol.K}$
62.36 $\dfrac{L.torr}{mol.K}=\dfrac{L.mmHg}{mol.K}$
8.314 $\dfrac{J}{mol.K}$
The ideal gas law is used like any other gas law, with attention paid to the unit and expression of the temperature in kelvin. However, the ideal gas law does not require a change in the conditions of a gas sample. The ideal gas law implies that if you know any three of the physical properties of a gas, you can calculate the fourth property.
Example $1$
A 4.22 mol sample of $\ce{Ar}$ has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?
Solution
The first step is to convert temperature to kelvins:
$34 + 273 = 307\, \ce{K} \nonumber$
Now we can substitute the conditions into the ideal gas law:
$(1.21atm)(V)=(4.22\, mol)\left(0.08205\dfrac{L.atm}{mol.K}\right)(307\, K)\nonumber$
The atm unit is in the numerator of both sides, so it cancels. On the right side of the equation, the mol and K units appear in the numerator and the denominator, so they cancel as well. The only unit remaining is L, which is the unit of volume that we are looking for. We isolate the volume variable by dividing both sides of the equation by 1.21:
$V=\dfrac{(4.22)(0.08205)(307)}{1.21}L\nonumber$
Then solving for volume, we get V = 87.9 L
Exercise $1$
A 0.0997 mol sample of $\ce{O2}$ has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?
Answer
3.94 L
Example $2$
At a given temperature, 0.00332 g of Hg in the gas phase has a pressure of 0.00120 mmHg and a volume of 435 L. What is its temperature?
Solution
We are not given the number of moles of Hg directly, but we are given a mass. We can use the molar mass of Hg to convert to the number of moles.
$0.00332\cancel{g\, Hg}\times \frac{1\, mol\, Hg}{200.59\cancel{g\, \, Hg}}=0.0000165\, mol=1.65\times 10^{-5}mol\nonumber$
Pressure is given in units of millimeters of mercury. We can either convert this to atmospheres or use the value of the ideal gas constant that includes the mmHg unit. We will take the second option. Substituting into the ideal gas law,
$(0.00332\, mm\, Hg)(435\, L)=(1.65\times 10^{-5}mol)(62.36\frac{L.mmHg}{mol.K})T\nonumber$
The mmHg, L, and mol units cancel, leaving the K unit, the unit of temperature. Isolating T on one side, we get
$T=\frac{(0.00332)(435)}{(1.65\times 10^{-5})(62.36)}K\nonumber$
Then solving for K, we get T = 1,404 K.
Exercise $2$
For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume?
Answer
8.21 L
The ideal gas law can also be used in stoichiometry problems.
Example $3$
What volume of $\ce{H2}$ is produced at 299 K and 1.07 atm when 55.8 g of $\ce{Zn}$ metal react with excess $\ce{HCl}$?
$\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}\nonumber$
Solution
Here we have a stoichiometry problem where we need to find the number of moles of H2 produced. Then we can use the ideal gas law, with the given temperature and pressure, to determine the volume of gas produced. First, the number of moles of H2 is calculated:
$55.8\cancel{g\, Zn}\times \frac{1\cancel{mol\, Zn}}{65.41\cancel{g\, Zn}}\times \dfrac{1\, mol\, H_{2}}{1\cancel{mol\, Zn}}=0.853\, H_{2}\nonumber$
Now that we know the number of moles of gas, we can use the ideal gas law to determine the volume, given the other conditions:
$(1.07atm)V=(0.853\, mol)\left(0.08205\dfrac{L.atm}{mol.K}\right)(299\, K)\nonumber$
All the units cancel except for L, for volume, which means V = 19.6 L
Exercise $3$
What pressure of $\ce{HCl}$ is generated if 3.44 g of $\ce{Cl2}$ are reacted in 4.55 L at 455 K?
$\ce{H2(g) + Cl2(g) → 2HCl(g)}\nonumber$
Answer
0.796 atm
It should be obvious by now that some physical properties of gases depend strongly on the conditions. What we need is a set of standard conditions so that properties of gases can be properly compared to each other. Standard Temperature and Pressure (STP) is defined as exactly 100 kPa of pressure (0.986 atm) and 273 K (0°C). For simplicity, we will use 1 atm as standard pressure. Defining STP allows us to more directly compare the properties of gases that differ from one another.
One property shared among gases is a molar volume. The molar volume is the volume of 1 mol of a gas. At STP, the molar volume of a gas can be easily determined by using the ideal gas law:
$(1\, atm)V=(1\, mol) \left(0.08205\dfrac{L.atm}{mol.K}\right)(273\, K)\nonumber$
All the units cancel except for L, the unit of volume. So V = 22.4 L
Note that we have not specified the identity of the gas; we have specified only that the pressure is 1 atm and the temperature is 273 K. This makes for a very useful approximation: any gas at STP has a volume of 22.4 L per mole of gas; that is, the molar volume at STP is 22.4 L/mol (Figure $1$). This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at STP. If the conditions are not at STP, a molar volume of 22.4 L/mol is not applicable. However, if the conditions are at STP, the combined gas law can be used to calculate what the volume of the gas would be if at STP; then the 22.4 L/mol molar volume can be used.
Example $4$
How many moles of $\ce{Ar}$ are present in 38.7 L at STP?
Solution
We can use the molar volume, 22.4 L/mol, as a conversion factor, but we need to reverse the fraction so that the L units cancel and mol units are introduced. It is a one-step conversion:
$38.7\, \cancel{L}\times \frac{1\, mol}{22.4\cancel{L}}=1.73\, mol\nonumber$
Exercise $4$
What volume does 4.87 mol of $\ce{Kr}$ have at STP?
Answer
109 L
Example $5$
What volume of $\ce{H2}$ is produced at STP when 55.8 g of $\ce{Zn}$ metal react with excess $\ce{HCl}$?
$\ce{Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)} \nonumber$
Solution
This is a stoichiometry problem with a twist: we need to use the molar volume of a gas at STP to determine the final answer. The first part of the calculation is the same as in a previous example:
$55.8\cancel{g\, Zn}\times \frac{1\cancel{mol\, Zn}}{65.41\cancel{g\, Zn}}\times \frac{1\, mol\, H_{2}}{1\cancel{mol\, Zn}}=0.853\, H_{2}\nonumber$
Now we can use the molar volume, 22.4 L/mol, because the gas is at STP:
$0.853\cancel{mol\, H_{2}}\times \frac{22.4\, L}{1\cancel{mol\, H_{2}}}=19.1\, L\, H_{2}\nonumber$
Alternatively, we could have applied the molar volume as a third conversion factor in the original stoichiometry calculation.
Exercise $5$
What volume of $\ce{HCl}$ is generated if 3.44 g of $\ce{Cl2}$ are reacted at STP?
$\ce{H2(g) + Cl2(g) → 2HCl(g)} \nonumber$
Answer
2.17 L
The ideal gas law can also be used to determine the density of gases. Density, recall, is defined as the mass of a substance divided by its volume:
$d=\dfrac{m}{V} \label{density}$
Assume that you have exactly 1 mol of a gas. If you know the identity of the gas, you can determine the molar mass of the substance. Using the ideal gas law, you can also determine the volume of that mole of gas, using whatever the temperature and pressure conditions are. Then you can calculate the density of the gas by using
$\text{density}=\dfrac{\text{molar mass}}{\text{molar volume}} \nonumber$
Example $6$
What is the density of $\ce{N2}$ at 25°C and 0.955 atm?
Solution
First, we must convert the temperature into kelvin:
$25 + 273 = 298\, \ce{K} \nonumber$
If we assume exactly 1 mol of $\ce{N2}$, then we know its mass: 28.0 g. Using the ideal gas law, we can calculate the volume:
$(0.955\, atm)V=(1\, mol)\left(0.08205\frac{L.atm}{mol.K}\right)(298\, K)\nonumber$
All the units cancel except for $L$, the unit of volume. So $V = 25.6\,\ce{L}$
Knowing the molar mass and the molar volume, we can determine the density of $\ce{N2}$ under these conditions using Equation \ref{density}:
$d=\frac{28.0\, g}{25.6\, L}=1.09\, g/L\nonumber$
Exercise $6$
What is the density of $\ce{CO2}$ at a pressure of 0.0079 atm and 227 K? (These are the approximate atmospheric conditions on Mars.)
Answer
0.019 g/L
Chemistry is Everywhere: Breathing
Breathing (more properly called respiration) is the process by which we draw air into our lungs so that our bodies can take up oxygen from the air. Let us apply the gas laws to breathing.
Start by considering pressure. We draw air into our lungs because the diaphragm, a muscle underneath the lungs, moves down to reduce pressure in the lungs, causing external air to rush in to fill the lower-pressure volume. We expel air by the diaphragm pushing against the lungs, increasing pressure inside the lungs and forcing the high-pressure air out. What are the pressure changes involved? A quarter of an atmosphere? A tenth of an atmosphere? Actually, under normal conditions, it's only 1 or 2 torr of pressure difference that makes us breathe in and out.
A normal breath is about 0.50 L. If room temperature is about 22°C, then the air has a temperature of about 295 K. With normal pressure being 1.0 atm, how many moles of air do we take in for every breath? The ideal gas law gives us an answer:
$(1.0\, atm)(0.50\, L)=n \left(0.08205\dfrac{L.atm}{mol.K}\right)(295\, K)\nonumber$
Solving for the number of moles, we get
$n = 0.021\, \ce{mol\, air} \nonumber$
This ends up being about 0.6 g of air per breath—not much, but enough to keep us alive.
Summary
• The ideal gas law relates the four independent physical properties of a gas at any time.
• The ideal gas law can be used in stoichiometry problems with chemical reactions that involve gases.
• Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases.
• At STP, gases have a volume of 22.4 L per mole.
• The ideal gas law can be used to determine the density of gases. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/06%3A_Gases/6.06%3A_The_Ideal_Gas_Law_and_Some_Applications.txt |
Learning Objective
• Learn Dalton's law of partial pressures.
One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.
In gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture continue to do that as well.) However, each gas has its own pressure. The partial pressure of a gas, $P_i$, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture.
Dalton's law of partial pressures states that the total pressure of a gas mixture, $P_{tot}$, is equal to the sum of the partial pressures of the components, $P_i$:
\begin{align} P_{tot} &=P_{1}+P_{2}+P_{3}+... \nonumber \[4pt] &=\sum_{i} P_{i} \label{dalton} \end{align}
where $i$ counts over all gases in mixture.
Although this law may seem trivial, it reinforces the idea that gases behave independently of each other.
Example $1$
A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container?
Solution
Dalton's law of partial pressures (Equation \ref{dalton}) states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:
Ptot = 2.33 atm + 0.77 atm = 3.10 atm
Exercise $1$
N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2?
Answer
152 torr
Example $2$
A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?
Solution
Because gases act independently of each other, we can determine the resulting final pressures using Boyle's law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle's law to determine the final pressure of H2:
(2.50 atm)(2.00 L) = P2(7.00 L)
Solving for P2, we get P2 = 0.714 atm = partial pressure of H2.
Now we do that same thing for the O2:
(1.90 atm)(5.00 L) = P2(7.00 L)P2 = 1.36 atm = partial pressure of O2
The total pressure is the sum of the two resulting partial pressures:
Ptot = 0.714 atm + 1.36 atm = 2.07 atm
Exercise $2$
If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the containers are opened, what is the resulting total pressure?
Answer
0.51 atm
One of the reasons we have to deal with Dalton's law of partial pressures is because gases are frequently collected by bubbling through water. As we will see in Chapter 10, liquids are constantly evaporating into a vapor until the vapor achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a vapor pressure. Table $1$ lists the vapor pressures of H2O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure is used; the term vapor pressure is reserved for the partial pressure of a vapor when the liquid is the normal phase under a given set of conditions.
Table $1$: Vapor Pressure of Water versus Temperature
Temperature (°C) Vapor Pressure (torr) Temperature (°C) Vapor Pressure (torr)
5 6.54 30 31.84
10 9.21 35 42.20
15 12.79 40 55.36
20 17.54 50 92.59
21 18.66 60 149.5
22 19.84 70 233.8
23 21.08 80 355.3
24 22.39 90 525.9
25 23.77 100 760.0
Any time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus the vapor pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.
Example $3$
Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2 gas were generated?
Solution
We need to take into account that the total pressure includes the vapor pressure of water. According to Table $1$, the vapor pressure of water at 22°C is 19.84 torr. According to Dalton's law of partial pressures (Equation \ref{dalton}), the total pressure equals the sum of the pressures of the individual gases, so
$733\, torr=P_{H_{2}}+P_{H_{2}O}=P_{H_{2}}+19.84\, torr\nonumber$
We solve by subtracting:
$P_{H_{2}}=713\, torr\nonumber$
Now we can use the ideal gas law to determine the number of moles (remembering to convert temperature to kelvins, making it 295 K):
$(713\, torr)(2.00\, L)=n \left(62.36\dfrac{L.atm}{mol.K}\right)(295\, K)\nonumber$
All the units cancel except for mol, which is what we are looking for.
Therefore $n = 0.0775\, mol\, \ce{H2}$ collected
Exercise $1$
$\ce{CO2}$, generated by the decomposition of $\ce{CaCO3}$, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of $\ce{CO2}$ were generated?
Answer
0.129 mol
Finally, we introduce a new unit that can be useful, especially for gases: the mole fraction. The ratio of the number of moles of a component in a mixture divided by the total number of moles in the sample, $\chi_i$, is the ratio of the number of moles of component i in a mixture divided by the total number of moles in the sample:
$\chi _{i}=\frac{\text{moles of component i}}{\text{total number of moles}}\nonumber$
($\chi$ is the lowercase Greek letter chi.) Note that mole fraction is not a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of $\ce{Ne}$. Converting both to moles, we get
$4.00\cancel{g\, He}\times \frac{1\, mol\, He}{4.00\cancel{g\, He}}=1.00\, mol\, He \nonumber$
and
$5.0\cancel{g\, Ne}\times \frac{1\, mol\, Ne}{20.0\cancel{g\, Ne}}=0.25\, mol\, Ne\nonumber$
The total number of moles is the sum of the two mole amounts:
$\text{total moles} = 1.00\, mol + 0.025 \,mol = 1.25\, mol \nonumber$
The mole fractions are simply the ratio of each mole amount and the total number of moles, 1.25 mol:
$\chi _{He}=\frac{1.00\cancel{mol}}{1.25\cancel{mol}}=0.800\nonumber$
$\chi _{Ne}=\frac{0.25\cancel{mol}}{1.25\cancel{mol}}=0.200\nonumber$
The sum of the mole fractions equals exactly 1.
$\chi _{He} + \chi _{Ne} = 0.800 + 0.200 =1 \nonumber$
For gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:
$\chi _{i}=\frac{P_{i}}{P_{tot}}\nonumber$
This expression allows us to determine mole fractions without calculating the moles of each component directly.
Example $4$
A container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What are the mole fractions of each component?
Solution
According to Dalton's law, the total pressure is the sum of the partial pressures:
$P_{tot} = 0.80\, atm + 0.60\, atm = 1.40\, atm \nonumber$
The mole fractions are the ratios of the partial pressure of each component and the total pressure:
$\chi _{He}=\frac{0.80\, atm}{1.40\, atm}=0.57\nonumber$
$\chi _{Ne}=\frac{0.60\, atm}{1.40\, atm}=0.43\nonumber$
Again, the sum of the mole fractions is exactly 1.
Exercise $4$
What are the mole fractions when 0.65 atm of O2 and 1.30 atm of N2 are mixed in a container?
Food and Drink Application: Carbonated Beverages
Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have $\ce{CO2}$ gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.
All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of $\ce{CO2}$ gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly-sealed package (usually a bottle or a can) and sold. When the container is opened, the $\ce{CO2}$ pressure is released, resulting in the well-known hiss of an opening container, and $\ce{CO2}$ bubbles come out of solution. This must be done with care: if the $\ce{CO2}$ comes out too violently, a mess can occur!
The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates $\ce{CO2}$ as a digestion product. This process is called fermentation. The overall reaction is
$\ce{C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(aq)}\nonumber$
When this process occurs in a closed container, the $\ce{CO2}$ produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of $\ce{CO2}$ gas.
Summary
• The pressure of a gas in a gas mixture is termed the partial pressure.
• Dalton's law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures.
• Collecting gases over water requires that we take the vapor pressure of water into account.
• Mole fraction is another way to express the amount of each component in a mixture. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/06%3A_Gases/6.07%3A_Gas_Mixtures.txt |
Energy is a very important quantity in science and the world around us. Although most of our energy ultimately comes from the sun, much of the energy we use on a daily basis is rooted in chemical reactions. The gasoline in your car, the electricity in your house, the food in your diet-all provide substances for chemical reactions to provide energy (gasoline, food) or are produced from chemical reactions (electricity, about 50% of which is generated by burning coal). As such, it is only natural that the study of chemistry involves energy.
• 7.1: Introduction
• 7.2: Energy
Energy is the ability to do work and uses the unit joule. The law of conservation of energy states that the total energy of an isolated system does not increase or decrease.
• 7.3: Work and Heat
Work can be defined as a gas changing volume against a constant external pressure. Heat is the transfer of energy due to temperature differences. Heat can be calculated in terms of mass, temperature change, and specific heat.
• 7.4: Enthalpy and Chemical Reactions
Every chemical reaction occurs with a concurrent change in energy. The change in enthalpy, a kind of energy, equals heat at constant pressure. Enthalpy changes can be expressed by using thermochemical equations. Enthalpy changes are measured by using calorimetry.
• 7.5: Stoichiometry Calculations Using Enthalpy
The energy change of a chemical reaction can be used in stoichiometry calculations.
• 7.6: Hess's Law
Hess's law allows us to combine reactions algebraically and then combine their enthalpy changes the same way.
• 7.7: Formation Reactions
A formation reaction is the formation of one mole of a substance from its constituent elements. Enthalpies of formation are used to determine the enthalpy change of any given reaction.
• 7.E: Energy and Chemistry (Exercises)
These are exercises and select solutions to accompany Chapter 7 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
07: Energy and Chemistry
It takes energy to launch a spaceship into space. If it takes 1 energy unit to warm 0.25 g of water by 1°C, then it takes over 15,100 energy units to put that 0.25 g of water into earth orbit. The most powerful engines designed to lift rockets into space were part of the Saturn V rocket, that was built by the National Aeronautics and Space Administration (NASA). The rocket had three stages, with the first stage having the capability to launch about 3.5 million kg of mass. About 2.3 million kg was the actual fuel for the first stage; rockets in space have the unpleasant task of having to take their own chemicals with them to provide thrust.
Having to carry its own fuel puts a lot of mass burden on an engine in space. This is why NASA is developing other types of engines to minimize fuel mass. An ion thruster uses xenon atoms that have had at least one electron removed from their atoms. The resulting ions can be accelerated by electric fields, causing a thrust. Because xenon atoms are very large for atoms, the thrusting efficiency is high even though the actual thrust is low. Because of this, ion engines are useful only in space.
Energy is a very important quantity in science and the world around us. Although most of our energy ultimately comes from the sun, much of the energy we use on a daily basis is rooted in chemical reactions. The gasoline in your car, the electricity in your house, the food in your diet—all provide substances for chemical reactions to provide energy (gasoline, food) or are produced from chemical reactions (electricity, about 50% of which is generated by burning coal). As such, it is only natural that the study of chemistry involves energy. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.01%3A_Introduction.txt |
Learning Objectives
• Define energy.
• Know the units of energy.
• Understand the law of conservation of energy.
Energy is the ability to do work. Think about it: when you have a lot of energy, you can do a lot of work; but if you're low on energy, you do not want to do much work. Work (w) itself is defined as a force (F) operating over a distance (Δx):
w = F × Δx
In SI, force has units of newtons (N), while distance has units of meters. Therefore, work has units of N·m. This compound unit is redefined as a Joule (J).
1 joule = 1 newton·meter1 J = 1 N·m
Because energy is the ability to do work, energy is also measured in joules. This is the primary unit of energy we will use here.
How much is 1 J? It is enough to warm up about one-fourth of a gram of water by 1°C. It takes about 12,000 J to warm a cup of coffee from room temperature to 50°C. So a joule is not a lot of energy. It will not be uncommon to measure energies in thousands of joules, so the kilojoule (kJ) is a common unit of energy, with 1 kJ equal to 1,000 J.
An older—but still common—unit of energy is the calorie. The calorie (cal) was originally defined in terms of warming up a given quantity of water. The modern definition of calorie equates it to joules:
1 cal = 4.184 J
The calorie is used when considering nutrition. Energy contents of foods are often expressed in calories. However, the calorie unit used for foods is actually the kilocalorie (kcal). Most foods indicate this by spelling the word with a capital C—Calorie. Figure $1$ - Calories on Food Labels, shows one example.
Example $1$
The label in Figure $1$ states that the serving has 38 Cal. How many joules is this?
Solution
We recognize that with a capital C, the Calories unit is actually kilocalories. To determine the number of joules, we convert first from kilocalories to calories (using the definition of the kilo- prefix) and then from calories to joules (using the relationship between calories and joules). So
$38\cancel{kcal}\times \frac{1000\cancel{cal}}{1\cancel{kcal}}\times \frac{4.184\, J}{1\cancel{cal}}=160,000\, J\nonumber$
Exercise $1$
A serving of breakfast cereal usually has 110 Cal. How many joules of energy is this?
Answer
460,000 J
In the study of energy, we use the term system to describe the part of the universe under study: a beaker, a flask, or a container whose contents are being observed and measured. An isolated system is a system that does not allow a transfer of energy or matter into or out of the system. A good approximation of an isolated system is a closed, insulated thermos-type bottle. The fact that the thermos-type bottle is closed keeps matter from moving in or out, and the fact that it is insulated keeps energy from moving in or out.
One of the fundamental ideas about the total energy of an isolated system is that it does not increase or decrease. When this happens to a quantity, we say that the quantity is conserved. The law of conservation of energy states that the total energy of an isolated system does not change. As a scientific law, this concept occupies the highest level of understanding we have about the natural universe.
Key Takeaways
• Energy is the ability to do work and uses the unit joule.
• The law of conservation of energy states that the total energy of an isolated system does not increase or decrease. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.02%3A_Energy.txt |
Learning Objectives
• Define a type of work in terms of pressure and volume.
• Define heat.
• Relate the amount of heat to a temperature change.
We have already defined work as a force acting through a distance. There are other equivalent definitions of work that are also important in chemistry.
When a certain volume of a gas expands, it works against an external pressure to expand (Figure $1$ - Volume versus Pressure). That is, the gas must perform work. Assuming that the external pressure Pext is constant, the amount of work done by the gas is given by the equation
$w = −P_{ext} × ΔV \nonumber$
where $ΔV$ is the change in volume of the gas. This term is always the final volume minus the initial volume,
$ΔV = V_{final} − V_{initial} \nonumber$
and can be positive or negative, depending on whether Vfinal is larger (is expanding) or smaller (is contracting) than Vinitial. The negative sign in the equation for work is important and implies that as volume expands (ΔV is positive), the gas in the system is losing energy as work. On the other hand, if the gas is contracting, ΔV is negative, and the two negative signs make the work positive, so energy is being added to the system.
Finally, let us consider units. Volume changes are usually expressed in units like liters, while pressures are usually expressed in atmospheres. When we use the equation to determine work, the unit for work comes out as liter·atmospheres, or L·atm. This is not a very common unit for work. However, there is a conversion factor between L·atm and the common unit of work, joules:
1 L·atm = 101.32 J
Using this conversion factor and the previous equation for work, we can calculate the work performed when a gas expands or contracts.
Example $1$
What is the work performed by a gas if it expands from 3.44 L to 6.19 L against a constant external pressure of 1.26 atm? Express the final answer in joules.
Solution
First we need to determine the change in volume, ΔV. A change is always the final value minus the initial value:
ΔV = VfinalVinitial = 6.19 L − 3.44 L = 2.75 L
Now we can use the definition of work to determine the work done:
w = −Pext · ΔV = −(1.26 atm)(2.75 L) = −3.47 L·atm
Now we construct a conversion factor from the relationship between liter·atmospheres and joules:
$-3.47\cancel{L.atm}\times \times \frac{101.32\, J}{1\cancel{L.atm}}=-351\, J\nonumber$
We limit the final answer to three significant figures, as appropriate.
Exercise $1$
What is the work performed when a gas expands from 0.66 L to 1.33 L against an external pressure of 0.775 atm?
Answer
−53 J
Heat is another aspect of energy. Heat is the transfer of energy from one body to another due to a difference in temperature. For example, when we touch something with our hands, we interpret that object as either hot or cold depending on how energy is transferred: If energy is transferred into your hands, the object feels hot. If energy is transferred from your hands to the object, your hands feel cold. Because heat is a measure of energy transfer, heat is also measured in joules.
For a given object, the amount of heat (q) involved is proportional to two things: the mass of the object ($m$) and the temperature change ($ΔT$) evoked by the energy transfer. We can write this mathematically as
$q\alpha \, m\times \Delta T\nonumber$
where ∝ means "is proportional to." To make a proportionality an equality, we include a proportionality constant. In this case, the proportionality constant is labeled c and is called the specific heat capacity, or, more succinctly, specific heat:
$q = mcΔT \nonumber$
where the mass, specific heat, and change in temperature are multiplied together. Specific heat is a measure of how much energy is needed to change the temperature of a substance; the larger the specific heat, the more energy is needed to change the temperature. The units for specific heat are
$\frac{J}{g.C}\, or\, \frac{J}{g.K}\nonumber$
depending on what the unit of $ΔT$ is. You may note a departure from the insistence that temperature be expressed in Kelvin. That is because a change in temperature has the same value whether the temperatures are expressed in degrees Celsius or kelvins.
Example $2$
Calculate the heat involved when 25.0 g of Fe increase temperature from 22°C to 76°C. The specific heat of Fe is 0.449 J/g·°C.
Solution
First we need to determine ΔT. A change is always the final value minus the initial value:
ΔT = 76°C − 22°C = 54°C
Now we can use the expression for q, substitute for all variables, and solve for heat:
$q=(25.0\, g)(0.449\frac{J}{g.^{\circ}C})(54^{\circ}C)=610\, J\nonumber$
Note how the g and °C units cancel, leaving J, a unit of heat. Also note that this value of q is inherently positive, meaning that energy is going into the system.
Exercise $2$
Calculate the heat involved when 76.5 g of Ag increase temperature from 17.8°C to 144.5°C. The specific heat of Ag is 0.233 J/g·°C.
Answer
2,260 J
As with any equation, when you know all but one variable in the expression for q, you can determine the remaining variable by using algebra.
Example $3$
It takes 5,408 J of heat to raise the temperature of 373 g of Hg by 104°C. What is the specific heat of Hg?
Solution
We can start with the equation for q, but now different values are given, and we need to solve for specific heat. Note that ΔT is given directly as 104°C. Substituting,
5,408 J = (373 g)c(104°C)
We divide both sides of the equation by 373 g and 104°C:
$c=\frac{5408\, J}{(373g)(104^{\circ}C)}\nonumber$
Combining the numbers and bringing together all the units, we get
$c=0.139\frac{J}{g.^{\circ}C}\nonumber$
Exercise $3$
Gold has a specific heat of 0.129 J/g·°C. If 1,377 J are needed to increase the temperature of a sample of gold by 99.9°C, what is the mass of the gold?
Answer
107 g
Table $1$: Specific Heats of Various Substances, lists the specific heats of some substances. Specific heat is a physical property of substances, so it is a characteristic of the substance. The general idea is that the lower the specific heat, the less energy is required to change the temperature of the substance by a certain amount.
Table $1$: Specific Heats of Various Substances.
Substance Specific Heat (J/g·°C)
water 4.184
iron 0.449
gold 0.129
mercury 0.139
aluminum 0.900
ethyl alcohol 2.419
magnesium 1.03
helium 5.171
oxygen 0.918
Key Takeaways
• Work can be defined as a gas changing volume against a constant external pressure.
• Heat is the transfer of energy due to temperature differences.
• Heat can be calculated in terms of mass, temperature change, and specific heat. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.03%3A_Work_and_Heat.txt |
Learning Objectives
• Define enthalpy.
• Properly express the enthalpy change of chemical reactions.
• Explain how enthalpy changes are measured experimentally.
Now that we have shown how energy, work, and heat are related, we are ready to consider energy changes in chemical reactions. A fundamental concept is that every chemical reaction occurs with a concurrent change in energy. Now we need to learn how to properly express these energy changes.
Our study of gases in Chapter 6, and our definition of work in Section 7.3, indicate that conditions like pressure, volume, and temperature affect the energy content of a system. What we need is a definition of energy that holds when some of these conditions are specified (somewhat similar to our definition of standard temperature and pressure in our study of gases). We define the enthalpy change H) as the heat of a process when pressure is held constant:
$\Delta H\equiv q\; at\; constant\; pressure\nonumber$
The letter H stands for "enthalpy," a kind of energy, while the Δ implies a change in the quantity. We will always be interested in the change in H, rather than the absolute value of H itself.
When a chemical reaction occurs, there is a characteristic change in enthalpy. The enthalpy change for a reaction is typically written after a balanced chemical equation and on the same line. For example, when two moles of hydrogen react with one mole of oxygen to make two moles of water, the characteristic enthalpy change is 570 kJ. We write the equation as
$\ce{2H2(g) + O2(g) → 2H2O(ℓ)}\nonumber$
with $ΔH = −570\, kJ$
A chemical equation that includes an enthalpy change is called a thermochemical equation. A thermochemical equation is assumed to refer to the equation in molar quantities, which means it must be interpreted in terms of moles, not individual molecules.
Example $1$
Write the thermochemical equation for the reaction of PCl3(g) with Cl2(g) to make PCl5(g), which has an enthalpy change of −88 kJ.
Solution
The thermochemical equation is
PCl3(g) + Cl2(g) → PCl5(g) ΔH = −88 kJ
Exercise $1$
Write the thermochemical equation for the reaction of N2(g) with O2(g) to make 2NO(g), which has an enthalpy change of 181 kJ.
Answer
N2(g) + O2(g) → 2NO(g) ΔH = 181 kJ
You may have noticed that the ΔH for a chemical reaction may be positive or negative. The number is assumed to be positive if it has no sign; a + sign can be added explicitly to avoid confusion. A chemical reaction that has a positive ΔH is said to be endothermic, while a chemical reaction that has a negative ΔH is said to be exothermic.
What does it mean if the ΔH of a process is positive? It means that the system in which the chemical reaction is occurring is gaining energy. If one considers the energy of a system as being represented as a height on a vertical energy plot, the enthalpy change that accompanies the reaction can be diagrammed as in part (a) of Figure $1$ Reaction Energy: the energy of the reactants has some energy, and the system increases its energy as it goes to products. The products are higher on the vertical scale than the reactants. Endothermic, then, implies that the system gains, or absorbs, energy.
An opposite situation exists for an exothermic process, as shown in part (b) of Figure $1$ - Reaction Energy. If the enthalpy change of a reaction is negative, the system is losing energy, so the products have less energy than the reactants, and the products are lower on the vertical energy scale than the reactants are. Exothermic, then, implies that the system loses, or gives off, energy.
Example $2$
Consider this thermochemical equation.
$\ce{2CO(g) + O2(g) → 2CO2(g)}\nonumber$
with ΔH = −565 kJ
Is it exothermic or endothermic? How much energy is given off or absorbed?
Solution
By definition, a chemical reaction that has a negative ΔH is exothermic, meaning that this much energy—in this case, 565 kJ—is given off by the reaction.
Exercise $2$
Consider this thermochemical equation.
CO2(g) + H2(g) → CO(g) + H2O(g) ΔH = 42 kJ
Is it exothermic or endothermic? How much energy is given off or absorbed?
Answer
Endothermic; 42 kJ are absorbed.
How are ΔH values measured experimentally? Actually, ΔH is not measured; q is measured. But the measurements are performed under conditions of constant pressure, so ΔH is equal to the q measured.
Experimentally, q is measured by taking advantage of the equation
$q = mcΔT\nonumber$
We pre-measure the mass of the chemicals in a system. Then we let the chemical reaction occur and measure the change in temperature (ΔT) of the system. If we know the specific heat of the materials in the system (typically we do), we can calculate q. That value of q is numerically equal to the ΔH of the process, which we can scale up to a molar scale. The container in which the system resides is typically insulated, so any energy change goes into changing the temperature of the system, rather than being leaked from the system. The container is referred to as a calorimeter, and the process of measuring changes in enthalpy is called calorimetry.
For example, suppose 4.0 g of NaOH, or 0.10 mol of NaOH, are dissolved to make 100.0 mL of aqueous solution; while 3.65 g of HCl, or 0.10 mol of HCl, are dissolved to make another 100.0 mL of aqueous solution. The two solutions are mixed in an insulated calorimeter, a thermometer is inserted, and the calorimeter is covered (See Figure $2$ - Calorimeters for an example setup). The thermometer measures the temperature change as the following chemical reaction occurs:
NaOH (aq) + HCl(aq) → NaCl(aq) + H2O(ℓ)
An observer notes that the temperature increases from 22.4°C to 29.1°C. Assuming that the heat capacities and densities of the solutions are the same as those of pure water, we now have the information we need to determine the enthalpy change of the chemical reaction. The total amount of solution is 200.0 mL, and with a density of 1.00 g/mL, we thus have 200.0 g of solution. Using the equation for q, we substitute for our experimental measurements and the specific heat of water (in Table $1$ of Section 7.3).
$q=(200.0\cancel{g})(4.184\frac{J}{\cancel{g}.\cancel{^{\circ}C}})(6.7\cancel{^{\circ}C})\nonumber$
Solving for q, we get
$q=5600\, J\equiv \Delta H\: for\: the\: reaction\nonumber$
The heat q is equal to the ΔH for the reaction because the chemical reaction occurs at constant pressure. However, the reaction is giving off this amount of energy, so the actual sign on ΔH is negative:
ΔH = −5,600 J for the reaction
Thus, we have the following thermochemical equation for the chemical reaction that occurred in the calorimeter:
$\frac{1}{10}NaOH(aq)+\frac{1}{10}HCl(aq)\rightarrow \frac{1}{10}NaCl(aq)+\frac{1}{10}H_{2}O(l)\; \Delta H=\, -\, 5600\: J\nonumber$
The 1/10 coefficients are present to remind us that we started with one-tenth of a mole of each reactant, so we make one-tenth of a mole of each product. Typically, however, we report thermochemical equations in terms of moles, not one-tenth of a mole. To scale up to molar quantities, we must multiply the coefficients by 10. However, when we do this, we get 10 times as much energy. Thus, we have
NaOH (aq) + HCl(aq) → NaCl(aq) + H2O(ℓ) ΔH = −56,000 J
The ΔH can be converted into kJ units, so our final thermochemical equation is
NaOH (aq) + HCl(aq) → NaCl(aq) + H2O(ℓ) ΔH = −56 kJ
We have just taken our experimental data from calorimetry and determined the enthalpy change of a chemical reaction. Similar measurements on other chemical reactions can determine the ΔH values of any chemical reaction you want to study.
Example $3$
A 100 mL solution of 0.25 mol of Ca2+(aq) was mixed with 0.50 mol of F(aq) ions, and CaF2 was precipitated:
Ca2+(aq) + 2F(aq) → CaF2(s)
The temperature of the solution increased by 10.5°C. What was the enthalpy change for the chemical reaction? What was the enthalpy change for the production of 1 mol of CaF2? Assume that the solution has the same density and specific heat as water.
Solution
Because we are given ΔT directly, we can determine the heat of the reaction, which is equal to ΔH:
$q=(100\cancel{g})(4.184\frac{J}{\cancel{g}.\cancel{^{\circ}C}})(10.5\cancel{^{\circ}C})\nonumber$
Solving for q, we get
q = 4,400 J
Therefore, ΔH = −4,400 J.
According to the stoichiometry of the reaction, exactly 0.25 mol of CaF2 will form, so this quantity of heat is for 0.25 mol. For 1 mol of CaF2, we need to scale up the heat by a factor of four:
q = 4,400 J × 4 = 17,600 J for 1 mol CaF2
On a molar basis, the change in enthalpy is
ΔH = −17,600 J = −17.6 kJ
Exercise $3$
In a calorimeter at constant pressure, 0.10 mol of CH4(g) and 0.20 mol of O2(g) are reacted.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
The reaction warms 750.0 g of H2O by 28.4°C. What is ΔH for the reaction on a molar scale?
Answer
−891 kJ
Key Takeaways
• Every chemical reaction occurs with a concurrent change in energy.
• The change in enthalpy equals heat at constant pressure.
• Enthalpy changes can be expressed by using thermochemical equations.
• Enthalpy changes are measured by using calorimetry. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.04%3A_Enthalpy_and_Chemical_Reactions.txt |
Learning Objective
• Perform stoichiometry calculations using energy changes from thermochemical equations.
Previously, we related quantities of one substance to another in a chemical equation by performing calculations that used the balanced chemical equation; the balanced chemical equation provided equivalents that we used to construct conversion factors. For example, in the balanced chemical equation
$\ce{2H2(g) + O2(g) → 2H2O(ℓ)} \nonumber$
we recognized the equivalents
$\ce{2 mol\, H2 ⇔ 1 mol\, O2 ⇔ 2 mol\, H2O} \nonumber$
where ⇔ is the mathematical symbol for "is equivalent to." In our thermochemical equation, however, we have another quantity-energy change:
$\ce{2H2(g) + O2(g) → 2H2O(ℓ)} \quad ΔH = −570\, kJ \nonumber$
This new quantity allows us to add another equivalent to our list:
$\ce{2 mol\, H2 ⇔ 1 mol \,O2 ⇔ 2 mol\, H2O ⇔ −570\, kJ} \nonumber$
That is, we can now add an energy amount to the equivalents—the enthalpy change of a balanced chemical reaction. This equivalence can also be used to construct conversion factors so that we can relate enthalpy change to amounts of substances reacted or produced.
Note that these equivalents address a concern. When an amount of energy is listed for a balanced chemical reaction, what amount(s) of reactants or products does it refer to? The answer is that it relates to the number of moles of the substance, as indicated by its coefficient in the balanced chemical reaction. Thus, 2 mol of $\ce{H2}$ are related to −570 kJ, while 1 mol of $\ce{O2}$ is related to −570 kJ. This is why the unit on the energy change is kJ, not kJ/mol.
For example, consider the thermochemical equation
$\ce{H2(g) + Cl2(g) → 2HCl(g)} \quad ΔH = −184.6\, kJ \nonumber$
The equivalencies for this thermochemical equation are
$\ce{1 mol\, H2 ⇔ 1 mol\, Cl2 ⇔ 2 mol \,HCl ⇔ −184.6 kJ} \nonumber$
Suppose we are asked how much energy is given off when 8.22 mol of $\ce{H2}$ react. We would construct a conversion factor between the number of moles of $\ce{H2}$ and the energy given off, −184.6 kJ:
$8.22\cancel{mol\, H_{2}}\times \frac{-184.6\, kJ}{1\cancel{mol\, H_{2}}}=-1520\, kJ\nonumber$
The negative sign means that this much energy is given off.
Example $1$
Given the thermochemical equation
$\ce{N2(g) + 3H2(g) → 2NH3(g)} \quad ΔH = −91.8\, kJ \nonumber$
how much energy is given off when 222.4 g of N2 reacts?
Solution
The balanced thermochemical equation relates the energy change to moles, not grams, so we first convert the amount of N2 to moles and then use the thermochemical equation to determine the energy change:
$222.4\cancel{g\, N_{2}}\times \frac{1\cancel{mol\, N_{2}}}{28.00\cancel{g\, N_{2}}}\times \frac{-91.8\, kJ}{1\cancel{mol\, N_{2}}}=-729\, kJ\nonumber$
Exercise $1$
Given the thermochemical equation
$\ce{N2(g) + 3H2(g) → 2NH3(g)} \quad ΔH = −91.8\, kJ \nonumber$
how much heat is given off when 1.00 g of H2 reacts?
Answer
−15.1 kJ
Like any stoichiometric quantity, we can start with energy and determine an amount, rather than the other way around.
Example $1$
Given the thermochemical equation
$\ce{N2(g) + O2(g) → 2NO(g)} \quad ΔH = 180.6\, kJ \nonumber$
if 558 kJ of energy are supplied, what mass of $\ce{NO}$ can be made?
Solution
This time, we start with an amount of energy
$558\cancel{kJ}\times \frac{2\cancel{mol\, NO}}{180.6\cancel{kJ}}\times \frac{30.0\, g\, NO}{1\cancel{mol\, NO}}=185\, g\, NO\nonumber$
Exercise $1$
How many grams of N2 will react if 100.0 kJ of energy are supplied?
$\ce{N2(g) + O2(g) → 2NO(g)} \quad ΔH = 180.6\, kJ \nonumber$
Answer
15.5 g
Chemistry is Everywhere: Welding with Chemical Reactions
One very energetic reaction is called the thermite reaction. Its classic reactants are aluminum metal and iron(III) oxide; the reaction produces iron metal and aluminum oxide:
$\ce{2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)} \quad ΔH = −850.2\, kJ \nonumber$
When properly done, the reaction gives off so much energy that the iron product comes off as a liquid. (Iron normally melts at 1,536°C.) If carefully directed, the liquid iron can fill spaces between two or more metal parts and, after it quickly cools, can weld the metal parts together.
Thermite reactions are used for this purpose even today. For civilian purposes, they are used to reweld broken locomotive axles that cannot be easily removed for repair. They are also used to weld railroad tracks together. Thermite reactions can also be used to separate thin pieces of metal if, for whatever reason, a torch doesn't work.
Key Takeaway
• The energy change of a chemical reaction can be used in stoichiometry calculations. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.05%3A_Stoichiometry_Calculations_Using_Enthalpy.txt |
Learning Objective
• Learn how to combine chemical equations and their enthalpy changes.
Now that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:
$\ce{2C(s) + O2(g) → 2CO(g)} \quad ΔH = ? \nonumber$
In reality, this is extremely difficult to do. Given the opportunity, carbon will react to make another compound, carbon dioxide:
$\ce{2C(s) + O2(g) → 2CO2(g)} \quad ΔH = −393.5 kJ\nonumber$
Is there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be canceled out (much like a spectator ion in ionic equations). For example, consider these two reactions:
\begin{align*} \ce{2C(s) + 2O2(g) &→ 2CO2(g)} \[4pt] \ce{2CO2(g) &→ 2CO(g) + O2(g)}\end{align*} \nonumber
If we added these two equations by combining all the reactants together and all the products together, we would get
$\ce{2C(s) + 2O2(g) + 2CO2(g) → 2CO2(g) + 2CO(g) + O2(g)}\nonumber$
We note that $\ce{2CO2(g)}$ appears on both sides of the arrow, so they cancel:
$\ce{2C(s) + 2O_{2}(g) + \cancel{2CO_{2}(g)}\rightarrow \cancel{2CO_{2}(g)}+2CO(g)+O_{2}(g)}\nonumber$
We also note that there are 2 mol of O2 on the reactant side, and 1 mol of O2 on the product side. We can cancel 1 mol of O2 from both sides:
$\ce{2C(s) + 2O_{2}(g)\rightarrow 2CO(g)+O_{2}(g)}\nonumber$
What do we have left?
$\ce{2C(s) + O2(g) → 2CO(g)}\nonumber$
This is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.
What about the enthalpy changes? Hess's law states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:
1. If a chemical reaction is reversed, the sign on $ΔH$ is changed.
2. If a multiple of a chemical reaction is taken, the same multiple of the $ΔH$ is taken as well.
What are the equations being combined? The first chemical equation is the combustion of C, which produces CO2:
$\ce{2C(s) + 2O2(g) → 2CO2(g)}\nonumber$
This reaction is two times the reaction to make $\ce{CO2}$ from $\ce{C(s)}$ and $\ce{O2(g)}$, whose enthalpy change is known:
$\ce{C(s) + O2(g) → CO2(g)} \quad ΔH = −393.5 kJ\nonumber$
According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:
$\ce{2C(s) + 2O2(g) → 2CO2(g)} \quad ΔH = −787.0 kJ\nonumber$
The second reaction in the combination is related to the combustion of CO(g):
$\ce{2CO(g) + O2(g) → 2CO2(g)} \quad ΔH = −566.0 kJ\nonumber$
The second reaction in our combination is the reverse of the combustion of CO. When we reverse the reaction, we change the sign on the ΔH:
$\ce{2CO2(g) → 2CO(g) + O2(g)} \quad ΔH = +566.0 kJ\nonumber$
Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the $ΔH$ values and add them:
$\begin{cases} 2C(s)+2O_{2}(g)\rightarrow 2CO_{2}(g)\; \; \; \; \; \; \Delta H=-787.0kJ\ \cancel{2CO_{2}(g)}\rightarrow 2CO(g))+\cancel{O_{2}(g)}\; \; \, \, \Delta H=+566.0kJ\ -------------------------\ 2C(s)+O_{2}(g)\rightarrow 2CO(g)\; \; \; \; \; \; \; \; \; \; \, \Delta H=-787.0+566.0=-221.0kJ\ \end{cases}\nonumber$
Hess's law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.
Example $1$
Determine the enthalpy change of
$\ce{C2H4 + 3O2 → 2CO2 + 2H2O} \quad ΔH = ?\nonumber$
from these reactions:
$\ce{C2H2 + H2 → C2H4} \quad ΔH = −174.5 kJ\nonumber$
$\ce{2C2H2 + 5O2 → 4CO2 + 2H2O} \quad ΔH = −1,692.2 kJ\nonumber$
$\ce{2CO2 + H2 → 2O2 + C2H2} \quad ΔH = −167.5 kJ\nonumber$
Solution
We will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C2H4 as a reactant, and only one reaction from our data has C2H4. However, it has C2H4 as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the ΔH:
$\ce{C2H4 → C2H2 + H2} \quad ΔH = +174.5 kJ\nonumber$
We need CO2 and H2O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all of our reactions are added):
$\ce{2C2H2 + 5O2 → 4CO2 + 2H2O} \quad ΔH = −1,692.2 kJ\nonumber$
We note that we now have 4 mol of CO2 as products; we need to get rid of 2 mol of CO2. The last reaction has 2CO2 as a reactant. Let us use it as written:
$\ce{2CO2 + H2 → 2O2 + C2H2} \quad ΔH = −167.5 kJ\nonumber$
We combine these three reactions, modified as stated:
What cancels? 2C2H2, H2, 2O2, and 2CO2. What is left is
$\ce{C2H4 + 3O2 → 2CO2 + 2H2O}\nonumber$
which is the reaction we are looking for. The $ΔH$ of this reaction is the sum of the three $ΔH$ values:
ΔH = +174.5 − 1,692.2 − 167.5 = −1,685.2 kJ
Exercise $1$
Given the thermochemical equations
$\ce{Pb + Cl2 → PbCl2} \quad ΔH = −223 kJ\nonumber$
$\ce{PbCl2 + Cl2 → PbCl4} \quad ΔH = −87 kJ\nonumber$
determine $ΔH$ for
$\ce{2PbCl2 → Pb + PbCl4}\nonumber$
Answer
+136 kJ
Key Takeaway
• Hess's law allows us to combine reactions algebraically and then combine their enthalpy changes the same way. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.06%3A_Hess%27s_Law.txt |
Learning Objectives
• Define a formation reaction and be able to recognize one.
• Use enthalpies of formation to determine the enthalpy of reaction.
Hess's law allows us to construct new chemical reactions and predict what their enthalpies of reaction will be. This is a very useful tool because it is not necessary to measure the enthalpy changes of every possible reaction. We need measure only the enthalpy changes of certain benchmark reactions, and then use these reactions to algebraically construct any possible reaction and combine the enthalpies of the benchmark reactions accordingly.
But what are the benchmark reactions? We need to have some agreed-on sets of reactions that provide the central data for any thermochemical equation.
Formation reactions are chemical reactions that form one mole of a substance from its constituent elements in their standard states. The term standard states means as a diatomic molecule, if that is how the element exists, and in the proper phase at normal temperatures (typically room temperature). The product is one mole of substance, which may require that coefficients on the reactant side be fractional (a change from our normal insistence that all coefficients be whole numbers). For example, the formation reaction for methane ($\ce{CH4}$) is
$\ce{C(s) + 2H2(g) → CH4(g)}\nonumber$
The formation reaction for carbon dioxide ($\ce{CO2}$) is
$\ce{C(s) + O2(g) → CO2(g)}\nonumber$
In both cases, one of the elements is a diatomic molecule because that is the standard state for that particular element. The formation reaction for $\ce{H2O}$:
$\ce{2H2(g) + O2(g) → 2H2O(ℓ)}\nonumber$
—is not in a standard state because the coefficient on the product is 2; for a proper formation reaction, only one mole of product is formed. Thus, we have to divide all coefficients by 2:
$\ce{H2(g) + 1/2O2(g) → H2O(ℓ)}\nonumber$
On a molecular scale, we are using half of an oxygen molecule, which may be problematic to visualize. However, on a molar level, it implies that we are reacting only half of a mole of oxygen molecules, which should be an easy concept for us to understand.
Example $1$
Which of the following are proper formation reactions?
1. $\ce{H2(g) + Cl2(g) → 2HCl(g)}$
2. $\ce{Si(s) + 2F2(g) → SiF4(g)}$
3. $\ce{CaO(s) + CO2 → CaCO3(s)}$
Solution
1. In this reaction, two moles of product are produced, so this is not a proper formation reaction.
2. In this reaction, one mole of a substance is produced from its elements in their standard states, so this is a proper formation reaction.
3. One mole of a substance is produced, but it is produced from two other compounds, not its elements. So this is not a proper formation reaction.
Exercise $1$
Is this a proper formation reaction? Explain why or why not.
$\ce{2Fe(s) + 3P(s) + 12O(g) → Fe2(PO4)3(s)} \nonumber$
Answer
This is not a proper formation reaction because oxygen is not written as a diatomic molecule.
Given the formula of any substance, you should be able to write the proper formation reaction for that substance.
Example $2$
Write formation reactions for each of the following.
1. $\ce{FeO(s)}$
2. $\ce{C2H6(g)}$
Solution
In both cases, there is one mole of the substance as product, and the coefficients of the reactants may have to be fractional to balance the reaction.
1. $\ce{Fe(s) + 1/2O2(g) → FeO(s)}$
2. $\ce{2C(s) + 3H2(g) → C2H6(g)}$
Exercise $2$
Write the equation for the formation of $\ce{CaCO3(s)}$.
Answer
$\ce{Ca(s) + C(s) + 3/2O2(g) → CaCO3(s)}\nonumber$
The enthalpy change for a formation reaction is called the enthalpy of formation and is given the symbol ΔHf. The subscript f is the clue that the reaction of interest is a formation reaction. Thus, for the formation of FeO(s),
$Fe(s)+\frac{1}{2}O_{2}(g)\rightarrow FeO(s)\; \; \; \Delta H=\Delta H_{f}=-272kJ/mol\nonumber$
Note that now we are using kJ/mol as the unit because it is understood that the enthalpy change is for one mole of substance. Note, too, by definition, that the enthalpy of formation of an element is exactly zero because making an element from an element is no change. For example,
H2(g) → H2(g) ΔHf = 0
Formation reactions and their enthalpies are important because these are the thermochemical data that are tabulated for any chemical reaction. Table $1$ - Enthalpies of Formation for Various Substances, lists some enthalpies of formation for a variety of substances; in some cases, however, phases can be important (e.g., for H2O).
It is easy to show that any general chemical equation can be written in terms of the formation reactions of its reactants and products, some of them reversed (which means the sign must change in accordance with Hess's law). For example, consider
2NO2(g) → N2O4(g)
We can write it in terms of the (reverse) formation reaction of NO2 and the formation reaction of N2O4:
$2\times \left [ NO_{2}(g) \rightarrow \frac{1}{2}N_{2}(g)+O_{2}(g)\right ]\; \; \; \; \Delta H=-2\times \Delta H_{r}\left [ NO_{2} \right ]=-2(33.1kJ)\ N_{2}(g)+2O_{2}(g)\rightarrow N_{2}O_{4}(g)\: \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \Delta H=\Delta H_{r}\left [ N_{2}O_{4} \right ]\; \; \; \; \; \; \, =9.1kJ\ ---------------------------------\ 2NO_{2}(g)\rightarrow N_{2}O_{4}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \Delta H=-57.1kJ\nonumber$
We must multiply the first reaction by 2 to get the correct overall balanced equation. We are simply using Hess's law in combining the ΔHf values of the formation reactions.
Table $1$ Enthalpies of Formation for Various Substances
Compound ΔHf (kJ/mol) Compound ΔHf (kJ/mol) Compound ΔHf (kJ/mol) Compound ΔHf (kJ/mol)
Ag(s) 0 Ca(s) 0 Hg2Cl2(s) −265.37 NaHCO3(s) −950.81
AgBr(s) −100.37 CaCl2(s) −795.80 I2(s) 0 NaN3(s) 21.71
AgCl(s) −127.01 CaCO3(s, arag) −1,207.1 K(s) 0 Na2CO3(s) −1,130.77
Al(s) 0 CaCO3(s, calc) −1,206.9 KBr(s) −393.8 Na2O(s) −417.98
Al2O3(s) −1,675.7 Cl2(g) 0 KCl(s) −436.5 Na2SO4(s) −331.64
Ar(g) 0 Cr(s) 0 KF(s) −567.3 Ne(g) 0
Au(s) 0 Cr2O3(s) −1,134.70 KI(s) −327.9 Ni(s) 0
BaSO4(s) −1,473.19 Cs(s) 0 Li(s) 0 O2(g) 0
Br2(ℓ) 0 Cu(s) 0 LiBr(s) −351.2 O3(g) 142.67
C(s, dia) 1.897 F2(g) 0 LiCl(s) −408.27 PH3(g) 22.89
C(s, gra) 0 Fe(s) 0 LiF(s) −616.0 Pb(s) 0
CCl4(ℓ) −128.4 Fe2(SO4)3(s) −2,583.00 LiI(s) −270.4 PbCl2(s) −359.41
CH2O(g) −115.90 Fe2O3(s) −825.5 Mg(s) 0 PbO2(s) −274.47
CH3COOH(ℓ) −483.52 Ga(s) 0 MgO(s) −601.60 PbSO4(s) −919.97
CH3OH(ℓ) −238.4 HBr(g) −36.29 NH3(g) −45.94 Pt(s) 0
CH4(g) −74.87 HCl(g) −92.31 NO(g) 90.29 S(s) 0
CO(g) −110.5 HF(g) −273.30 NO2(g) 33.10 SO2(g) −296.81
CO2(g) −393.51 HI(g) 26.5 N2(g) 0 SO3(g) −395.77
C2H5OH(ℓ) −277.0 HNO2(g) −76.73 N2O(g) 82.05 SO3(ℓ) −438
C2H6(g) −83.8 HNO3(g) −134.31 N2O4(g) 9.08 Si(s) 0
C6H12(ℓ) −157.7 H2(g) 0 N2O5(g) 11.30 U(s) 0
C6H12O6(s) −1277 H2O(g) −241.8 Na(s) 0 UF6(s) −2,197.0
C6H14(ℓ) −198.7 H2O(ℓ) −285.83 NaBr(s) −361.1 UO2(s) −1,085.0
C6H5CH3(ℓ) 12.0 H2O(s) −292.72 NaCl(s) −385.9 Xe(g) 0
C6H6(ℓ) 48.95 He(g) 0 NaF(s) −576.6 Zn(s) 0
C10H8(s) 77.0 Hg(ℓ) 0 NaI(s) −287.8 ZnCl2(s) −415.05
C12H22O11(s) −2,221.2
Sources: National Institute of Standards and Technology's Chemistry WebBook(opens in new window); D. R. Lide, ed., CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008); J. A. Dean, ed., Lange's Handbook of Chemistry, 14th ed. (New York: McGraw-Hill, 1992).
Example $3$
Show that the reaction
$\ce{Fe2O3(s) + 3SO3(g) → Fe2(SO4)3(s)} \nonumber \nonumber$
can be written as a combination of formation reactions.
Solution
There will be three formation reactions. The one for the products will be written as a formation reaction, while the ones for the reactants will be written in reverse. Furthermore, the formation reaction for SO3 will be multiplied by 3 because there are three moles of SO3 in the balanced chemical equation. The formation reactions are as follows:
$Fe_{2}O_{3}(s)\rightarrow 2Fe(s)+\frac{3}{2}O_{2}(g)\ 3\times \left [ SO_{3}(g)\rightarrow S(s)+\frac{3}{2}O_{2}(g) \right ]\nonumber$
$\ce{2Fe(s) + 3S(s) + 6O2(g) → Fe2(SO4)3(s)}\nonumber$
When these three equations are combined and simplified, the overall reaction is
$\ce{Fe2O3(s) + 3SO3(s) → Fe2(SO4)3(s)}\nonumber \nonumber$
Exercise $3$
Write the formation reactions that will yield
$\ce{2SO2(g) + O2(g) → 2SO3(g).} \nonumber \nonumber$
Answer
$2\times \left [ SO_{2}(g)\rightarrow S(s)+O_{2}(g) \right ]\ 2\times \left [ S(s)+\frac{3}{2} O_{2}(g)\rightarrow 2SO_{3}(g)\right ] \nonumber \nonumber$
Now that formation reactions have been established as the major type of thermochemical reaction being examined in this chapter, is it necessary to write all of the formation reactions when the aim is to determine the enthalpy change of any random chemical reaction? No. There is an easier way. You may have noticed in all of our examples that signs are changed on the enthalpies of formation of the reactants, and signs are not changed on the enthalpies of formation of the products. We also multiply the enthalpies of formation of any substance by its coefficient—technically, even when it is just 1. This allows us to make the following statement: the enthalpy change of any chemical reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In mathematical terms,
$\Delta H_{rxn}=\sum n_{p}\Delta H_{f,p}-\sum n_{r}\Delta H_{f,r}\nonumber$
where np and nr are the number of moles of products and reactants, respectively (even if they are just 1 mol), and ΔHf,p and ΔHf,r are the enthalpies of formation of the product and reactant species, respectively. This products-minus-reactants scheme is very useful in determining the enthalpy change of any chemical reaction, if the enthalpy of formation data are available. Because the mol units cancel when multiplying the amount by the enthalpy of formation, the enthalpy change of the chemical reaction has units of energy (joules or kilojoules) only.
Example $4$
Use the products-minus-reactants approach to determine the enthalpy of reaction for
$2HBr(g)+Cl_{2}(g)\rightarrow 2HCl(g)+Br_{2}(l)\ \Delta H_{f}\; -36.3\; \; \; \; \; \; \; \; 0\; \; \; \; \; \; \; \; \; -92.3\; \; \; \; \; \; \; \; \; \; \; \; 0\; \; \; \; \; kJ/mol\nonumber$
Solution
The enthalpies of formation are multiplied by the number of moles of each substance in the chemical equation, and the total enthalpy of formation for reactants is subtracted from the total enthalpy of formation of the products:
$\Delta H_{rxn}=\left [ \left ( 2\cancel{mol} \right )\left ( -92.3kJ/\cancel{mol} \right )+\left ( 1\cancel{mol} \right ) \left ( 0kJ/\cancel{mol} \right )\right ]-\left [ \left ( 2\cancel{mol} \right )\left ( -36.3kJ/\cancel{mol} \right )+\left ( 1\cancel{mol} \right ) \left ( 0kJ/\cancel{mol} \right )\right ]\nonumber$
All the mol units cancel. Multiplying and combining all the values, we get
ΔHrxn = −112.0 kJ
Exercise $4$
What is the enthalpy of reaction for this chemical equation?
$CO(g)\; \; \; +\; \; H_{2}O(l)\rightarrow \;CO_{2}(g)+H_{2}(g)\ \Delta H_{f}\; -110.5\; -285.8\; \; \; \; \; \; -393.5\; \; \; \; \; \; \; 0\; \; \; \; \; kJ/mol \nonumber \nonumber$
Answer
+2.8 kJ
Food and Drink Application: Calories and Nutrition
Section 7.2 mentioned the connection between the calorie unit and nutrition: the calorie is the common unit of energy used in nutrition, but we actually consider the kilocalorie (spelled Calorie with a capital C). A daily diet of 2,000 Cal is actually 2,000,000 cal, or over 8,000,000 J, of energy.
Nutritionists typically generalize the Calorie content of a food by separating it into the three main components: proteins, carbohydrates, and fats. The general rule of thumb is as follows:
Table with two columns and three rows. The first (left) column is labeled "If the food is..." and the second (right) column two is labeled "It has this energy content...". Underneath the first column, in the rows are different food components. Underneath the second column, in the rows are the calories for the different corresponding food in the left rows.
If the food is... It has this energy content...
protein 4 Cal/g
carbohydrate 4 Cal/g
fat 9 Cal/g
This table is very useful. Assuming a 2,000 Cal daily diet, if our diet consists solely of proteins and carbohydrates, we need only about 500 g of food for sustenance—a little more than a pound. If our diet consists solely of fats, we need only about 220 g of food—less than a half pound. Of course, most of us have a mixture of proteins, carbohydrates, and fats in our diets. Water has no caloric value in the diet, so any water in the diet is calorically useless. (However, it is important for hydration; also, many forms of water in our diet are highly flavored and sweetened, which bring other nutritional issues to bear.)
When your body works, it uses calories provided by the diet as its energy source. If we eat more calories than our body uses, we gain weight—about 1 lb of weight for every additional 3,500 Cal we ingest. Similarly, if we want to lose weight, we need to expend an extra 3,500 Cal than we ingest to lose 1 lb of weight. No fancy or fad diets are needed; maintaining an ideal body weight is a straightforward matter of thermochemistry—pure and simple.
Key Takeaways
• A formation reaction is the formation of one mole of a substance from its constituent elements.
• Enthalpies of formation are used to determine the enthalpy change of any given reaction. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.07%3A_Formation_Reactions.txt |
Additional Exercises
1. What is the work when 124 mL of gas contract to 72.0 mL under an external pressure of 822 torr?
2. What is the work when 2,345 mL of gas contract to 887 mL under an external pressure of 348 torr?
• A 3.77 L volume of gas is exposed to an external pressure of 1.67 atm. As the gas contracts, 156 J of work are added to the gas. What is the final volume of the gas?
• A 457 mL volume of gas contracts when 773 torr of external pressure act on it. If 27.4 J of work are added to the gas, what is its final volume?
• What is the heat when 1,744 g of Hg increase in temperature by 334°C? Express your final answer in kJ.
• What is the heat when 13.66 kg of Fe cool by 622°C? Express your final answer in kJ.
• What is final temperature when a 45.6 g sample of Al at 87.3°C gains 188 J of heat?
• What is final temperature when 967 g of Au at 557°C lose 559 J of heat?
• Plants take CO2 and H2O and make glucose (C6H12O6) and O2. Write a balanced thermochemical equation for this process. Use data in Table \(1\) of Section 7.7
• Exercise 9 described the formation of glucose in plants, which take in CO2 and H2O and give off O2. Is this process exothermic or endothermic? If exothermic, where does the energy go? If endothermic, where does the energy come from?
• The basic reaction in the refining of aluminum is to take Al2O3(s) and turn it into Al(s) and O2(g). Write the balanced thermochemical equation for this process. Use data in Table \(1\) of Section 7.7
• Is the enthalpy change of the reaction
H2O(ℓ) → H2O(g)
zero or nonzero? Use data in Table \(1\) of Section 7.7
• What mass of H2O can be heated from 22°C to 80°C in the combustion of 1 mol of CH4? You will need the balanced thermochemical equation for the combustion of CH4. Use data in Table \(1\) of Section 7.7
• What mass of H2O can be heated from 22°C to 80°C in the combustion of 1 mol of C2H6? You will need the balanced thermochemical equation for the combustion of C2H6. Use data in Table \(1\) of Section 7.7
• What is the enthalpy change for the unknown reaction?
Pb(s) + Cl2(g) → PbCl2(s) ΔH = −359 kJPbCl2(s) + Cl2(g) → PbCl4(ℓ) ΔH = ?Pb(s) + 2Cl2(g) → PbCl4(ℓ) ΔH = −329 kJ
• What is the enthalpy change for the unknown reaction?
P(s) + 3/2Br2(ℓ) → PBr3(ℓ) ΔH = −185 kJPI3(s) → P(s) + 3/2I2(s) ΔH = ?PI3(s) + 3/2Br2(ℓ) → PBr3(ℓ) + 3/2I2(s) ΔH = −139 kJ
• What is the ΔH for this reaction? The label gra means graphite, and the label dia means diamond. What does your answer mean?
C(s, gra) → C(s, dia)
• Without consulting any tables, determine the ΔH for this reaction. Explain your answer.
H2O(ℓ, 25°C) → H2O(ℓ, 25°C)
• Answers
1. 5.70 J
2.
3. 4.69 L
4.
5. 80.97 kJ
6.
7. 91.9°C
8.
9. 6CO2(g) + 6H2O(ℓ) → C6H12O6(s) + 6O2(g) ΔH = 2,799 kJ
10.
11. 2Al2O3(s) → 4Al(s) + 3O2(g) ΔH = 3351.4 kJ
12.
13. 3,668 g
14.
15. ΔH = 30 kJ
16.
17. ΔH = 1.897 kJ; the reaction is endothermic. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/07%3A_Energy_and_Chemistry/7.E%3A_Energy_and_Chemistry_%28Exercises%29.txt |
Atoms act the way that they do because of their structure. We already know that atoms are composed of protons, neutrons, and electrons. Protons and neutrons are located in the nucleus, and electrons orbit around the nucleus. But knowing structural details is key to understanding why atoms react the way they do. Virtually everything known about atoms ultimately comes from light. Before the composition of atoms (especially electrons) can be understood, the properties of light need to be understood.
• 8.1: Prelude to Electronic Structure
A startling conclusion of modern science is that electrons also act as waves. However, the wavelength of electrons is much, much shorter—about 0.5 to 1 nm. This allows electron microscopes to magnify 600–700 times more than light microscopes. This allows us to see even smaller features in a world that is invisible to the naked eye.
• 8.2: Light
Light acts like a wave, with a frequency and a wavelength. The frequency and wavelength of light are related by the speed of light, a constant. Light acts like a particle of energy, whose value is related to the frequency of light.
• 8.3: Quantum Numbers for Electrons
Electrons in atoms have quantized energies. The state of electrons in atoms is described by four quantum numbers.
• 8.4: Organization of Electrons in Atoms
The Pauli exclusion principle limits the number of electrons in the subshells and shells. Electrons in larger atoms fill shells and subshells in a regular pattern that can be followed. Electron configurations are a shorthand method of indicating what subshells electrons occupy in atoms. Abbreviated electron configurations are a simpler way of representing electron configurations for larger atoms. Exceptions to the strict filling of subshells with electrons occur.
• 8.5: Electronic Structure and the Periodic Table
The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table.
• 8.6: Periodic Trends
Certain properties—notably atomic radius, IE, and EA—can be qualitatively understood by the positions of the elements on the periodic table.
• 8.E: Electronic Structure (Exercises)
These are exercises and select solutions to accompany Chapter 8 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
08: Electronic Structure
Normal light microscopes can magnify objects up to about 1,500 times. Electron microscopes can magnify objects up to 1,000,000 times. Why can electron microscopes magnify images so much? A microscope's resolution depends on the wavelength of light used. The smaller the wavelength, the more a microscope can magnify. Light is a wave, and as such, it has a wavelength associated with it. The wavelength of visible light, which is detected by the eyes, varies from about 700 nm to about 400 nm.
A startling conclusions of modern science is that electrons also act as waves. However, the wavelength of electrons is much, much shorter—about 0.5 to 1 nm. This allows electron microscopes to magnify 600–700 times more than light microscopes. This allows us to see even smaller features in a world that is invisible to the naked eye. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.01%3A_Prelude_to_Electronic_Structure.txt |
Learning Objectives
• Describe light with its frequency and wavelength.
• Describe light as a particle of energy.
What is known as light is more properly called electromagnetic radiation. We know from experiments that light acts as a wave. As such, it can be described as having a frequency and a wavelength. The wavelength of light is the distance between corresponding points in two adjacent light cycles, and the frequency of light is the number of cycles of light that pass a given point in one second. Wavelength is typically represented by λ, the lowercase Greek letter lambda, while frequency is represented by ν, the lowercase Greek letter nu (although it looks like a Roman "vee," it is actually the Greek equivalent of the letter "en"). Wavelength has units of length (meters, centimeters, etc.), while frequency has units of per second, written as s−1 and sometimes called a hertz (Hz). Figure $1$ - Characteristics of Light Waves, shows how these two characteristics are defined.
One property of waves is that their speed is equal to their wavelength times their frequency. That means we have
$speed=\lambda \nu\nonumber$
For light, however, speed is actually a universal constant when light is traveling through a vacuum (or, to a very good approximation, air). The measured speed of light (c) in a vacuum is 2.9979 × 108 m/s, or about 3.00 × 108 m/s. Thus, we have
$c=\lambda \nu\nonumber$
Because the speed of light is a constant, the wavelength and the frequency of light are related to each other: as one increases, the other decreases and vice versa. We can use this equation to calculate what one property of light has to be when given the other property.
Example $1$
What is the frequency of light if its wavelength is 5.55 × 10−7 m?
Solution
We use the equation that relates the wavelength and frequency of light with its speed. We have
$3.00\times 10^{8}m/s=\left ( 5.55\times 10^{-7} m\right )\nu\nonumber$
We divide both sides of the equation by 5.55 × 10−7 m and get
$\nu =5.41\times 10^{14}s^{-1}\nonumber$
Note how the m units cancel, leaving s in the denominator. A unit in a denominator is indicated by a −1 power, s−1, and read as "per second."
Exercise $1$
What is the wavelength of light if its frequency is 1.55 × 1010 s−1?
Answer
0.0194 m, or 19.4 mm
Light also behaves like a package of energy. It turns out that for light, the energy of the "package" of energy is proportional to its frequency. (For most waves, energy is proportional to wave amplitude, or the height of the wave.) The mathematical equation that relates the energy (E) of light to its frequency is
$E=h\nu\nonumber$
where ν is the frequency of the light, and h is a constant called Planck's constant. Its value is 6.626 × 10−34 J·s—a very small number that is another fundamental constant of our universe, like the speed of light. The units on Planck's constant may look unusual, but these units are required so that the algebra works out.
Example $2$
What is the energy of light if its frequency is 1.55 × 1010 s−1?
Solution
Using the formula for the energy of light, we have
E = (6.626 × 10−34 J·s)(1.55 × 1010 s−1)
Seconds are in the numerator and the denominator, so they cancel, leaving us with joules, the unit of energy. So
E = 1.03 × 10−23 J
This is an extremely small amount of energy—but this is for only one light wave.
Exercise $2$
What is the frequency of a light wave if its energy is 4.156 × 10−20 J?
Answer
6.27 × 1013 s−1
Because a light wave behaves like a little particle of energy, light waves have a particle-type name: the photon. It is not uncommon to hear light described as photons.
Wavelengths, frequencies, and energies of light span a wide range; the entire range of possible values for light is called the electromagnetic spectrum. We are mostly familiar with visible light, which is light having a wavelength range between about 400 nm and 700 nm. Light can have much longer and much shorter wavelengths than this, with corresponding variations in frequency and energy. Figure $2$ shows the entire electromagnetic spectrum and how certain regions of the spectrum are labeled. You may already be familiar with some of these regions; they are all light—with different frequencies, wavelengths, and energies.
Summary
Light acts like a wave, with a frequency and a wavelength. The frequency and wavelength of light are related by the speed of light, a constant. Light acts like a particle of energy, whose value is related to the frequency of light. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.02%3A_Light.txt |
Learning Objectives
• Explain what spectra are.
• Learn the quantum numbers that are assigned to electrons.
There are two fundamental ways of generating light: either heat an object up to be so hot that it glows, or pass an electrical current through a sample of matter (usually a gas). Incandescent lights and fluorescent lights generate light via these two methods, respectively.
A hot object gives off a continuum of light. We notice this when the visible portion of the electromagnetic spectrum is passed through a prism: the prism separates light into its constituent colors, and all colors are present in a continuous rainbow (Figure $\PageIndex{1a}$ - Prisms and Light). This image is known as a continuous spectrum. However, when electricity is passed through a gas and light is emitted and this light is passed though a prism, we see only certain lines of light in the image (Figure $\PageIndex{1b}$). This image is called a line spectrum. It turns out that every element has its own unique, characteristic line spectrum.
Why does the light emitted from an electrically excited gas have only certain colors, while light given off by hot objects has a continuous spectrum? For a long time, it was not well explained. Particularly simple was the spectrum of hydrogen gas, which could be described easily by an equation; no other element has a spectrum that is so predictable (Figure $2$ - Hydrogen Spectrum). Late nineteenth century scientists found that the positions of the lines obeyed a pattern given by the equation
$\frac{1}{\lambda }=(109,700cm^{-1})\left ( \frac{1}{4} -\frac{1}{n^{2}}\right )\nonumber$
where n = 3, 4, 5, 6, etc. However, they could not explain why this was so.
In 1913, the Danish scientist Niels Bohr suggested a reason why the hydrogen atom spectrum looked this way. He suggested that the electron in a hydrogen atom could not have any random energy, having only certain fixed values of energy that were indexed by the number n (the same n in the equation above and now called a quantum number). Quantities that have certain specific values are quantized values. Bohr suggested that the energy of the electron in hydrogen was quantized because it was in a specific orbit. Because the energies of the electron can have only certain values, the changes in energies can have only certain values (somewhat similar to a staircase—not only are the stair steps set at specific heights, but the height between steps is fixed). Finally, Bohr suggested that the energy of light emitted from electrified hydrogen gas was equal to the energy difference of the electron's energy states:
$E_{light}=h\nu =\Delta E_{electron}\nonumber$
This means that only certain frequencies (and thus, certain wavelengths) of light are emitted. Figure $3$ - Bohr's Model of the Hydrogen Atom, shows a model of the hydrogen atom based on Bohr's ideas.
Bohr's ideas were useful, but were applicable only to the hydrogen atom. However, later researchers generalized Bohr's ideas into a new theory called quantum mechanics, which explains the behavior of electrons as if they were acting as a wave, not as particles. Quantum mechanics predicts two major things: quantized energies for electrons of all atoms (not just hydrogen) and an organization of electrons within atoms. Electrons are no longer thought of as being randomly distributed around a nucleus or restricted to certain orbits (in that regard, Bohr was wrong). Instead, electrons are collected into groups and subgroups that explain much about the chemical behavior of the atom.
In the quantum-mechanical model of an atom, the state of an electron is described by four quantum numbers, not just the one predicted by Bohr. The first quantum number is called the principal quantum number(n). The principal quantum number largely determines the energy of an electron. Electrons in the same atom that have the same principal quantum number are said to occupy an electron shell of the atom. The principal quantum number can be any nonzero positive integer: 1, 2, 3, 4,….
Within a shell, there may be multiple possible values of the next quantum number, the angular momentum quantum number (). The ℓ quantum number has a minor effect on the energy of the electron but also affects the spatial distribution of the electron in three-dimensional space—that is, the shape of an electron's distribution in space. The value of the ℓ quantum number can be any integer between 0 and n − 1: ℓ = 0, 1, 2,…, n − 1.
Thus, for a given value of n, there are different possible values of ℓ:
Table with two columns and four rows. The first (left) column is labeled "If n equals" and the second (right) column is labeled "ℓ can be”. Underneath the columns in the rows are different values. All of the rows have numerical values.
If n equals ℓ can be
1 0
2 0 or 1
3 0, 1, or 2
4 0, 1, 2, or 3
and so forth. Electrons within a shell that have the same value of ℓ are said to occupy a subshell in the atom. Commonly, instead of referring to the numerical value of ℓ, a letter represents the value of ℓ (to help distinguish it from the principal quantum number):
Table with two columns and four rows. The first (left) column is labeled "If ℓ equals" and the second (right) column is labeled "The letter is”. The rows underneath the left column are numbers and the rows underneath the right column are letters.
If ℓ equals The letter is
0 s
1 p
2 d
3 f
The next quantum number is called the magnetic quantum number (m). For any value of ℓ, there are 2ℓ + 1 possible values of m, ranging from −ℓ to ℓ:
$−ℓ \lt mℓ \lt ℓ\nonumber$
or
$|m_ℓ|\lt ℓ\nonumber$
The following explicitly lists the possible values of m for the possible values of ℓ:
Table with two columns and four rows. The first (left) column is labeled "If ℓ equals" and the second (right) column is labeled "The mℓ values can be”. The rows underneath the left column are positive numbers and the rows underneath the right column are negative numbers.
If ℓ equals The m values can be
0 0
1 −1, 0, or 1
2 −2, −1, 0, 1, or 2
3 −3, −2, −1, 0, 1, 2, or 3
The particular value of m dictates the orientation of an electron's distribution in space. When ℓ is zero, m can be only zero, so there is only one possible orientation. When ℓ is 1, there are three possible orientations for an electron's distribution. When ℓ is 2, there are five possible orientations of electron distribution. This goes on and on for other values of ℓ, but we need not consider any higher values of ℓ here. Each value of m designates a certain orbital. Thus, there is only one orbital when ℓ is zero, three orbitals when ℓ is 1, five orbitals when ℓ is 2, and so forth. The m quantum number has no effect on the energy of an electron unless the electrons are subjected to a magnetic field—hence its name.
The ℓ quantum number dictates the general shape of electron distribution in space (Figure $4$ - Electron Orbitals). Any s orbital is spherically symmetric (Figure $\PageIndex{4a}$ - Electron Orbitals), and there is only one orbital in any s subshell. Any p orbital has a two-lobed, dumbbell-like shape (Figure $\PageIndex{4b}$ - Electron Orbitals); because there are three of them, we normally represent them as pointing along the x-, y-, and z-axes of Cartesian space. The d orbitals are four-lobed rosettes (Figure $\PageIndex{4c}$ - Electron Orbitals) and they are oriented differently in space (the one labeled $dz2$ has two lobes and a torus instead of four lobes, but it is equivalent to the other orbitals). When there is more than one possible value of m, each orbital is labeled with one of the possible values. It should be noted that the diagrams in Figure $4$ are estimates of the electron distribution in space, not surfaces electrons are fixed on.
The final quantum number is the spin quantum number (ms). Electrons and other subatomic particles behave as if they are spinning (we cannot tell if they really are, but they behave as if they are). Electrons themselves have two possible spin states, and because of mathematics, they are assigned the quantum numbers +1/2 and −1/2. These are the only two possible choices for the spin quantum number of an electron.
Example $1$
Of the set of quantum numbers {n, ℓ, m, ms}, which are possible and which are not allowed?
1. {3, 2, 1, +1/2}
2. {2, 2, 0, −1/2}
3. {3, −1, 0, +1/2}
Solution
1. The principal quantum number n must be an integer, as it is here. The quantum number ℓ must be less than n, which it is. The m quantum number must be between −ℓ and ℓ, which it is. The spin quantum number is +1/2, which is allowed. Because this set of quantum numbers follows all restrictions, it is possible.
2. The quantum number n is an integer, but the quantum number ℓ must be less than n, which it is not. Thus, this is not an allowed set of quantum numbers.
3. The principal quantum number n is an integer, but ℓ is not allowed to be negative. Therefore, this is not an allowed set of quantum numbers.
Exercise $1$
Of the set of quantum numbers {n, ℓ, m, ms}, which are possible and which are not allowed?
1. {4, 2, −2, 1}
2. {3, 1, 0, −1/2}
Answers
1. Spin must be either +1/2 or −1/2, so this set of quantum number is not allowed.
2. allowed
Chemistry is Everywhere: Neon Lights
A neon light is basically an electrified tube with a small amount of gas in it. Electricity excites electrons in the gas atoms, which then give off light as the electrons go back into a lower energy state. However, many so-called "neon" lights do not contain neon!
Although we know now that a gas discharge gives off only certain colors of light, without a prism or other component to separate the individual light colors, we see a composite of all the colors emitted. It is not unusual for a certain color to predominate. True neon lights, with neon gas in them, have a reddish-orange light due to the large amount of red-, orange-, and yellow-colored light emitted. However, if krypton is used instead of neon, a whitish light is emitted, while using argon yields a blue-purple light. A light filled with nitrogen gas glows purple, as does a helium lamp. Other gases—and mixtures of gases—emit other colors of light. Ironically, despite its importance in the development of modern electronic theory, hydrogen lamps emit little visible light and are rarely used for illumination purposes.
Key Takeaways
Electrons in atoms have quantized energies. The state of electrons in atoms is described by four quantum numbers. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.03%3A_Quantum_Numbers_for_Electrons.txt |
Learning Objectives
• Learn how electrons are organized within atoms.
• Represent the organization of electrons by an electron configuration.
Now that you know that electrons have quantum numbers, how are they arranged in atoms? The key to understanding electronic arrangement is summarized in the Pauli exclusion principle: no two electrons in an atom can have the same set of four quantum numbers. This dramatically limits the number of electrons that can exist in a shell or a subshell.
Electrons are typically organized around an atom by starting at the lowest possible quantum numbers first, which are the shells-subshells with lower energies. Consider H, an atom with a single electron only. Under normal conditions, the single electron would go into the n = 1 shell, which has only a single s subshell with one orbital (because m can equal only 0). The convention is to label the shell-subshell combination with the number of the shell and the letter that represents the subshell. Thus, the electron goes in the 1s shell-subshell combination. It is usually not necessary to specify the m or ms quantum numbers, but for the H atom, the electron has m = 0 (the only possible value) and an ms of either +1/2 or −1/2.
The He atom has two electrons. The second electron can also go into the 1s shell-subshell combination, but only if its spin quantum number is different from the first electron's spin quantum number. Thus, the sets of quantum numbers for the two electrons are {1, 0, 0, +1/2} and {1, 0, 0, −1/2}. Notice that the overall set is different for the two electrons, as required by the Pauli exclusion principle.
The next atom is Li, with three electrons. However, now the Pauli exclusion principle implies that we cannot put that electron in the 1s shell-subshell because no matter how we try, this third electron would have the same set of four quantum numbers as one of the first two electrons. So this third electron must be assigned to a different shell-subshell combination. However, the n = 1 shell doesn't have another subshell; it is restricted to having just ℓ = 0, or an s subshell. Therefore, this third electron has to be assigned to the n = 2 shell, which has an s (ℓ = 0) subshell and a p (ℓ = 1) subshell. Again, we usually start with the lowest quantum number, so this third electron is assigned to the 2s shell-subshell combination of quantum numbers.
The Pauli exclusion principle has the net effect of limiting the number of electrons that can be assigned a shell-subshell combination of quantum numbers. For example, in any s subshell, no matter what the shell number, there can be a maximum of only two electrons. Once the s subshell is filled up, any additional electrons must go to another subshell in the shell (if it exists), or to a higher-numbered shell. A similar analysis shows that a p subshell can hold a maximum of six electrons. A d subshell can hold a maximum of 10 electrons, while an f subshell can have a maximum of 14 electrons. By limiting subshells to these maxima, we can distribute the available electrons to their shells and subshells.
Example \(1\): Carbon Atoms
How would the six electrons for C be assigned to the n and ℓ quantum numbers?
Solution
The first two electrons go into the 1s shell-subshell combination. Two additional electrons can go into the 2s shell-subshell, but now this subshell is filled with the maximum number of electrons. The n = 2 shell also has a p subshell, so the remaining two electrons can go into the 2p subshell. The 2p subshell is not completely full because it can hold a maximum of six electrons.
Exercise \(1\): Sodium Atoms
How would the 11 electrons for Na be assigned to the n and ℓ quantum numbers?
Answer
two 1s electrons, two 2s electrons, six 2p electrons, and one 3s electron
Now that we see how electrons are partitioned among the shells and subshells, we need a more concise way of communicating this partitioning. Chemists use an electron configuration to represent the organization of electrons in shells and subshells in an atom. An electron configuration simply lists the shell and subshell labels, with a right superscript giving the number of electrons in that subshell. The shells and subshells are listed in the order of filling.
For example, an H atom has a single electron in the 1s subshell. Its electron configuration is
H: 1s1
The He atom has two electrons in the 1s subshell. Its electron configuration is
He: 1s2
The three electrons for Li are arranged in the 1s subshell (two electrons) and the 2s subshell (one electron). The electron configuration of Li is
Li: 1s22s1
The Be atom has four electrons, two in the 1s subshell and two in the 2s subshell. Its electron configuration is
Be: 1s22s2
Now that the 2s subshell is filled, electrons in larger atoms must go into the 2p subshell, which can hold a maximum of six electrons. The next six elements progressively fill up the 2p subshell:
• B: 1s22s22p1
• C: 1s22s22p2
• N: 1s22s22p3
• O: 1s22s22p4
• F: 1s22s22p5
• Ne: 1s22s22p6
Now that the 2p subshell is filled (all possible subshells in the n = 2 shell), the next electron for the next-larger atom must go into the n = 3 shell, s subshell.
Example \(2\): Sodium
What is the electron configuration for Na, which has 11 electrons?
Solution
The first two electrons occupy the 1s subshell. The next two occupy the 2s subshell, while the next six electrons occupy the 2p subshell. This gives us 10 electrons so far, with 1 electron left. This last electron goes into the n = 3 shell, s subshell. Thus, the electron configuration of Na is 1s22s22p63s1.
Exercise \(2\): Magnesium
What is the electron configuration for Mg, which has 12 electrons?
Answer
1s22s22p63s2
For larger atoms, the electron arrangement becomes more complicated. This is because after the 3p subshell is filled, filling the 4s subshell first actually leads to a lesser overall energy than filling the 3d subshell. Recall that while the principal quantum number largely dictates the energy of an electron, the angular momentum quantum number also has an impact on energy; by the time we get to the 3d and 4s subshells, we see overlap in the filling of the shells. Thus, after the 3p subshell is completely filled (which occurs for Ar), the next electron for K occupies the 4s subshell, not the 3d subshell:
K: 1s22s22p63s23p64s1,
not
K: 1s22s22p63s23p63d1
For larger and larger atoms, the order of filling the shells and subshells seems to become even more complicated. There are some useful ways to remember the order, like that shown in Figure \(1\). If you follow the arrows in order, they pass through the subshells in the order that they are filled with electrons in larger atoms. Initially, the order is the same as the expected shell-subshell order, but for larger atoms, there is some shifting around of the principal quantum numbers. However, Figure \(1\) gives a valid ordering of filling subshells with electrons for most atoms.
Example \(3\): Tin
What is the predicted electron configuration for Sn, which has 50 electrons?
Solution
We will follow the chart in Figure 8.4., until we can accommodate 50 electrons in the subshells in the proper order: Sn: 1s22s22p63s23p64s23d104p65s24d105p2
Verify by adding the superscripts, which indicate the number of electrons: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 2 = 50, so we have placed all 50 electrons in subshells in the proper order.
Exercise \(3\): Barium
What is the electron configuration for Ba, which has 56 electrons?
Answer
1s22s22p63s23p64s23d104p65s24d105p66s2
As the previous example demonstrated, electron configurations can get fairly long. An abbreviated electron configuration uses one of the elements from the last column of the periodic table, which contains the noble gases, to represent the core of electrons up to that element. Then, the remaining electrons are listed explicitly. For example, the abbreviated electron configuration for Li, which has three electrons, would be
Li: [He]2s1
where [He] represents the two-electron core that is equivalent to He's electron configuration. The square brackets represent the electron configuration of a noble gas. This is not much of an abbreviation. However, consider the abbreviated electron configuration for W, which has 74 electrons:
W: [Xe]6s24f145d4
This is a significant simplification over an explicit listing of all 74 electrons. So for larger elements, the abbreviated electron configuration can be a very useful shorthand.
Example \(4\): Phosphorus
What is the abbreviated electron configuration for P, which has 15 electrons?
Solution
With 15 electrons, the electron configuration of P is
P: 1s22s22p63s23p3
The first immediate noble gas is Ne, which has an electron configuration of 1s22s22p6. Using the electron configuration of Ne to represent the first 10 electrons, the abbreviated electron configuration of P is
P: [Ne]3s23p3
Exercise \(4\): Rubidium
What is the abbreviated electron configuration for Rb, which has 37 electrons?
Answer
[Kr]5s1
There are some exceptions to the rigorous filling of subshells by electrons. In many cases, an electron goes from a higher-numbered shell to a lower-numbered, but later-filled, subshell to fill the later-filled subshell. One example is Ag. With 47 electrons, its electron configuration is predicted to be
Ag: [Kr]5s24d9
However, experiments have shown that the electron configuration is actually
Ag: [Kr]5s14d10
This, then, qualifies as an exception to our expectations. At this point, you do not need to memorize the exceptions; but if you come across one, understand that it is an exception to the normal rules of filling subshells with electrons, which can happen.
Summary
The Pauli exclusion principle limits the number of electrons in the subshells and shells. Electrons in larger atoms fill shells and subshells in a regular pattern that can be followed. Electron configurations are a shorthand method of indicating what subshells electrons occupy in atoms. Abbreviated electron configurations are a simpler way of representing electron configurations for larger atoms. Exceptions to the strict filling of subshells with electrons do occur. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.04%3A_Organization_of_Electrons_in_Atoms.txt |
Learning Objectives
• Relate the electron configurations of the elements to the shape of the periodic table.
• Determine the expected electron configuration of an element by its place on the periodic table.
In Chapter 3, we introduced the periodic table as a tool for organizing the known chemical elements. A periodic table is shown in Figure \(1\). The elements are listed by atomic number (the number of protons in the nucleus), and elements with similar chemical properties are grouped together in columns.
Why does the periodic table have the structure it does? The answer is rather simple, if you understand electron configurations, the shape of the periodic table mimics the filling of the subshells with electrons.
The shape of the periodic table mimics the filling of the subshells with electrons.
Let us start with H and He. Their electron configurations are 1s1 and 1s2, respectively; with He, the n = 1 shell is filled. These two elements make up the first row of the periodic table (Figure \(2\))
The next two electrons, for Li and Be, would go into the 2s subshell. Figure \(3\) shows that these two elements are adjacent on the periodic table.
For the next six elements, the 2p subshell is being occupied with electrons. On the right side of the periodic table, these six elements (B through Ne) are grouped together (Figure \(4\)).
The next subshell to be filled is the 3s subshell. The elements coinciding this subshell being filled, Na and Mg, are back on the left side of the periodic table (Figure \(5\)).
Next, the 3p subshell is filled with the next six elements (Figure \(6\)).
Instead of filling the 3d subshell next, electrons go into the 4s subshell (Figure \(7\)).
After the 4s subshell is filled, the 3d subshell is filled with up to 10 electrons. This explains the section of 10 elements in the middle of the periodic table (Figure \(8\)).
And so forth. As we go across the columns of the periodic table, the overall shape of the table outlines how the electrons are occupying the shells and subshells.
The first two columns on the left side of the periodic table are where the s subshells are being occupied. Because of this, the first two rows of the periodic table are labeled the s block. Similarly, the p block is the right-most six columns of the periodic table, the d block is the middle 10 columns of the periodic table, while the f block is the 14-column section that is normally depicted as detached from the main body of the periodic table. It could be part of the main body, but then the periodic table would be rather long and cumbersome. Figure \(9\) shows the blocks of the periodic table.
The electrons in the highest-numbered shell, plus any electrons in the last unfilled subshell, are called valence electrons; the highest-numbered shell is called the valence shell. (The inner electrons are called core electrons.) The valence electrons largely control the chemistry of an atom. If we look at just the valence shell's electron configuration, we find that in each column, the valence shell's electron configuration is the same. For example, take the elements in the first column of the periodic table: H, Li, Na, K, Rb, and Cs. Their electron configurations (abbreviated for the larger atoms) are as follows, with the valence shell electron configuration highlighted:
Table with two columns and 6 rows. The first column on the left has various elements in the rows underneath. The second column on the right has the different corresponding electron configurations for the specified element in the rows underneath.
H: 1s1
Li: 1s22s1
Na: [Ne]3s1
K: [Ar]4s1
Rb: [Kr]5s1
Cs: [Xe]6s1
They all have a similar electron configuration in their valence shells: a single s electron. Because much of the chemistry of an element is influenced by valence electrons, we would expect that these elements would have similar chemistry—and they do. The organization of electrons in atoms explains not only the shape of the periodic table, but also the fact that elements in the same column of the periodic table have similar chemistry.
The same concept applies to the other columns of the periodic table. Elements in each column have the same valence shell electron configurations, and the elements have some similar chemical properties. This is strictly true for all elements in the s and p blocks. In the d and f blocks, because there are exceptions to the order of filling of subshells with electrons, similar valence shells are not absolute in these blocks. However, many similarities do exist in these blocks, so a similarity in chemical properties is expected.
Similarity of valence shell electron configuration implies that we can determine the electron configuration of an atom solely by its position on the periodic table. Consider Se, as shown in Figure \(10\). It is in the fourth column of the p block. This means that its electron configuration should end in a p4 electron configuration. Indeed, the electron configuration of Se is [Ar]4s23d104p4, as expected.
Example \(1\)
From the element's position on the periodic table, predict the valence shell electron configuration for each atom (Figure \(11\)).
1. Ca
2. Sn
Solution
1. Ca is located in the second column of the s block. We would expect that its electron configuration should end with s2. Calcium's electron configuration is [Ar]4s2.
2. Sn is located in the second column of the p block, so we expect that its electron configuration would end in p2. Tin's electron configuration is [Kr]5s24d105p2.
Exercise \(1\)
From the element's position on the periodic table, predict the valence shell electron configuration for each atom. Figure \(11\).
1. Ti
2. Cl
Answer a
[Ar]4s23d2
Answer b
[Ne]3s23p5
Food and Drink Application: Artificial Colors
The color of objects comes from a different mechanism than the colors of neon and other discharge lights. Although colored lights produce their colors, objects are colored because they preferentially reflect a certain color from the white light that shines on them. A red tomato, for example, is bright red because it reflects red light while absorbing all the other colors of the rainbow.
Many foods, such as tomatoes, are highly colored; in fact, the common statement "you eat with your eyes first" is an implicit recognition that the visual appeal of food is just as important as its taste. But what about processed foods?
Many processed foods have food colorings added to them. There are two types of food colorings: natural and artificial. Natural food colorings include caramelized sugar for brown; annatto, turmeric, and saffron for various shades of orange or yellow; betanin from beets for purple; and even carmine, a deep red dye that is extracted from the cochineal, a small insect that is a parasite on cacti in Central and South America. (That's right—you may be eating bug juice!)
Some colorings are artificial. In the United States, the Food and Drug Administration currently approves only seven compounds as artificial colorings in food, beverages, and cosmetics:
• FD&C Blue #1: Brilliant Blue FCF
• FD&C Blue #2: Indigotine
• FD&C Green #3: Fast Green FCF
• RD&C Red #3: Erythrosine
• FD&C Red #40: Allura Red AC
• FD&C Yellow #5: Tartrazine
• FD&C Yellow #6: Sunset Yellow FCF
Lower-numbered colors are no longer on the market or have been removed for various reasons. Typically, these artificial colorings are large molecules that absorb certain colors of light very strongly, making them useful even at very low concentrations in foods and cosmetics. Even at such low amounts, some critics claim that a small portion of the population (especially children) is sensitive to artificial colorings and urge that their use be curtailed or halted. However, formal studies of artificial colorings and their effects on behavior have been inconclusive or contradictory. Many people continue to enjoy processed foods with artificial coloring (like those shown in the accompanying figure).
Summary
The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.05%3A_Electronic_Structure_and_the_Periodic_Table.txt |
Learning Objective
• Be able to state how certain properties of atoms vary based on their relative position on the periodic table.
One reason the periodic table is so useful is that its structure allows us to qualitatively determine how some properties of the elements vary according to their position on the periodic table. The variations of properties according to positions on the periodic table are called periodic trends. There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row, or down an entire column of the periodic table.
The first periodic trend we will consider is atomic radius. The atomic radius is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals.
As you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. This trend can be summarized as follows:
$as\downarrow PT,atomic\; radius \uparrow\nonumber$
where PT stands for periodic table. Going across a row on the periodic table, left to right, the trend is different. This is because although the valence shell maintains the same principal quantum number, the number of protons—and hence the nuclear charge—is increasing as you go across the row. The increasing positive charge casts a tighter grip on the valence electrons, so as you go across the periodic table, the atomic radii decrease. Again, we can summarize this trend as follows:
$as\rightarrow PT,atomic\; radius \downarrow\nonumber$
Figure $1$ shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius.
Example $1$: Atomic Radii
Referring only to a periodic table and not to Figure $1$, which atom is larger in each pair?
1. Si or S
2. S or Te
Solution
1. Si is to the left of S on the periodic table, so it is larger because as you go across the row, the atoms get smaller.
2. S is above Te on the periodic table, so Te is larger because as you go down the column, the atoms get larger.
Exercise $1$: Atomic Radii
Referring only to a periodic table and not to Figure $1$, which atom is smaller, Ca or Br?
Answer
Br
Ionization energy (IE) is the amount of energy required to remove an electron from an atom in the gas phase:
$A(g)\rightarrow A^{+}(g)+e^{-}\; \; \; \; \; \Delta H\equiv IE\nonumber$
IE is usually expressed in kJ/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. Thus,
$as\downarrow PT,\; IE\downarrow\nonumber$
However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron. As a result, IE increases:
$as\rightarrow PT,\; IE\uparrow\nonumber$
Figure $2$ shows values of IE versus position on the periodic table. Again, the trend isn't absolute, but the general trends going across and down the periodic table should be obvious.
IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with):
• First Ionization Energy (IE1):
$A(g) → A^+(g) + e^-\nonumber$
• Second Ionization Energy (IE2):
$A^{+}(g) → A^{2+}(g) + e^-\nonumber$
• Third Ionization Energy (IE3):
$A^{2+}(g) → A^{3+}(g) + e^-\nonumber$
and so forth.
Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration is 1s22s22p63s2:
First Ionization Energy (IE1) = 738 kJ/mol:
$Mg(g) → Mg^{+}(g) + e^−\nonumber$
Second Ionization Energy (IE2) = 1,450 kJ/mol:
$Mg^+(g) → Mg^{2+}(g) + e^−\nonumber$
Third Ionization Energy (IE3) = 7,734 kJ/mol:
$Mg^{2+}(g) → Mg^{3+}(g) + e^− \nonumber$
The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over five times the previous one. Why is it so much larger? Because the first two electrons are removed from the 3s subshell, but the third electron has to be removed from the n = 2 shell (specifically, the 2p subshell, which is lower in energy than the n = 3 shell). Thus, it takes much more energy than just overcoming a larger ionic charge would suggest. Trends like this one demonstrate that electrons within atoms are organized in groups.
Example $2$: Ionization Energies
Which atom in each pair has the larger first ionization energy?
1. Ca or Sr
2. K or K+
Solution
1. Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE.
2. Because K+ has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K+ to be removed comes from another shell.
Exercise $2$: Ionization Energies
Which atom has the lower ionization energy, C or F?
Answer
C
The opposite of IE is described by electron affinity (EA), which is the energy change when a gas-phase atom accepts an electron:
$A(g)+e^{-}\rightarrow A^{-}(g)\; \; \; \; \; \Delta H\equiv EA\nonumber$
EA is also usually expressed in kJ/mol. EA also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously. Generally, as you go across the periodic table, EA increases its magnitude:
$as\rightarrow PT,\; EA\uparrow\nonumber$
There is not a definitive trend as you go down the periodic table; sometimes EA increases, sometimes it decreases. Figure $3$ shows EA values versus position on the periodic table for the s- and p-block elements. The trend isn't absolute, especially considering the large positive EA values for the second column. However, the general trend going across the periodic table should be obvious.
Example $3$: Electron Affinities
Predict which atom in each pair will have the highest magnitude of Electron Affinity.
1. C or F
2. Na or S
Solution
1. C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA.
2. Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA.
Exercise $3$: Electron Affinities
Predict which atom will have the highest magnitude of Electron Affinity: As or Br.
Br
Summary
Certain properties—notably atomic radius, ionization energies, and electron affinitiescan be qualitatively understood by the positions of the elements on the periodic table. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.06%3A_Periodic_Trends.txt |
8.1: Light
Q8.1.1
Describe the characteristics of a light wave.
Q8.1.1
Light has a wavelength and a frequency.
Q8.1.2
What is a characteristic of a particle of light?
Q8.1.3
What is the frequency of light if its wavelength is 7.33 × 10−5 m?
4.09 × 1012 s−1
Q8.1.4
What is the frequency of light if its wavelength is 1.226 m?
Q8.1.5
What is the frequency of light if its wavelength is 733 nm?
4.09 × 1014 s−1
Q8.1.6
What is the frequency of light if its wavelength is 8.528 cm?
Q8.1.7
What is the wavelength of light if its frequency is 8.19 × 1014 s−1?
3.66 × 10−7 m
Q8.1.8
What is the wavelength of light if its frequency is 3.66 × 106 s−1?
Q8.1.9
What is the wavelength of light if its frequency is 1.009 × 106 Hz?
297 m
Q8.1.10
What is the wavelength of light if its frequency is 3.79 × 10−3 Hz?
Q8.1.11
What is the energy of a photon if its frequency is 5.55 × 1013 s−1?
3.68 × 10−20 J
Q8.1.12
What is the energy of a photon if its frequency is 2.06 × 1018 s−1?
Q8.1.13
What is the energy of a photon if its wavelength is 5.88 × 10−4 m?
Q8.1.14
What is the energy of a photon if its wavelength is 1.888 × 102 m?
1.053 × 10−27 J
8.2: Quantum Numbers for Electrons
Q8.2.1
Differentiate between a continuous spectrum and a line spectrum.
Q8.2.1
A continuous spectrum is a range of light frequencies or wavelengths; a line spectrum shows only certain frequencies or wavelengths.
Q8.2.2
Under what circumstances is a continuous spectrum formed? Under what circumstances is a line spectrum formed?
Q8.2.3
What is the wavelength of light from the hydrogen atom spectrum when n = 3?
Q8.2.3
6.56 × 10−7 m, or 656 nm
Q8.2.4
What is the wavelength of light from the hydrogen atom spectrum when n = 5?
Q8.2.5
What are the restrictions on the principal quantum number?
Q8.2.5
The principal quantum number is restricted to being a positive whole number.
Q8.2.6
What are the restrictions on the angular momentum quantum number?
Q8.2.7
What are the restrictions on the magnetic quantum number?
Q8.2.7
The absolute value of m must be less than or equal to ℓ: |m| ≤ ℓ.
Q8.2.8
What are the restrictions on the spin quantum number?
Q8.2.9
What are the possible values for ℓ when n = 5?
Q8.2.9
ℓ can be 0, 1, 2, 3, or 4.
Q8.2.10
What are the possible values for ℓ when n = 1?
Q8.2.11
What are the possible values for m when ℓ = 3?
Q8.2.11
m can be −3, −2, −1, 0, 1, 2, or 3.
Q8.2.12
What are the possible values for m when ℓ = 6?
Q8.2.13
Describe the shape of an s orbital.
Q8.2.13
An s orbital is spherical in shape.
Q8.2.14
Describe the shape of a p orbital.
Q8.2.15
Which of these sets of quantum numbers is allowed? If it is not, explain why.
1. {5, 2, −1, −1/2}
2. {3, −1, −1, −1/2}
Q8.2.15
1. Because |m| must be less than ℓ, this set of quantum numbers is not allowed.
2. allowed
Q8.2.16
Which of these sets of quantum numbers is allowed? If it is not, explain why.
1. {4, 1, −2, +1/2}
2. {2, 0, 0, −1/2}
8.3: Organization of Electrons in Atoms
Q8.3.1
Give two possible sets of four quantum numbers for the electron in an H atom.
Q8.3.1
{1, 0, 0, 1/2} and [1, 0, 0, −1/2}
Q8.3.2
Give the possible sets of four quantum numbers for the electrons in a Li atom.
Q8.3.3
How many subshells are completely filled with electrons for Na? How many subshells are unfilled?
Q8.3.3
Three subshells (1s, 2s, 2p) are completely filled, and one shell (3s) is partially filled.
Q8.3.4
How many subshells are completely filled with electrons for Mg? How many subshells are unfilled?
Q8.3.5
What is the maximum number of electrons in the entire n = 2 shell?
8 electrons
Q8.3.6
What is the maximum number of electrons in the entire n = 4 shell?
Q8.3.7
Write the complete electron configuration for each atom.
1. Si, 14 electrons
2. Sc, 21 electrons
Q8.3.7
1. 1s22s22p63s23p2
2. 1s22s22p63s23p64s23d1
Q8.3.8
Write the complete electron configuration for each atom.
1. Br, 35 electrons
2. Be, 4 electrons
Q8.3.9
Write the complete electron configuration for each atom.
1. Cd, 48 electrons
2. Mg, 12 electrons
Q8.3.9
1. 1s22s22p63s23p64s23d104p65s24d10
2. 1s22s22p63s2
Q8.3.10
Write the complete electron configuration for each atom.
1. Cs, 55 electrons
2. Ar, 18 electrons
Q8.3.11
Write the abbreviated electron configuration for each atom in Exercise 7.
1. [Ne]3s23p2
2. [Ar]4s23d1
Q8.3.12
Write the abbreviated electron configuration for each atom in Exercise 8.
Q8.3.13
Write the abbreviated electron configuration for each atom in Exercise 9.
1. [Kr]5s24d10
2. [Ne]3s2
Q8.3.14
Write the abbreviated electron configuration for each atom in Exercise 10.
8.4: Electronic Structure and the Periodic Table
Q8.4.1
Where on the periodic table are s subshells being occupied by electrons?
Q8.4.1
the first two columns
Q8.4.2
Where on the periodic table are d subshells being occupied by electrons?
Q8.4.3
In what block is Ra found?
the s block
Q8.4.4
In what block is Br found?
Q8.4.5
What are the valence shell electron configurations of the elements in the second column of the periodic table?
ns2
Q8.4.6
What are the valence shell electron configurations of the elements in the next-to-last column of the periodic table?
Q8.4.7
What are the valence shell electron configurations of the elements in the first column of the p block?
ns2np1
Q8.4.8
What are the valence shell electron configurations of the elements in the last column of the p block?
Q8.4.9
From the element’s position on the periodic table, predict the electron configuration of each atom.
1. Sr
2. S
Q8.4.9
1. 1s22s22p63s23p64s23d104p65s2
2. 1s22s22p63s23p4
Q8.4.10
From the element’s position on the periodic table, predict the electron configuration of each atom.
1. Fe
2. Ba
Q8.4.11
From the element’s position on the periodic table, predict the electron configuration of each atom.
1. V
2. Ar
Q8.4.11
1. 1s22s22p63s23p64s23d3
2. 1s22s22p63s23p6
Q8.4.12
From the element’s position on the periodic table, predict the electron configuration of each atom.
1. Cl
2. K
Q8.4.13
From the element’s position on the periodic table, predict the electron configuration of each atom.
1. Ge
2. C
Q8.4.13
1. 1s22s22p63s23p64s23d104p2
2. 1s22s22p2
Q8.4.14
From the element’s position on the periodic table, predict the electron configuration of each atom.
1. Mg
2. I
Q8.5.1
Write a chemical equation with an IE energy change.
Q8.5.1
Na(g) → Na+(g) + e ΔH = IE (answers will vary)
Q8.5.2
Write a chemical equation with an EA energy change.
Q8.5.3
State the trends in atomic radii as you go across and down the periodic table.
Q8.5.3
As you go across, atomic radii decrease; as you go down, atomic radii increase.
Q8.5.4
State the trends in IE as you go across and down the periodic table.
Q8.5.5
Which atom of each pair is larger?
1. Na or Cs
2. N or Bi
1. Cs
2. Bi
Q8.5.6
Which atom of each pair is larger?
1. C or Ge
2. Be or Ba
Q8.5.7
Which atom of each pair is larger?
1. K or Cl
2. Ba or Bi
1. K
2. Ba
Q8.5.8
Which atom of each pair is larger?
1. Si or S
2. H or He
Q8.5.9
Which atom has the higher IE?
1. Na or S
2. Ge or Br
1. S
2. Br
Q8.5.10
Which atom has the higher IE?
1. C or Ne
2. Rb or I
Q8.5.11
Which atom has the higher IE?
1. Li or Cs
2. Se or O
1. Li
2. O
Q8.5.12
Which atom has the higher IE?
1. Al or Ga
2. F or I
Q8.5.13
A third-row element has the following successive IEs: 738; 1,450; 7,734; and 10,550 kJ/mol. Identify the element.
Mg
Q8.5.14
A third-row element has the following successive IEs: 1,012; 1,903; 2,912; 4,940; 6,270; and 21,300 kJ/mol. Identify the element.
Q8.5.15
For which successive IE is there a large jump in IE for Ca?
Q8.5.15
The third IE shows a large jump in Ca.
Q8.5.16
For which successive IE is there a large jump in IE for Al?
Q8.5.17
Which atom has the greater magnitude of EA?
1. C or F
2. Al or Cl
1. F
2. Cl
Q8.5.18
Which atom has the greater magnitude of EA?
1. K or Br
2. Mg or S
QExtra.1
What is the frequency of light if its wavelength is 1.00 m?
3.00 × 108 s−1
QExtra.2
What is the wavelength of light if its frequency is 1.00 s−1?
QExtra.3
What is the energy of a photon if its wavelength is 1.00 meter?
1.99 × 10−22 J
QExtra.4
What is the energy of a photon if its frequency is 1.00 s−1?
QExtra.5
If visible light is defined by the wavelength limits of 400 nm and 700 nm, what is the energy range for visible light photons?
SExtra.5
4.97 × 10−19 J to 2.84 × 10−19 J
QExtra.6
Domestic microwave ovens use microwaves that have a wavelength of 122 mm. What is the energy of one photon of this microwave?
QExtra.7
Use the equation for the wavelengths of the lines of light in the H atom spectrum to calculate the wavelength of light emitted when n is 7 and 8.
SExtra.7
3.97 × 10−7 m and 3.89 × 10−7 m, respectively
QExtra.8
Use the equation for the wavelengths of the lines of light in the H atom spectrum to calculate the wavelengths of light emitted when n is 5 and 6.
QExtra.9
Make a table of all the possible values of the four quantum numbers when the principal quantum number n = 5.
SExtra.9
n m ms
5 0 0 1/2 or −1/2
5 1 −1, 0, 1 1/2 or −1/2
5 2 −2, −1, 0, 1, 2 1/2 or −1/2
5 3 −3, −2, −1, 0, 1, 2, 3 1/2 or −1/2
5 4 −4, −3, −2, −1, 0, 1, 2, 3, 4 1/2 or −1/2
QExtra.10
Make a table of all the possible values of m and ms when ℓ = 4. What is the lowest value of the principal quantum number for this to occur?
1. Predict the electron configurations of Sc through Zn.
2. From a source of actual electron configurations, determine how many exceptions there are from your predictions in part a.
SExtra.10
1. The electron configurations are predicted to end in 3d1, 3d2, 3d3, 3d4, 3d5, 3d6, 3d7, 3d8, 3d9, and 3d10.
2. Cr and Cu are exceptions.
QExtra.11
1. Predict the electron configurations of Ga through Kr.
2. From a source of actual electron configurations, determine how many exceptions there are from your predictions in part a.
QExtra.12
Recently, Russian chemists reported experimental evidence of element 117. Use the periodic table to predict its valence shell electron configuration.
SExtra.12
Element 117’s valence shell electron configuration should be 7s27p5.
QExtra.13
Bi (atomic number 83) is used in some stomach discomfort relievers. Using its place on the periodic table, predict its valence shell electron configuration.
QExtra.14
Which atom has a higher ionization energy (IE), O or P?
O
QExtra.15
Which atom has a higher IE, F or As?
QExtra.16
Which atom has a smaller radius, As or Cl?
Cl
QExtra.17
Which atom has a smaller radius, K or F?
QExtra.18
How many IEs does an H atom have? Write the chemical reactions for the successive ionizations.
SExtra.18
H has only one IE: H → H+ + e
QExtra.19
How many IEs does a Be atom have? Write the chemical reactions for the successive ionizations.
QExtra.20
Based on what you know of electrical charges, do you expect Na+ to be larger or smaller than Na?
smaller
QExtra.21
Based on what you know of electrical charges, do you expect Cl to be larger or smaller than Cl? | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/08%3A_Electronic_Structure/8.E%3A_Electronic_Structure_%28Exercises%29.txt |
How do atoms make compounds? Typically they join together in such a way that they lose their identities as elements and adopt a new identity as a compound. These joins are called chemical bonds. But how do atoms join together? Ultimately, it all comes down to electrons. Before we discuss how electrons interact, we need to introduce a tool to simply illustrate electrons in an atom.
• 9.1: Prelude to Chemical Bonds
Diamond is the hardest natural material known on Earth. Yet diamond is just pure carbon. What is special about this element that makes diamond so hard? Bonds. Chemical bonds.
• 9.2: Lewis Electron Dot Diagrams
Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol. Lewis electron dot diagrams for ions have less (for cations) or more (for anions) dots than the corresponding atom.
• 9.3: Electron Transfer - Ionic Bonds
The tendency to form species that have eight electrons in the valence shell is called the octet rule. The attraction of oppositely charged ions caused by electron transfer is called an ionic bond. The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions.
• 9.4: Covalent Bonds
Covalent bonds are formed when atoms share electrons. Lewis electron dot diagrams can be drawn to illustrate covalent bond formation. Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules.
• 9.5: Other Aspects of Covalent Bonds
Covalent bonds can be nonpolar or polar, depending on the electronegativities of the atoms involved. Covalent bonds can be broken if energy is added to a molecule. The formation of covalent bonds is accompanied by energy given off. Covalent bond energies can be used to estimate the enthalpy changes of chemical reactions.
• 9.6: Violations of the Octet Rule
There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules.
• 9.7: Molecular Shapes
The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.
• 9.E: Chemical Bonds (Exercises)
These are exercises and select solutions to accompany Chapter 9 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
09: Chemical Bonds
Diamond is the hardest natural material known on Earth. Yet diamond is just pure carbon. What is special about this element that makes diamond so hard? Bonds. Chemical bonds.
In a perfect diamond crystal, each C atom makes four connections—bonds—to four other C atoms in a three-dimensional matrix. Four is the greatest number of bonds that is commonly made by atoms, so C atoms maximize their interactions with other atoms. This three-dimensional array of connections extends throughout the diamond crystal, making it essentially one large molecule. Breaking a diamond means breaking every bond at once. Also, the bonds are moderately strong. There are stronger interactions known, but the carbon-carbon connection is fairly strong itself. Not only does a person have to break many connections at once, but the bonds are also strong connections from the start.
Diamond is the hardest known natural substance and is composed solely of the element carbon. (CC SA-BY 3.0; Mario Sarto).
There are other substances that have bonding arrangements similar to diamond. Silicon dioxide and boron nitride have some similarities, but neither of them comes close to the ultimate hardness of diamond. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.01%3A_Prelude_to_Chemical_Bonds.txt |
Learning Objective
• Draw a Lewis electron dot diagram for an atom or a monatomic ion.
In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful.
A Lewis electron dot diagram (or electron dot diagram, or a Lewis diagram, or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (The order in which the positions are used does not matter.) For example, the Lewis electron dot diagram for hydrogen is simply
$\mathbf{H}\mathbf{\cdot}\nonumber$
Because the side is not important, the Lewis electron dot diagram could also be drawn as follows:
$\mathbf{\dot{H}}\; \; or\; \mathbf{\cdot}\mathbf{H}\; \; \; or\; \; \; \mathbf{\underset{.}H}\nonumber$
The electron dot diagram for helium, with two valence electrons, is as follows:
$\mathbf{He}\mathbf{:}\nonumber$
By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1s22s1, so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used:
$\mathbf{Li}\mathbf{\cdot}\nonumber$
Beryllium has two valence electrons in its 2s shell, so its electron dot diagram is like that of helium:
$\mathbf{Be}\mathbf{:}\nonumber$
The next atom is boron. Its valence electron shell is 2s22p1, so it has three valence electrons. The third electron will go on another side of the symbol:
$\mathbf{\dot{Be}}\mathbf{:}\nonumber$
Again, it does not matter on which sides of the symbol the electron dots are positioned.
For carbon, there are four valence electrons, two in the 2s subshell and two in the 2p subshell. As usual, we will draw two dots together on one side, to represent the 2s electrons. However, conventionally, we draw the dots for the two p electrons on different sides. As such, the electron dot diagram for carbon is as follows:
$\mathbf{\cdot \dot{C}}\mathbf{:}\nonumber$
With N, which has three p electrons, we put a single dot on each of the three remaining sides:
$\mathbf{\cdot}\mathbf{\dot{\underset{.}N}}\mathbf{:}\nonumber$
For oxygen, which has four p electrons, we now have to start doubling up on the dots on one other side of the symbol. When doubling up electrons, make sure that a side has no more than two electrons.
$\mathbf{\cdot}\mathbf{\ddot{\underset{.}O}}\mathbf{:}\nonumber$
Fluorine and neon have seven and eight dots, respectively:
$\mathbf{:}\mathbf{\ddot{\underset{.}F}}\mathbf{:}\nonumber$
$\mathbf{:}\mathbf{\ddot{\underset{.\: .}Ne}}\mathbf{:}\nonumber$
With the next element, sodium, the process starts over with a single electron because sodium has a single electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we see that the Lewis electron dot diagrams of atoms will never have more than eight dots around the atomic symbol.
Example $1$
What is the Lewis electron dot diagram for each element?
1. aluminum
2. selenium
Solution
1. The valence electron configuration for aluminum is 3s23p1. So it would have three dots around the symbol for aluminum, two of them paired to represent the 3s electrons:
$\dot{Al:} \nonumber \nonumber$
1. The valence electron configuration for selenium is 4s24p4. In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows:
$\mathbf{\cdot }\mathbf{\dot{\underset{.\: .}Se}}\mathbf{:}\nonumber \nonumber$
Exercise $1$
What is the Lewis electron dot diagram for each element?
1. phosphorus
2. argon
Answer
$\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:}\nonumber \nonumber$
$\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:}\nonumber \nonumber$
Elements in the same column of the periodic table have similar Lewis electron dot diagrams because they have the same valence shell electron configuration. The electron dot diagrams for the first column of elements are as follows:
$\mathbf{H\: \cdot }\; \; \; \mathbf{Li\: \cdot }\; \; \; \mathbf{Na\: \cdot }\; \; \; \mathbf{K\: \cdot }\; \; \; \mathbf{Rb\: \cdot }\; \; \; \mathbf{Cs\: \cdot }\; \; \;\nonumber$
Monatomic ions are atoms that have either lost (for cations) or gained (for anions) electrons. Electron dot diagrams for ions are the same as for atoms, except that some electrons have been removed for cations, while some electrons have been added for anions. Thus, in comparing the electron configurations and electron dot diagrams for the Na atom and the Na+ ion, we note that the Na atom has a single valence electron in its Lewis diagram, while the Na+ ion has lost that one valence electron:
$\text{Lewis dot diagram}: \quad \mathbf{Na\: \cdot }\; \; \; \; \; Na^{+}\nonumber$
$\text{Electron configuration}: \quad \left [ Ne\right]3s^{1}\; \; \; \; \left [ Ne \right ]\nonumber$
Technically, the valence shell of the Na+ ion is now the n = 2 shell, which has eight electrons in it. So why do we not put eight dots around Na+? Conventionally, when we show electron dot diagrams for ions, we show the original valence shell of the atom, which in this case is the n = 3 shell and empty in the Na+ ion.
In making cations, electrons are first lost from the highest numbered shell, not necessarily the last subshell filled. For example, in going from the neutral Fe atom to the Fe2+ ion, the Fe atom loses its two 4s electrons first, not its 3d electrons, despite the fact that the 3d subshell is the last subshell being filled. Thus, we have
$\text{Lewis dot diagram}: \quad \mathbf{Fe\: :}\; \; \; \; \; Fe^{2+}\nonumber$
$\text{Electron configuration}: \quad \left [ Ar\right]4s^{2}3d^{6}\; \; \; \; \left [ Ar \right ]3d^{6}\nonumber$
Anions have extra electrons when compared to the original atom. Here is a comparison of the Cl atom with the Cl ion:
$\text{Lewis dot diagram}: \quad \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\cdot }\; \; \; \; \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{:}^{-}\nonumber$
$\text{Electron configuration}: \quad \left [ Ne\right]3s^{2}3p^{5}\; \; \; \; \left [ Ne \right ]3s^{2}3p^{6}\nonumber$
Example $2$
What is the Lewis electron dot diagram for each ion?
1. Ca2+
2. O2−
Solution
1. Having lost its two original valence electrons, the Lewis electron dot diagram is simply: Ca2+
2. The O2− ion has gained two electrons in its valence shell, so its Lewis electron dot diagram is as follows:
$\mathbf{:}\mathbf{\ddot{\underset{.\: .}O}}\mathbf{:}^{2-}\nonumber \nonumber$
Exercise $2$
The electronic configuration of the thallium ion is 6s24f145d10. What is the Lewis electron dot diagram for the Tl+ ion?
Answer
$\mathbf{Tl:}^{+}\nonumber \nonumber$
Summary
• Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol.
• Lewis electron dot diagrams for ions have less (for cations) or more (for anions) dots than the corresponding atom. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.02%3A_Lewis_Electron_Dot_Diagrams.txt |
Learning Objectives
• State the octet rule.
• Define ionic bond.
• Demonstrate electron transfer between atoms to form ionic bonds.
In Section 9.2, we saw how ions are formed by losing electrons to make cations, or by gaining electrons to form anions. The astute reader may have noticed something: many of the ions that form have eight electrons in their valence shell. Either atoms gain enough electrons to have eight electrons in the valence shell and become the appropriately charged anion, or they lose the electrons in their original valence shell. In the case of electron loss, the lower shell, now the valence shell, has eight electrons in it; so the atom becomes positively charged. For whatever reason, having eight electrons in a valence shell is a particularly energetically stable arrangement of electrons. The trend that atoms like to have eight electrons in their valence shell is called the octet rule. When atoms form compounds, the octet rule is not always satisfied for all atoms at all times, but it is a very good rule of thumb for understanding the kinds of bonding arrangements that atoms can make.
It is not impossible to violate the octet rule. Consider sodium: in its elemental form, it has one valence electron and is stable. It is rather reactive, however, and does not require a lot of energy to remove that electron to make the Na+ ion. We could remove another electron by adding even more energy to the ion to make the Na2+ ion. However, that requires much more energy than is normally available in chemical reactions, so sodium stops at a 1+ charge after losing a single electron. It turns out that the Na+ ion has a complete octet in its new valence shell, the n = 2 shell, which satisfies the octet rule. The octet rule is a result of trends in energies and is useful in explaining why atoms form the ions that they do.
Consider an Na atom in the presence of a Cl atom. The two atoms have these Lewis electron dot diagrams and electron configurations:
$\mathbf{Na\, \cdot }\; \; \; \; \; \; \; \; \; \; \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}\nonumber$
$\left [ Ne \right ]3s^{1}\; \; \; \; \left [ Ne \right ]3s^{2}3p^{5}\nonumber$
For the Na atom to obtain an octet, it must lose an electron; for the Cl atom to obtain an octet, it must gain an electron. An electron transfers from the Na atom to the Cl atom:
$\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}\nonumber$
resulting in two ions—the Na+ ion and the Cl ion:
$\mathbf{Na\, \cdot }^{+}\; \; \; \; \; \; \; \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\nonumber$
$\left [ Ne \right ]\; \; \; \; \; \left [ Ne \right ]3s^{2}3p^{6}\nonumber$
Both species now have complete octets, and the electron shells are energetically stable. From basic physics, we know that opposite charges attract. This is what happens to the Na+ and Cl ions:
$\mathbf{Na\, \cdot }^{+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\rightarrow Na^{+}Cl^{-}\; \; or\; \; NaCl\nonumber$
where we have written the final formula (the formula for sodium chloride) as per the convention for ionic compounds, without listing the charges explicitly. The attraction between oppositely charged ions is called an ionic bond, and it is one of the main types of chemical bonds in chemistry. Ionic bonds are caused by electrons transferring from one atom to another.
In electron transfer, the number of electrons lost must equal the number of electrons gained. We saw this in the formation of NaCl. A similar process occurs between Mg atoms and O atoms, except in this case two electrons are transferred:
The two ions each have octets as their valence shell, and the two oppositely charged particles attract, making an ionic bond:
$\mathbf{Mg\,}^{2+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}O}}\mathbf{\: :}^{2-}\; \; \; \; \; Mg^{2+}O^{2-}\; or\; MgO\nonumber$
Remember, in the final formula for the ionic compound, we do not write the charges on the ions.
What about when an Na atom interacts with an O atom? The O atom needs two electrons to complete its valence octet, but the Na atom supplies only one electron:
$\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.}O}}\mathbf{\: :}\nonumber$
The O atom still does not have an octet of electrons. What we need is a second Na atom to donate a second electron to the O atom:
These three ions attract each other to form an overall neutrally charged ionic compound, which we write as Na2O. The need for the number of electrons lost to be equal to the number of electrons gained explains why ionic compounds have the ratio of cations to anions that they do. This is also required by the law of conservation of matter.
Example $1$
With arrows, illustrate the transfer of electrons to form calcium chloride from Ca atoms and Cl atoms.
Solution
A Ca atom has two valence electrons, while a Cl atom has seven electrons. A Cl atom needs only one more to complete its octet, while Ca atoms have two electrons to lose. We need two Cl atoms to accept the two electrons from one Ca atom. The transfer process is as follows:
The oppositely charged ions attract one another to make CaCl2.
Exercise $1$
With arrows, illustrate the transfer of electrons to form potassium sulfide from K atoms and S atoms.
Answer
The strength of ionic bonding depends on two major characteristics: the magnitude of the charges and the size of the ion. The greater the magnitude of the charge, the stronger the ionic bond. The smaller the ion, the stronger the ionic bond (because a smaller ion size allows the ions to get closer together). The measured strength of ionic bonding is called the lattice energy. Some lattice energies are given in Table $1$ - Lattice Energies of Some Ionic Compounds.
Table $1$: Lattice Energies of Some Ionic Compounds
Compound Lattice Energy (kJ/mol)
LiF 1,036
LiCl 853
NaCl 786
NaBr 747
MgF2 2,957
Na2O 2,481
MgO 3,791
Chemistry is Everywhere: Salt
The element sodium (part [a] in the accompanying figure) is a very reactive metal; given the opportunity, it will react with the sweat on your hands and form sodium hydroxide, which is a very corrosive substance. The element chlorine (part [b] in the accompanying figure) is a pale yellow, corrosive gas that should not be inhaled due to its poisonous nature. Bring these two hazardous substances together, however, and they react to make the ionic compound sodium chloride (part [c] in the accompanying figure), known simply as salt.
Salt is necessary for life. Na+ ions are one of the main ions in the human body and are necessary to regulate the fluid balance in the body. Cl ions are necessary for proper nerve function and respiration. Both of these ions are supplied by salt. The taste of salt is one of the fundamental tastes; salt is probably the most ancient flavoring known, and one of the few rocks we eat.
The health effects of too much salt are still under debate, although a 2010 report by the US Department of Agriculture concluded that "excessive sodium intake…raises blood pressure, a well-accepted and extraordinarily common risk factor for stroke, coronary heart disease, and kidney disease" (US Department of Agriculture Committee for Nutrition Policy and Promotion, Report of the Dietary Guidelines Advisory Committee on the Dietary Guidelines for Americans, accessed January 5, 2010). It is clear that most people ingest more salt than their bodies need, and most nutritionists recommend curbing salt intake. Curiously, people who suffer from low salt (called hyponatria) do so not because they ingest too little salt, but because they drink too much water. Endurance athletes and others involved in extended strenuous exercise need to watch their water intake so that their body's salt content is not diluted to dangerous levels.
Summary
• The tendency to form species that have eight electrons in the valence shell is called the octet rule.
• The attraction of oppositely charged ions caused by electron transfer is called an ionic bond.
• The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.03%3A_Electron_Transfer_-_Ionic_Bonds.txt |
Learning Objectives
• Define covalent bond.
• Illustrate covalent bond formation with Lewis electron dot diagrams.
Ionic bonding typically occurs when it is easy for one atom to lose one or more electrons, and for another atom to gain one or more electrons. However, some atoms will not give up or gain electrons easily. Yet they still participate in compound formation. How? There is another mechanism for obtaining a complete valence shell: sharing electrons. When electrons are shared between two atoms, they form a covalent bond.
Let us illustrate a covalent bond by using H atoms, with the understanding that H atoms need only two electrons to fill the 1s subshell. Each H atom starts with a single electron in its valence shell:
$\mathbf{H\, \cdot }\; \; \; \; \; \mathbf{\cdot \: H}\nonumber$
The two H atoms can share their electrons:
$\mathbf{H}\: \mathbf{: H}\nonumber$
We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom's valence shell:
Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. (This explains why hydrogen is one of the diatomic elements.) For simplicity's sake, it is not unusual to represent the covalent bond with a dash, instead of with two dots:
H–H
Because two atoms are sharing one pair of electrons, this covalent bond is called a single bond. As another example, consider fluorine. F atoms have seven electrons in their valence shell:
These two atoms can do the same thing that the H atoms did; they share their unpaired electrons to make a covalent bond.
Note that each F atom has a complete octet around it now:
We can also write this using a dash to represent the shared electron pair:
There are two different types of electrons in the fluorine diatomic molecule. The bonding electron pair makes the covalent bond. Each F atom has three other pairs of electrons that do not participate in the bonding; they are called lone pair electrons. Each F atom has one bonding pair and three lone pairs of electrons.
Covalent bonds can be made between different elements as well. One example is HF. Each atom starts out with an odd number of electrons in its valence shell:
The two atoms can share their unpaired electrons to make a covalent bond:
Note that the H atom has a full valence shell with two electrons, while the F atom has a complete octet of electrons.
Example $1$
Use Lewis electron dot diagrams to illustrate the covalent bond formation in HBr.
Solution
HBr is very similar to HF, except that it has Br instead of F. The atoms are as follows:
The two atoms can share their unpaired electron:
Exercise $1$
Use Lewis electron dot diagrams to illustrate the covalent bond formation in Cl2.
Answer
More than two atoms can participate in covalent bonding, although any given covalent bond will be between two atoms only. Consider H and O atoms:
The H and O atoms can share an electron to form a covalent bond:
The H atom has a complete valence shell. However, the O atom has only seven electrons around it, which is not a complete octet. This can be fixed by including a second H atom, whose single electron will make a second covalent bond with the O atom:
(It does not matter what side the second H atom is positioned on.) Now the O atom has a complete octet around it, and each H atom has two electrons, filling its valence shell. This is how a water molecule, H2O, is made.
Example $2$
Use a Lewis electron dot diagram to show the covalent bonding in NH3.
Solution
The N atom has the following Lewis electron dot diagram:
It has three unpaired electrons, each of which can make a covalent bond by sharing electrons with an H atom. The electron dot diagram of NH3 is as follows:
Exercise $2$
Use a Lewis electron dot diagram to show the covalent bonding in PCl3.
Answer
There is a simple set of steps for determining the Lewis electron dot diagram of a simple molecule. First, you must identify the central atom and the surrounding atoms. The central atom is the atom in the center of the molecule, while the surrounding atoms are the atoms making bonds to the central atom. The central atom is usually written first in the formula of the compound (H2O is the notable exception). After the central and surrounding atoms have been identified, follow these steps:
1. Count the total number of valence electrons. Add extra if the species has negative charges and remove some for every positive charge on the species.
2. Write the central atom and surround it with the surrounding atoms.
3. Put a pair of electrons between the central atom and each surrounding atom.
4. Complete the octets around the surrounding atoms (except for H).
5. Put remaining electrons, if any, around the central atom.
6. Check that every atom has a full valence shell.
Let us try these steps to determine the electron dot diagram for BF4. The B atom is the central atom, and the F atoms are the surrounding atoms. There is a negative sign on the species, so we have an extra electron to consider.
1. Count the total number of electrons. B has 3, each F has 7, and there is one extra electron: 3 + 7 + 7 + 7 + 7 + 1 = 32.
2. Write the central atom surrounded by surrounding atoms.
3. Put a pair of electrons between the central atom and each surrounding atom. This uses up eight electrons, so we have 32 − 8 = 24 electrons left.
4. Complete the octets around the surrounding atoms (except for H). This uses up 24 more electrons, leaving 24 − 24 = 0 electrons left.
5. Put remaining electrons, if any, around the central atom. There are no additional electrons to add to the central atom.
6. Check. The B atom has eight electrons around it, as does each F atom. Each atom has a complete octet. This is a good Lewis electron dot diagram for BF4.
Sometimes, however, these steps do not work. If we were to follow these steps for the compound formaldehyde (CH2O), we would get the following:
The H and O atoms have the proper number of electrons, but the C atom only has six electrons around it, not the eight electrons for an octet. How do we fix this?
We fix this by recognizing that two atoms can share more than one pair of electrons. In the case of CH2O, the O and C atoms share two pairs of electrons, with the following Lewis electron dot diagram as a result:
By circling the electrons around each atom, we can now see that the O and C atoms have octets, while each H atom has two electrons:
Each valence shell is full, so this is an acceptable Lewis electron dot diagram. If we were to use lines to represent the bonds, we would use two lines between the C and O atoms:
The bond between the C and O atoms is a double bond and represents two bonding pairs of electrons between the atoms. If the rules for drawing Lewis electron dot diagrams do not work as written, a double bond may be required.
Example $3$: Carbon Dioxide
What is the proper Lewis electron dot diagram for CO2?
Solution
The central atom is a C atom, with O atoms as surrounding atoms. We have a total of 4 + 6 + 6 = 16 valence electrons. Following the rules for Lewis electron dot diagrams for compounds gives us
The O atoms have complete octets around them, but the C atom only has four electrons around it. The way to solve this dilemma is to make a double bond between carbon and each O atom:
Each O atom still has eight electrons around it, but now the C atom also has a complete octet. This is an acceptable Lewis electron dot diagram for CO2.
Exercise $3$
What is the proper Lewis electron dot diagram for carbonyl sulfide (COS)?
Answer
It is also possible to have a triple bond, in which there are three pairs of electrons between two atoms. Good examples of this are elemental nitrogen (N2) and acetylene (C2H2):
Acetylene is an interesting example of a molecule with two central atoms, which are both C atoms. Polyatomic ions are bonded together with covalent bonds. Because they are ions, however, they participate in ionic bonding with other ions. So both major types of bonding can occur at the same time.
Food and Drink Application: Vitamins and Minerals
Vitamins are nutrients that our bodies need in small amounts but cannot synthesize; therefore, they must be obtained from the diet. The word vitamin comes from "vital amine" because it was once thought that all these compounds had an amine group (NH2) in it. This is not actually true, but the name stuck anyway.
All vitamins are covalently bonded molecules. Most of them are commonly named with a letter, although all of them also have formal chemical names. Thus vitamin A is also called retinol, vitamin C is called ascorbic acid, and vitamin E is called tocopherol. There is no single vitamin B; there is a group of substances called the B complex vitamins that are all water soluble and participate in cell metabolism. If a diet is lacking in a vitamin, diseases such as scurvy or rickets develop. Luckily, all vitamins are available as supplements, so any dietary deficiency in a vitamin can be easily corrected.
A mineral is any chemical element other than carbon, hydrogen, oxygen, or nitrogen that is needed by the body. Minerals that the body needs in quantity include sodium, potassium, magnesium, calcium, phosphorus, sulfur, and chlorine. Essential minerals that the body needs in tiny quantities (so-called trace elements) include manganese, iron, cobalt, nickel, copper, zinc, molybdenum, selenium, and iodine. Minerals are also obtained from the diet. Interestingly, most minerals are consumed in ionic form, rather than as elements or from covalent molecules. Like vitamins, most minerals are available in pill form, so any deficiency can be compensated for by taking supplements.
Summary
• Covalent bonds are formed when atoms share electrons.
• Lewis electron dot diagrams can be drawn to illustrate covalent bond formation.
• Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.04%3A_Covalent_Bonds.txt |
Learning Objectives
• Describe a nonpolar bond and a polar bond.
• Use electronegativity to determine whether a bond between two elements will be nonpolar covalent, polar covalent, or ionic.
• Describe the bond energy of a covalent bond.
Consider the H2 molecule:
$\mathbf{H}\: \mathbf{: H}\nonumber$
Because the nuclei of each H atom contain protons, the electrons in the bond are attracted to the nuclei (opposite charges attract). But because the two atoms involved in the covalent bond are both H atoms, each nucleus attracts the electrons by the same amount. Thus the electron pair is equally shared by the two atoms. The equal sharing of electrons in a covalent bond is called a nonpolar covalent bond.
Now consider the $HF$ molecule:
There are two different atoms involved in the covalent bond. The H atom has one proton in its nucleus that is attracting the bonding pair of electrons. However, the F atom has nine protons in its nucleus, with nine times the attraction of the H atom. The F atom attracts the electrons so much more strongly that the electrons remain closer to the F atom than to the H atom; the electrons are no longer equally balanced between the two nuclei. Instead of representing the HF molecule as
it may be more appropriate to draw the covalent bond as
with the electrons in the bond being nearer to the F atom than the H atom. Because the electrons in the bond are nearer to the F atom, this side of the molecule takes on a partial negative charge, which is represented by δ− (δ is the lowercase Greek letter delta). The other side of the molecule, the H atom, adopts a partial positive charge, which is represented by δ+:
A covalent bond between different atoms that attract the shared electrons by different amounts, and cause an imbalance of electron distribution is called a polar covalent bond.
Technically, any covalent bond between two different elements is polar. However, the degree of polarity is important. A covalent bond between two different elements may be so slightly unbalanced that the bond is, essentially, nonpolar. A bond may be so polar that an electron actually transfers from one atom to another, forming a true ionic bond. How do we judge the degree of polarity? Scientists have devised a scale called electronegativity, a scale for judging how strongly atoms of any element attract electrons. Electronegativity is a unitless number; the higher the number, the more an atom attracts electrons. A common scale for electronegativity is shown in Figure $1$.
The polarity of a covalent bond can be judged by determining the difference of the electronegativities of the two atoms involved in the covalent bond, as summarized in the following table:
Table with two columns and four rows. The first column on the left has different values underneath in the row. The second column on the right side has the corresponding bond type for the values underneath in the rows.
Electronegativity Difference Bond Type
0 nonpolar covalent
0–0.4 slightly polar covalent
0.4–1.9 definitely polar covalent
>1.9 likely ionic
Example $1$
What is the polarity of each bond?
1. C–H
2. O–H
Solution
Using Figure $1$, we can calculate the difference of the electronegativities of the atoms involved in the bond.
1. For the C–H bond, the difference in the electronegativities is 2.5 − 2.1 = 0.4. Thus we predict that this bond will be slightly polar covalent.
2. For the O–H bond, the difference in electronegativities is 3.5 − 2.1 = 1.4, so we predict that this bond will be definitely polar covalent.
Exercise $1$
What is the polarity of each bond?
1. Rb–F
2. P–Cl
Answer a
likely ionic
Answer b
polar covalent
The polarity of a covalent bond can have significant influence on the properties of the substance. If the overall molecule is polar, the substance may have a higher melting point and boiling point than expected; also, it may or may not be soluble in various other substances, such as water or hexane.
It should be obvious that covalent bonds are stable because molecules exist. However, they can be broken if enough energy is supplied to a molecule. For most covalent bonds between any two given atoms, a certain amount of energy must be supplied. Although the exact amount of energy depends on the molecule, the approximate amount of energy to be supplied is similar if the atoms in the bond are the same. The approximate amount of energy needed to break a covalent bond is called the bond energy of the covalent bond. Table $1$, lists the bond energies of some covalent bonds.
Table $1$: Bond Energies of Covalent Bonds
Bond Energy (kJ/mol) Bond Energy (kJ/mol)
C–C 348 N–N 163
C=C 611 N=N 418
C≡C 837 N≡N 946
C–O 351 N–H 389
C=O 799 O–O 146
C–Cl 328 O=O 498
C–H 414 O–H 463
F–F 159 S–H 339
H–Cl 431 S=O 523
H–F 569 Si–H 293
H–H 436 Si–O 368
A few trends are obvious from Table $1$. For bonds that involve the same two elements, a double bond is stronger than a single bond, and a triple bond is stronger than a double bond. The energies of multiple bonds are not exact multiples of the single bond energy; for carbon-carbon bonds, the energy increases somewhat less than double or triple the C–C bond energy, while for nitrogen-nitrogen bonds the bond energy increases at a rate greater than the multiple of the N–N single bond energy. The bond energies in Table $1$ are average values; the exact value of the covalent bond energy will vary slightly among molecules with these bonds, but should be close to these values.
To be broken, covalent bonds always require energy; that is, covalent bond breaking is always an endothermic process. Thus the ΔH for this process is positive:
Molecule–O–H → Molecule–O + H
with ΔH ≈ +463 kJ/mol
However, when making a covalent bond, energy is always given off; covalent bond making is always an exothermic process. Thus ΔH for this process is negative:
Molecule–S + H → Molecule–S–H
with ΔH ≈ −339 kJ/mol
Bond energies can be used to estimate the energy change of a chemical reaction. When bonds are broken in the reactants, the energy change for this process is endothermic. When bonds are formed in the products, the energy change for this process is exothermic. We combine the positive energy change with the negative energy change to estimate the overall energy change of the reaction. For example, in
2H2 + O2 → 2H2O
we can draw Lewis electron dot diagrams for each substance to see what bonds are broken and what bonds are formed:
(The lone electron pairs on the O atoms are omitted for clarity.) We are breaking two H–H bonds and one O–O double bond and forming four O–H single bonds. The energy required for breaking the bonds is as follows:
2 H–H bonds: 2(+436 kJ/mol)
1 O=O bond: +498 kJ/mol
Total: +1,370 kJ/mol
The energy given off when the four O–H bonds are made is as follows:
4 O–H bonds: 4(−463 kJ/mol)
Total: −1,852 kJ/mol
Combining these two numbers:
+1,370 kJ/mol + (−1,852 kJ/mol)
Net Change: −482 kJ/mol ≈ ΔH
The actual ΔH is −572 kJ/mol; we are off by about 16%. Although not ideal, a 16% difference is reasonable because we used estimated, not exact, bond energies.
Example $1$
Estimate the energy change of this reaction.
Solution
Here, we are breaking a C–C double bond and an H–H single bond and making a C–C single bond and two C–H single bonds. Bond breaking is endothermic, while bond making is exothermic. For the bond breaking:
1 C=C: +611 kJ/mol
1 H–H: +436 kJ/mol
Total: +1,047 kJ/mol
For the bond making:
1 C–C: −348 kJ/mol
2 C–H: 2(−414 kJ/mol)
Total −1,176 kJ/mol
Overall, the energy change is +1,047 + (−1,176) = −129 kJ/mol.
Exercise $1$
Estimate the energy change of this reaction.
Summary
• Covalent bonds can be nonpolar or polar, depending on the electronegativities of the atoms involved.
• Covalent bonds can be broken if energy is added to a molecule.
• The formation of covalent bonds is accompanied by energy given off.
• Covalent bond energies can be used to estimate the enthalpy changes of chemical reactions. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.05%3A_Other_Aspects_of_Covalent_Bonds.txt |
Learning Objective
• Recognize the three major types of violations of the octet rule.
As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations.
There are three violations to the octet rule. Odd-electron molecules are the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are \(\ce{NO}\), \(\ce{NO2}\), and \(\ce{ClO2}\). The Lewis electron dot diagram for \(\ce{NO}\) is as follows:
Although the \(\ce{O}\) atom has an octet of electrons, the \(\ce{N}\) atom has only seven electrons in its valence shell. Although \(\ce{NO}\) is a stable compound, it is very chemically reactive, as are most other odd-electron compounds.
Electron-deficient molecules are the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell:
Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the \(\ce{B}\) atom. A well-known example is \(\ce{BF3}\):
The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is \(\ce{PF5}\). The only reasonable Lewis electron dot diagram for this compound has the \(\ce{P}\) atom making five covalent bonds:
Formally, the \(\ce{P}\) atom has 10 electrons in its valence shell.
Example \(1\)
Identify each violation to the octet rule by drawing a Lewis electron dot diagram.
1. \(\ce{ClO}\)
2. \(\ce{SF6}\)
Solution
1. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows:
1. In \(\ce{SF6}\), the central \(\ce{S}\) atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows:
Exercise \(1\): Xenon Difluoride
Identify the violation to the octet rule in \(\ce{XeF2}\) by drawing a Lewis electron dot diagram.
Answer
The Xe atom has an expanded valence shell with more than eight electrons around it.
.
Summary
There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules.
9.07: Molecular Shapes
Learning Objective
• Determine the shape of simple molecules.
Molecules have shapes. There is an abundance of experimental evidence to that effect, from their physical properties to their chemical reactivity. Small molecules—molecules with a single central atom—have shapes that can be easily predicted. The basis of molecular shapes is called valence shell electron pair repulsion (VSEPR). VSEPR says that electron pairs, being composed of negatively charged particles, repel each other to get as far away from each other as possible. VSEPR makes a distinction between electron group geometry, which expresses how electron groups (bonds and nonbonding electron pairs) are arranged, and molecular geometry, which expresses how the atoms in a molecule are arranged. However, the two geometries are related.
There are two types of electron groups: any type of bond—single, double, or triple—and lone electron pairs. When applying VSEPR to simple molecules, the first thing to do is to count the number of electron groups around the central atom. Remember that a multiple bond counts as only one electron group.
Any molecule with only two atoms is linear. A molecule whose central atom contains only two electron groups orients those two groups as far apart from each other as possible—180° apart. When the two electron groups are 180° apart, the atoms attached to those electron groups are also 180° apart, so the overall molecular shape is linear. Examples include BeH2 and CO2:
A molecule with three electron groups orients the three groups as far apart as possible. They adopt the positions of an equilateral triangle—120° apart and in a plane. The shape of such molecules is trigonal planar. An example is BF3:
Some substances have a trigonal planar electron group distribution, but have atoms bonded to only two of the three electron groups. An example is GeF2:
From an electron group geometry perspective, GeF2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. This shape is called bent or angular.
A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron, as shown in Figure \(1\): Tetrahedral Geometry. If there are four atoms attached to these electron groups, then the molecular shape is also tetrahedral. Methane (CH4) is an example.
This diagram of CH4 illustrates the standard convention of displaying a three-dimensional molecule on a two-dimensional surface. The straight lines are in the plane of the page, the solid wedged line is coming out of the plane toward the reader, and the dashed wedged line is going out of the plane away from the reader.
NH3 is an example of a molecule whose central atom has four electron groups, but only three of them are bonded to surrounding atoms.
Although the electron groups are oriented in the shape of a tetrahedron, from a molecular geometry perspective, the shape of NH3 is trigonal pyramidal.
H2O is an example of a molecule whose central atom has four electron groups, but only two of them are bonded to surrounding atoms.
Although the electron groups are oriented in the shape of a tetrahedron, the shape of the molecule is bent or angular. A molecule with four electron groups about the central atom, but only one electron group bonded to another atom is linear because there are only two atoms in the molecule.
Double or triple bonds count as a single electron group. CH2O has the following Lewis electron dot diagram.
The central C atom has three electron groups around it because the double bond counts as one electron group. The three electron groups repel each other to adopt a trigonal planar shape:
(The lone electron pairs on the O atom are omitted for clarity.) The molecule will not be a perfect equilateral triangle because the C–O double bond is different from the two C–H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule.
Example \(1\)
What is the approximate shape of each molecule?
1. \(\ce{PCl3}\)
2. \(\ce{NOF}\)
Solution
The first step is to draw the Lewis electron dot diagram of the molecule.
a:
For PCl3, the electron dot diagram is as follows:
The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal.
b:
The electron dot diagram for NOF is as follows:
The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent.
Exercise \(1\)
What is the approximate molecular shape of \(\ce{CH2Cl2}\)?
Answer
Tetrahedral
Table \(1\) summarizes the shapes of molecules based on their number of electron groups and surrounding atoms.
Table \(1\): Summary of Molecular Shapes
Number of Electron Groups on Central Atom Number of Surrounding Atoms Molecular Shape
any 1 linear
2 2 linear
3 3 trigonal planar
3 2 bent
4 4 tetrahedral
4 3 trigonal pyramidal
4 2 bent
Summary
The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.06%3A_Violations_of_the_Octet_Rule.txt |
9.2: Lewis Electron Dot Diagrams
1. Explain why the first two dots in a Lewis electron dot diagram are drawn on the same side of the atomic symbol.
2. Is it necessary for the first dot around an atomic symbol to go on a particular side of the atomic symbol?
3. What column of the periodic table has Lewis electron dot diagrams with two electrons?
4. What column of the periodic table has Lewis electron dot diagrams that have six electrons in them?
5. Draw the Lewis electron dot diagram for each element.
1. strontium
2. silicon
6. Draw the Lewis electron dot diagram for each element.
1. krypton
2. sulfur
7. Draw the Lewis electron dot diagram for each element.
1. titanium
2. phosphorus
8. Draw the Lewis electron dot diagram for each element.
1. bromine
2. gallium
9. Draw the Lewis electron dot diagram for each ion.
1. Mg2+
2. S2−
10. Draw the Lewis electron dot diagram for each ion.
1. In+
2. Br
11. Draw the Lewis electron dot diagram for each ion.
1. Fe2+
2. N3−
12. Draw the Lewis electron dot diagram for each ion.
1. H+
2. H
Answers
1. The first two electrons in a valence shell are s electrons, which are paired.
2. the second column of the periodic table
1. Mg2+
1. Fe2+
9.3: Electron Transfer - Ionic Bonds
1. Comment on the possible formation of the K2+ ion. Why is its formation unlikely?
2. Comment on the possible formation of the Cl2− ion. Why is its formation unlikely?
3. How many electrons does a Ba atom have to lose to have a complete octet in its valence shell?
4. How many electrons does a Pb atom have to lose to have a complete octet in its valence shell?
5. How many electrons does an Se atom have to gain to have a complete octet in its valence shell?
6. How many electrons does an N atom have to gain to have a complete octet in its valence shell?
7. With arrows, illustrate the transfer of electrons to form potassium chloride from K atoms and Cl atoms.
8. With arrows, illustrate the transfer of electrons to form magnesium sulfide from Mg atoms and S atoms.
9. With arrows, illustrate the transfer of electrons to form scandium fluoride from Sc atoms and F atoms.
10. With arrows, illustrate the transfer of electrons to form rubidium phosphide from Rb atoms and P atoms.
11. Which ionic compound has the higher lattice energy—KI or MgO? Why?
12. Which ionic compound has the higher lattice energy—KI or LiF? Why?
1. Which ionic compound has the higher lattice energy—BaS or MgO? Why?
Answers
1. The K2+ ion is unlikely to form because the K+ ion already satisfies the octet rule and is rather stable.
2. two
3. two
4. MgO because the ions have a higher magnitude charge
5. MgO because the ions are smaller
9.4: Covalent Bonds
1. How many electrons will be in the valence shell of H atoms when it makes a covalent bond?
2. How many electrons will be in the valence shell of non-H atoms when they make covalent bonds?
3. What is the Lewis electron dot diagram of I2? Circle the electrons around each atom to verify that each valence shell is filled.
4. What is the Lewis electron dot diagram of H2S? Circle the electrons around each atom to verify that each valence shell is filled.
5. What is the Lewis electron dot diagram of NCl3? Circle the electrons around each atom to verify that each valence shell is filled.
6. What is the Lewis electron dot diagram of SiF4? Circle the electrons around each atom to verify that each valence shell is filled.
7. Draw the Lewis electron dot diagram for each substance.
1. SF2
2. BH4
8. Draw the Lewis electron dot diagram for each substance.
1. PI3
2. OH
9. Draw the Lewis electron dot diagram for each substance.
1. GeH4
2. ClF
10. Draw the Lewis electron dot diagram for each substance.
1. AsF3
2. NH4+
11. Draw the Lewis electron dot diagram for each substance. Double or triple bonds may be needed.
1. SiO2
2. C2H4 (assume two central atoms)
12. Draw the Lewis electron dot diagram for each substance. Double or triple bonds may be needed.
1. CN
2. C2Cl2 (assume two central atoms)
13. Draw the Lewis electron dot diagram for each substance. Double or triple bonds may be needed.
1. CS2
2. NH2CONH2 (assume that the N and C atoms are the central atoms)
14. Draw the Lewis electron dot diagram for each substance. Double or triple bonds may be needed.
1. POCl
2. HCOOH (assume that the C atom and one O atom are the central atoms)
Answers
1. two
9.5: Other Aspects of Covalent Bonds
1. Give an example of a nonpolar covalent bond. How do you know it is nonpolar?
2. Give an example of a polar covalent bond. How do you know it is polar?
3. How do you know which side of a polar bond has the partial negative charge? Identify the negatively charged side of each polar bond.
1. H–Cl
2. H–S
4. How do you know which side of a polar bond has the partial positive charge? Identify the positively charged side of each polar bond.
1. H–Cl
2. N-F
5. Label the bond between the given atoms as nonpolar covalent, slightly polar covalent, definitely polar covalent, or likely ionic.
1. H and C
2. C and F
3. K and F
6. Label the bond between the given atoms as nonpolar covalent, slightly polar covalent, definitely polar covalent, or likely ionic.
1. S and Cl
2. P and O
3. Cs and O
7. Which covalent bond is stronger—a C–C bond or a C–H bond?
8. Which covalent bond is stronger—an O–O double bond or an N–N double bond?
9. Estimate the enthalpy change for this reaction: N2 + 3H2 → 2NH3 .Start by drawing the Lewis electron dot diagrams for each substance.
10. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance: HN=NH + 2H2 → 2NH3
11. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance: CH4 + 2O2 → CO2 + 2H2O
12. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance: 4NH3 + 3O2 → 2N2 + 6H2O
Answers
1. H–H; it is nonpolar because the two atoms have the same electronegativities (answers will vary).
1. Cl side
2. S side
1. slightly polar covalent
2. definitely polar covalent
3. likely ionic
2. C–H bond
3. −80 kJ
4. −798 kJ
9.6: Violations of the Octet Rule
1. Why can an odd-electron molecule not satisfy the octet rule?
2. Why can an atom in the second row of the periodic table not form expanded valence shell molecules?
3. Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule.
1. NO2
2. XeF4
4. Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule.
1. BCl3
2. ClO2
5. Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule.
1. POF3
2. ClF3
6. Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule.
1. SF4
2. BeH2
Answers
1. There is no way all electrons can be paired if there are an odd number of them.
9.7: Molecular Shapes
1. What is the basic premise behind VSEPR?
2. What is the difference between the electron group geometry and the molecular geometry?
3. Identify the electron group geometry and the molecular geometry of each molecule.
1. H2S
2. POCl3
4. Identify the electron group geometry and the molecular geometry of each molecule.
1. CS2
2. H2S
5. Identify the electron group geometry and the molecular geometry of each molecule.
1. HCN
2. CCl4
6. Identify the electron group geometry and the molecular geometry of each molecule.
1. BI3
2. PH3
7. What is the geometry of each species?
1. CN
2. PO43−
8. What is the geometry of each species?
1. PO33−
2. NO3
9. What is the geometry of each species?
1. COF2
2. C2Cl2 (both C atoms are central atoms and are bonded to each other)
10. What is the geometry of each species?
1. CO32−
2. N2H4 (both N atoms are central atoms and are bonded to each other)
Answers
1. Electron pairs repel each other.
1. electron group geometry: tetrahedral; molecular geometry: bent
2. electron group geometry: tetrahedral; molecular geometry: tetrahedral
1. electron group geometry: linear; molecular geometry: linear
2. electron group geometry: tetrahedral; molecular geometry: tetrahedral
1. linear
2. tetrahedral
1. trigonal planar
2. linear and linear about each central atom
9.8 Additional Exercises
1. Explain why iron and copper have the same Lewis electron dot diagram when they have different numbers of electrons.
2. Name two ions with the same Lewis electron dot diagram as the Cl ion.
3. Based on the known trends, what ionic compound from the first column of the periodic table and the next-to-last column of the periodic table should have the highest lattice energy?
4. Based on the known trends, what ionic compound from the first column of the periodic table and the next-to-last column of the periodic table should have the lowest lattice energy?
5. P2 is not a stable form of phosphorus, but if it were, what would be its likely Lewis electron dot diagram?
6. Se2 is not a stable form of selenium, but if it were, what would be its likely Lewis electron dot diagram?
7. What are the Lewis electron dot diagrams of SO2, SO3, and SO42−?
8. What are the Lewis electron dot diagrams of PO33− and PO43−?
9. Which bond do you expect to be more polar—an O–H bond or an N–H bond?
10. Which bond do you expect to be more polar—an O–F bond or an S–O bond?
11. Use bond energies to estimate the energy change of this reaction.
C3H8 + 5O2 → 3CO2 + 4H2O
12. Use bond energies to estimate the energy change of this reaction.
N2H4 + O2 → N2 + 2H2O
13. Ethylene (C2H4) has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule.
14. Hydrogen peroxide (H2O2) has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule.
Answers
1. Iron has d electrons that typically are not shown on Lewis electron dot diagrams.
2. LiF
3. It would be like N2:
4. an O–H bond
5. −2,000 kJ
6. trigonal planar about both central C atoms | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/09%3A_Chemical_Bonds/9.E%3A_Chemical_Bonds_%28Exercises%29.txt |
Chapter 6 discussed the properties of gases. In this chapter, properties of liquids and solids are considered. As a review, the Table below lists some general properties of the three phases of matter.
Phase Shape Density Compressibility
Gas fills entire container low high
Liquid fills a container from bottom to top high low
Solid rigid high low
• 10.1: Prelude to Solids and Liquids
Liquids flow when a small force is placed on them, even if only very slowly. Solids may deform under a small force, but they return to their original shape when the force is relaxed. This is how glass behaves: it goes back to its original shape (unless it breaks under the applied force). Observers also point out that telescopes with glass lenses to focus light still do so even decades after manufacture—a circumstance that would not be so if the lens were liquid and flowed.
• 10.2: Intermolecular Forces
All substances experience dispersion forces between their particles. Substances that are polar experience dipole-dipole interactions. Substances with covalent bonds between an H atom and N, O, or F atoms experience hydrogen bonding. The preferred phase of a substance depends on the strength of the intermolecular force and the energy of the particles.
• 10.3: Phase Transitions - Melting, Boiling, and Subliming
Phase changes can occur between any two phases of matter. All phase changes occur with a simultaneous change in energy. All phase changes are isothermal.
• 10.4: Properties of Liquids
All liquids evaporate. If volume is limited, evaporation eventually reaches a dynamic equilibrium, and a constant vapor pressure is maintained. All liquids experience surface tension, an imbalance of forces at the surface of the liquid. All liquids experience capillary action, demonstrating either capillary rise or capillary depression in the presence of other substances.
• 10.5: Solids
Solids can be divided into amorphous solids and crystalline solids. Crystalline solids can be ionic, molecular, covalent network, or metallic.
• 10.E: Solids and Liquids (Exercises)
These are exercises and select solutions to accompany Chapter 10 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
10: Solids and Liquids
There is an urban legend that glass is an extremely thick liquid rather than a solid, even at room temperature. Proponents claim that old windows are thicker at the bottom than at the top, suggesting that the glass flowed down over time. Unfortunately, the proponents of this idea have no credible evidence that this is true, as old windows were likely not subject to the stricter manufacturing standards that exist today. Also, when mounting a piece of glass that has an obviously variable thickness, it makes structural sense to put the thicker part at the bottom, where it will support the object better.
Liquids flow when a small force is placed on them, even if only very slowly. Solids may deform under a small force, but they return to their original shape when the force is relaxed. This is how glass behaves: it goes back to its original shape (unless it breaks under the applied force). Observers also point out that telescopes with glass lenses to focus light still do so even decades after manufacture—a circumstance that would not be so if the lens were liquid and flowed. Thus, glass is a solid at room temperature. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/10%3A_Solids_and_Liquids/10.01%3A_Prelude_to_Solids_and_Liquids.txt |
Learning Objective
• Relate phase to intermolecular forces.
Why does a substance have the phase that it does? The preferred phase of a substance at a given set of conditions is a balance between the energy of the particles and intermolecular forces (or intermolecular interactions) between the particles. If the forces between particles are strong enough, the substance is a liquid or, if stronger, a solid. If the forces between particles are weak and sufficient energy is present, the particles separate from each other, so the gas phase is the preferred phase. The energy of the particles is mostly determined by temperature, so temperature is the main variable that determines what phase is stable at any given point.
What forces define intermolecular interactions? There are several. A force present in all substances with electrons is the dispersion force (sometimes called the London dispersion force, after the physicist Fritz London, who first described this force in the early 1900s). This interaction is caused by the instantaneous position of an electron in a molecule, which temporarily makes that point of the molecule negatively charged and the rest of the molecule positively charged. In an instant, the electron is now somewhere else, but the fleeting imbalance of electric charge in the molecule allows molecules to interact with each other. As you might expect, the greater the number of electrons in a species, the stronger the dispersion force; this partially explains why smaller molecules are gases and larger molecules are liquids and solids at the same temperature. (Mass is a factor as well.)
Molecules with a permanent dipole moment experience dipole-dipole interactions, which are generally stronger than dispersion forces if all other things are equal. The oppositely charged ends of a polar molecule, which have partial charges on them, attract each other (Figure \(1\)). Thus, a polar molecule such as CH2Cl2 has a significantly higher boiling point (313 K, or 40°C) than a nonpolar molecule like CF4 (145 K, or −128°C), even though it has a lower molar mass (85 g/mol vs. 88 g/mol).
An unusually strong form of dipole-dipole interaction is called hydrogen bonding. Hydrogen bonding is found in molecules with an H atom bonded to an N atom, an O atom, or an F atom. Such covalent bonds are very polar, and the dipole-dipole interaction between these bonds in two or more molecules is strong enough to create a new category of intermolecular force. Hydrogen bonding is the reason water has unusual properties. For such a small molecule (its molar mass is only 18 g/mol), H2O has relatively high melting and boiling points. Its boiling point is 373 K (100°C), while the boiling point of a similar molecule, H2S, is 233 K (−60°C). This is because H2O molecules experience hydrogen bonding, while H2S molecules do not. This strong attraction between H2O molecules requires additional energy to separate the molecules in the condensed phase, so its boiling point is higher than would be expected. Hydrogen bonding is also responsible for water's ability as a solvent, its high heat capacity, and its ability to expand when freezing; the molecules line up in such a way that there is extra space between the molecules, increasing its volume in the solid state (Figure \(2\)).
Example \(1\)
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
1. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
2. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
3. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise \(1\)
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answers
1. hydrogen bonding
2. dipole-dipole interactions
The preferred phase a substance adopts can change with temperature. At low temperatures, most substances are solids (only helium is predicted to be a liquid at absolute zero). As the temperature increases, those substances with very weak intermolecular forces become gases directly (in a process called sublimation, which will be discussed in Section 10.3). Substances with weak interactions can become liquids as the temperature increases. As the temperature increases even more, the individual particles will have so much energy that the intermolecular forces are overcome, so the particles separate from each other, and the substance becomes a gas (assuming that their chemical bonds are not so weak that the compound decomposes from the high temperature). Although it is difficult to predict the temperature ranges for which solid, liquid, or gas is the preferred phase for any random substance, all substances progress from solid to liquid to gas (in that order) as temperature increases.
Summary
• All substances experience dispersion forces between their particles.
• Substances that are polar experience dipole-dipole interactions.
• Substances with covalent bonds between an H atom and N, O, or F atoms experience hydrogen bonding.
• The preferred phase of a substance depends on the strength of the intermolecular force and the energy of the particles. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/10%3A_Solids_and_Liquids/10.02%3A_Intermolecular_Forces.txt |
Learning Objective
• Describe what happens during a phase change.
• Calculate the energy change needed for a phase change.
Substances can change phase—often because of a temperature change. At low temperatures, most substances are solid; as the temperature increases, they become liquid; at higher temperatures still, they become gaseous.
The process of a solid becoming a liquid is called melting (an older term that you may see sometimes is fusion). The opposite process, a liquid becoming a solid, is called solidification. For any pure substance, the temperature at which melting occurs—known as the melting point—is a characteristic of that substance. It requires energy for a solid to melt into a liquid. Every pure substance has a certain amount of energy it needs to change from a solid to a liquid. This amount is called the enthalpy of fusion (or heat of fusion) of the substance, represented as ΔHfus. Some ΔHfus values are listed in Table $1$; it is assumed that these values are for the melting point of the substance. Note that the unit of ΔHfus is kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The ΔHfus is always tabulated as a positive number. However, it can be used for both the melting and the solidification processes, minding that melting is always endothermic (so ΔH will be positive), while solidification is always exothermic (so ΔH will be negative).
Table $1$: Enthalpies of Fusion for Various Substances
Substance (Melting Point) ΔHfus (kJ/mol)
Water (0°C) 6.01
Aluminum (660°C) 10.7
Benzene (5.5°C) 9.95
Ethanol (−114.3°C) 5.02
Mercury (−38.8°C) 2.29
Example $1$
What is the energy change when 45.7 g of $\ce{H2O}$ melt at 0°C?
Solution
The $ΔH_{fus}$ of $\ce{H2O}$ is 6.01 kJ/mol. However, our quantity is given in units of grams, not moles, so the first step is to convert grams to moles using the molar mass of $\ce{H_2O}$, which is 18.0 g/mol. Then we can use $ΔH_{fus}$ as a conversion factor. Because the substance is melting, the process is endothermic, so the energy change will have a positive sign.
$45.7\cancel{g\: H_{2}O}\times \frac{1\cancel{mol\: H_{2}O}}{18.0\cancel{g}}\times \frac{6.01kJ}{\cancel{mol}}=15.3\,kJ \nonumber \nonumber$
Without a sign, the number is assumed to be positive.
Exercise $1$
What is the energy change when 108 g of $\ce{C6H6}$ freeze at 5.5°C?
Answer
−13.8 kJ
During melting, energy goes exclusively to changing the phase of a substance; it does not go into changing the temperature of a substance. Hence melting is an isothermal process because a substance stays at the same temperature. Only when all of a substance is melted does additional energy go to changing its temperature.
What happens when a solid becomes a liquid? In a solid, individual particles are stuck in place because the intermolecular forces cannot be overcome by the energy of the particles. When more energy is supplied (e.g., by raising the temperature), there comes a point at which the particles have enough energy to move around, but not enough energy to separate. This is the liquid phase: particles are still in contact, but are able to move around each other. This explains why liquids can assume the shape of their containers: the particles move around and, under the influence of gravity, fill the lowest volume possible (unless the liquid is in a zero-gravity environment—see Figure $1$.
The phase change between a liquid and a gas has some similarities to the phase change between a solid and a liquid. At a certain temperature, the particles in a liquid have enough energy to become a gas. The process of a liquid becoming a gas is called boiling (or vaporization), while the process of a gas becoming a liquid is called condensation. However, unlike the solid/liquid conversion process, the liquid/gas conversion process is noticeably affected by the surrounding pressure on the liquid because gases are strongly affected by pressure. This means that the temperature at which a liquid becomes a gas, the boiling point, can change with surrounding pressure. Therefore, we define the normal boiling point as the temperature at which a liquid changes to a gas when the surrounding pressure is exactly 1 atm, or 760 torr. Unless otherwise specified, it is assumed that a boiling point is for 1 atm of pressure.
Like the solid/liquid phase change, the liquid/gas phase change involves energy. The amount of energy required to convert a liquid to a gas is called the enthalpy of vaporization (or heat of vaporization), represented as ΔHvap. Some ΔHvap values are listed in Table $2$; it is assumed that these values are for the normal boiling point temperature of the substance, which is also given in the table. The unit for ΔHvap is also kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The ΔHvap is also always tabulated as a positive number. It can be used for both the boiling and the condensation processes as long as you keep in mind that boiling is always endothermic (so ΔH will be positive), while condensation is always exothermic (so ΔH will be negative).
Table $1$: Enthalpies of Vaporization for Various Substances
Substance (Normal Boiling Point) ΔHvap (kJ/mol)
Water (100°C) 40.68
Bromine (59.5°C) 15.4
Benzene (80.1°C) 30.8
Ethanol (78.3°C) 38.6
Mercury (357°C) 59.23
Example $2$
What is the energy change when 66.7 g of Br2(g) condense to a liquid at 59.5°C?
Solution
The ΔHvap of Br2 is 15.4 kJ/mol. Even though this is a condensation process, we can still use the numerical value of ΔHvap as long as we realize that we must take energy out, so the ΔH value will be negative. To determine the magnitude of the energy change, we must first convert the amount of Br2 to moles. Then we can use ΔHvap as a conversion factor.
$66.7\cancel{g\: Br_{2}}\times \frac{1\cancel{mol\: Br_{2}}}{159.8\cancel{g}}\times \frac{15.4kJ}{\cancel{mol}}=6.43\,kJ \nonumber \nonumber$
Because the process is exothermic, the actual value will be negative: ΔH = −6.43 kJ.
Exercise $2$
What is the energy change when 822 g of $\ce{C2H5OH(ℓ)}$ boil at its normal boiling point of 78.3°C?
Answer
689 kJ
As with melting, the energy in boiling goes exclusively to changing the phase of a substance; it does not go into changing the temperature of a substance. So boiling is also an isothermal process. Only when all of a substance has boiled does any additional energy go to changing its temperature.
What happens when a liquid becomes a gas? We have already established that a liquid is composed of particles in contact with each other. When a liquid becomes a gas, the particles separate from each other, with each particle going its own way in space. This is how gases tend to fill their containers. Indeed, in the gas phase most of the volume is empty space; only about 1/1,000th of the volume is actually taken up by matter (Figure $1$). It is this property of gases that explains why they can be compressed, a fact that is considered in Chapter 6.
Under some circumstances, the solid phase can transition directly to the gas phase without going through a liquid phase, and a gas can directly become a solid. The solid-to-gas change is called sublimation, while the reverse process is called deposition. Sublimation is isothermal, like the other phase changes. There is a measurable energy change during sublimation—this energy change is called the enthalpy of sublimation, represented as ΔHsub. The relationship between the ΔHsub and the other enthalpy changes is as follows:
$ΔH_{sub} = ΔH_{fus} + ΔH_{vap}\nonumber$
As such, ΔHsub is not always tabulated because it can be simply calculated from ΔHfus and ΔHvap.
There are several common examples of sublimation. A well-known product, dry ice, is actually solid CO2. Dry ice is dry because it sublimes, with the solid bypassing the liquid phase and going straight to the gas phase. The sublimation occurs at temperature of −77°C, so it must be handled with caution. If you have ever noticed that ice cubes in a freezer tend to get smaller over time, it is because the solid water is very slowly subliming. "Freezer burn" isn't actually a burn; it occurs when certain foods, such as meats, slowly lose solid water content because of sublimation. The food is still good, but looks unappetizing. Reducing the temperature of a freezer will slow the sublimation of solid water.
Chemical equations can be used to represent a phase change. In such cases, it is crucial to use phase labels on the substances. For example, the chemical equation for the melting of ice to make liquid water is as follows:
$H_2O(s) → H_2O(ℓ)\nonumber$
No chemical change is taking place; however, a physical change is taking place.
Summary
• Phase changes can occur between any two phases of matter.
• All phase changes occur with a simultaneous change in energy.
• All phase changes are isothermal. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/10%3A_Solids_and_Liquids/10.03%3A_Phase_Transitions_-_Melting_Boiling_and_Subliming.txt |
Learning Objective
• Define the vapor pressure of liquids.
• Explain the origin of both surface tension and capillary action.
There are some properties that all liquids, including water, have. All liquids have a certain portion of particles with enough energy to enter the gas phase, and if these particles are at the surface of the liquid, they do so (Figure \(1\)). The formation of a gas from a liquid at temperatures below the boiling point is called evaporation. At these temperatures, the material in the gas phase is called vapor, rather than gas; the term gas is reserved for when the gas phase is the stable phase.
If the available volume is large enough, eventually all the liquid will become vapor. But if the available volume is not enough, eventually some of the vapor particles will reenter the liquid phase (Figure \(2\) Equilibrium). At some point, the number of particles entering the vapor phase will equal the number of particles leaving the vapor phase, so there is no net change in the amount of vapor in the system. We say that the system is at equilibrium. The partial pressure of the vapor at equilibrium is called the vapor pressure of the liquid.
Understand that the liquid has not stopped evaporating. The reverse process—condensation—is occurring as fast as evaporation, so there is no net change in the amount of vapor in the system. The term dynamic equilibrium represents a situation in which a process still occurs, but the opposite process also occurs at the same rate, so that there is no net change in the system.
The vapor pressure for a substance is dependent on the temperature of the substance; as the temperature increases, so does the vapor pressure. Figure \(3\) - Plots of Vapor Pressure versus Temperature for Several Liquids, is a plot of vapor pressure versus temperature for several liquids. Having defined vapor pressure, we can also redefine the boiling point of a liquid: the temperature at which the vapor pressure of a liquid equals the surrounding environmental pressure. The normal vapor pressure, then, is the temperature at which the vapor pressure is 760 torr, or exactly 1 atm. Thus boiling points vary with surrounding pressure, a fact that can have large implications on cooking foods at lower- or higher-than-normal elevations. Atmospheric pressure varies significantly with altitude.
Example \(1\)
Use Figure \(3\) to estimate the boiling point of water at 500 torr, which is the approximate atmospheric pressure at the top of Mount Everest.
Solution
See the accompanying figure. Five hundred torr is between 400 and 600, so we extend a line from that point on the y-axis across to the curve for water and then drop it down to the x-axis to read the associated temperature. It looks like the point on the water vapor pressure curve corresponds to a temperature of about 90°C, so we conclude that the boiling point of water at 500 torr is 90°C.
By reading the graph properly, you can estimate the boiling point of a liquid at different temperatures.
Exercise \(1\)
Use Figure \(3\) to estimate the boiling point of ethanol at 400 torr.
Answer
about 65°C
The vapor pressure curve for water is not exactly zero at the melting point 0°C. Even ice has a vapor pressure—that is why it sublimes over time. However, the vapor pressures of solids are typically much lower than that of liquids. At −1°C, the vapor pressure of ice is 4.2 torr. At a freezer temperature of 0°F (−17°C), the vapor pressure of ice is only 1.0 torr; so-called deep freezers can get down to −23°C, where the vapor pressure of ice is only 0.6 torr.
All liquids share some other properties as well. Surface tension is an effect caused by an imbalance of forces on the atoms at the surface of a liquid, as shown in Figure \(4\). The blue particle in the bulk of the liquid experiences intermolecular forces from all around, as illustrated by the arrows. However, the yellow particle on the surface does not experience any forces above it because there are no particles above it. This leads to an imbalance of forces, called surface tension.
Surface tension is responsible for several well-known behaviors of liquids, including water. Liquids with high surface tension tend to bead up when present in small amounts (Figure \(5\)).
Surface tension causes liquids to form spheres in free fall or zero gravity. Surface tension is also responsible for the fact that small insects can "walk" on water. Because of surface tension, it takes energy to break the surface of a liquid, and if an object (such as an insect) is light enough, there is not enough force due to gravity for the object to break through the surface, so the object stays on top of the water (Figure \(6\)). Carefully done, this phenomenon can also be illustrated with a thin razor blade or a paper clip.
The fact that small droplets of water bead up on surfaces does not mean that water—or any other liquid—does not interact with other substances. Sometimes the attraction can be very strong. Adhesion is the tendency of a substance to interact with other substances because of intermolecular forces, while cohesion is the tendency of a substance to interact with itself. If cohesive forces within a liquid are stronger than adhesive forces between a liquid and another substance, then the liquid tends to keep to itself; it will bead up. However, if adhesive forces between a liquid and another substance are stronger than cohesive forces, then the liquid will spread out over the other substance, trying to maximize the interface between the other substance and the liquid. It is said that the liquid wets the other substance.
Adhesion and cohesion are important for other phenomena as well. In particular, if adhesive forces are strong, then when a liquid is introduced to a small-diameter tube of another substance, the liquid will move up or down in the tube, as if ignoring gravity. Because tiny tubes are called capillaries, this phenomenon is called capillary action. For example, one type of capillary action, capillary rise, is seen when water or water-based liquids rise up in thin glass tubes (like the capillaries sometimes used in blood tests), forming an upwardly curved surface called a meniscus. Capillary action is also responsible for the "wicking" effect that towels and sponges use to dry wet objects; the matting of fibers forms tiny capillaries that have good adhesion with water. Cotton is a good material for this; polyester and other synthetic fabrics do not display similar capillary action, which is why you seldom find rayon bath towels. A similar effect is observed with liquid fuels or melted wax and their wicks. Capillary action is thought to be at least partially responsible for transporting water from the roots to the tops of trees, even tall ones.
On the other hand, some liquids have stronger cohesive forces than adhesive forces. In this case, in the presence of a capillary, the liquid is forced down from its surface; this is an example of a type of capillary action called capillary depression. In this case, the meniscus curves downward. Mercury has very strong cohesive forces; when a capillary is placed in a pool of mercury, the surface of the mercury liquid is depressed (Figure \(7\)).
Chemistry is Everywhere: Waxing a Car
Responsible car owners are encouraged to wax their cars regularly. In addition to making the car look nicer, it also helps protect the surface, especially if the surface is metal. Why?
The answer has to do with cohesion and adhesion (and, to a lesser extent, rust). Water is an important factor in the rusting of iron, sometimes used extensively in outer car bodies. Keeping water away from the metal is one way to minimize rusting. A coat of paint helps with this. However, dirty or scratched paint can attract water, and adhesive forces will allow the water to wet the surface, maximizing its contact with the metal and promoting rust.
Wax is composed of long hydrocarbon molecules that do not interact well with water. (Hydrocarbons are compounds with C and H atoms; for more information on hydrocarbons, see Chapter 16). That is, a thin layer of wax will not be wetted by water. A freshly waxed car has low adhesive forces with water, so water beads up on the surface, as a consequence of its cohesion and surface tension. This minimizes the contact between water and metal, thus minimizing rust.
Summary
• All liquids evaporate.
• If volume is limited, evaporation eventually reaches a dynamic equilibrium, and a constant vapor pressure is maintained.
• All liquids experience surface tension, an imbalance of forces at the surface of the liquid.
• All liquids experience capillary action, demonstrating either capillary rise or capillary depression in the presence of other substances. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/10%3A_Solids_and_Liquids/10.04%3A_Properties_of_Liquids.txt |
Learning Objective
• Describe the general properties of a solid.
• Describe the six different types of solids.
A solid is like a liquid in that particles are in contact with one another. Solids are unlike liquids in that the intermolecular forces are strong enough to hold the particles in place. At low enough temperatures, all substances are solids (helium is the lone exception), but the temperature at which the solid state becomes the stable phase varies widely among substances, from 20 K (−253°C) for hydrogen to over 3,900 K (3,600°C) for carbon.
The solid phase has several characteristics. First, solids maintain their shape. They do not fill their entire containers like gases do, and they do not adopt the shape of their containers like liquids do. They cannot be easily compressed like gases can, and they have relatively high densities.
Solids may also demonstrate a variety of properties. For example, many metals can be beaten into thin sheets or drawn into wires, while compounds such as NaCl will shatter if they are struck. Some metals, such as sodium and potassium, are rather soft, while others, such as diamond, are very hard and can easily scratch other substances. Appearances differ as well: most metals are shiny and silvery, but sulfur (a nonmetal) is yellow, and ionic compounds can take on a rainbow of colors. Solid metals conduct electricity and heat, while ionic solids do not. Many solids are opaque, but some are transparent. Some dissolve in water, but some do not. Figure \(1\), shows two solids that exemplify the similar and dissimilar properties of solids.
Solids can have a wide variety of physical properties. We will review the different types of solids and the bonding that gives them their properties.
First, we must distinguish between two general types of solids. An amorphous solid is a solid with no long-term structure or repetition. Examples include glass and many plastics, both of which are composed of long chains of molecules with no order from one molecule to the next. A crystalline solid is a solid that has a regular, repeating three-dimensional structure. A crystal of NaCl (Figure \(1\)) is one example: at the atomic level, NaCl is composed of a regular three-dimensional array of Na+ ions and Cl ions.
There is only one type of amorphous solid. However, there are several different types of crystalline solids, depending on the identity of the units that compose the crystal.
An ionic solid is a crystalline solid composed of ions (even if the ions are polyatomic). NaCl is an example of an ionic solid (Figure \(2\) - An Ionic Solid). The Na+ ions and Cl ions alternate in three dimensions, repeating a pattern that goes on throughout the sample. The ions are held together by the attraction of opposite charges—a very strong force. Hence most ionic solids have relatively high melting points; for example, the melting point of NaCl is 801°C. Ionic solids are typically very brittle. To break them, the very strong ionic attractions need to be broken; a displacement of only about 1 × 10−10 m will move ions next to ions of the same charge, which results in repulsion. Ionic solids do not conduct electricity in their solid state; however, in the liquid state and when dissolved in some solvent, they do conduct electricity. This fact originally promoted the idea that some substances exist as ionic particles.
A molecular solid is a crystalline solid whose components are covalently bonded molecules. Many molecular substances, especially when carefully solidified from the liquid state, form solids where the molecules line up with a regular fashion similar to an ionic crystal, but they are composed of molecules instead of ions. Because the intermolecular forces between molecules are typically less strong than in ionic solids, molecular solids typically melt at lower temperatures and are softer than ionic solids. Ice is an example of a molecular solid. In the solid state, the molecules line up in a regular pattern (Figure \(3\)). Some very large molecules, such as biological molecules, will form crystals only if they are very carefully solidified from the liquid state or, more often, from a dissolved state; otherwise, they will form amorphous solids.
Some solids are composed of atoms of one or more elements that are covalently bonded together in a seemingly never-ending fashion. Such solids are called covalent network solids. Each piece of the substance is essentially one huge molecule, as the covalent bonding in the crystal extends throughout the entire crystal. The two most commonly known covalent network solids are carbon in its diamond form and silicon dioxide (SiO2). Figure \(4\) - Covalent Network Solids, shows the bonding in a covalent network solid. Generally, covalent network solids are poor conductors of electricity, although their ability to conduct heat is variable: diamond is one of the most thermally conductive substances known, while SiO2 is about 100 times less thermally conductive. Most covalent network solids are very hard, as exemplified by diamond, which is the hardest known substance. Covalent network solids have high melting points by virtue of their network of covalent bonds, all of which would have to be broken for them to transform into a liquid. Indeed, covalent network solids are among the highest-melting substances known: the melting point of diamond is over 3,500°C, while the melting point of SiO2 is around 1,650°C. These characteristics are explained by the network of covalent bonds throughout the sample.
A metallic solid is a solid with the characteristic properties of a metal: shiny and silvery in color and a good conductor of heat and electricity. A metallic solid can also be hammered into sheets and pulled into wires. A metallic solid exhibits metallic bonding, a type of intermolecular interaction caused by the sharing of the s valence electrons by all atoms in the sample. It is the sharing of these valence electrons that explains the ability of metals to conduct electricity and heat well. It is also relatively easy for metals to lose these valence electrons, which explains why metallic elements usually form cations when they make compounds.
Example \(1\)
Predict the type of crystal exhibited by each solid.
1. MgO
2. Ag
3. CO2
Solution
1. A combination of a metal and a nonmetal makes an ionic compound, so MgO would exist as ionic crystals in the solid state.
2. Silver is a metal, so it would exist as a metallic solid in the solid state.
3. CO2 is a covalently bonded molecular compound. In the solid state, it would form molecular crystals. (You can actually see the crystals in dry ice with the naked eye.)
Exercise \(1\)
Predict the type of crystal exhibited by each solid.
1. I2
2. Ca(NO3)2
Answers
1. molecular crystals
2. ionic crystals
Food and Drink Application: The Rocks We Eat
The foods and beverages we eat and drink all have different phases: solid, liquid, and gas. (How do we ingest gases? Carbonated beverages have gas, which sometimes cause a person to belch.) However, among the solids we eat, three in particular are, or are produced from, rocks. Yes, rocks!
The first one is NaCl, or common salt. Salt is the only solid that we ingest that is actually mined as a rock (hence the term rock salt; it really is a rock). Salt provides both Na+ ions and Cl ions, both of which are necessary for good health. Salt preserves food, a function that was much more important before the days of modern food preparation and storage. The fact that saltiness is one of the major tastes the tongue can detect suggests a strong evolutionary link between ingesting salt and survival. There is some concern today that there is too much salt in the diet; it is estimated that the average person consumes at least three times as much salt daily as is necessary for proper bodily function.
The other two rocks we eat are related: sodium bicarbonate (NaHCO3) and sodium carbonate (Na2CO3). However, we do not mine these substances directly from the ground; we mine trona, whose chemical formula is Na3H(CO3)2. This substance is dissolved in water and treated with CO2 gas to make either Na2CO3 or NaHCO3. Another process, called the Solvay process, is also used to make Na2CO3. In the Solvay process, NH3 and CO2 are added to solutions of NaCl to make NaHCO3 and NH4Cl; the NaHCO3 precipitates and is heated to produce Na2CO3. Either way, we get these two products from the ground (i.e., rocks).
NaHCO3 is also known as baking soda, which is used in many baked goods. Na2CO3 is used in foods to regulate the acid balance. It is also used in laundry (where it is called washing soda) to interact with other ions in water that tend to reduce detergent efficiency.
Summary
• Solids can be divided into amorphous solids and crystalline solids.
• Crystalline solids can be ionic, molecular, covalent network, or metallic.
10.E: Solids and Liquids (Exercises)
Additional Exercises
1. All other things being equal, rank the intermolecular forces in order of increasing strength.
2. Which subatomic particles (protons, neutrons, electrons) are most responsible for intermolecular forces? Explain your answer.
• Can a molecule experience more than one intermolecular force at the same time? Why or why not?
• Of the properties boiling point, structure of the solid phase, and molar mass, which are influenced by hydrogen bonding? Explain your answer.
• How many grams of solid water can be melted with 1.55 kJ of energy?
• How many grams of Hg can be vaporized using 29,330 J of energy?
• Another way to minimize freezer burn is to wrap food tightly before freezing. Why would this minimize freezer burn?
• The ΔHsub of naphthalene (C10H8) is 72.6 kJ/mol. What energy is needed to sublime 100.0 g of C10H8?
• Which do you think would have a higher surface tension, liquid neon or liquid krypton? Explain your answer.
• Under what condition would a liquid not show either capillary rise or capillary depression?
• Answers
1. dispersion forces < dipole-dipole interactions < hydrogen bonding < ionic bonding
2.
3. Yes, but one intermolecular force usually dominates.
4.
5. 4.64 g
6.
7. Water in the vapor phase has no space to evaporate into.
8.
9. liquid krypton because it would have stronger dispersion forces | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/10%3A_Solids_and_Liquids/10.05%3A_Solids.txt |
A solution is a homogeneous mixture-a mixture of two or more substances that are so intimately mixed that the mixture behaves in many ways like a single substance. Many chemical reactions occur when the reactants are dissolved in solution. In this chapter, we will introduce concepts that are applicable to solutions and the chemical reactions that occur in them.
• 11.1: Prelude to Solutions
A solution is a homogeneous mixture-a mixture of two or more substances that are so intimately mixed that the mixture behaves in many ways like a single substance. Many chemical reactions occur when the reactants are dissolved in solution. In this chapter, we will introduce concepts that are applicable to solutions and the chemical reactions that occur in them.
• 11.2: Definitions
Solutions are composed of a solvent (major component) and a solute (minor component). Concentration is the expression of the amount of solute in a given amount of solvent and can be described by several qualitative terms. Solubility is a specific amount of solute that can dissolve in a given amount of solvent. "Like dissolves like" is a useful rule for deciding if a solute will be soluble in a solvent.
• 11.3: Quantitative Units of Concentration
Quantitative units of concentration include molarity, molality, mass percentage, parts per thousand, parts per million, and parts per billion.
• 11.4: Dilutions and Concentrations
Calculate the new concentration or volume for a dilution or concentration of a solution.
• 11.5: Concentrations as Conversion Factors
Concentration units can be used as conversion factors.
• 11.6: Colligative Properties of Solutions
Colligative properties depend only on the number of dissolved particles (that is, the concentration), not their identity. Raoult's law is concerned with the vapor pressure depression of solutions. The boiling points of solutions are always higher, and the freezing points always lower, than those of the pure solvent.
• 11.7: Colligative Properties of Ionic Solutes
For ionic solutes, the calculation of colligative properties must include the fact that the solutes separate into multiple particles when they dissolve. The equations for calculating colligative properties of solutions of ionic solvents include the van't Hoff factor, i.
• 11.E: Solutions (Exercises)
These are exercises and select solutions to accompany Chapter 11 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
11: Solutions
More than 70% of the earth's surface is covered by a very important solution—seawater. It is likely that without seawater, no life would exist on Earth.
At its simplest, seawater is mostly H2O. However, about 3.5% of seawater is dissolved solids—mostly NaCl—but other ions as well. Table \(1\) lists the percentage by mass of the various ions in seawater. Because it is highly likely that life on Earth originated in the oceans, it should not be surprising that many bodily fluids resemble seawater, especially blood. Table \(1\) also lists the percentage by mass of ions in a typical sample of blood.
Table \(1\): Percentage by Mass of Ions in Seawater and Blood
Ion Percentage in Seawater Percentage in Blood
Na+ 2.36 0.322
Cl 1.94 0.366
Mg2+ 0.13 0.002
SO42 0.09 -
K+ 0.04 0.016
Ca2+ 0.04 0.0096
HCO3 0.002 0.165
HPO42, H2PO4 - 0.01
Most ions are more abundant in seawater than they are in blood, with some notable exceptions. There is far more hydrogen carbonate ion (HCO3) in blood than in seawater; indeed, it is the third most common ion in blood. This difference is significant because the HCO3 ion and some related species [CO32, CO2(aq)] have an important role in controlling the acid-base properties of blood. Although there is a negligible amount of the two hydrogen phosphate ions (HPO42 and H2PO4) in seawater, there is a small amount in blood, where these ions affect acid-base properties. Another notable difference is that blood has a negligible amount of the sulfate ion (SO42), but this ion is present in seawater. Gold is present in seawater—but only a tiny amount. A current estimate of the amount of gold is about 1 part per every 1 × 1013 parts of seawater, which makes the extraction of gold from seawater unfeasible. However, it does mean that there are about 1.4 × 1014 g of gold in the world's oceans!
A solution is a homogeneous mixture: a mixture of two or more substances that are so intimately mixed that the mixture behaves in many ways like a single substance. Many chemical reactions occur when the reactants are dissolved in solution. In this chapter, we will introduce concepts that are applicable to solutions and the chemical reactions that occur in them. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.01%3A_Prelude_to_Solutions.txt |
Learning Objective
• Learn some terminology involving solutions.
• Recognize which terminology is qualitative and which terminology is quantitative.
• Explain why certain substances dissolve in other substances.
The major component of a solution is called the solvent. The minor component of a solution is called the solute. "Major" and "minor" indicate which component has the greater or lesser presence by mass or moles, respectively. Sometimes this becomes confusing, especially when considering substances with very different molar masses. We will confine the discussion here to solutions for which the major component and the minor component are obvious.
Solutions exist for every possible phase of the solute and the solvent. Salt water, for example, is a solution of solid NaCl in liquid water; soda water is a solution of gaseous CO2 in liquid water, while air is a solution of a gaseous solute (O2) in a gaseous solvent (N2). In all cases, however, the overall phase of the solution is the same phase as the solvent.
Example \(1\): Sugar Water
A solution is made by dissolving 1.00 g of sucrose (C12H22O11) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution.
Solution
Either by mass or by moles, the obvious minor component is sucrose, so it is the solute. Water—the majority component—is the solvent. The fact that the resulting solution is the same phase as water also suggests that water is the solvent.
Exercise \(1\)
A solution is made by dissolving 3.33 g of HCl(g) in 40.0 g of liquid methyl alcohol (CH3OH). Identify the solvent and solute in the resulting solution.
Answer
solute: HCl(g); solvent: CH3OH
One important concept of solutions is in defining how much solute is dissolved in a given amount of solvent. This concept is called concentration. Various words are used to describe the relative amounts of solute. Dilute describes a solution that has very little solute, while concentrated describes a solution that has a lot of solute. One problem is that these terms are qualitative; they describe more or less but not exactly how much.
In most cases, only a certain maximum amount of solute can be dissolved in a given amount of solvent. This maximum amount is called the solubility of the solute. It is usually expressed in terms of the amount of solute that can dissolve in 100 g of the solvent at a given temperature. Table \(1\) lists the solubilities of some simple ionic compounds. These solubilities vary widely: NaCl can dissolve up to 31.6 g per 100 g of H2O, while AgCl can dissolve only 0.00019 g per 100 g of H2O.
Table \(1\): Solubilities of Some Ionic Compounds
Solute Solubility (g per 100 g of H2O at 25°C)
AgCl 0.00019
CaCO3 0.0006
KBr 70.7
NaCl 36.1
NaNO3 94.6
When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute. When less than the maximum amount of solute is dissolved in a given amount of solute, the solution is unsaturated. These terms are also qualitative terms because each solute has its own solubility. A solution of 0.00019 g of AgCl per 100 g of H2O may be saturated, but with so little solute dissolved, it is also rather dilute. A solution of 36.1 g of NaCl in 100 g of H2O is also saturated but rather concentrated. Ideally, we need more precise ways of specifying the amount of solute in a solution. We will introduce such ways in Section 11.3.
In some circumstances, it is possible to dissolve more than the maximum amount of a solute in a solution. Usually, this happens by heating the solvent, dissolving more solute than would normally dissolve at regular temperatures, and letting the solution cool down slowly and carefully. Such solutions are called supersaturated solutions and are not stable; given an opportunity (such as dropping a crystal of solute in the solution), the excess solute will precipitate from the solution.
It should be obvious that some solutes dissolve in certain solvents but not others. NaCl, for example, dissolves in water but not in vegetable oil. Beeswax dissolves in liquid hexane but not water. What is it that makes a solute soluble in some solvents but not others?
The answer is intermolecular interactions. The intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding (as described in Chapter 10). From experimental studies, it has been determined that if molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl—a very polar substance because it is composed of ions—dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane but not in polar water. This concept leads to the general rule that "like dissolves like" for predicting whether a solute is soluble in a given solvent. However, this is a general rule, not an absolute statement, so it must be applied with care.
Example \(2\): Polar and Nonpolar Solvents
Would I2 be more soluble in CCl4 or H2O? Explain your answer.
Solution
I2 is nonpolar. Of the two solvents, CCl4 is nonpolar and H2O is polar, so I2 would be expected to be more soluble in CCl4.
Exercise \(2\)
Would C3H7OH be more soluble in CCl4 or H2O? Explain your answer.
Answer
H2O because both experience hydrogen bonding
Summary
• Solutions are composed of a solvent (major component) and a solute (minor component).
• Concentration is the expression of the amount of solute in a given amount of solvent and can be described by several qualitative terms.
• Solubility is a specific amount of solute that can dissolve in a given amount of solvent.
• "Like dissolves like" is a useful rule for deciding if a solute will be soluble in a solvent. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.02%3A_Definitions.txt |
Learning Objective
• Learn to determine specific concentrations with several common units.
Rather than qualitative terms (Section 11.2 - Definitions), we need quantitative ways to express the amount of solute in a solution; that is, we need specific units of concentration. In this section, we will introduce several common and useful units of concentration.
Molarity (M) is defined as the number of moles of solute divided by the number of liters of solution:
$molarity \: =\: \frac{moles\: of\: solute}{liters\: of\: solution}\nonumber$
which can be simplified as
$M\: =\: \frac{mol}{L},\; or\; mol/L\nonumber$
As with any mathematical equation, if you know any two quantities, you can calculate the third, unknown, quantity.
For example, suppose you have 0.500 L of solution that has 0.24 mol of NaOH dissolved in it. The concentration of the solution can be calculated as follows:
$molarity \: =\: \frac{0.24\: mol\: NaOH}{0.500L}=0.48\, M\; NaOH\nonumber$
The concentration of the solution is 0.48 M, which is spoken as "zero point forty-eight molarity" or "zero point forty-eight molar." If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol):
$22.4\cancel{gHCl}\times \frac{1\: mol\: Hcl}{36.5\cancel{gHCl}}=0.614\, M\; HCl\nonumber$
Now we can use the definition of molarity to determine a concentration:
$M \: =\: \frac{0.614\: mol\: HCl}{1.56L}=0.394\, M\nonumber$
Example $1$
What is the molarity of a solution made when 32.7 g of NaOH are dissolved to make 445 mL of solution?
Solution
To use the definition of molarity, both quantities must be converted to the proper units. First, convert the volume units from milliliters to liters:
$445\cancel{mL}\times \frac{1\: L}{1000\cancel{mL}}=0.445\, L\nonumber$
Now we convert the amount of solute to moles, using the molar mass of NaOH, which is 40.0 g/mol:
$32.7\cancel{gNaOH}\times \frac{1\: mol\: NaOH}{40.0\cancel{gNaOH}}=0.818\, mol\: NaOH\nonumber$
Now we can use the definition of molarity to determine the molar concentration:
$M \: =\: \frac{0.818\: mol\: NaOH}{0.445L}=1.84\, M\: NaOH\nonumber$
Exercise $1$
What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution?
Answer
1.57 M
The definition of molarity can be used to determine the amount of solute or the volume of solution, if the other information is given. Example 4 illustrates this situation.
Example $2$
How many moles of solute are present in 0.108 L of a 0.887 M NaCl solution?
Solution
We know the volume and the molarity; we can use the definition of molarity to mathematically solve for the amount in moles. Substituting the quantities into the definition of molarity:
$0.887\, M \: =\: \frac{mol\: NaCl}{0.108L}\nonumber$
We multiply the 0.108 L over to the other side of the equation and multiply the units together; "molarity × liters" equals moles, according to the definition of molarity. So
Exercise $2$
How many moles of solute are present in 225 mL of a 1.44 M CaCl2 solution?
Answer
0.324 mol
If you need to determine volume, remember the rule that the unknown quantity must be by itself and in the numerator to determine the correct answer. Thus, rearrangement of the definition of molarity is required.
Example $3$
What volume of a 2.33 M NaNO3 solution is needed to obtain 0.222 mol of solute?
Solution
Using the definition of molarity, we have
$2.33\, M \: =\: \frac{0.222mol}{L}\nonumber$
To solve for the number of liters, we bring the 2.33 M over to the right into the denominator, and the number of liters over to the left in the numerator. We now have
$L \: =\: \frac{0.222mol}{2.33\, M}\nonumber$
Dividing, the volume is 0.0953 L = 95.3 mL.
Exercise $3$
What volume of a 0.570 M K2SO4 solution is needed to obtain 0.872 mol of solute?
Answer
1.53 L
A similar unit of concentration is molality (m), which is defined as the number of moles of solute per kilogram of solvent, not per liter of solution:
$molality\: =\: \frac{moles\: solute}{kilograms\: solvent}\nonumber$
Mathematical manipulation of molality is the same as with molarity.
Another way to specify an amount is percentage composition by mass (or mass percentage, % m/m). It is defined as follows:
$\%m/m\: =\: \frac{mass\: of\: solute}{mass\: of\: entire\: sample}\times 100\%\nonumber$
It is not uncommon to see this unit used on commercial products (see Figure $1$ - Concentration in Commercial Applications).
Example $4$
What is the mass percentage of Fe in a piece of metal with 87.9 g of Fe in a 113 g sample?
Solution
Using the definition of mass percentage, we have
$\%m/m\: =\: \frac{87.9\, g\, Fe}{113\, g\, sample}\times 100\%=77.8\%\, Fe\nonumber$
Related concentration units are parts per thousand (ppth), parts per million (ppm) and parts per billion (ppb). Parts per thousand is defined as follows:
$ppth\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1000\nonumber$
There are similar definitions for parts per million and parts per billion:
$ppm\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1,000,000\nonumber$
and
$ppb\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1,000,000,000\nonumber$
Each unit is used for progressively lower and lower concentrations. The two masses must be expressed in the same unit of mass, so conversions may be necessary.
Example $5$
If there are 0.6 g of Pb present in 277 g of solution, what is the Pb concentration in parts per thousand?
Solution
Use the definition of parts per thousand to determine the concentration. Substituting
$\frac{0.6g\, Pb}{277g\, solution}\times 1000=2.17\, ppth\nonumber$
Exercise $5$
If there are 0.551 mg of As in 348 g of solution, what is the As concentration in ppm?
Answer
1.58 ppm
As with molarity and molality, algebraic rearrangements may be necessary to answer certain questions.
Example $6$
The concentration of Cl ion in a sample of H2O is 15.0 ppm. What mass of Cl ion is present in 240.0 mL of H2O, which has a density of 1.00 g/mL?
Solution
First, use the density of H2O to determine the mass of the sample:
$240.0\cancel{mL}\times \frac{1.00\: g}{\cancel{mL}}=240.0\, g\nonumber$
Now we can use the definition of ppm:
$15.0\, ppm\: =\: \frac{mass\: of\: solute}{240.0\: g\: solution}\times 1,000,000\nonumber$
Rearranging to solve for the mass of solute,
$mass\: solute =\: \frac{(15.0\, ppm)(240.0\: g\: solution)}{1,000,000}=0.0036g=3.6\, mg\nonumber$
Exercise $6$
The concentration of Fe3+ ion in a sample of H2O is 335.0 ppm. What mass of Fe3+ ion is present in 3,450 mL of H2O, which has a density of 1.00 g/mL?
Answer
1.16 g
For ionic solutions, we need to differentiate between the concentration of the salt versus the concentration of each individual ion. Because the ions in ionic compounds go their own way when a compound is dissolved in a solution, the resulting concentration of the ion may be different from the concentration of the complete salt. For example, if 1 M NaCl were prepared, the solution could also be described as a solution of 1 M Na+(aq) and 1 M Cl(aq) because there is one Na+ ion and one Cl ion per formula unit of the salt. However, if the solution were 1 M CaCl2, there are two Cl(aq) ions for every formula unit dissolved, so the concentration of Cl(aq) would be 2 M, not 1 M.
In addition, the total ion concentration is the sum of the individual ion concentrations. Thus for the 1 M NaCl, the total ion concentration is 2 M; for the 1 M CaCl2, the total ion concentration is 3 M.
Key Takeaway
• Quantitative units of concentration include molarity, molality, mass percentage, parts per thousand (ppth), parts per million (ppm), and parts per billion (ppb). | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.03%3A_Quantitative_Units_of_Concentration.txt |
Learning Objective
• Learn how to dilute and concentrate solutions.
Often, a worker will need to change the concentration of a solution by changing the amount of solvent. Dilution is the addition of solvent, which decreases the concentration of the solute in the solution. Concentration is the removal of solvent, which increases the concentration of the solute in the solution. (Do not confuse the two uses of the word concentration here!)
In both dilution and concentration, the amount of solute stays the same. This gives us a way to calculate what the new solution volume must be for the desired concentration of solute. From the definition of molarity,
$\text{molarity} = \dfrac{\text{moles of solute}}{\text{liters of solution}}\nonumber$
we can solve for the number of moles of solute:
moles of solute = (molarity)(liters of solution)
A simpler way of writing this is to use M to represent molarity and V to represent volume. So the equation becomes
moles of solute = MV
Because this quantity does not change before and after the change in concentration, the product MV must be the same before and after the concentration change. Using numbers to represent the initial and final conditions, we have
$M_1V_1 = M_2V_2\nonumber$
as the dilution equation. The volumes must be expressed in the same units. Note that this equation gives only the initial and final conditions, not the amount of the change. The amount of change is determined by subtraction.
Example $1$
If 25.0 mL of a 2.19 M solution are diluted to 72.8 mL, what is the final concentration?
Solution
It does not matter which set of conditions is labeled 1 or 2, as long as the conditions are paired together properly. Using the dilution equation, we have
(2.19 M)(25.0 mL) = M2(72.8 mL)
Solving for the second concentration (noting that the milliliter units cancel),
M2 = 0.752 M
The concentration of the solution has decreased. In going from 25.0 mL to 72.8 mL, 72.8 − 25.0 = 47.8 mL of solvent must be added.
Exercise $1$
A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?
Answer
135.4 mL
Concentrating solutions involves removing solvent. Usually this is done by evaporating or boiling, assuming that the heat of boiling does not affect the solute. The dilution equation is used in these circumstances as well.
Chemistry is Everywhere: Preparing IV Solutions
In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl?
Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation.
If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then it is easy to calculate how much stock solution to use:
(10%)V1 = (0.50%)(100 mL)V1 = 5 mL
Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough even for medical purposes.
Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be useless, harmful, or even fatal!
Summary
• Molarity and volume are used to determine dilutions and concentrations of solutions. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.04%3A_Dilutions_and_Concentrations.txt |
Learning Objective
• Apply concentration units as conversion factors.
Concentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.
A simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition. For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol/L, we can use this second expression for the concentration as a conversion factor:
$0.108\cancel{L\, NaCl}\times \frac{0.887\, mol\, NaCl}{\cancel{L\, NaCl}}=0.0958\, mol\, NaCl\nonumber$
(There is an understood 1 in the denominator of the conversion factor.) If we used the definition approach, we get the same answer, but now we are using conversion factor skills. Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor.
Example $1$
Using concentration as a conversion factor, how many liters of 2.35 M CuSO4 are needed to obtain 4.88 mol of CuSO4?
Solution
This is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out:
$4.88\cancel{mol\, CuSO_{4}}\times \frac{1\, L}{2.35\cancel{mol}}=2.08\, L\, of \, solution\nonumber$
Exercise $1$
Using concentration as a conversion factor, how many liters of 0.0444 M CH2O are needed to obtain 0.0773 mol of CH2O?
Answer:
1.74 L
Of course, once quantities in moles are available, another conversion can give the mass of the substance, using molar mass as a conversion factor.
Example $2$
What mass of solute is present in 0.765 L of 1.93 M NaOH?
Solution
This is a two-step conversion, first using concentration as a conversion factor to determine the number of moles and then the molar mass of NaOH (40.0 g/mol) to convert to mass:
$0.765\cancel{L}\times \frac{1.93\cancel{mol\, NaOH}}{\cancel{L\, solution}}\times \frac{40.0g\, NaOH}{1\cancel{mol\, NaOH}}=59.1\, g\, NaOH\nonumber$
Exercise $2$
What mass of solute is present in 1.08 L of 0.0578 M H2SO4?
Answer
6.12 g
More complex stoichiometry problems using balanced chemical reactions can also use concentrations as conversion factors. For example, suppose the following equation represents a chemical reaction:
$\ce{2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq)}\nonumber$
If we wanted to know what volume of 0.555 M CaCl2 would react with 1.25 mol of AgNO3, we first use the balanced chemical equation to determine the number of moles of CaCl2 that would react and then use concentration to convert to liters of solution:
$1.25\cancel{mol\, AgNO_{3}}\times \frac{1\cancel{mol\, CaCl_{2}}}{2\cancel{mol\, AgNO_{3}}}\times \frac{1L\, solution}{0.555\cancel{mol\, CaCl_{2}}}=1.13\, L\, CaCl_{2}\nonumber$
This can be extended by starting with the mass of one reactant, instead of moles of a reactant.
Example $3$
What volume of 0.0995 M Al(NO3)3 will react with 3.66 g of Ag according to the following chemical equation?
$\ce{3Ag(s) + Al(NO3)3(aq) → 3AgNO3 + Al(s)}\nonumber$
Solution
Here, we first must convert the mass of Ag to moles before using the balanced chemical equation and then the definition of molarity as a conversion factor:
$3.66\cancel{g\, Ag}\times \frac{1\cancel{mol\, Ag}}{107.97\cancel{g\, Ag}}\times \frac{1\cancel{mol\, Al(NO_{3})_{3}}}{3\cancel{mol\, Ag}}\times \frac{1L\, solution}{0.0995\cancel{mol\, Al(NO_{3})_{3}}}=0.114\, L\nonumber$
The strikeouts show how the units cancel.
Exercise $3$
What volume of 0.512 M NaOH will react with 17.9 g of H2C2O4(s) according to the following chemical equation?
$\ce{H2C2O4(s) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(ℓ)}\nonumber$
Answer:
0.777 L
We can extend our skills even further by recognizing that we can relate quantities of one solution to quantities of another solution. Knowing the volume and concentration of a solution containing one reactant, we can determine how much of another solution of another reactant will be needed using the balanced chemical equation.
Example $4$
A student takes a precisely measured sample, called an aliquot, of 10.00 mL of a solution of FeCl3. The student carefully adds 0.1074 M Na2C2O4 until all the Fe3+(aq) has precipitated as Fe2(C2O4)3(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na2C2O4 solution was added to completely precipitate the Fe3+(aq). What was the concentration of the FeCl3 in the original solution? (A precisely measured experiment like this, which is meant to determine the amount of a substance in a sample, is called a titration.) The balanced chemical equation is as follows:
$\ce{2FeCl3(aq) + 3Na2C2O4(aq) → Fe2(C2O4)3(s) + 6NaCl(aq)}\nonumber$
Solution
First we need to determine the number of moles of Na2C2O4 that reacted. We will convert the volume to liters and then use the concentration of the solution as a conversion factor:
$9.04\cancel{mL}\times \frac{1\cancel{L}}{1000\cancel{mL}}\times \frac{0.1074mol\, Na_{2}C_{2}O_{4}}{\cancel{L}}=0.000971\, mol\, Na_{2}C_{2}O_{4}\nonumber$
Now we will use the balanced chemical equation to determine the number of moles of Fe3+(aq) that were present in the initial aliquot:
$0.000971\cancel{mol\, Na_{2}C_{2}O_{4}}\times \frac{2mol\, FeCl_{3}}{3\cancel{molNa_{2}C_{2}O_{4}}}=0.000647mol\, FeCl_{3}\nonumber$
Then we determine the concentration of FeCl3 in the original solution. Converting 10.00 mL into liters (0.01000 L), we use the definition of molarity directly:
$M=\frac{mol}{L}=\frac{0.000647mol\, FeCl_{3}}{0.01000L}=0.0647M\, FeCl_{3}\nonumber$
Exercise $4$
A student titrates 25.00 mL of H3PO4 with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H3PO4?
$\ce{H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O}\nonumber$
Answer:
0.0711 M
We have used molarity exclusively as the concentration of interest, but that will not always be the case. The next example shows a different concentration unit being used.
Example $5$
H2O2 is used to determine the amount of Mn according to this balanced chemical equation:
$\ce{2MnO4^{-}(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 5O2(g) + 8H2O(ℓ)}\nonumber$
What mass of 3.00% m/m H2O2 solution is needed to react with 0.355 mol of MnO4(aq)?
Solution
Because we are given an initial amount in moles, all we need to do is use the balanced chemical equation to determine the number of moles of H2O2 and then convert to find the mass of H2O2. Knowing that the H2O2 solution is 3.00% by mass, we can determine the mass of solution needed:
$0.355\cancel{mol\, MnO_{4}^{-}}\times \frac{5\cancel{mol\, H_{2}O_{2}}}{2\cancel{molMnO_{4}^{-}}}\times \frac{34.02\cancel{g\, H_{2}O_{2}}}{\cancel{mol\, H_{2}O_{2}}}\times \frac{100g\; solution}{3\cancel{g\, H_{2}O_{2}}}=1006g\; sol\nonumber$
The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H2O2, and the third conversion factor comes from the definition of percentage concentration by mass.
Exercise $5$
Use the balanced chemical reaction for MnO4 and H2O2 to determine what mass of O2 is produced if 258 g of 3.00% m/m H2O2 is reacted with MnO4.
Answer
7.28 g
Summary
Know how to apply concentration units as conversion factors. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.05%3A_Concentrations_as_Conversion_Factors.txt |
the solvent. However, some of the properties of solutions differ from pure solvents in measurable and predictable ways. The differences are proportional to the fraction that the particles occupy in the . These properties are called ; the word comes from the Greek word meaning "related to the number," implying that these properties are related to the number of particles, not their identities. unit. The of the $i^{th}$ component in a , $\chi_i$, is the number of moles of that component divided by the total number of moles in the sample: is always a number between 0 and 1 (inclusive) and has no units; it is just a number. fractions of all substances in a mixture equals 1. Thus the of $\ce{C6H6}$ in Example $1$ could be calculated by evaluating the definition of a second time, or—because there are only two substances in this particular mixture—we can subtract the of the $\ce{C10H8}$ from 1 to get the of $\ce{C6H6}$. unit has been introduced, the first colligative property can be considered. As mentioned in Chapter 10, all pure liquids have a characteristic in equilibrium with the liquid phase, the of which is dependent on temperature. Solutions, however, have a lower than the pure solvent has, and the amount of lowering is dependent on the fraction of particles, as long as the itself does not have a significant (the term is used to describe such solutes). This colligative property is called (or ). The actual of the can be calculated as follows: }\] of the , $\chi_{solv}$ is the of the solvent particles, and $P^{*}_{solv}$ is the of the pure solvent at that temperature (which is data that must be provided). Equation \ref{Raoult's } is known as (the approximate pronunciation is ). is rationalized by presuming that particles take positions at the surface in place of solvent particles, so not as many solvent particles can evaporate. are related to as expressed in . As a review, recall the definition of : } =\frac{\text{moles of }}{\text{kilograms of solvent}}\nonumber \] of a with a nonvolatile is depressed compared to that of the pure solvent, it requires a higher temperature for the 's to reach 1.00 atm (760 ). Recall that this is the definition of the : the temperature at which the of the liquid equals 1.00 atm. As such, the of the is higher than that of the pure solvent. This property is called . of the and is called the which is a characteristic of the solvent. Several constants (as well as boiling point temperatures) are listed in Table $1$. . is higher than the boiling point of the pure solvent, but the opposite occurs with the freezing point. The freezing point of a is lower than the freezing point of the pure solvent. Think of this by assuming that particles interfere with solvent particles coming together to make a solid, so it takes a lower temperature to get the solvent particles to solidify. This is called . is similar to the equation for the : of the and is called the which is also a characteristic of the solvent only. Several constants (as well as freezing point temperatures) are listed in Table $2$. is one colligative property that we use in everyday life. Many antifreezes used in automobile radiators use solutions that have a lower freezing point than normal so that automobile engines can operate at subfreezing temperatures. We also take advantage of when we sprinkle various compounds on ice to thaw it in the winter for safety (Figure $1$). The compounds make solutions that have a lower freezing point, so rather than forming slippery ice, any ice is liquefied and runs off, leaving a safer pavement behind. is a thin membrane that will pass certain small but not others. A thin sheet of cellophane, for example, acts as a . Consider the system in Figure $2$. of a is easy to calculate: of a , $M$ is the of the , $R$ is the constant, and $T$ is the absolute temperature. This equation is reminiscent of the discussed previously. is important in biological systems because cell walls are semipermeable membranes. In particular, when a person is receiving intravenous (IV) fluids, the of the fluid needs to be approximately the same as blood serum; otherwise there may be negative consequences. Figure $3$ shows three red blood cells: is also the reason you should not drink seawater if you're stranded in a lifeboat on an ocean; seawater has a higher than most of the fluids in your body. You drink the water, but ingesting it will pull water out of your cells as osmosis works to dilute the seawater. Ironically, your cells will die of thirst, and you will also die. (It is OK to drink the water if you are stranded on a body of freshwater, at least from an perspective.) is also thought to be important in getting water to the tops of tall trees, in addition to capillary action. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.06%3A_Colligative_Properties_of_Solutions.txt |
Learning Objective
• Determine the colligative properties of solutions of ionic solutes.
Previously, we considered the colligative properties of solutions with molecular solutes. What about solutions with ionic solutes? Do they exhibit colligative properties?
There is a complicating factor: ionic solutes separate into ions when they dissolve. This increases the total number of particles dissolved in solution and increases the impact on the resulting colligative property. Historically, this greater-than-expected impact on colligative properties was one main piece of evidence for ionic compounds separating into ions (increased electrical conductivity was another piece of evidence).
For example, when NaCl dissolves, it separates into two ions:
$\ce{NaCl(s) → Na^{+}(aq) + Cl^{-}(aq)}\nonumber$
This means that a 1 M solution of NaCl actually has a net particle concentration of 2 M. The observed colligative property will then be twice as large as expected for a 1 M solution.
It is easy to incorporate this concept into our equations to calculate the respective colligative property. We define the van 't Hoff factor ($i$) as the number of particles each solute formula unit breaks apart into when it dissolves. Previously, we have always tacitly assumed that the van 't Hoff factor is simply 1. But for some ionic compounds, $i$ is not 1, as shown in Table $1$.
Table $1$ Ideal van 't Hoff Factors for Ionic Compounds
Compound i
NaCl 2
KBr 2
LiNO3 2
CaCl2 3
Mg(C2H3O2)2 3
FeCl3 4
Al2(SO4)3 5
The ideal van 't Hoff factor is equal to the number of ions that form when an ionic compound dissolves.
Example $1$
Predict the van 't Hoff factor for Sr(OH)2.
Solution
When Sr(OH)2 dissolves, it separates into one Sr2+ ion and two OH ions:
$\ce{Sr(OH)2 \rightarrow Sr^{2+}(aq) + 2OH^{−}(aq)} \nonumber \nonumber$
Because it breaks up into three ions, its van 't Hoff factor is 3.
Exercise $1$
What is the van 't Hoff factor for Fe(NO3)3?
Answer
4
It is the "ideal" van 't Hoff factor because this is what we expect from the ionic formula. However, this factor is usually correct only for dilute solutions (solutions less than 0.001 M). At concentrations greater than 0.001 M, there are enough interactions between ions of opposite charge that the net concentration of the ions is less than expected—sometimes significantly. The actual van 't Hoff factor is thus less than the ideal one. Here, we will use ideal van 't Hoff factors.
Revised equations to calculate the effect of ionization are then easily produced:
ΔTb = imKb
ΔTf = imKg
Π = iMRT
where all variables have been previously defined. To calculate vapor pressure depression according to Raoult's law, the mole fraction of solvent particles must be recalculated to take into account the increased number of particles formed on ionization.
Example $2$
Determine the freezing point of a 1.77 m solution of NaCl in H2O.
Solution
For NaCl, we need to remember to include the van 't Hoff factor, which is 2. Otherwise, the calculation of the freezing point is straightforward:
$ΔT_f = (2)(1.77\, m)(1.86\,°C/m) = 6.58°C\nonumber$
This represents the change in the freezing point, which is decreasing. So we have to subtract this change from the normal freezing point of water, 0.00°C:
0.00 − 6.58 = −6.58°C
Exercise $2$
Determine the boiling point of a 0.887 m solution of CaCl2 in H2O.
Answer
101.36 °C
Food and Drink App: Salting Pasta Cooking Water
When cooking dried pasta, many recipes call for salting the water before cooking the pasta. Some argue—with colligative properties on their side—that adding salt to the water raises the boiling point, thus cooking the pasta faster. Is there any truth to this?
To judge the veracity of this claim, we can calculate how much salt should be added to the water to raise the boiling temperature by 1.0°C, with the presumption that dried pasta cooks noticeably faster at 101°C than at 100°C (although a 1° difference may make only a negligible change in cooking times). We can calculate the molality that the water should have:
1.0°C = m(0.512°C/m)m = 1.95
We have ignored the van 't Hoff factor in our estimation because this obviously is not a dilute solution. Let us further assume that we are using 4 L of water (which is very close to 4 qt, which in turn equals 1 gal). Because 4 L of water is about 4 kg (it is actually slightly less at 100°C), we can determine how much salt (NaCl) to add:
$4\cancel{kg\, H_{2}O}\times \frac{1.95\cancel{mol\, NaCl}}{\cancel{kg\, H_{2}O}}\times \frac{58.5g\, NaCl}{\cancel{1\, mol\, NaCl}}=456.3g\, NaCl\nonumber$
This is just over 1 lb of salt and is equivalent to nearly 1 cup in the kitchen. In your experience, do you add almost a cup of salt to a pot of water to make pasta? Certainly not! A few pinches, perhaps one-fourth of a teaspoon, but not almost a cup! It is obvious that the little amount of salt that most people add to their pasta water is not going to significantly raise the boiling point of the water.
So why do people add some salt to boiling water? There are several possible reasons, the most obvious of which is taste: adding salt adds a little bit of salt flavor to the pasta. It cannot be much because most of the salt remains in the water, not in the cooked pasta. However, it may be enough to detect with our taste buds. The other obvious reason is habit; recipes tell us to add salt, so we do, even if there is little scientific or culinary reason to do so.
Summary
For ionic solutes, the calculation of colligative properties must include the fact that the solutes separate into multiple particles when they dissolve. The equations for calculating colligative properties of solutions of ionic solvents include the van 't Hoff factor, i. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.07%3A_Colligative_Properties_of_Ionic_Solutes.txt |
Additional Exercises
1. One brand of ethyl alcohol (Everclear) is 95% ethyl alcohol, with the remaining 5% being water. What is the solvent and what is the solute of this solution?
2. Give an example of each type of solution from your own experience.
1. A solution composed of a gas solute in a liquid solvent.
2. A solution composed of a solid solute in a liquid solvent.
3. A solution composed of a liquid solute in a liquid solvent.
4. A solution composed of a solid solute in a solid solvent. (Hint: usually such solutions are made as liquids and then solidified.)
3. Differentiate between the terms saturated and concentrated.
• Differentiate between the terms unsaturated and dilute.
• What mass of FeCl2 is present in 445 mL of 0.0812 M FeCl2 solution?
• What mass of SO2 is present in 26.8 L of 1.22 M SO2 solution?
• What volume of 0.225 M Ca(OH)2 solution is needed to deliver 100.0 g of Ca(OH)2?
• What volume of 12.0 M HCl solution is needed to obtain exactly 1.000 kg of HCl?
• The World Health Organization recommends that the maximum fluoride ion concentration in drinking water is 1.0 ppm. Assuming water has the maximum concentration, if an average person drinks 1,920 mL of water per day, how many milligrams of fluoride ion are being ingested?
• For sanitary reasons, water in pools should be chlorinated to a maximum level of 3.0 ppm. In a typical 5,000 gal pool that contains 21,200 kg of water, what mass of chlorine must be added to obtain this concentration?
• Given its notoriety, you might think that uranium is very rare, but it is present at about 2–4 ppm of the earth's crust, which is more abundant than silver or mercury. If the earth's crust is estimated to have a mass of 8.50 × 1020 kg, what range of mass is thought to be uranium in the crust?
• Chromium is thought to be an ultratrace element, with about 8.9 ng present in a human body. If the average body mass is 75.0 kg, what is the concentration of chromium in the body in pptr?
• What mass of 3.00% H2O2 solution is needed to produce 35.7 g of O2(g) at 295 K at 1.05 atm pressure?
2H2O2(aq) → 2H2O(ℓ) + O2(g)
• What volume of pool water is needed to generate 1.000 L of Cl2(g) at standard temperature and pressure if the pool contains 4.0 ppm HOCl and the water is slightly acidic? The chemical reaction is as follows:
HOCl(aq) + HCl(aq) → H2O(ℓ) + Cl2(g)
Assume the pool water has a density of 1.00 g/mL.
• A 0.500 m solution of MgCl2 has a freezing point of −2.60°C. What is the true van 't Hoff factor of this ionic compound? Why is it less than the ideal value?
• The osmotic pressure of a 0.050 M LiCl solution at 25.0°C is 2.26 atm. What is the true van 't Hoff factor of this ionic compound? Why is it less than the ideal value?
• Order these solutions in order of increasing boiling point, assuming an ideal van 't Hoff factor for each: 0.10 m C6H12O6, 0.06 m NaCl, 0.4 m Au(NO3)3, and 0.4 m Al2(SO4)3.
• Order these solutions in order of decreasing osmotic pressure, assuming an ideal van 't Hoff factor: 0.1 M HCl, 0.1 M CaCl2, 0.05 M MgBr2, and 0.07 M Ga(C2H3O2)3
• Answers
1. solvent: ethyl alcohol; solute: water
2.
3. Saturated means all the possible solute that can dissolve is dissolved, whereas concentrated implies that a lot of solute is dissolved.
4.
5. 4.58 g
6.
7. 6.00 L
8.
9. 1.92 mg
10.
11. 1.7 × 1015 to 3.4 × 1015 kg
12.
13. 2,530 g
14.
15. 2.80; it is less than 3 because not all ions behave as independent particles.
16.
17. 0.10 m C6H12O6 < 0.06 m NaCl < 0.4 m Au(NO3)3 < 0.4 m Al2(SO4)3 | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/11%3A_Solutions/11.E%3A_Solutions_%28Exercises%29.txt |
Acids and bases are important classes of chemical compounds. They are part of the foods and beverages we ingest, they are present in medicines and other consumer products, and they are prevalent in the world around us. In this chapter, we will focus on acids and bases and their chemistry.
• 12.1: Introduction
Certain household chemicals, such as some brands of cleaner, can be very concentrated bases, which makes them among the most potentially hazardous substances found around the home. If spilled on the skin, these strong caustic compounds can immediately remove H+ ions from the flesh, resulting in chemical burns. Compare this to the fact that we purposefully ingest substances such as citrus fruits, vinegar, and wine—all of which contain acids.
• 12.2: Arrhenius Acids and Bases
An Arrhenius acid is a compound that increases the H+ ion concentration in aqueous solution. An Arrhenius base is a compound that increases the OH− ion concentration in aqueous solution. The reaction between an Arrhenius acid and an Arrhenius base is called neutralization and results in the formation of water and a salt.
• 12.3: Brønsted-Lowry Acids and Bases
A Brønsted-Lowry acid is a proton donor; a Brønsted-Lowry base is a proton acceptor. Acid-base reactions include two sets of conjugate acid-base pairs.
• 12.4: Acid-Base Titrations
A titration is the quantitative reaction of an acid and a base. Indicators are used to show that all of the analyte has reacted with the titrant.
• 12.5: Strong and Weak Acids and Bases and their Salts
Strong acids and bases are 100% ionized in aqueous solution. Weak acids and bases are less than 100% ionized in aqueous solution. Salts of weak acids or bases can affect the acidity or basicity of their aqueous solutions.
• 12.6: Autoionization of Water
In any aqueous solution at room temperature, the product of $[H^+]$ and $[OH^−]$ equals $1.0 \times 10^{−14}$.
• 12.7: The pH Scale
pH is a logarithmic function of [H+]. [H+] can be calculated directly from pH. pOH is related to pH and can be easily calculated from pH.
• 12.8: Buffers
A buffer is a solution that resists sudden changes in pH.
• 12.E: Acids and Bases (Exercises)
These are exercises and select solutions to accompany Chapter 12 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
Thumbnail: Acids and Bases with litmus paper (CC BY-NC 4.0 (license of source has changed); CK-12 via CK-12(opens in new window)).
12: Acids and Bases
Formerly there were rather campy science-fiction television shows in which the hero was always being threatened with death by being plunged into a vat of boiling acid: "Mwa ha ha, Buck Rogers [or whatever the hero's name was], prepare to meet your doom by being dropped into a vat of boiling acid!" (The hero always escapes, of course.) This may have been interesting drama, but not very good chemistry. If the villain knew his/her/its science, the hero would have been dropped into a vat of boiling base.
Recall that the active component of a classic acid is the H+ ion, while the active part of a classic base is the OH ion. Both ions are related to water in that all H+ ion needs to become a water molecule is an OH ion, while all an OH ion needs to become water is an H+ ion. Consider the relative masses involved: an ion of mass 1 needs an ion of mass 17 to make water, while an ion of mass 17 needs an ion of mass 1 to make water. Which process do you think will be easier?
In fact, bases are more potentially dangerous than acids because it is much easier for an OH ion to rip off an H+ ion from surrounding matter than it is for an H+ ion to rip off an OH ion. Certain household chemicals, such as some brands of cleaner, can be very concentrated bases, which makes them among the most potentially hazardous substances found around the home. If spilled on the skin, these strong caustic compounds can immediately remove H+ ions from the flesh, resulting in chemical burns. Compare this to the fact that we occasionally purposefully ingest substances such as citrus fruits, vinegar, and wine—all of which contain acids. (Of course, some parts of the body, such as the eyes, are extremely sensitive to acids as well as bases.) It seems that our bodies are more capable of dealing with acids than with bases.
A note to all of the villains out there...get your chemistry right if you want to be successful! | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.01%3A_Introduction.txt |
Learning Objective
• Identify an Arrhenius acid and an Arrhenius base.
• Write the chemical reaction between an Arrhenius acid and an Arrhenius base.
Historically, the first chemical definition of an acid and a base was put forward by Svante Arrhenius, a Swedish chemist, in 1884. An Arrhenius acid is a compound that increases the H+ ion concentration in an aqueous solution. The H+ ion is just a bare proton, and it is rather clear that bare protons are not floating around in an aqueous solution. Instead, chemistry has defined the hydronium ion (H3O+) as the actual chemical species that represents an H+ ion. H+ ions and H3O+ ions are often considered interchangeable when writing chemical equations (although a properly balanced chemical equation should also include the additional H2O). Classic Arrhenius acids can be considered ionic compounds in which H+ is the cation. Table \(1\) lists some Arrhenius acids and their names.
Table \(1\) Some Arrhenius Acids
Formula Name
HC2H3O2 (also written CH3COOH) acetic acid
HClO3 chloric acid
HCl hydrochloric acid
HBr hydrobromic acid
HI hydriodic acid
HF hydrofluoric acid
HNO3 nitric acid
H2C2O4 oxalic acid
HClO4 perchloric acid
H3PO4 phosphoric acid
H2SO4 sulfuric acid
H2SO3 sulfurous acid
An Arrhenius base is a compound that increases the OH ion concentration in aqueous solution. Ionic compounds of the OH ion are classic Arrhenius bases.
Example \(1\)
Identify each compound as an Arrhenius acid, an Arrhenius base, or neither.
1. HNO3
2. CH3OH
3. Mg(OH)2
Solution
1. This compound is an ionic compound between H+ ions and NO3 ions, so it is an Arrhenius acid.
2. Although this formula has an OH in it, we do not recognize the remaining part of the molecule as a cation. It is neither an acid nor a base. (In fact, it is the formula for methanol, an organic compound.)
3. This formula also has an OH in it, but this time we recognize that the magnesium is present as Mg2+ cations. As such, this is an ionic compound of the OH ion and is an Arrhenius base.
Exercise \(1\)
Identify each compound as an Arrhenius acid, an Arrhenius base, or neither.
1. KOH
2. H2SO4
3. C2H6
Answer
1. Arrhenius base
2. Arrhenius acid
3. neither
Acids have some properties in common. They turn litmus, a plant extract, red. They react with some metals to give off H2 gas. They react with carbonate and hydrogen carbonate salts to give off CO2 gas. Acids that are ingested typically have a sour, sharp taste. (The name acid comes from the Latin word acidus, meaning "sour.") Bases also have some properties in common. They are slippery to the touch, turn litmus blue, and have a bitter flavor if ingested.
Acids and bases have another property: they react with each other to make water and an ionic compound called a salt. A salt, in chemistry, is any ionic compound made by combining an acid with a base. A reaction between an acid and a base is called a neutralization reaction and can be represented as:
acid + base → H2O + salt
The stoichiometry of the balanced chemical equation depends on the number of H+ ions in the acid and the number of OH ions in the base.
Example \(2\)
Write the balanced chemical equation for the neutralization reaction between H2SO4 and KOH. What is the name of the salt that is formed?
Solution
The general reaction is as follows:
H2SO4 + KOH → H2O + salt
Because the acid has two H+ ions in its formula, we need two OH ions to react with it, making two H2O molecules as product. The remaining ions, K+ and SO42, make the salt potassium sulfate (K2SO4). The balanced chemical reaction is as follows:
H2SO4 + 2KOH → 2H2O + K2SO4
Exercise \(2\)
Write the balanced chemical equation for the neutralization reaction between HCl and Mg(OH)2. What is the name of the salt that is formed?
Answer
2HCl + Mg(OH)2 → 2H2O + MgCl2; magnesium chloride
Key Takeaways
• An Arrhenius acid is a compound that increases the H+ ion concentration in aqueous solution.
• An Arrhenius base is a compound that increases the OH ion concentration in aqueous solution.
• The reaction between an Arrhenius acid and an Arrhenius base is called neutralization and results in the formation of water and a salt. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.02%3A_Arrhenius_Acids_and_Bases.txt |
Learning Objectives
• Identify a Brønsted-Lowry acid and a Brønsted-Lowry base.
• Identify conjugate acid-base pairs in an acid-base reaction.
The Arrhenius definition of acid and base is limited to aqueous (that is, water) solutions. Although this is useful because water is a common solvent, it is limited to the relationship between the H+ ion and the OH ion. What would be useful is a general definition more applicable to other chemical reactions and, importantly, independent of H2O.
In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).
It is easy to see that the Brønsted-Lowry definition covers the Arrhenius definition of acids and bases. Consider the prototypical Arrhenius acid-base reaction:
$\underset{acid}{H^{+}(aq)}+\underset{base}{OH^{-}(aq)}\rightarrow H_{2}O\, (l)\nonumber$
Acid species and base species are marked. The proton, however, is (by definition) a proton donor (labeled PD), while the OH ion is acting as the proton acceptor (labeled PA):
$\underset{PD}{H^{+}(aq)}+\underset{PA}{OH^{-}(aq)}\rightarrow H_{2}O\, (l)\nonumber$
The proton donor is a Brønsted-Lowry acid, and the proton acceptor is the Brønsted-Lowry base:
$\underset{BL\, acid}{H^{+}(aq)}+\underset{BL\, base}{OH^{-}(aq)}\rightarrow H_{2}O\, (l)\nonumber$
Thus H+ is an acid by both definitions, and OH is a base by both definitions.
Ammonia (NH3) is a base even though it does not contain OH ions in its formula. Instead, it generates OH ions as the product of a proton-transfer reaction with H2O molecules; NH3 acts like a Brønsted-Lowry base, and H2O acts like a Brønsted-Lowry acid:
A reaction with water is called hydrolysis; we say that NH3 hydrolyzes to make NH4+ ions and OH ions.
Even the dissolving of an Arrhenius acid in water can be considered a Brønsted-Lowry acid-base reaction. Consider the process of dissolving HCl(g) in water to make an aqueous solution of hydrochloric acid. The process can be written as follows:
$\ce{HCl(g) + H2O(ℓ) → H3O+(aq) + Cl^{-}(aq)} \nonumber \nonumber$
HCl(g) is the proton donor and therefore a Brønsted-Lowry acid, while H2O is the proton acceptor and a Brønsted-Lowry base. These two examples show that H2O can act as both a proton donor and a proton acceptor, depending on what other substance is in the chemical reaction. A substance that can act as a proton donor or a proton acceptor is called amphiprotic. Water is probably the most common amphiprotic substance we will encounter, but other substances are also amphiprotic.
Example $1$
Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.
$\ce{C6H5OH + NH2^{-} -> C6H5O^{-} + NH3} \nonumber \nonumber$
Solution
The C6H5OH molecule is losing an H+; it is the proton donor and the Brønsted-Lowry acid. The NH2 ion (called the amide ion) is accepting the H+ ion to become NH3, so it is the Brønsted-Lowry base.
Exercise $1$: Aluminum Ions in Solution
Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.
$\ce{Al(H2O)6^{3+} + H2O -> Al(H2O)5(OH)^{2+} + H3O^{+}} \nonumber \nonumber$
Answer
Brønsted-Lowry acid: Al(H2O)63+; Brønsted-Lowry base: H2O
In the reaction between NH3 and H2O,
the chemical reaction does not go to completion; rather, the reverse process occurs as well, and eventually the two processes cancel out any additional change. At this point, we say the chemical reaction is at equilibrium. Both processes still occur, but any net change by one process is countered by the same net change of the other process; it is a dynamic, rather than a static, equilibrium. Because both reactions are occurring, it makes sense to use a double arrow instead of a single arrow:
What do you notice about the reverse reaction? The NH4+ ion is donating a proton to the OH ion, which is accepting it. This means that the NH4+ ion is acting as the proton donor, or Brønsted-Lowry acid, while the OH ion, the proton acceptor, is acting as a Brønsted-Lowry base. The reverse reaction is also a Brønsted-Lowry acid base reaction:
BL bases. NH4+ and H2O are the BL acids. lt-chem-64081" style="width: 750px; height: 173px;" width="750px" height="173px" src="/@api/deki/files/92035/3765cabac9591a0fb3dd5878f56075e2.jpg" data-quail-id="174">
This means that both reactions are acid-base reactions by the Brønsted-Lowry definition. If you consider the species in this chemical reaction, two sets of similar species exist on both sides. Within each set, the two species differ by a proton in their formulas, and one member of the set is a Brønsted-Lowry acid, while the other member is a Brønsted-Lowry base. These sets are marked here:
The two sets—NH3/NH4+ and H2O/OH—are called conjugate acid-base pairs. We say that NH4+ is the conjugate acid of NH3, OH is the conjugate base of H2O, and so forth. Every Brønsted-Lowry acid-base reaction can be labeled with two conjugate acid-base pairs.
Example $2$
Identify the conjugate acid-base pairs in this equilibrium.
$(CH_{3})_{3}N+H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+}+OH^{-} \nonumber \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base. The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Exercise $2$
Identify the conjugate acid-base pairs in this equilibrium.
$NH_{2}^{-}+H_{2}O\rightleftharpoons NH_{3}+OH^{-} \nonumber \nonumber$
Answer
H2O (acid) and OH (base); NH2 (base) and NH3 (acid)
Chemistry is Everywhere: Household Acids and Bases
Many household products are acids or bases. For example, the owner of a swimming pool may use muriatic acid to clean the pool. Muriatic acid is another name for HCl(aq). In Section 4.6, vinegar was mentioned as a dilute solution of acetic acid [HC2H3O2(aq)]. In a medicine chest, one may find a bottle of vitamin C tablets; the chemical name of vitamin C is ascorbic acid (HC6H7O6).
One of the more familiar household bases is NH3, which is found in numerous cleaning products. NH3 is a base because it increases the OH ion concentration by reacting with H2O:
NH3(aq) + H2O(ℓ) → NH4+(aq) + OH(aq)
Many soaps are also slightly basic because they contain compounds that act as Brønsted-Lowry bases, accepting protons from H2O and forming excess OH ions. This is one explanation for why soap solutions are slippery.
Perhaps the most dangerous household chemical is the lye-based drain cleaner. Lye is a common name for NaOH, although it is also used as a synonym for KOH. Lye is an extremely caustic chemical that can react with grease, hair, food particles, and other substances that may build up and clog a water pipe. Unfortunately, lye can also attack body tissues and other substances in our bodies. Thus when we use lye-based drain cleaners, we must be very careful not to touch any of the solid drain cleaner or spill the water it was poured into. Safer, non-lye drain cleaners (like the one in the accompanying figure) use peroxide compounds to react on the materials in the clog and clear the drain.
Key Takeaways
• A Brønsted-Lowry acid is a proton donor; a Brønsted-Lowry base is a proton acceptor.
• Acid-base reactions include two sets of conjugate acid-base pairs. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.03%3A_Brnsted-Lowry_Acids_and_Bases.txt |
Learning Objectives
• Describe a titration experiment.
• Explain what an indicator does.
• Perform a titration calculation correctly.
The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a titration. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions.
In a titration, one reagent has a known concentration or amount, while the other reagent has an unknown concentration or amount. Typically, the known reagent (the titrant) is added to the unknown quantity and is dissolved in solution. The unknown amount of substance (the analyte) may or may not be dissolved in solution (but usually is). The titrant is added to the analyte using a precisely calibrated volumetric delivery tube called a burette (also spelled buret; see Figure $1$). The burette has markings to determine how much volume of solution has been added to the analyte. When the reaction is complete, it is said to be at the equivalence point; the number of moles of titrant can be calculated from the concentration and the volume, and the balanced chemical equation can be used to determine the number of moles (and then concentration or mass) of the unknown reactant.
For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted:
# mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl
We also have the balanced chemical reaction between HCl and NaOH:
HCl + NaOH → NaCl + H2O
So we can construct a conversion factor to convert to number of moles of NaOH reacted:
$0.002766\cancel{mol\, HCl}\times \frac{1\, mol\, NaOH}{1\cancel{mol\, HCl}}=0.002766\, mol\, NaOH\nonumber$
Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol):
$0.002766\cancel{mol\, HCl}\times \frac{40.00\, g\, NaOH}{1\cancel{mol\, HCl}}=0.1106\, g\, NaOH\nonumber$
This type of calculation is performed as part of a titration.
Example $1$
What mass of Ca(OH)2 is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO3? The balanced chemical equation is as follows:
2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
Solution
In liters, the volume is 0.04402 L. We calculate the number of moles of titrant:
# moles HNO3 = (0.04402 L)(0.0885 M) = 0.00390 mol HNO3
Using the balanced chemical equation, we can determine the number of moles of Ca(OH)2 present in the analyte:
$0.00390\cancel{mol\, HNO_{3}}\times \frac{1\, mol\, Ca(OH)_{2}}{2\cancel{mol\, HNO_{3}}}=0.00195\, mol\, Ca(OH)_{2}\nonumber$
Then we convert this to a mass using the molar mass of Ca(OH)2:
$0.00195\cancel{mol\, Ca(OH)_{2}}\times \frac{74.1\, g\, Ca(OH)_{2}}{\cancel{mol\, Ca(OH)_{2}}}=0.144\, g\, Ca(OH)_{2}\nonumber$
Exercise $1$
What mass of H2C2O4 is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows:
$\ce{H2C2O4 + 2NaOH → Na2C2O4 + 2H2O} \nonumber$
Answer
0.182 g
How does one know if a reaction is at its equivalence point? Usually, the person performing the titration adds a small amount of an indicator, a substance that changes color depending on the acidity or basicity of the solution. Because different indicators change colors at different levels of acidity, choosing the correct one is important in performing an accurate titration.
Summary
A titration is the quantitative reaction of an acid and a base. Indicators are used to show that all of the analyte has reacted with the titrant. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.04%3A_Acid-Base_Titrations.txt |
Learning Objectives
• Define a strong and a weak acid and base.
• Recognize an acid or a base as strong or weak.
• Determine if a salt produces an acidic or a basic solution.
Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way. Consider $\ce{HCl(aq)}$. When $\ce{HCl}$ is dissolved in $\ce{H2O}$, it completely dissociates into $\ce{H^{+}(aq)}$ and $\ce{Cl^{−}(aq)}$ ions; all the $\ce{HCl}$ molecules become ions:
$\ce{HCl\overset{100\%}{\rightarrow}H^{+}(aq)+Cl^{-}(aq)}\nonumber$
Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. Acetic acid ($\ce{HC2H3O2}$) is an example of a weak acid:
$\ce{HC2H3O2 \overset{\sim 5\%}{\longrightarrow} H^{+}(aq) + C2H3O2^{-}(aq)} \nonumber$
Because this reaction does not go 100% to completion, it is more appropriate to write it as a reversible reaction:
$\ce{HC2H3O2 \rightleftharpoons H^{+}(aq) + C2H3O2^{-}(aq)} \nonumber$
As it turns out, there are very few strong acids, which are given in Table $1$. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid.
Table $1$: Strong Acids and Bases
Acids Bases
HCl LiOH
HBr NaOH
HI KOH
HNO3 RbOH
H2SO4 CsOH
HClO3 Mg(OH)2
HClO4 Ca(OH)2
Sr(OH)2
Ba(OH)2
The issue is similar with bases: a strong base is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a weak base. There are very few strong bases (Table $1$); any base not listed is a weak base. All strong bases are OH compounds. So a base dependent on some other mechanism, such as NH3 (which does not contain OH ions as part of its formula), will be a weak base.
Example $1$
Identify each acid or base as strong or weak.
1. HCl
2. Mg(OH)2
3. C5H5N
Solution
1. Because HCl is listed in Table $1$, it is a strong acid.
2. Because Mg(OH)2 is listed in Table $1$, it is a strong base.
3. The nitrogen in C5H5N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base.
Exercise $1$
Identify each acid or base as strong or weak.
1. RbOH
2. HNO2
Answer a
strong base
Answer b
weak acid
Example $2$
Write the balanced chemical equation for the dissociation of Ca(OH)2 and indicate whether it proceeds 100% to products or not.
Solution
This is an ionic compound of Ca2+ ions and OH ions. When an ionic compound dissolves, it separates into its constituent ions:
$\ce{Ca(OH)2 → Ca^{2+}(aq) + 2OH^{-}(aq)} \nonumber \nonumber$
Because Ca(OH)2 is listed in Table $1$, this reaction proceeds 100% to products.
Exercise $2$
Write the balanced chemical equation for the dissociation of hydrazoic acid (HN3) and indicate whether it proceeds 100% to products or not.
Answer
The reaction is as follows:
$\ce{HN3 → H+(aq) + N3^{-}(aq)} \nonumber \nonumber$
It does not proceed 100% to products because hydrazoic acid is not a strong acid.
Certain salts will also affect the acidity or basicity of aqueous solutions because some of the ions will undergo hydrolysis, just like NH3 does, to make a basic solution. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyze, while salts with ions that are part of weak acids or bases will hydrolyze.
Consider NaCl. When it dissolves in an aqueous solution, it separates into Na+ ions and Cl ions:
$\ce{NaCl → Na+(aq) + Cl^{−}(aq)}\nonumber$
Will the Na+(aq) ion hydrolyze? If it does, it will interact with the OH ion to make NaOH:
$\ce{Na+(aq) + H2O → NaOH + H+(aq)}\nonumber$
However, NaOH is a strong base, which means that it is 100% ionized in solution:
$\ce{NaOH → Na^+(aq) + OH^{−}(aq)}\nonumber$
The free OH(aq) ion reacts with the H+(aq) ion to remake a water molecule:
$\ce{H+(aq) + OH^{−}(aq) → H2O}\nonumber$
The net result? There is no change, so there is no effect on the acidity or basicity of the solution from the Na+(aq) ion. What about the Cl ion? Will it hydrolyze? If it does, it will take an H+ ion from a water molecule:
$\ce{Cl^{−}(aq) + H2O → HCl + OH^{−}}\nonumber$
However, HCl is a strong acid, which means that it is 100% ionized in solution:
$\ce{HCl → H+(aq) + Cl^{−}(aq)}\nonumber$
The free H+(aq) ion reacts with the OH(aq) ion to remake a water molecule:
$\ce{H+(aq) + OH^{−}(aq) → H2O}\nonumber$
The net result? There is no change, so there is no effect on the acidity or basicity of the solution from the Cl(aq) ion. Because neither ion in NaCl affects the acidity or basicity of the solution, NaCl is an example of a neutral salt. A neutral salt is any ionic compound that does not affect the acidity or basicity of its aqueous solution.
Things change, however, when we consider a salt like NaC2H3O2. We already know that the Na+ ion won't affect the acidity of the solution. What about the acetate ion? If it hydrolyzes, it will take an H+ from a water molecule:
$\ce{C2H3O2^{−}(aq) + H2O \rightleftharpoons HC2H3O2 + OH−(aq)}\nonumber$
Does this happen? Yes, it does. Why? Because HC2H3O2 is a weak acid. Any chance a weak acid has to form, it will (the same with a weak base). As some C2H3O2 ions hydrolyze with H2O to make the molecular weak acid, OH ions are produced. OH ions make solutions basic. Thus NaC2H3O2 solutions are slightly basic, so such a salt is called a basic salt.
There are also salts whose aqueous solutions are slightly acidic. NH4Cl is an example. When NH4Cl is dissolved in H2O, it separates into NH4+ ions and Cl ions. We have already seen that the Cl ion does not hydrolyze. However, the NH4+ ion will:
$\ce{NH4+(aq) + H2O \rightleftharpoons NH3(aq) + H3O+(aq)}\nonumber$
Recall from Section 12.2, that H3O+ ion is the hydronium ion, the more chemically proper way to represent the H+ ion. This is the classic acid species in solution, so a solution of NH4+(aq) ions is slightly acidic. NH4Cl is an example of an acid salt. The molecule NH3 is a weak base, and it will form when it can, just like a weak acid will form when it can.
So there are two general rules:
1. If an ion derives from a strong acid or base, it will not affect the acidity of the solution.
2. If an ion derives from a weak acid, it will make the solution basic; if an ion derives from a weak base, it will make the solution acidic.
Example $3$
Identify each salt as acidic, basic, or neutral.
1. KCl
2. KNO2
3. NH4Br
Solution
1. The ions from KCl derive from a strong acid (HCl) and a strong base (KOH). Therefore, neither ion will affect the acidity of the solution, so KCl is a neutral salt.
2. Although the K+ ion derives from a strong base (KOH), the NO2 ion derives from a weak acid (HNO2). Therefore the solution will be basic, and KNO2 is a basic salt.
3. Although the Br ions derive from a strong acid (HBr), the NH4+ ion derives from a weak base (NH3), so the solution will be acidic, and NH4Br is an acidic salt.
Exercise $3$
Identify each salt as acidic, basic, or neutral.
1. (C5H5NH)Cl
2. Na2SO3
Answer a
acidic
Answer b
basic
Some salts are composed of ions that come from both weak acids and weak bases. The overall effect on an aqueous solution depends on which ion exerts more influence on the overall acidity. We will not consider such salts here.
Summary
• Strong acids and bases are 100% ionized in aqueous solution.
• Weak acids and bases are less than 100% ionized in aqueous solution.
• Salts of weak acids or bases can affect the acidity or basicity of their aqueous solutions. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.05%3A_Strong_and_Weak_Acids_and_Bases_and_their_Salts.txt |
Learning Objectives
• Describe the autoionization of water.
• Calculate the concentrations of H+ and OH in solutions, knowing the other concentration.
We have already seen that H2O can act as an acid or a base:
NH3 + H2O → NH4+ + OH (H2O acts as an acid)
HCl + H2O → H3O+ + Cl (H2O acts as a base)
It may not be surprising to learn, then, that within any given sample of water, some H2O molecules are acting as acids, and other H2O molecules are acting as bases. The chemical equation is as follows:
H2O + H2O → H3O+ + OH
This occurs only to a very small degree: only about 6 in 108 H2O molecules are participating in this process, which is called the autoionization of water. At this level, the concentration of both H+(aq) and OH(aq) in a sample of pure H2O is about 1.0 × 10−7 M. If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have
[H+] = [OH] = 1.0 × 10−7 M
for any sample of pure water because H2O can act as both an acid and a base. The product of these two concentrations is 1.0 × 10−14:
[H+] × [OH] = (1.0 × 10−7)(1.0 × 10−7) = 1.0 × 10−14
In acids, the concentration of H+(aq)—[H+]—is greater than 1.0 × 10−7 M, while for bases the concentration of OH(aq)—[OH]—is greater than 1.0 × 10−7 M. However, the product of the two concentrations—[H+][OH]—is always equal to 1.0 × 10−14, no matter whether the aqueous solution is an acid, a base, or neutral:
[H+][OH] = 1.0 × 10−14
This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted Kw:
Kw = [H+][OH] = 1.0 × 10−14
This means that if you know [H+] for a solution, you can calculate what [OH] has to be for the product to equal 1.0 × 10−14, or if you know [OH], you can calculate [H+]. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of Kw.
Example $1$
What is [OH] of an aqueous solution if [H+] is 1.0 × 10−4 M?
Solution
Using the expression and known value for Kw,
Kw = [H+][OH] = 1.0 × 10−14 = (1.0 × 10−4)[OH]
We solve by dividing both sides of the equation by 1.0 × 10−4:
$\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber$
It is assumed that the concentration unit is molarity, so [OH] is 1.0 × 10−10 M.
Exercise $1$
What is [H+] of an aqueous solution if [OH] is 1.0 × 10−9 M?
Answer
1.0 × 10−5 M
When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H+ or OH ions in the formula unit because [H+] or [OH] may not be the same as the concentration of the acid or base itself.
Example $2$
What is [H+] in a 0.0044 M solution of Ca(OH)2?
Solution
We begin by determining [OH]. The concentration of the solute is 0.0044 M, but because Ca(OH)2 is a strong base, there are two OH ions in solution for every formula unit dissolved, so the actual [OH] is two times this, or 2 × 0.0044 M = 0.0088 M. Now we can use the Kw expression:
[H+][OH] = 1.0 × 10−14 = [H+](0.0088 M)
Divide both sides by 0.0088:
$\left [ H^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M\nonumber$
[H+] has decreased significantly in this basic solution.
Exercise $2$
What is [OH] in a 0.00032 M solution of H2SO4? (Hint: assume both H+ ions ionize.)
Answer
1.6 × 10−11 M
For strong acids and bases, [H+] and [OH] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionized by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionization would need to be known before we can determine [H+] and [OH].
Example $3$
A 0.0788 M solution of HC2H3O2 is 3.0% ionized into H+ ions and C2H3O2 ions. What are [H+] and [OH] for this solution?
Solution
Because the acid is only 3.0% ionized, we can determine [H+] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form:
[H+] = 0.030 × 0.0788 = 0.00236 M
With this [H+], then [OH] can be calculated as follows:
$\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{0.00236}=4.2\times 10^{-12}M\nonumber$
This is about 30 times higher than would be expected for a strong acid of the same concentration.
Exercise $3$
A 0.0222 M solution of pyridine (C5H5N) is 0.44% ionized into pyridinium ions (C5H5NH+) and OH ions. What are [OH] and [H+] for this solution?
Answer
[OH] = 9.77 × 10−5 M; [H+] = 1.02 × 10−10 M
Summary
In any aqueous solution, the product of [H+] and [OH−] equals $1.0 \times 10^{−14}$ (at room temperature).
12.07: The pH Scale
Learning Objectives
• Define pH.
• Determine the pH of acidic and basic solutions.
As we have seen, [H+] and [OH] values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions.
pH is a logarithmic function of [H+]:
pH = −log[H+]
pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on [H+], we can summarize as follows:
• If pH < 7, then the solution is acidic.
• If pH = 7, then the solution is neutral.
• If pH > 7, then the solution is basic.
This is known as the pH scale and is the range of values from 0 to 14 that describes the acidity or basicity of a solution. You can use pH to quickly determine whether a given aqueous solution is acidic, basic, or neutral.
Example \(1\)
Label each solution as acidic, basic, or neutral based only on the stated pH.
1. milk of magnesia, pH = 10.5
2. pure water, pH = 7
3. wine, pH = 3.0
Solution
1. With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)2.)
2. Pure water, with a pH of 7, is neutral.
3. With a pH of less than 7, wine is acidic.
Exercise \(1\)
Identify each substance as acidic, basic, or neutral based only on the stated pH.
1. human blood, pH = 7.4
2. household ammonia, pH = 11.0
3. cherries, pH = 3.6
Answers
1. basic
2. basic
3. acidic
Table \(1\) gives the typical pH values of some common substances. Note that several food items are on the list, and most of them are acidic.
Table \(1\) Typical pH Values of Various Substances*
Substance pH
stomach acid 1.7
lemon juice 2.2
vinegar 2.9
soda 3.0
wine 3.5
coffee, black 5.0
milk 6.9
pure water 7.0
blood 7.4
seawater 8.5
milk of magnesia 10.5
ammonia solution 12.5
1.0 M NaOH 14.0
*Actual values may vary depending on conditions
pH is a logarithmic scale. A solution that has a pH of 1.0 has 10 times the [H+] as a solution with a pH of 2.0, which in turn has 10 times the [H+] as a solution with a pH of 3.0 and so forth.
Using the definition of pH, it is also possible to calculate [H+] (and [OH]) from pH and vice versa. The general formula for determining [H+] from pH is as follows:
[H+] = 10−pH
Key Takeaways
• pH is a logarithmic function of [H+].
• [H+] can be calculated directly from pH.
• pOH is related to pH and can be easily calculated from pH. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.06%3A_Autoionization_of_Water.txt |
Learning Objectives
• Define buffer.
• Correctly identify the two components of a buffer.
As indicated in Section 12.5, weak acids are relatively common, even in the foods we eat. But we occasionally encounter a strong acid or base, such as stomach acid, which has a strongly acidic pH of 1.7. By definition, strong acids and bases can produce a relatively large amount of H+ or OH ions and consequently have marked chemical activities. In addition, very small amounts of strong acids and bases can change the pH of a solution very quickly. If 1 mL of stomach acid [approximated as 0.1 M HCl(aq)] were added to the bloodstream and no correcting mechanism were present, the pH of the blood would decrease from about 7.4 to about 4.7—a pH that is not conducive to continued living. Fortunately, the body has a mechanism for minimizing such dramatic pH changes.
This mechanism involves a buffer, a solution that resists dramatic changes in pH. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus a salt derived from that weak acid, or a weak base plus a salt of that weak base. For example, a buffer can be composed of dissolved HC2H3O2 (a weak acid) and NaC2H3O2 (the salt derived from that weak acid). Another example of a buffer is a solution containing NH3 (a weak base) and NH4Cl (a salt derived from that weak base).
Let us use an HC2H3O2/NaC2H3O2 buffer to demonstrate how buffers work. If a strong base—a source of OH(aq) ions—is added to the buffer solution, those OH ions will react with the HC2H3O2 in an acid-base reaction:
$\ce{HC2H3O2(aq) + OH^{-}(aq) \rightarrow H2O(ℓ) + C2H3O^{-}2(aq)} \label{Eq1}$
Rather than changing the pH dramatically by making the solution basic, the added OH ions react to make H2O, so the pH does not change much.
If a strong acid—a source of H+ ions—is added to the buffer solution, the H+ ions will react with the anion from the salt. Because HC2H3O2 is a weak acid, it is not ionized much. This means that if lots of H+ ions and C2H3O2 ions are present in the same solution, they will come together to make HC2H3O2:
$\ce{H^{+}(aq) + C2H3O^{−}2(aq) \rightarrow HC2H3O2(aq)} \label{Eq2}$
Rather than changing the pH dramatically and making the solution acidic, the added H+ ions react to make molecules of a weak acid. Figure $1$ illustrates both actions of a buffer.
Buffers made from weak bases and salts of weak bases act similarly. For example, in a buffer containing NH3 and NH4Cl, NH3 molecules can react with any excess H+ ions introduced by strong acids:
$\ce{NH3(aq) + H^{+}(aq) \rightarrow NH^{+}4(aq)} \label{Eq3}$
while the ammonium ion ($\ce{NH4^{+}(aq)}$) can react with any hydroxide ions introduced by strong bases:
$\ce{NH^{+}4(aq) + OH^{-}(aq) \rightarrow NH3(aq) + H2O(ℓ)} \label{Eq4}$
Example $1$
Which combinations of compounds can make a buffer solution?
1. HCHO2 and NaCHO2
2. HCl and NaCl
3. CH3NH2 and CH3NH3Cl
4. NH3 and NaOH
Solution
1. HCHO2 is formic acid, a weak acid, while NaCHO2 is the salt made from the anion of the weak acid (the formate ion [CHO2]). The combination of these two solutes would make a buffer solution.
2. HCl is a strong acid, not a weak acid, so the combination of these two solutes would not make a buffer solution.
3. CH3NH2 is methylamine, which is like NH3 with one of its H atoms substituted with a CH3 group. Because it is not listed in Table $1$, we can assume that it is a weak base. The compound CH3NH3Cl is a salt made from that weak base, so the combination of these two solutes would make a buffer solution.
4. NH3 is a weak base, but NaOH is a strong base. The combination of these two solutes would not make a buffer solution.
Exercise $1$
Which combinations of compounds can make a buffer solution?
1. NaHCO3 and NaCl
2. H3PO4 and NaH2PO4
3. NH3 and (NH4)3PO4
4. NaOH and NaCl
Answer a
Yes.
Answer b
No. Need a weak acid or base and a salt of its conjugate base or acid.
Answer c
Yes.
Answer d
No. Need a weak base or acid.
Buffers work well only for limited amounts of added strong acid or base. Once either solute is completely reacted, the solution is no longer a buffer, and rapid changes in pH may occur. We say that a buffer has a certain capacity. Buffers that have more initial solute dissolved within them have larger capacities, as might be expected.
Human blood has a buffering system to minimize extreme changes in pH. One buffer in blood is based on the presence of HCO3 and H2CO3 [the second compound is another way to write CO2(aq)]. With this buffer present, even if some stomach acid were to find its way directly into the bloodstream, the change in the pH of blood would be minimal. Inside many of the body's cells, there is a buffering system based on phosphate ions.
Food and Drink Application: the Acid that Eases Pain
Although medicines are not exactly "food and drink," we do ingest them, so let's take a look at an acid that is probably the most common medicine: acetylsalicylic acid, also known as aspirin. Aspirin is well known as a pain reliever and antipyretic (fever reducer).
The structure of aspirin is shown in the accompanying figure. The acid part is circled; it is the H atom in that part that can be donated as aspirin acts as a Brønsted-Lowry acid. Because it is not given in Table $1$, acetylsalicylic acid is a weak acid. However, it is still an acid, and given that some people consume relatively large amounts of aspirin daily, its acidic nature can cause problems in the stomach lining, despite the stomach's defenses against its own stomach acid.
Because the acid properties of aspirin may be problematic, many aspirin brands offer a "buffered aspirin" form of the medicine. In these cases, the aspirin also contains a buffering agent—usually MgO—that regulates the acidity of the aspirin to minimize its acidic side effects.
As useful and common as aspirin is, it was formally marketed as a drug starting in 1899. The US Food and Drug Administration (FDA), the governmental agency charged with overseeing and approving drugs in the United States, wasn't formed until 1906. Some have argued that if the FDA had been formed before aspirin was introduced, aspirin may never have gotten approval due to its potential for side effects—gastrointestinal bleeding, ringing in the ears, Reye's syndrome (a liver problem), and some allergic reactions. However, recently aspirin has been touted for its effects in lessening heart attacks and strokes, so it is likely that aspirin will remain on the market.
Summary
A buffer is a solution that resists sudden changes in pH.
12.E: Acids and Bases (Exercises)
Additional Exercises
1. Write the balanced chemical equation between Zn metal and HCl(aq). The other product is ZnCl2.
2. Write the neutralization reaction in which ZnCl2, also found in Exercise 1, is the salt product.
3. Why isn't an oxide compound like CaO considered a salt? (Hint: what acid-base combination would be needed to make it if it were a salt?)
4. Metal oxides are considered basic because they react with H2O to form OH compounds. Write the chemical equation for a reaction that forms a base when CaO is combined with H2O.
5. Write the balanced chemical equation between aluminum hydroxide and sulfuric acid.
6. Write the balanced chemical equation between phosphoric acid and barium hydroxide.
7. Write the equation for the chemical reaction that occurs when caffeine (C8H10N4O2) acts as a Brønsted-Lowry base.
8. Citric acid (C6H8O7) is the acid found in citrus fruits. It can lose a maximum of three H+ ions in the presence of a base. Write the chemical equations for citric acid acting stepwise as a Brønsted-Lowry acid.
9. Can an amphiprotic substance be a strong acid and a strong base at the same time? Explain your answer.
10. Can an amphiprotic substance be a weak acid and a weak base at the same time? If so, explain why and give an example.
11. Under what conditions will the equivalence point of a titration be slightly acidic?
12. Under what conditions will the equivalence point of a titration be slightly basic?
13. Write the chemical equation for the autoionization of NH3.
14. Write the chemical equation for the autoionization of HF.
15. What is the pOH range for an acidic solution?
16. What is the pOH range for a basic solution?
17. The concentration of commercial HCl is about 12 M. What is its pH and pOH?
18. The concentration of concentrated H2SO4 is about 18 M. Assuming only one H+ comes off the H2SO4 molecule, what is its pH and pOH? What would the pH and pOH be if the second H+ were also ionized?
Answers
1. Zn + 2HCl → ZnCl2 + H2
2.
3. The O2− ion would come from H2O, which is not considered a classic acid in the Arrhenius sense.
4.
5. 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
6.
7. C8H10N4O2 + H2O → C8H10N4O2H+ + OH; the H+ ion attaches to one of the N atoms in the caffeine molecule.
8.
9. As a strong acid or base, an amphiprotic substance reacts 100% as an acid or a base, so it cannot be a base or an acid at the same time.
10.
11. if the salt produced is an acidic salt
12.
13. NH3 + NH3 → NH4+ + NH2
14.
15. pOH > 7
16.
17. pH = −1.08; pOH = 15.08 | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/12%3A_Acids_and_Bases/12.08%3A_Buffers.txt |
So far in this text, when presented a chemical reaction, it has been implicitly assumed that the reaction goes to completion. Indeed, previous stoichiometric calculations were based on this; when asked how much of a product is produced when so much of a reactant reacts, it was assumed that all of a reactant reacts. However, this is usually not the case; many reactions do not go to completion, and many chemists have to deal with that. In this chapter, we will study this phenomenon and see ways in which we can affect the extent of chemical reactions.
• 13.1: Prelude to Chemical Equilibrium
More chemical reactions come to an equilibrium. The actual position of the equilibrium—whether it favors the reactants or the products—is characteristic of a chemical reaction; it is difficult to see just by looking at the balanced chemical equation. But chemistry has tools to help explain the equilibrium of chemical reactions—the focus of this chapter.
• 13.2: Chemical Equilibrium
Chemical reactions eventually reach equilibrium, a point at which forward and reverse reactions balance each other's progress. Chemical equilibria are dynamic: the chemical reactions are always occurring; they just cancel each other's progress.
• 13.3: The Equilibrium Constant
Every chemical equilibrium can be characterized by an equilibrium constant, known as Keq. The Keq and KP expressions are formulated as amounts of products divided by amounts of reactants; each amount (either a concentration or a pressure) is raised to the power of its coefficient in the balanced chemical equation. Solids and liquids do not appear in the expression for the equilibrium constant.
• 13.4: Shifting Equilibria - Le Chatelier's Principle
Le Chatelier's principle addresses how an equilibrium shifts when the conditions of an equilibrium are changed. The direction of shift can be predicted for changes in concentrations, temperature, or pressure. Catalysts do not affect the position of an equilibrium; they help reactions achieve equilibrium faster.
• 13.5: Calculating Equilibrium Constant Values
• 13.6: Some Special Types of Equilibria
• 13.7: End-of-Chapter Material
These are exercises and select solutions to accompany Chapter 13 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
13: Chemical Equilibrium
Imagine you are stranded in a rowboat in the middle of the ocean. Suddenly, your boat springs a small leak, and you need to bail out water. You grab a bucket and begin to bail. After a few minutes, your efforts against the leak keep the water to only about half an inch, but any further bailing doesn't change the water level; the leak brings in as much water as you bail out.
You are at a dynamics equilibrium. Two opposing processes have reached the same speed, and there is no more overall change in the process.
Chemical reactions are like that as well. Most of them come to an equilibrium. The actual position of the equilibrium—whether it favors the reactants or the products—is characteristic of a chemical reaction; it is difficult to see just by looking at the balanced chemical equation. But chemistry has tools to help you understand the equilibrium of chemical reactions—the focus of this chapter.
13.02: Chemical Equilibrium
Learning Objectives
• Define chemical equilibrium.
• Recognize chemical equilibrium as a dynamic process.
Consider the following reaction occurring in a closed container (so that no material can go in or out):
H2 + I2 → 2HI
This is simply the reaction between elemental hydrogen and elemental iodine to make hydrogen iodide. The way the equation is written, we are led to believe that the reaction goes to completion, that all the H2 and the I2 react to make HI.
However, this is not the case. The reverse chemical reaction is also taking place:
2HI → H2 + I2
It acts to undo what the first reaction does. Eventually, the reverse reaction proceeds so quickly that it matches the speed of the forward reaction. When that happens, any continued overall reaction stops: the reaction has reached chemical equilibrium (sometimes just spoken as equilibrium; plural equilibria), the point at which the forward and reverse processes balance each other's progress.
Because two opposing processes are occurring at once, it is conventional to represent an equilibrium using a double arrow, like this:
$H_{2}+I_{2}\rightleftharpoons 2HI\nonumber$
The double arrow implies that the reaction is going in both directions. Note that the reaction must still be balanced.
Example $1$
Write the equilibrium equation that exists between calcium carbonate as a reactant and calcium oxide and carbon dioxide as products.
Solution
As this is an equilibrium situation, a double arrow is used. The equilibrium equation is written as follows:
$CaCO_{3}+\rightleftharpoons CaO+CO_{2}\nonumber$
Exercise $1$
Write the equilibrium equation between elemental hydrogen and elemental oxygen as reactants and water as the product.
Answer
$2H_{2}+O_{2}+\rightleftharpoons 2H_{2}O\nonumber$
One thing to note about equilibrium is that the reactions do not stop; both the forward reaction and the reverse reaction continue to occur. They both occur at the same rate, so any overall change by one reaction is canceled by the reverse reaction. We say that chemical equilibrium is dynamic, rather than static. Also, because both reactions are occurring simultaneously, the equilibrium can be written backward. For example, representing an equilibrium as
$H_{2}+I_{2}\rightleftharpoons 2HI\nonumber$
is the same thing as representing the same equilibrium as
$2HI\rightleftharpoons H_{2}+I_{2}\nonumber$
The reaction must be at equilibrium for this to be the case, however.
Key Takeaways
• Chemical reactions eventually reach equilibrium, a point at which forward and reverse reactions balance each other's progress.
• Chemical equilibria are dynamic: the chemical reactions are always occurring; they just cancel each other's progress. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/13%3A_Chemical_Equilibrium/13.01%3A_Prelude_to_Chemical_Equilibrium.txt |
Learning Objectives
• Explain the importance of the equilibrium constant.
• Construct an equilibrium constant expression for a chemical reaction.
In the mid 1860s, Norwegian scientists C. M. Guldberg and P. Waage noted a peculiar relationship between the amounts of reactants and products in an equilibrium. No matter how many reactants they started with, a certain ratio of reactants and products was achieved at equilibrium. Today, we call this observation the law of mass action. It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution,
$aA+bB\rightleftharpoons cC+dD\nonumber$
the equilibrium constant, also known as Keq, is defined by the following expression:
$K_{eq}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\nonumber$
where [A] is the molar concentration of species A at equilibrium, and so forth. The coefficients a, b, c, and d in the chemical equation become exponents in the expression for Keq. The Keq is a characteristic numerical value for a given reaction at a given temperature; that is, each chemical reaction has its own characteristic Keq. The concentration of each reactant and product in a chemical reaction at equilibrium is related; the concentrations cannot be random values, but they depend on each other. The numerator of the expression for Keq has the concentrations of every product (however many products there are), while the denominator of the expression for Keq has the concentrations of every reactant, leading to the common products over reactants definition for the Keq.
Let us consider a simple example. Suppose we have this equilibrium:
$A\rightleftharpoons B\nonumber$
There is one reactant, one product, and the coefficients on each are just 1 (assumed, not written). The Keq expression for this equilibrium is
$K_{eq}=\frac{[B]}{[A]}\nonumber$
(Exponents of 1 on each concentration are understood.) Suppose the numerical value of Keq for this chemical reaction is 2.0. If [B] = 4.0 M, then [A] must equal 2.0 M so that the value of the fraction equals 2.0:
$K_{eq}=\frac{[B]}{[A]}=\frac{4.0}{2.0}=2.0\nonumber$
By convention, the units are understood to be M and are omitted from the Keq expression. Suppose [B] were 6.0 M. For the Keq value to remain constant (it is, after all, called the equilibrium constant), then [A] would have to be 3.0 M at equilibrium:
$K_{eq}=\frac{[B]}{[A]}=\frac{6.0}{3.0}=2.0\nonumber$
If [A] were not equal to 3.0 M, the reaction would not be at equilibrium, and a net reaction would occur until that ratio was indeed 2.0. At that point, the reaction is at equilibrium, and any net change would cease. (Recall, however, that the forward and reverse reactions do not stop because chemical equilibrium is dynamic.)
The issue is the same with more complex expressions for the Keq; only the mathematics become more complex. Generally speaking, given a value for the Keq and all but one concentration at equilibrium, the missing concentration can be calculated.
Example $1$
Given the following reaction:
$\ce{H2 + I2 <=> 2HI}$
If the equilibrium [HI] is 0.75 M and the equilibrium [H2] is 0.20 M, what is the equilibrium [I2] if the Keq is 0.40?
Solution
We start by writing the Keq expression. Using the products over reactants approach, the Keq expression is as follows:
$K_{eq}=\frac{[HI]^{2}}{[H_{2}][I_{2}]}\nonumber$
Note that [HI] is squared because of the coefficient 2 in the balanced chemical equation. Substituting for the equilibrium [H2] and [HI] and for the given value of Keq:
$0.40=\frac{(0.75)^{2}}{(0.20)[I_{2}]}\nonumber$
To solve for [I2], we have to do some algebraic rearrangement: divide the 0.40 into both sides of the equation and multiply both sides of the equation by [I2]. This brings [I2] into the numerator of the left side and the 0.40 into the denominator of the right side:
$[I_{2}]=\frac{(0.75)^{2}}{(0.20)(0.40)}\nonumber$
Solving,
[I2] = 7.0 M
The concentration unit is assumed to be molarity. This value for [I2] can be easily verified by substituting 0.75, 0.20, and 7.0 into the expression for Keq and evaluating: you should get 0.40, the numerical value of Keq (and you do).
Exercise $1$
Given the following reaction:
$\ce{H2 + I2 <=> 2HI}$
If the equilibrium [HI] is 0.060 M and the equilibrium [I2] is 0.90 M, what is the equilibrium [H2] if the Keq is 0.40?
Answer
0.010 M
In some types of equilibrium problems, square roots, cube roots, or even higher roots need to be analyzed to determine a final answer. Make sure you know how to perform such operations on your calculator; if you do not know, ask your instructor for assistance.
Example $2$
The following reaction is at equilibrium:
$\ce{N2 + 3H2 <=> 2NH3}$
The Keq at a particular temperature is 13.7. If the equilibrium [N2] is 1.88 M and the equilibrium [NH3] is 6.62 M, what is the equilibrium [H2]?
Solution
We start by writing the Keq expression from the balanced chemical equation:
$K_{eq}=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\nonumber$
Substituting for the known equilibrium concentrations and the Keq, this becomes
$13.7=\frac{(6.62)^{2}}{(1.88)[H_{2}]^{3}}\nonumber$
Rearranging algebraically and then evaluating the numerical expression, we get
$[H_{2}]^{3}=\frac{(6.62)^{2}}{(1.88)(13.7)}=1.7015219754\nonumber$
To solve for [H2], we need to take the cube root of the equation. Performing this operation, we get
[H2] = 1.19 M
You should verify that this is correct using your own calculator to confirm that you know how to do a cube root correctly.
Exercise $2$
The following reaction is at equilibrium:
$\ce{N2 + 3H2 <=> 2NH3}$
The Keq at a particular temperature is 13.7. If the equilibrium [N2] is 0.055 M and the equilibrium [H2] is 1.62 M, what is the equilibrium [NH3]?
Answer
1.79 M
The Keq was defined earlier in terms of concentrations. For gas-phase reactions, the Keq can also be defined in terms of the partial pressures of the reactants and products, Pi. For the gas-phase reaction
$aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)\nonumber$
the pressure-based equilibrium constant, KP, is defined as follows:
$K_{P}=\frac{P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}}\nonumber$
where PA is the partial pressure of substance A at equilibrium in atmospheres, and so forth. As with the concentration-based equilibrium constant, the units are omitted when substituting into the expression for KP.
Example $3$
What is the KP for this reaction, given the equilibrium partial pressures of 0.664 atm for NO2 and 1.09 for N2O4?
$\ce{2NO2(g) <=> N2O4(g)}$
Solution
Write the KP expression for this reaction:
$K_{P}=\frac{P_{N_{2}O_{4}}}{P_{NO_{2}}^{2}}\nonumber$
Then substitute the equilibrium partial pressures into the expression and evaluate:
$K_{P}=\frac{(1.09)}{(0.664)^{2}}=2.47\nonumber$
Exercise $3$
What is the KP for this reaction, given the equilibrium partial pressures of 0.44 atm for H2, 0.22 atm for Cl2, and 2.98 atm for HCl?
$\ce{H2 + Cl2 <=> 2HCl}$
Answer
91.7
There is a simple relationship between Keq (based on concentration units) and KP (based on pressure units):
$K_{P}=K_{eq}\cdot \left ( RT\right )^{\Delta n}\nonumber$
where R is the ideal gas law constant (in units of L·atm/mol·K), T is the absolute temperature, and Δn is the change in the number of moles of gas in the balanced chemical equation, defined as ngas,prodsngas,rcts. Note that this equation implies that if the number of moles of gas are the same in reactants and products, Keq = KP.
Example $4$
What is the KP at 25°C for this reaction if the Keq is 4.2 × 10−2?
$\ce{N2 + 3H2 <=> 2NH3}$
Solution
Before we use the relevant equation, we need to do two things: convert the temperature to kelvins and determine Δn. Converting the temperature is easy:
T = 25 + 273 = 298 K
To determine the change in the number of moles of gas, take the number of moles of gaseous products and subtract the number of moles of gaseous reactants. There are 2 mol of gas as product and 4 mol of gas of reactant:
Δn = 2 − 4 = −2 mol
Note that Δn is negative. Now we can substitute into our equation, using R = 0.08205 L·atm/mol·K. The units are omitted for clarity:
KP = (4.2 × 10−2)(0.08205)(298)−2
Solving,
KP = 7.0 × 10−5
.Exercise $4$
What is the KP at 25°C for this reaction if the Keq is 98.3?,-
$\ce{I2(g) <=> 2I(g)}$
Answer
2.40 × 103
Finally, we recognize that many chemical reactions involve substances in the solid or liquid phases. For example, a particular chemical reaction is represented as follows:
$\ce{2NaHCO3(s) <=> Na2CO3(s) + CO2(g) + H2O(l)}$
This chemical equation includes all three phases of matter. This kind of equilibrium is called a heterogeneous equilibrium because there is more than one phase present.
The rule for heterogeneous equilibria is as follows: Do not include the concentrations of pure solids and pure liquids in Keq expressions. Only partial pressures for gas-phase substances or concentrations in solutions are included in the expressions of equilibrium constants. As such, the equilibrium constant expression for this reaction would simply be
$K_{P}=P_{CO_{2}}\nonumber$
because the two solids and one liquid would not appear in the expression.
Key Takeaways
• Every chemical equilibrium can be characterized by an equilibrium constant, known as Keq.
• The Keq and KP expressions are formulated as amounts of products divided by amounts of reactants; each amount (either a concentration or a pressure) is raised to the power of its coefficient in the balanced chemical equation.
• Solids and liquids do not appear in the expression for the equilibrium constant. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/13%3A_Chemical_Equilibrium/13.03%3A_The_Equilibrium_Constant.txt |
Learning Objectives
• Define Le Chatelier's principle.
• Predict the direction of shift for an equilibrium under stress.
Once equilibrium is established, the reaction is over, right? Not exactly. An experimenter has some ability to affect the equilibrium.
Chemical equilibria can be shifted by changing the conditions that the system experiences. We say that we "stress" the equilibrium. When we stress the equilibrium, the chemical reaction is no longer at equilibrium, and the reaction starts to move back toward equilibrium in such a way as to decrease the stress. The formal statement is called Le Chatelier's principle: If an equilibrium is stressed, then the reaction shifts to reduce the stress.
There are several ways to stress an equilibrium. One way is to add or remove a product or a reactant in a chemical reaction at equilibrium. When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.
Example $1$
Given this reaction at equilibrium:
$N_{2}+3H_{2}\rightleftharpoons 2NH_{3}\nonumber$
In which direction—toward reactants or toward products-—does the reaction shift if the equilibrium is stressed by each change?
1. H2 is added.
2. NH3 is added.
3. NH3 is removed.
Solution
1. If H2 is added, there is now more reactant, so the reaction will shift toward products to reduce the added H2.
2. If NH3 is added, there is now more product, so the reaction will shift toward reactants to reduce the added NH3.
3. If NH3 is removed, there is now less product, so the reaction will shift toward products to replace the product removed.
Exercise $1$
Given this reaction at equilibrium:
$CO(g)+Br_{2}(g)\rightleftharpoons COBr_{2}(g)\nonumber$
In which direction—toward reactants or toward products—does the reaction shift if the equilibrium is stressed by each change?
1. Br2 is removed.
2. COBr2 is added.
Answers
1. toward reactants
2. toward reactants
It is worth noting that when reactants or products are added or removed, the value of the Keq does not change. The chemical reaction simply shifts, in a predictable fashion, to reestablish concentrations so that the Keq expression reverts to the correct value.
How does an equilibrium react to a change in pressure? Pressure changes do not markedly affect the solid or liquid phases. However, pressure strongly impacts the gas phase. Le Chatelier's principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas. If the number of moles of gas is the same on both sides of the reaction, pressure has no effect.
Example $2$
What is the effect on this equilibrium if pressure is increased?
$N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)\nonumber$
Solution
According to Le Chatelier's principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts toward the products side.
Exercise $2$
What is the effect on this equilibrium if pressure is decreased?
$3O_{2}(g)\rightleftharpoons 2O_{3}(g)\nonumber$
Answer
Reaction shifts toward reactants.
What is the effect of temperature changes on an equilibrium? It depends on whether the reaction is endothermic or exothermic. Recall that endothermic means that energy is absorbed by a chemical reaction, while exothermic means that energy is given off by the reaction. As such, energy can be thought of as a reactant or a product, respectively, of a reaction:
endothermic: energy + reactants → products
exothermic: reactants → products + energy
Because temperature is a measure of the energy of the system, increasing temperature can be thought of as adding energy. The reaction will react as if a reactant or a product is being added and will act accordingly by shifting to the other side. For example, if the temperature is increased for an endothermic reaction, essentially a reactant is being added, so the equilibrium shifts toward products. Decreasing the temperature is equivalent to decreasing a reactant (for endothermic reactions) or a product (for exothermic reactions), and the equilibrium shifts accordingly.
Example $3$
Predict the effect of increasing the temperature on this equilibrium.
$PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}+60kJ\nonumber$
Solution
Because energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts back toward reactants.
Exercise $3$
Predict the effect of decreasing the temperature on this equilibrium.
$N_{2}O_{4}+57kJ\rightleftharpoons 2NO_{2}\nonumber$
Answer
Equilibrium shifts toward reactants.
In the case of temperature, the value of the equilibrium has changed because the Keq is dependent on temperature. That is why equilibria shift with changes in temperature.
A catalyst is a substance that increases the speed of a reaction. Overall, a catalyst is not a reactant and is not used up, but it still affects how fast a reaction proceeds. However, a catalyst does not affect the extent or position of a reaction at equilibrium. It helps a reaction achieve equilibrium faster.
Chemistry is Everywhere: Equilibria in the Garden
Hydrangeas are common flowering plants around the world. Although many hydrangeas are white, there is one common species (Hydrangea macrophylla) whose flowers can be either red or blue, as shown in the accompanying figure. How is it that a plant can have different colored flowers like this?
Interestingly, the color of the flowers is due to the acidity of the soil that the hydrangea is planted in. An astute gardener can adjust the pH of the soil and actually change the color of the flowers. However, it is not the H+ or OH ions that affect the color of the flowers. Rather, it is the presence of aluminum that causes the color change.
The solubility of aluminum in soil, and the ability of plants to absorb it, is dependent upon the acidity of the soil. If the soil is relatively acidic, the aluminum is more soluble, and plants can absorb it more easily. Under these conditions, hydrangea flowers are blue, as Al ions interact with anthocyanin pigments in the plant. In more basic soils, aluminum is less soluble, and under these conditions the hydrangea flowers are red. Gardeners who change the pH of their soils to change the color of their hydrangea flowers are therefore employing Le Chatelier's principle: the amount of acid in the soil changes the equilibrium of aluminum solubility, which in turn affects the color of the flowers.
Key Takeaways
• Le Chatelier's principle addresses how an equilibrium shifts when the conditions of an equilibrium are changed.
• The direction of shift can be predicted for changes in concentrations, temperature, or pressure.
• Catalysts do not affect the position of an equilibrium; they help reactions achieve equilibrium faster.
Exercise $1$
1. Define Le Chatelier's principle.
2. What is meant by a stress? What are some of the ways an equilibrium can be stressed?
3. Given this equilibrium, predict the direction of shift for each stress. $H_{2}(g)+I_{2}(s)+53kJ\rightleftharpoons 2HI(g)\nonumber$
1. decreased temperature
2. increased pressure
3. removal of HI
4. Given this equilibrium, predict the direction of shift for each stress. $H_{2}(g)+F_{2}(g)\rightleftharpoons 2HF(g)+546kJ\nonumber$
1. increased temperature
2. addition of H2
3. decreased pressure
5. Given this equilibrium, predict the direction of shift for each stress. $2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)+196kJ\nonumber$
1. removal of SO3
2. addition of O2
3. decreased temperature
6. Given this equilibrium, predict the direction of shift for each stress. $CO_{2}(g)+C(s)+171kJ\rightleftharpoons 2CO(g)\nonumber$
1. addition of CO
2. increased pressure
3. addition of a catalyst
7. The synthesis of NH3 uses this chemical reaction. $N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)+92kJ\nonumber$
Identify three stresses that can be imposed on the equilibrium to maximize the amount of NH3.
8. The synthesis of CaCO3 uses this chemical reaction. $CaO(s)+CO_{2}(g)\rightleftharpoons CaCO_{3}(s)+180kJ\nonumber$
Identify three stresses that can be imposed on the equilibrium to maximize the amount of CaCO3.
Answers
1. When an equilibrium is stressed, the equilibrium shifts to minimize that stress.
2.
1. toward reactants
• toward reactants
• toward products
•
1. toward products
2. toward products
3. toward products
•
• increased pressure, decreased temperature, removal of NH3 | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/13%3A_Chemical_Equilibrium/13.04%3A_Shifting_Equilibria_-_Le_Chatelier%27s_Principle.txt |
Learning Objective
• Calculate equilibrium concentrations from the values of the initial amounts and the Keq.
There are some circumstances in which, given some initial amounts and the Keq, you will have to determine the concentrations of all species when equilibrium is achieved. Such calculations are not difficult to do, especially if a consistent approach is applied. We will consider such an approach here.
Suppose we have this simple equilibrium. Its associated Keq is 4.0, and the initial concentration of each reactant is 1.0 M:
$\underset{1.0M}{H_{2}(g)}+\underset{1.0M}{Cl_{2}(g)}\rightleftharpoons 2HCl(g)\; \; \; \; \; K_{eq}=4.0\nonumber$
Because we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. But by how much will it proceed? We do not know, so let us assign it a variable. Let us assume that x M H2 reacts as the reaction goes to equilibrium. This means that at equilibrium, we have (1.0 − x) M H2 left over.
According to the balanced chemical equation, H2 and Cl2 react in a 1:1 ratio. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that if x M H2 reacts, x M Cl2 reacts as well. If we start with 1.0 M Cl2 at the beginning and we react x M, we have (1.0 − x) M Cl2 left at equilibrium.
How much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H2 (or Cl2) that reacts (from the coefficients in the balanced chemical equation), so if we lose x M H2, we gain 2x M HCl. So now we know the equilibrium concentrations of our species:
$\underset{(1.0-x)M}{H_{2}(g)}+\underset{(1.0-x)M}{Cl_{2}(g)}\rightleftharpoons \underset{2xM}{2HCl(g)}\; \; \; \; \; K_{eq}=4.0\nonumber$
We can substitute these concentrations into the Keq expression for this reaction and combine it with the known value of Keq:
$K_{eq}=\frac{[HCl]^{2}}{[H_{2}][Cl_{2}]}=\frac{(2x)^{2}}{(1-x)(1-x)}=4.0\nonumber$
This is an equation in one variable, so we should be able to solve for the unknown value. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well:
$\frac{(2x)^{2}}{(1-x)^{2}}=4.0\nonumber$
The fraction is a perfect square, as is the 4.0 on the right. So we can take the square root of both sides:
$\frac{(2x)}{(1-x)}=2.0\nonumber$
Now we rearrange and solve (be sure you can follow each step):
$2x=2.0-2.0x\ 4x=2.0\ x=0.50\nonumber$
Now we have to remind ourselves what x is—the amount of H2 and Cl2 that reacted—and that 2x is the equilibrium [HCl]. To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 − x and 2x to get the equilibrium concentrations of our species:
1.0 − x = 1.0 − 0.50 = 0.50 M = [H2] = [Cl2]2x = 2(0.50) = 1.0 M = [HCl]
The units are assumed to be molarity. To check, we simply substitute these concentrations and verify that we get the numerical value of the Keq, in this case 4.0:
$\frac{(1.0)^{2}}{(0.50)(0.50)}=4.0\nonumber$
We formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium. The initial values go in the first row of the chart. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. However, the change values must be in the proper stoichiometric ratio as indicated by the balanced chemical equation. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. The expressions in the equilibrium row are substituted into the Keq expression, which yields an algebraic equation that we try to solve.
The ICE chart for the above example would look like this:
H2(g) + Cl2(g) $⇄$ 2HCl(g) Keq = 4.0
I 1.0 1.0 0
C x x +2x
E 1.0 − x 1.0 − x +2x
Substituting the last row into the expression for the Keq yields
$K_{eq}=\frac{[HCl]^{2}}{[H_{2}][Cl_{2}]}=\frac{(2x)^{2}}{(1-x)(1-x)}=4.0\nonumber$
which, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. The ICE chart is a more formalized way to do these types of problems. The + sign is included explicitly in the change row of the ICE chart to avoid any confusion.
Sometimes when an ICE chart is set up and the Keq expression is constructed, a more complex algebraic equation will result. One of the more common equations has an x2 term in it and is called a quadratic equation. There will be two values possible for the unknown x, and for a quadratic equation with the general formula ax2 + bx + c = 0 (where a, b, and c are the coefficients of the quadratic equation), the two possible values are as follows:
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\nonumber$
One value of x is the + sign used in the numerator, and the other value of x is the − sign used in the numerator. In this case, one value of x typically makes no sense as an answer and can be discarded as physically impossible, leaving only one possible value and the resulting set of concentrations. Example 9 illustrates this.
Example $1$:
Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.
$\underset{0.55M}{COI_{2}(g)}\rightleftharpoons \underset{0}{CO(g)}+\underset{0}{I_{2}(g)}\; \; \; \; \; K_{eq}=0.00088\nonumber$
Solution
The ICE chart is set up like this. First, the initial values:
COI2(g) $⇄$ CO(g) + I2(g)
I 0.55 0 0
C
E
Some of the COI2 will be lost, but how much? We do not know, so we represent it by the variable x. So x M COI2 will be lost, and for each COI2 that is lost, x M CO and x M I2 will be produced. These expressions go into the change row:
COI2(g) $⇄$ CO(g) + I2(g)
I 0.55 0 0
C x +x +x
E
At equilibrium, the resulting concentrations will be a combination of the initial amount and the changes:
COI2(g) $⇄$ CO(g) + I2(g)
I 0.55 0 0
C x +x +x
E 0.55 − x +x +x
The expressions in the equilibrium row go into the Keq expression:
$K_{eq}=\frac{[CO][I_{2}]}{[COI_{2}]}=0.00088=\frac{(x)(x)}{(0.55-x))}\nonumber$
We rearrange this into a quadratic equation that equals 0:
0.000484 − 0.00088x = x2x2 + 0.00088x − 0.000484 = 0
Now we use the quadratic equation to solve for the two possible values of x:
$x=\frac{-0.00088\pm \sqrt{(0.00088)^{2}-4(1)(-0.000484)}}{2(1)}\nonumber$
Evaluate for both signs in the numerator—first the + sign and then the − sign:
x = 0.0216 or x = −0.0224
Because x is the final concentration of both CO and I2, it cannot be negative, so we discount the second numerical answer as impossible. Thus x = 0.0216.
Going back to determine the final concentrations using the expressions in the E row of our ICE chart, we have
[COI2] = 0.55 − x = 0.55 − 0.0216 = 0.53 M[CO] = x = 0.0216 M[I2] = x = 0.0216 M
You can verify that these numbers are correct by substituting them into the Keq expression and evaluating and comparing to the known Keq value.
Exercise $1$
Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.
$\underset{0.075M}{N_{2}H_{2}(g)}\rightleftharpoons \underset{0}{N_{2}(g)}+\underset{0}{H_{2}(g)}\; \; \; \; \; K_{eq}=0.052\nonumber$
Answer
The completed ICE chart is as follows:
N2H2(g) $⇄$ N2(g) + H2(g)
I 0.075 0 0
C x +x +x
E 0.075 − x +x +x
Solving for x gives the equilibrium concentrations as [N2H2] = 0.033 M; [N2] = 0.042 M; and [H2] = 0.042 M
Key Takeaway
• An ICE chart is a convenient way to determine equilibrium concentrations from starting amounts. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/13%3A_Chemical_Equilibrium/13.05%3A_Calculating_Equilibrium_Constant_Values.txt |
Learning Objective
• Identify several special chemical equilibria and construct their Ka expressions.
In one sense, all chemical equilibria are treated the same. However, there are several classes of reactions that are noteworthy because of either the identities of the reactants and products, or the form of the Keq expression.
Weak Acids and Bases
In Chapter 12 - Acids and Bases, we noted how some acids and bases are strong and some are weak. If an acid or base is strong, it is ionized 100% in H2O. HCl(aq) is an example of a strong acid:
$HCl(aq)=\overset{100\%}{\rightarrow}H^{+}(aq)+Cl^{-}(aq)\nonumber$
However, if an acid or base is weak, it may dissolve in H2O, but does not ionize completely. This means that there is an equilibrium between the unionized acid or base and the ionized form. HC2H3O2 is an example of a weak acid:
$HC_{2}H_{3}O_{2}(aq)\rightleftharpoons H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq)\nonumber$
HC2H3O2 is soluble in H2O (in fact, it is the acid in vinegar), so the reactant concentration will appear in the equilibrium constant expression. But not all the molecules separate into ions. This is the case for all weak acids and bases.
An acid dissociation constant, Ka, is the equilibrium constant for the dissociation of a weak acid into ions. Note the a subscript on the K; it implies that the substance is acting as an acid. The larger Ka is, the stronger the acid is. Table $1$ - Acid Dissociation Constants for Some Weak Acids, lists several acid dissociation constants. Keep in mind that they are just equilibrium constants.
Table, two columns and 7 rows. The first column on the right has different acids in the rows underneath. The second column on the right side has the acid dissociation content for the corresponding acid in the rows underneath.
Acid Ka
HC2H3O2 1.8 × 10−5
HClO2 1.1 × 10−2
H2PO4 6.2 × 10−8
HCN 6.2 × 10−10
HF 6.3 × 10−4
HNO2 5.6 × 10−4
H3PO4 7.5 × 10−3
Table $1$ Acid Dissociation Constants for Some Weak Acids
Note also that the acid dissociation constant refers to one H+ ion coming off of the initial reactant. Thus the acid dissociation constant for H3PO4 refers to this equilibrium:
$H_{3}PO_{4}(aq)\rightleftharpoons H^{+}(aq)+H_{2}PO_{4}^{-}(aq)\; \; \; \; \; K_{a}=7.5\times 10^{-3}\nonumber$
The H2PO4 ion, called the dihydrogen phosphate ion, is also a weak acid with its own acid dissociation constant:
$H_{2}PO_{4}^{-}(aq)\rightleftharpoons H^{+}(aq)+HPO_{4}^{2-}(aq)\; \; \; \; \; K_{a}=6.2\times 10^{-8}\nonumber$
Thus, for so-called polyprotic acids, each H+ ion comes off in sequence; each H+ ion that ionizes does so with its own characteristic Ka.
Example $1$
Write the equilibrium equation and the Ka expression for HSO4 acting as a weak acid.
Solution
HSO4 acts as a weak acid by separating into an H+ ion and an SO42 ion:
$HSO_{4}^{-}(aq)\rightleftharpoons H^{+}(aq)+SO_{4}^{2-}(aq)\nonumber$
The Ka is written just like any other equilibrium constant, in terms of the concentrations of products divided by concentrations of reactants:
$K_{a}=\frac{[H^{+}][SO_{4}^2-]}{[HSO_{4}^{-}]}\nonumber$
Exercise $1$
Write the equilibrium equation and the Ka expression for HPO42 acting as a weak acid.
Answer
$HPO_{4}^{2-}(aq)\rightleftharpoons H^{+}(aq)+PO_{4}^{3-}(aq)\; \; \; \; \; K_{a}=\frac{[H^{+}][PO_{4}^{3-}]}{[HPO_{4}^{2-}]}\nonumber$
The Ka is used in equilibrium constant problems just like other equilibrium constants are. However, in some cases, we can simplify the mathematics if the numerical value of the Ka is small, much smaller than the concentration of the acid itself. Example 11 illustrates this.
Example $2$
What is the pH of a 1.00 M solution of HC2H3O2? The Ka of HC2H3O2 is 1.8 × 10−5.
Solution
This is a two-part problem. We need to determine [H+] and then use the definition of pH to determine the pH of the solution. For the first part, we can use an ICE chart:
Solutions to Example 13.6.2
HC2H3O2(aq) $⇄$ H+(g) + C2H3O2(g)
I 1.00 0 0
C x +x +x
E 1.00 − x +x +x
We now construct the Ka expression, substituting the concentrations from the equilibrium row in the ICE chart:
$K_{a}=\frac{[H^{+}][C_{2}H_{3}O_{2}^{-}]}{[HC_{2}H_{3}O_{2}]}=\frac{(x)(x)}{(1.00-x)}=1.8\times 10^{-5}\nonumber$
Here is where a useful approximation comes in: at 1.8 × 10−5, HC2H3O2 will not ionize very much, so we expect that the value of x will be small. It should be so small that in the denominator of the fraction, the term (1.00 − x) will likely be very close to 1.00. As such, we would introduce very little error if we simply neglect the x in that term, making it equal to 1.00:
(1.00 − x) ≈ 1.00 for small values of x
This simplifies the mathematical expression we need to solve:
$\frac{(x)(x)}{1.00}=1.8\times 10^{-5}\nonumber$
This is much easier to solve than a more complete quadratic equation. The new equation to solve becomes
x2 = 1.8 × 10−5
Taking the square root of both sides,
x = 4.2 × 10−3
Because x is the equilibrium concentration of H+ and C2H3O2, we thus have
[H+] = 4.2 × 10−3 M
Notice that we are justified by neglecting the x in the denominator; it truly is small compared to 1.00. Now we can determine the pH of the solution:
pH = −log[H+] = −log(4.2 × 10−3) = 2.38
Exercise $2$
What is the pH of a 0.500 M solution of HCN? The Ka of HCN is 6.2 × 10−10.
Answer
4.75
Weak bases also have dissociation constants, labeled Kb (the b subscript stands for base). However, values of Kb are rarely tabulated because there is a simple relationship between the Kb of a base and the Ka of its conjugate acid:
Ka × Kb = 1.0 × 10−14
Thus it is simple to calculate the Kb of a base from the Ka of its conjugate acid.
Example $3$
What is the value of Kb for C2H3O2, which can accept a proton and act as a base?
Solution
To determine the Kb for C2H3O2, we need to know the Ka of its conjugate acid. The conjugate acid of C2H3O2 is HC2H3O2. The Ka for HC2H3O2 is in Table $1$ "Acid Dissociation Constants for Some Weak Acids" and is 1.8 × 10−5. Using the mathematical relationship between Ka and Kb:
(1.8 × 10−5)Kb = 1.0 × 10−14
Solving,
$K_{b}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.6\times 10^{-10}\nonumber$
Exercise $3$
What is the value of Kb for PO43, which can accept a proton and act as a base? The Ka for HPO42 is 2.2 × 10−13.
Answer
4.5 × 10−2
Autoionization of water
In Chapter 12, "Acids and Bases," we introduced the autoionization of water—the idea that water can act as a proton donor and proton acceptor simultaneously. Because water is not a strong acid (Table 12.5.1 - Strong Acids and Bases), it must be a weak acid, which means that its behavior as an acid must be described as an equilibrium. That equilibrium is as follows:
$H_{2}O(l)+H_{2}O(l)\rightleftharpoons H_{3}O^{+}(aq)+OH^{-}(aq)\nonumber$
The equilibrium constant includes [H3O+] and [OH] but not [H2O(ℓ)] because it is a pure liquid. Hence the expression does not have any terms in its denominator:
K = [H3O+][OH] ≡ Kw = 1.0 × 10−14
This is the same Kw that was introduced in Chapter 12 and the same 1.0 × 10−14 that appears in the relationship between the Ka and the Kb of a conjugate acid-base pair. In fact, we can rewrite this relationship as follows:
Ka × Kb = Kw
Insoluble Compounds
In Chapter 4, section 4.3: "Types of Chemical Reactions - Single and Double Displacement Reactions," the concept of soluble and insoluble compounds was introduced. Solubility rules were presented that allow a person to predict whether certain simple ionic compounds will or will not dissolve.
Describing a substance as soluble or insoluble is a bit misleading because virtually all substances are soluble; they are just soluble to different extents. In particular, for ionic compounds, what we typically describe as an insoluble compound can actually be ever so slightly soluble; an equilibrium is quickly established between the solid compound and the ions that do form in solution. Thus, the hypothetical compound MX does in fact dissolve but only very slightly. That means we can write an equilibrium for it:
$MX(s)\rightleftharpoons M^{+}(aq)+X^{-}(aq)\nonumber$
The equilibrium constant for a compound normally considered insoluble is called a solubility product constant and is labeled Ksp (with the subscript sp, meaning "solubility product"). Because the reactant is a solid, its concentration does not appear in the Ksp expression, so like Kw, expressions for Ksp do not have denominators. For example, the chemical equation and the expression for the Ksp for AgCl, normally considered insoluble, are as follows:
$AgCl(s)\rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)\; \; \; \; \; K_{sp}=[Ag^{+}][Cl^{-}]\nonumber$
Table $2$ - Solubility Product Constants for Slightly Soluble Ionic Compounds, lists some values of the Ksp for slightly soluble ionic compounds.
Table, two columns and 8 rows. The first column on the right has different compounds in the rows underneath. The second column on the right side has the acid dissociation content for the corresponding compounds in the rows underneath.
Compound Ksp
BaSO4 1.1 × 10−10
Ca(OH)2 5.0 × 10−6
Ca3(PO4)2 2.1 × 10−33
Mg(OH)2 5.6 × 10−12
HgI2 2.9 × 10−29
AgCl 1.8 × 10−10
AgI 8.5 × 10−17
Ag2SO4 1.5 × 10−5
Table $2$ Solubility Product Constants for Slightly Soluble Ionic Compounds.
Example $4$
Write the Ksp expression for Ca3(PO4)2.
Solution
Recall that when an ionic compound dissolves, it separates into its individual ions. For Ca3(PO4)2, the ionization reaction is as follows:
$Ca_{3}(PO_{4})_{2}(s)\rightleftharpoons 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber$
Hence the Ksp expression is
Ksp = [Ca2+]3[PO43]2
Exercise $4$
Write the Ksp expression Ag2SO4.
Answer
Ksp = [Ag+]2[SO42]
Equilibrium problems involving the Ksp can also be done, and they are usually more straightforward than other equilibrium problems because there is no denominator in the Ksp expression. Care must be taken, however, in completing the ICE chart and evaluating exponential expressions.
Example $5$
What are [Ag+] and [Cl] in a saturated solution of AgCl? The Ksp of AgCl is 1.8 × 10−10.
Solution
The chemical equation for the dissolving of AgCl is
$AgCl(s)\rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)\nonumber$
The Ksp expression is as follows:
Ksp = [Ag+][Cl]
So the ICE chart for the equilibrium is as follows:
Solutions to Example 13.6.5
AgCl(s) $⇄$ Ag+(aq) + Cl(aq)
I 0 0
C x +x +x
E +x +x
Notice that we have little in the column under AgCl except the stoichiometry of the change; we do not need to know its initial or equilibrium concentrations because its concentration does not appear in the Ksp expression. Substituting the equilibrium values into the expression:
(x)(x) = 1.8 × 10−10
Solving,
x2 = 1.8 × 1010x = 1.3 × 10−5
Thus [Ag+] and [Cl] are both 1.3 × 10−5 M.
Exercise $5$
What are [Ba2+] and [SO42] in a saturated solution of BaSO4? The Ksp of BaSO4 is 1.1 × 10−10.
Answer
1.0 × 10−5 M
Example $6$
What are [Ca2+] and [PO43] in a saturated solution of Ca3(PO4)2? The Ksp of Ca3(PO4)2 is 2.1 × 10−33.
Solution
This is similar to Example 14, but the ICE chart is much different because of the number of ions formed.
Solutions to Example 13.6.6
Ca3(PO4)2(s) $⇄$ 3Ca2+(aq) + 2PO43(aq)
I 0 0
C x +3x +2x
E +3x +2x
For every unit of Ca3(PO4)2 that dissolves, three Ca2+ ions and two PO43 ions are formed. The expression for the Ksp is also different:
Ksp = [Ca2+]3[PO43]2 = 2.1 × 10−33
Now when we substitute the unknown concentrations into the expression, we get
(3x)3(2x)2 = 2.1 × 10−33
When we raise each expression inside parentheses to the proper power, remember that the power affects everything inside the parentheses, including the number. So,
(27x3)(4x2) = 2.1 × 10−33
Simplifying,
108x5 = 2.1 × 10−33
Dividing both sides of the equation by 108, we get
x5 = 1.9 × 10−35
Now we take the fifth root of both sides of the equation (be sure you know how to do this on your calculator):
x = 1.1 × 10−7
We are not done yet. We still need to determine the concentrations of the ions. According to the ICE chart, [Ca2+] is 3x, not x. So
[Ca2+] = 3x = 3 × 1.1 × 10−7 = 3.3 × 10−7 M
[PO43] is 2x, so
[PO43] = 2x = 2 × 1.1 × 10−7 = 2.2 × 10−7 M
Exercise $6$
What are [Mg2+] and [OH] in a saturated solution of Mg(OH)2? The Ksp of Mg(OH)2 is 5.6 × 10−12.
Answer
[Mg2+] = 1.1 × 10−4 M; [OH] = 2.2 × 10−4 M
Food and Drink Application: Solids in Your Wine Bottle
People who drink wine from bottles (as opposed to boxes) will occasionally notice some insoluble materials in the wine, either crusting the bottle, stuck to the cork, or suspended in the liquid wine itself. The accompanying figure shows a cork encrusted with colored crystals. What are these crystals?
One of the acids in wine is tartaric acid (H2C4H4O6). Like the other acids in wine (citric and malic acids, among others), tartaric acid imparts a slight tartness to the wine. Tartaric acid is rather soluble in H2O, dissolving over 130 g of the acid in only 100 g of H2O. However, the potassium salt of singly ionized tartaric acid, potassium hydrogen tartrate (KHC4H4O6; also known as potassium bitartrate and better known in the kitchen as cream of tartar), has a solubility of only 6 g per 100 g of H2O. Thus, over time, wine stored at cool temperatures will slowly precipitate potassium hydrogen tartrate. The crystals precipitate in the wine or grow on the insides of the wine bottle and, if the bottle is stored on its side, on the bottom of the cork. The color of the crystals comes from pigments in the wine; pure potassium hydrogen tartrate is clear in its crystalline form, but in powder form it is white.
The crystals are harmless to ingest; indeed, cream of tartar is used as an ingredient in cooking. However, most wine drinkers do not like to chew their wine, so if tartrate crystals are present in a wine, the wine is usually filtered or decanted to remove the crystals. Tartrate crystals are almost exclusively in red wines; white and rose wines do not have as much tartaric acid in them.
Key Takeaway
• Equilibrium constants exist for certain groups of equilibria, such as weak acids, weak bases, the autoionization of water, and slightly soluble salts.
13.07: End-of-Chapter Material
Additional Exercises
1. What is the relationship between the Ksp expressions for a chemical reaction and its reverse chemical reaction?
2. What is the relationship between the Kw value for H2O and its reverse chemical reaction?
3. For the equilibrium $PCl_{3}(g)+Cl^{2+}(g)\rightleftharpoons PCl_{5}(g)+60kJ\nonumber$
list four stresses that serve to increase the amount of PCl5.
4. For the equilibrium $N_{2}O_{4}+57kJ\rightleftharpoons 2NO_{2}\nonumber$
list four stresses that serve to increase the amount of NO2.
5. Does a very large Keq favor the reactants or the products? Explain your answer.
6. Is the Keq for reactions that favor reactants large or small? Explain your answer.
7. Show that Ka × Kb = Kw by determining the expressions for these two reactions and multiplying them together. $HX(aq)\rightleftharpoons H^{+}(aq)+X^{-}(aq)\ X^{+}(aq)+H_{2}O(l)\rightleftharpoons HX(aq)+OH^{-}(aq)\nonumber$
8. Is the conjugate base of a strong acid weak or strong? Explain your answer.
9. What is the solubility in moles per liter of AgCl? Use data from Table $2$ - Solubility Product Constants for Slightly Soluble Ionic Compounds.
10. What is the solubility in moles per liter of Ca(OH)2? Use data from Table $2$ - Solubility Product Constants for Slightly Soluble Ionic Compounds.
11. Under what conditions is Keq = KP?
12. Under what conditions is Keq > KP when the temperature is 298 K?
13. What is the pH of a saturated solution of Mg(OH)2? Use data from Table $2$ - Solubility Product Constants for Slightly Soluble Ionic Compounds.
14. What are the pH and the pOH of a saturated solution of Fe(OH)3? The Ksp of Fe(OH)3 is 2.8 × 10−39.
1. For a salt that has the general formula MX, an ICE chart shows that the Ksp is equal to x2, where x is the concentration of the cation. What is the appropriate formula for the Ksp of a salt that has a general formula of MX2?
2. Referring to Exercise 15, what is the appropriate formula for the Ksp of a salt that has a general formula of M2X3 if the concentration of the cation is defined as 2x, rather than x?
3. Consider a saturated solution of PbBr2(s). If [Pb2+] is 1.33 × 10−5 M, find each of the following.
1. [Br]
2. the Ksp of PbBr2(s)
4. Consider a saturated solution of Pb3(PO4)2(s). If [Pb2+] is 7.34 × 10−14 M, find each of the following.
1. [PO43]
2. the Ksp of Pb3(PO4)2(s)
Answers
1. They are reciprocals of each other.
2.
3. increase the pressure; decrease the temperature; add PCl3; add Cl2; remove PCl5
4.
1. favor products because the numerator of the ratio for the Keq is larger than the denominator
2.
3. $K_{a}\times K_{b}=\frac{[H^{+}][X]}{[HX]}\times \frac{[HX][OH]}{[X]}=[H^{+}][OH^{-}]=K_{W}\nonumber$
4.
5. 1.3 × 10−5 mol/L
6.
7. Keq = KP when the number of moles of gas on both sides of the reaction is the same.
8.
9. 10.35
10.
11. 4x3
12.
1. 2.66 × 10−5 M
2. 9.41 × 10−15 | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/13%3A_Chemical_Equilibrium/13.06%3A_Some_Special_Types_of_Equilibria.txt |
One important type of chemical reaction is the oxidation-reduction reaction, also known as the redox reaction. Although we introduced redox reactions in Section 4.7, it is worth reviewing some basic concepts.
• 14.1: Introduction to Oxidation and Reduction
Most of us are familiar with rusty iron: metal that has a dark red-brown scale that falls off an object, ultimately weakening it. Although we usually attribute rusting exclusively to iron, this process occurs with many materials. The more formal term for rusting is corrosion. Corrosion is an example of the type of chemical reaction discussed in this chapter. Although we usually think of corrosion as bad, the reaction it typifies can actually be put to good use.
• 14.2: Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another. Oxidation numbers are used to keep track of electrons in atoms. There are rules for assigning oxidation numbers to atoms. Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number (gain of electrons).
• 14.3: Balancing Redox Reactions
Redox reactions can be balanced by inspection or by the half reaction method. A solvent may participate in redox reactions; in aqueous solutions, H2O, H+, and OH− may be reactants or products.
• 14.4: Applications of Redox Reactions - Voltaic Cells
A voltaic cell produces electricity as a redox reaction occurs. The voltage of a voltaic cell can be determined by the reduction potentials of the half reactions. Voltaic cells are fashioned into batteries, which are a convenient source of electricity.
• 14.5: Electrolysis
Electrolysis is the forcing of a nonspontaneous redox reaction to occur by the introduction of electricity into a cell from an outside source. Electrolysis is used to isolate elements and electroplate objects.
• 14.E: Oxidation-Reduction Reaction (Exercises)
These are exercises and select solutions to accompany Chapter 14 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
Thumbnail: Copper from a wire is displaced by silver in a silver nitrate solution it is dipped into, and solid silver precipitates out. (CC SA-BY 3.0 au; Toby Hudson via Wikipedia).
14: Oxidation and Reduction
Most of us are familiar with rusty iron: metal that has a dark red-brown scale that falls off an object, ultimately weakening it. Although we usually attribute rusting exclusively to iron, this process occurs with many materials. The more formal term for rusting is corrosion. Corrosion is defined as the disintegration of a material due to chemical reactions with other substances in the environment. In many cases, oxygen in the air causes the disintegration. Corrosion is not uniformly destructive. Although the corrosion of iron is generally considered bad, the corrosion of aluminum and copper forms a protective barrier on the surface of the metal, protecting it from further reaction with the environment.
Having said that, it has been estimated that as much as 5% of expenditures in the United States apply to fixing problems caused by corrosion. The replacement of structures built with iron, steel, aluminum, and concrete must be performed regularly to keep these structures safe. As an example of what might happen, consider the story of the Silver Bridge on US Interstate 35, connecting West Virginia and Ohio. On December 15, 1967, the 39-year-old bridge collapsed, killing 46 people. The ultimate cause of the collapse was determined to be corrosion of a suspension chain on the Ohio side of the bridge.
Corrosion is an example of the type of chemical reaction discussed in this chapter. Although we usually think of corrosion as bad, the reaction it typifies can actually be put to good use. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/14%3A_Oxidation_and_Reduction/14.01%3A_Introduction_to_Oxidation_and_Reduction.txt |
Learning Objectives
• Define oxidation and reduction.
• Assign oxidation numbers to atoms in simple compounds.
• Recognize a reaction as an oxidation-reduction reaction.
Consider this chemical reaction:
$\ce{Mg(s) + Cl2(g) → MgCl2}\nonumber$
The reactants are two electrically neutral elements; they have the same number of electrons as protons. The product, however, is ionic; it is composed of Mg2+ and Cl ions. Somehow, the individual Mg atoms lose two electrons to make the Mg2+ ion, while the Cl atoms gain an electron to become Cl ions. This reaction involves the transfer of electrons between atoms.
The process of losing and gaining electrons occurs simultaneously. However, mentally we can separate the two processes. Oxidation is defined as the loss of one or more electrons by an atom. Reduction is defined as the gain of one or more electrons by an atom. So oxidation and reduction always occur together; it is only mentally that we can separate them. Chemical reactions that involve the transfer of electrons are called oxidation-reduction (or redox) reactions.
Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate.
Assigning Oxidation Numbers
The rules for assigning oxidation numbers to atoms are as follows:
1. Atoms in their elemental state are assigned an oxidation number of 0.
2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges.
3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion [H], in which case rule 2 prevails).
4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral).
Here are some examples for practice. In H2, both H atoms have an oxidation number of 0 by rule 1. In MgCl2, magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1 by rule 2. In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H2O2) has an oxidation number of +1, while each O atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO2. Each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.
Example $1$
Assign oxidation numbers to the atoms in each substance.
1. Cl2
2. GeO2
3. Ca(NO3)2
Solution
1. Cl2 is the elemental form of chlorine. Rule 1 states each atom has an oxidation number of 0.
2. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4.
3. Ca(NO3)2 can be separated into two parts: the Ca2+ ion and the NO3 ion. Considering these separately, the Ca2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3 ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation x + 3(−2) = −1
where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for x,
x + (−6) = −1x = +5
Thus the oxidation number on the N atom in the NO3 ion is +5.
Exercise $1$: Phosphoric Acid
Assign oxidation numbers to the atoms in H3PO4.
Answer
H: +1; O: −2; P: +5
All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively.
Example $2$
Identify what is being oxidized and reduced in this redox reaction.
$\ce{2Na + Br2 → 2NaBr} \nonumber \nonumber$
Solution
Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br ions have an oxidation number of −1.
$2\underset{0}{Na}+\underset{0}{Br_{2}}\rightarrow 2\underset{+1 -1}{NaBr} \nonumber \nonumber$
Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced:
Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom).
Exercise $2$
Identify what is being oxidized and reduced in this redox reaction.
$\ce{C + O2 → CO2}\nonumber \nonumber$
Answer
C is being oxidized from 0 to +4; O is being reduced from 0 to −2.
Oxidation reactions can become quite complex, as attested by the following redox reaction:
$6H^{+}(aq)+2\underset{+7}{MnO_{4}^{-}}(aq)+5\underset{-1}{H_{2}O_{2}}(l)\rightarrow 2\underset{+2}{Mn^{2+}}(aq)+5\underset{0}{O_{2}}(g)+8H_{2}O(l)\nonumber$
To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation. Eventually, we will need to learn techniques for writing correct (i.e., balanced) redox reactions.
Food and Drink Application: Fortifying Food with Iron
Iron is an essential mineral in our diet; iron-containing compounds like the heme protein in hemoglobin could not function without it. Most biological iron has the form of the Fe2+ ion; iron with other oxidation numbers is almost inconsequential in human biology (although the body does contain an enzyme to reduce Fe3+ to Fe2+, so Fe3+ must have some biological significance, albeit minor). To ensure that we ingest enough iron, many foods are enriched with iron. Although Fe2+ compounds are the most logical substances to use, some foods use "reduced iron" as an ingredient (bread and breakfast cereals are the most well-known examples). Reduced iron is simply iron metal; iron is added as a fine metallic powder. The metallic iron is oxidized to Fe2+ in the digestive system and then absorbed by the body, but the question remains: Why are we ingesting metallic iron? Why not just use Fe2+ salts as an additive?
Although it is difficult to establish conclusive reasons, a search of scientific and medical literature suggests a few reasons. One reason is that fine iron filings do not affect the taste of the product. The size of the iron powder (several dozen micrometers) is not noticeable when chewing iron-supplemented foods, and the tongue does not detect any changes in flavor that can be detected when using Fe2+ salts. Fe2+ compounds can affect other properties of foodstuffs during preparation and cooking, like dough pliability, yeast growth, and color. Finally, of the common iron substances that might be used, metallic iron is the least expensive. These factors appear to be among the reasons why metallic iron is the supplement of choice in some foods.
Key Takeaways
• Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another.
• Oxidation numbers are used to keep track of electrons in atoms.
• There are rules for assigning oxidation numbers to atoms.
• Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number (gain of electrons). | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/14%3A_Oxidation_and_Reduction/14.02%3A_Oxidation-Reduction_Reactions.txt |
Learning Objectives
• Learn to balance simple redox reactions by inspection.
• Learn to balance complex redox reactions by the half reaction method.
• Use the solvent, or parts of it, as a reactant or a product in balancing a redox reaction.
Balancing simple redox reactions can be a straightforward matter of going back and forth between products and reactants. For example, in the redox reaction of Na and Cl2:
$\ce{Na + Cl2 → NaCl}\nonumber$
it should be immediately clear that the Cl atoms are not balanced. We can fix this by putting the coefficient 2 in front of the product:
$\ce{Na + Cl2 → 2NaCl}\nonumber$
However, now the sodium is unbalanced. This can be fixed by including the coefficient 2 in front of the Na reactant:
$\ce{2Na + Cl2 → 2NaCl}\nonumber$
This reaction is now balanced. That was fairly straightforward; we say that we are able to balance the reaction by inspection. Many simple redox reactions can be balanced by inspection.
Example $1$
Balance this redox reaction by inspection:
SO2 + O2 → SO3
Solution
There is one S atom on both sides of the equation, so the sulfur is balanced. However, the reactant side has four O atoms while the product side has three. Clearly we need more O atoms on the product side, so let us start by including the coefficient 2 on the SO3:
SO2 + O2 → 2SO3
This now gives us six O atoms on the product side, and it also imbalances the S atoms. We can balance both the elements by adding coefficient 2 on the SO2 on the reactant side:
2SO2 + O2 → 2SO3
This gives us two S atoms on both sides and a total of six O atoms on both sides of the chemical equation. This redox reaction is now balanced.
Exercise $1$
Balance this redox reaction by inspection:
Al + O2 → Al2O3
Answer
4Al + 3O2 → 2Al2O3
The first thing you should do when encountering an unbalanced redox reaction is to try to balance it by inspection.
Some redox reactions are not easily balanced by inspection. Consider this redox reaction:
Al + Ag+ → Al3+ + Ag
At first glance, this equation seems balanced: there is one Ag atom on both sides and one Al atom on both sides. However, if you look at the total charge on each side, there is a charge imbalance: the reactant side has a total charge of 1+, while the product side has a total charge of 3+. Something is amiss with this chemical equation; despite the equal number of atoms on each side, it is not balanced.
A fundamental point about redox reactions that has not arisen previously is that the total number of electrons being lost must equal the total number of electrons being gained for a redox reaction to be balanced. This is not the case for the aluminum and silver reaction: the Al atom loses three electrons to become the Al3+ ion, while the Ag+ ion gains only one electron to become elemental silver.
To balance this, we will write each oxidation and reduction reaction separately, listing the number of electrons explicitly in each. Individually, the oxidation and reduction reactions are called half reactions. We will then take multiples of each reaction until the number of electrons on each side cancels completely and combine the half reactions into an overall reaction, which should then be balanced. This method of balancing redox reactions is called the half reaction method. (There are other ways of balancing redox reactions, but this is the only one that will be used in this text. The reason for this will be seen in 14.4: Applications of Redox Reactions - Voltaic Cells of this chapter.)
The oxidation half reaction involves aluminum, which is being oxidized:
Al → Al3+
This half reaction is not completely balanced because the overall charges on each side are not equal. When an Al atom is oxidized to Al3+, it loses three electrons. We can write these electrons explicitly as products:
Al → Al3+ + 3e
Now this half reaction is balanced in terms of both atoms and charges.
The reduction half reaction involves silver:
Ag+ → Ag
The overall charge is not balanced on both sides. But we can fix this by adding one electron to the reactant side because the Ag+ ion must accept one electron to become the neutral Ag atom:
Ag+ + e → Ag
This half reaction is now also balanced.
When combining the two half reactions into a balanced chemical equation, the key is that the total number of electrons must cancel, so the number of electrons lost by atoms are equal to the number of electrons gained by other atoms. This may require we multiply one or both half reaction(s) by an integer to make the number of electrons on each side equal. With three electrons as products and one as a reactant, the least common multiple of these two numbers is three. We can use a single aluminum reaction, but must take three times the silver reaction:
Al → Al3+ + 3e3 × [Ag+ + e → Ag]
The 3 on the second reaction is distributed to all species in the reaction:
Al → Al3+ + 3e3Ag+ + 3e → 3Ag
Now the two half reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. The same species on opposite sides of the arrow can be canceled:
$Al+3Ag^{+}+3e^{-}\rightarrow Al^{3+}+3Ag+3e^{-}\nonumber$
The net balanced redox reaction is as follows:
Al + 3Ag+ → Al3+ + 3Ag
There is still only one Al atom on each side of the chemical equation, but there are now three Ag atoms, and the total charge on each side of the equation is the same (3+ for both sides). This redox reaction is balanced. It took more effort to use the half reaction method than by inspection, but the correct balanced redox reaction was obtained.
Example $2$
Balance this redox reaction by using the half reaction method:
Fe2+ + Cr → Fe + Cr3+
Solution
We start by writing the two half reactions. Chromium is being oxidized, and iron is being reduced:
Cr → Cr3+ oxidation Fe2+ → Fe reduction
Then we include the appropriate number of electrons on the proper side to balance the charges for each reaction:
Cr → Cr3+ + 3eFe2+ + 2e → Fe
The first reaction involves three electrons, while the second reaction involves two electrons. The least common multiple of these two numbers is six, so to get six electrons in each reaction we need to double the first reaction and triple the second one:
2 × [Cr → Cr3+ + 3e] = 2Cr → 2Cr3+ + 6e3 × [Fe2+ + 2e → Fe] = 3Fe2+ + 6e → 3Fe
We can combine the two final reactions, noting that the electrons cancel:
$2Cr+3Fe^{2+}+6e^{-}\rightarrow 2Cr^{3+}+3Fe+6e^{-}\nonumber$
The overall, balanced redox reaction is
2Cr + 3Fe2+ → 2Cr3+ + 3Fe
Exercise $2$
Balance this redox reaction by using the half reaction method:
O2− + F2 → O2 + F
Answer
2O2 + 2F2 → O2 + 4F
Many redox reactions occur in aqueous solution—in water. Because of this, in many cases H2O or a fragment of an H2O molecule (H+ or OH, in particular) can participate in the redox reaction. As such, we need to learn how to incorporate the solvent into a balanced redox equation.
Consider the following oxidation half reaction in aqueous solution, which has one Cr atom on each side:
Cr3+ → CrO4
Here, the Cr atom is going from the +3 to the +7 oxidation state. To do this, the Cr atom must lose four electrons. Let us start by listing the four electrons as products:
Cr3+ → CrO4 + 4e
But where do the O atoms come from? They come from water molecules or a common fragment of a water molecule that contains an O atom: the OH ion. When we balance this half reaction, we should feel free to include either of these species in the reaction to balance the elements. Let us use H2O to balance the O atoms; we need to include four water molecules to balance the four O atoms in the products:
4H2O + Cr3+ → CrO4 + 4e
This balances the O atoms, but now introduces hydrogen to the reaction. We can balance the H atoms by adding an H+ ion, which is another fragment of the water molecule. We need to add eight H+ ions to the product side:
4H2O + Cr3+ → CrO4 + 4e + 8H+
The Cr atoms are balanced, the O atoms are balanced, and the H atoms are balanced; if we check the total charge on both sides of the chemical equation, they are the same (3+, in this case). This half reaction is now balanced, using water molecules and parts of water molecules as reactants and products.
Reduction reactions can be balanced in a similar fashion. When oxidation and reduction half reactions are individually balanced, they can be combined in the same fashion as before: by taking multiples of each half reaction as necessary to cancel all electrons. Other species, such as H+, OH, and H2O, may also have to be canceled in the final balanced reaction.
Unless otherwise noted, it does not matter if you add H2O or OH as a source of O atoms, although a reaction may specify acidic solution or basic solution as a hint of what species to use or what species to avoid. OH ions are not very common in acidic solutions, so they should be avoided in those circumstances.
Example $3$
Balance this redox reaction. Assume a basic solution.
MnO2 + CrO3 → Mn + CrO4
Solution
We start by separating the oxidation and reduction processes so we can balance each half reaction separately. The oxidation reaction is as follows:
CrO3 → CrO4
The Cr atom is going from a +5 to a +7 oxidation state and loses two electrons in the process. We add those two electrons to the product side:
CrO3 → CrO4 + 2e
Now we must balance the O atoms. Because the solution is basic, we should use OH rather than H2O:
OH + CrO3 → CrO4 + 2e
We have introduced H atoms as part of the reactants; we can balance them by adding H+ as products:
OH + CrO3 → CrO4 + 2e + H+
If we check the atoms and the overall charge on both sides, we see that this reaction is balanced. However, if the reaction is occurring in a basic solution, it is unlikely that H+ ions will be present in quantity. The way to address this is to add an additional OH ion to each side of the equation:
OH + CrO3 + OH → CrO4 + 2e + H+ + OH
The two OH ions on the left side can be grouped together as 2OH. On the right side, the H+ and OH ions can be grouped into an H2O molecule:
2OH + CrO3 → CrO4 + 2e + H2O
This is a more appropriate form for a basic solution.
Now we balance the reduction reaction:
MnO2 → Mn
The Mn atom is going from +4 to 0 in oxidation number, which requires a gain of four electrons:
4e + MnO2 → Mn
Then we balance the O atoms and then the H atoms:
4e + MnO2 → Mn + 2OH2H+ + 4e + MnO2 → Mn + 2OH
We add two OH ions to each side to eliminate the H+ ion in the reactants; the reactant species combine to make two water molecules, and the number of OH ions in the product increases to four:
2H2O + 4e + MnO2 → Mn + 4OH
This reaction is balanced for a basic solution.
Now we combine the two balanced half reactions. The oxidation reaction has two electrons, while the reduction reaction has four. The least common multiple of these two numbers is four, so we multiply the oxidation reaction by 2 so that the electrons are balanced:
2 × [2OH + CrO3 → CrO4 + 2e + H2O]2H2O + 4e + MnO2 → Mn + 4OH
Combining these two equations results in the following equation:
4OH + 2CrO3 + 2H2O + 4e + MnO2 → 2CrO4 + 4e + 2H2O + Mn + 4OH
The four electrons cancel. So do the two H2O molecules and the four OH ions. What remains is
2CrO3 + MnO2 → 2CrO4 + Mn
which is our final balanced redox reaction.
Exercise $3$
Balance this redox reaction. Assume a basic solution.
Cl + MnO4 → MnO2 + ClO3
Answer
H2O + Cl + 2MnO4 → 2MnO2 + ClO3 + 2OH
Key Takeaways
• Redox reactions can be balanced by inspection or by the half reaction method.
• A solvent may participate in redox reactions; in aqueous solutions, H2O, H+, and OH may be reactants or products. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/14%3A_Oxidation_and_Reduction/14.03%3A_Balancing_Redox_Reactions.txt |
Learning Objectives
• Learn the parts of a voltaic cell.
• Combine half reactions to determine the voltage of a voltaic cell.
• Understand how voltaic cells are used as batteries.
Consider this redox reaction:
$\ce{Zn + Cu^{2+} → Zn^{2+} + Cu}\nonumber$
If you were to mix zinc metal and copper ions in a container, this reaction would proceed by itself; we say that this reaction is spontaneous.
Suppose, however, we set up this reaction in a way depicted in Fig 14.4.1 - A Redox Reaction in Which the Two Half Reactions Are Physically Separated. Zinc and zinc ions are on one side of the system, while copper and copper ions are on the other side of the system. The two parts are connected with a wire.
Even though the two half reactions are physically separated, a spontaneous redox reaction still occurs. However, in this case, the electrons transfer through the wire connecting the two half reactions; that is, this setup becomes a source of electricity. Useful work can be extracted from the electrons as they transfer from one side to the other. For example, a light bulb can be lit, or a motor can be operated. The apparatus as a whole, which allows useful electrical work to be extracted from a redox reaction, is called a voltaic (galvanic) cell.
Each individual system that contains a half reaction is called a half cell. The half cell that contains the oxidation reaction is called the anode, while the half cell that contains the reduction reaction is called the cathode. The cathode and anode collectively are the electrodes of the voltaic cell. Because electrons are coming from the anode, the anode is considered the negative electrode of the cell, while the cathode is considered the positive electrode of the cell. Finally, because electrons are moving from one half cell to the other, a charge imbalance builds up as the reaction proceeds. To counter that, a salt bridge is used; the salt bridge contains a solution of some ionic compound whose ions migrate to either side of the voltaic cell to maintain the charge balance.
The tendency for electrons to go from one half cell to another is called the voltage of the voltaic cell, represented by E. Sometimes the term potential is used to represent the voltage of a cell. Voltage is expressed in volts (V). The voltage of a voltaic cell is determined by the difference in the tendencies of the individual half cells and is characteristic of a given redox reaction when concentrations are specific (1.0 M for dissolved species and 1.0 atm for gases). Because the voltage of a redox reaction is determined by the difference of the tendencies of the individual half reactions, absolute voltages are unnecessary; only relative voltages of each half reaction are needed. The relative voltage of each half cell is represented as E1/2 and is based on the standard that the E1/2 for the reaction
$H^{+}+e^{-}\rightarrow \frac{1}{2}H_{2}\nonumber$
is assigned to be exactly 0.000 V under standard conditions of pressure and concentration. Table $1$ - Standard Reduction Potentials of Half Reactions, lists some relative E1/2 values for some half reactions. Note that all half reactions are listed as reduction reactions, so these values are called the standard reduction potentials of each half reaction.
Reduction Half Reaction E1/2 (V)
F2 + 2e → 2F 2.87
Ce4+ + e → Ce3+ 1.61
MnO4 + 8H+ + 5e → Mn2+ + 4H2O 1.51
Cl2 + 2e → 2Cl 1.36
O2 + 4H+ + 4e → 2H2O 1.23
Br2 + 2e → 2Br 1.06
NO3 + 4H+ + 3eNO + 2H2O 0.96
Ag+ + e → Ag 0.80
Fe3+ + e → Fe2+ 0.77
I2 + 2e → 2I 0.54
Cu2+ + 2e → Cu 0.34
AgCl + e → Ag + Cl 0.222
Sn4+ + 2e → Sn2+ 0.15
2H+ + 2e → H2 0.000
Pb2+ + 2e → Pb −0.126
Ni2+ + 2e → Ni −0.25
Cr3+ + e → Cr2+ −0.41
Fe2+ + 2e → Fe −0.44
Cr3+ + 3e → Cr −0.74
Zn2+ + 2e → Zn −0.76
Cr2+ + 2e → Cr −0.91
Ba2+ + 2e → Ba −1.57
Al3++ 3e → Al −1.66
Mg2+ + 2e → Mg −2.37
Na+ + e → Na −2.714
Li+ + e → Li −3.045
Table $1$ Standard Reduction Potentials of Half Reactions
The above table lists only reduction reactions, but a redox reaction has a reduction and an oxidation. To make the oxidation reaction, simply reverse the reduction reaction in the above table and change the sign on the E1/2 value. If the reduction potential is negative, make the voltage for the oxidation positive; if the reduction potential is positive, make the voltage for the oxidation negative.
Example $1$
What is the value of E1/2 for this half reaction?
Ag + Cl → AgCl + e
Solution
The given reaction is the reverse of this reaction:
AgCl + e → Ag + Cl E1/2 = 0.222 V
Therefore, the E1/2 of the given reaction is −0.222 V.
Exercise $1$
What is the value of E1/2 for this half reaction?
Na → Na+ + e
Answer
2.714 V
To determine the overall voltage of a particular voltaic cell, simply combine the voltages of the oxidation and reduction half reactions. Even if you need to take a multiple of a half reaction for the electrons to cancel, do not take the multiple of the E1/2. Use the values directly as is from Table $1$ - Standard Reduction Potentials of Half Reactions.
Spontaneous redox reactions have positive overall voltages. If the voltage of the reaction as written is negative, it is not spontaneous in that direction. Rather, the reverse reaction is the spontaneous redox reaction.
Example $2$
What is the voltage of a voltaic cell based on this reaction? Is the reaction spontaneous as written?
2NO3 + 8H+ + 3Cu → 2NO + 4H2O + 3Cu2+
Solution
The overall redox reaction is formed from these two half reactions:
NO3 + 4H+ + 3e → NO + 2H2O E1/2 = 0.96 VCu2+ + 2 e → Cu E1/2 = 0.34 V
The second reaction is reversed in the overall redox reaction, so its voltage changes sign from the reduction reaction:
Cu → Cu2+ + 2 e E1/2 = −0.34 V
To obtain the voltage of the voltaic cell based on the overall reaction, we simply combine the two voltages of the half reactions:
E = 0.96 + (−0.34) = 0.62 V
Because the overall voltage is positive, the reaction is spontaneous as written.
Exercise $2$
What is the voltage of a voltaic cell based on this reaction? Is the reaction spontaneous as written?
5Ni + 2MnO4 + 16H+ → 3Mn2+ + 8H2O + 5Ni2+
Answer
1.76 V; spontaneous
Technically, any redox reaction can be set up to make a voltaic cell. In modern society, however, only certain redox reactions are put to practical use. A portable voltaic cell that generates electricity to power devices for our convenience is called a battery. All batteries are based on redox reactions.
The first battery (called a "voltaic pile") was constructed by the Italian scientist Alessandro Volta in 1800 and was based on the copper/zinc reaction depicted in Figure $1$ - A Redox Reaction in Which the Two Half Reactions Are Physically Separated. Unfortunately, it was messy, requiring quantities of copper and zinc salts dissolved in water. In 1866, the French scientist Georges Leclanché invented the dry cell, a precursor to today's modern battery. A schematic of a dry cell is shown in Figure $2$ - Dry Cell. The zinc case and the central carbon rod serve as the anode and cathode, respectively. The other reactants are combined into a moist paste that minimizes free liquid, so the battery is less messy (hence the name dry cell).
The actual redox reaction is complex but can be represented by the following redox reaction:
Zn + 2MnO2 + 2NH4+ → Zn2+ + Mn2O3 + 2NH3 + H2O
A dry cell has a voltage of about 1.56 V. While common and useful, dry cells have relatively short lifetimes and contain acidic components. They also cannot be recharged, so they are one-use only. Batteries that can be used only once are called primary batteries.
In the late 1950s, Lewis Urry of the Eveready Battery Company in Ohio invented the alkaline battery (still marketed today under the trade name Energizer). Alkaline batteries are similar to dry cells, but they use a basic moist paste rather than an acidic one. Moreover, the net amount of base does not change during the course of the redox reaction. The overall redox reaction is as follows:
Zn + 2MnO2 → ZnO + Mn2O3
Alkaline batteries have the advantage of being longer lasting and holding their voltage better—about 1.54 V—throughout their lifetime.
A common type of battery, especially with the increased popularity of personal electronic devices, is the button battery (Figure $3$ - Button Battery). A button battery is a small battery that can power small electronic devices; the batteries can be as small as 5 mm across. Two popular redox reactions used for button batteries are the alkaline dry-cell reaction and a silver oxide-based reaction:
Zn + Ag2O → ZnO + 2Ag
Some button batteries use a lithium-based redox reaction, typified by this anode reaction:
Li → Li+ + e E1/2 = 3.045 V
The actual redox reaction depends on the composition of the cathode and is variable depending on voltage. Lithium batteries can also be used for applications that require more energy, such as portable computers and electric vehicles. Some lithium-based batteries are rechargeable and can be used over and over again; such batteries are called secondary batteries.
An important secondary battery is the lead storage battery, shown in the Figure $4$ - Lead Storage Batteries. The lead storage battery is based on this redox reaction:
Pb + PbO2 + 4H+ + SO42 → 2PbSO4 + 2H2O
The redox reaction produces about 2 V, but it is typical to tie several individual batteries together to generate a larger voltage. The lead storage battery has the distinction that the product of both half reactions is PbSO4, which as a solid that accumulates on the many plates within each cell. The lead storage battery is a secondary battery, as it can be recharged and reused many times. Because it is based on lead, these batteries are rather heavy. They should also be recycled when replaced so that potentially dangerous lead does not escape into the environment. Because of their characteristics, lead storage batteries are used to start large engines in automobiles, boats, and airplanes.
Chemistry is Everywhere: Fuel Cells
A fuel cell is a type of battery in which reactants flow continuously into a specialized reaction chamber, and products flow out continuously while electrons are extracted from the reaction. Because all reactions in a fuel cell consist of a fuel and an oxidizer undergoing a redox reaction, an introduction of fuel cells is at home in a discussion of redox chemistry.
By far the most common fuel cell reaction is based on hydrogen and oxygen:
2H2 + O2 → 2H2O E = 1.23 V under standard conditions
However, fuel cells typically do not work under standard nor even optimal conditions, so they typically generate about 0.6–0.7 V. In this fuel cell, the only two products are water and electricity, so the fuel cell not only does not create pollution, but also makes a by-product that in some environments is a valuable commodity (water). Other fuels can be used besides hydrogen; fuel cells have been developed that work on methane, methyl alcohol, ethyl alcohol, carbon-rich materials, and even magnesium metal.
Hydrogen-based fuel cells were and are used to provide electricity for manned space vehicles, partly because their only chemical product is water, which could be used for drinking. However, there has been a recent resurgence in interest in fuel cells because of their potential use in electric cars. Most electric cars run on conventional batteries, which can be very heavy and expensive to replace. It is thought that fuel cells, rather than conventional batteries, might be better sources of electricity for automobiles.
Several current barriers to fuel cell use in electric cars include capacity, cost, and overall energy efficiency. The 2008 Honda FCX, the first production model of a vehicle powered with a fuel cell, can hold 4.1 kg (just under 9 lb) of highly pressured H2 gas and has a range of 450 km (280 mi). It costs about $120,000–$140,000 to build, making the vehicle beyond the ability of most people to own. Finally, it always requires more energy to produce elemental hydrogen as a fuel than can be extracted from hydrogen as a fuel. As such, hydrogen is described as an energy carrier (like electricity) rather than an energy source (like oil and gas). This distinction points out a fundamental argument against fuel cells as a "better" power source.
The 2008 Honda FCX was the first production car to use a fuel cell as a power source. Nonetheless, the car is in very limited service because of its need for relatively large quantities of elemental hydrogen as fuel.
The limitations notwithstanding, there is a lot of interest in fuel cell research. If ways can be found to circumvent their current limitations, fuel cells may become more and more common as power sources.
Key Takeaways
• A voltaic cell produces electricity as a redox reaction occurs.
• The voltage of a voltaic cell can be determined by the reduction potentials of the half reactions.
• Voltaic cells are fashioned into batteries, which are a convenient source of electricity. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/14%3A_Oxidation_and_Reduction/14.04%3A_Applications_of_Redox_Reactions_-_Voltaic_Cells.txt |
Learning Objectives
• Describe electrolysis from a perspective of redox reactions.
• Give examples of electrolysis applications.
Up to this point, we have considered redox reactions for processes that are spontaneous. When set up as a voltaic cell or battery, such reactions can be used as a source of electricity. However, it is possible to go in the other direction. By forcing electricity into a cell, we can make a redox reaction occur that normally would not be spontaneous. Under these circumstances, the cell is called an electrolytic cell, and the process that occurs in the cell is called electrolysis (Figure $1$).
Electrolysis has many applications. For example, if NaCl is melted at about 800°C in an electrolytic cell and an electric current is passed through it, elemental sodium will appear at the cathode and elemental chlorine will appear at the anode as the following two reactions occur:
$Na^+ + e^− \rightarrow Na\nonumber$
$2 Cl^− → Cl_2 + 2e^−\nonumber$
Normally we expect elemental sodium and chlorine to react spontaneously to make NaCl. However, by using an input of electricity, we can force the opposite reaction to occur and generate the elements. Lithium, potassium, and magnesium can also be isolated from compounds by electrolysis.
Another element that is isolated by electrolysis is aluminum. Aluminum formerly was a difficult metal to isolate in its elemental form; in fact, the top of the Washington Monument has a 2.8 kg cap of aluminum metal, which at the time (1884) was the largest piece of elemental aluminum ever isolated. However, in 1886, the American Charles Hall and the Frenchman Paul Héroult almost simultaneously worked out an electrolytic process for isolating aluminum from bauxite—an ore of aluminum whose chemical formula is AlOx(OH)3 − 2x . The basic reactions are as follows:
$Al^{3+} + 3e^− \rightarrow Al_2O^{2−} \rightarrow O_2 + 4e^−\nonumber$
With the development of the Hall-Héroult process, the price of aluminum dropped by a factor of over 200, and aluminum metal became common. So much elemental aluminum is produced in the United States each year that it has been estimated that the electrolysis of aluminum uses 5% of all the electricity in the country. (Recycling aluminum requires about 1/70th the energy of refining aluminum from ore, which illustrates the tremendous energy savings that recycling provides.)
Another application of electrolysis is electroplating, which is the deposition of a thin layer of metal on an object for protective or decorative purposes (Figure $2$). Essentially, a metal object is connected to the cathode of an electrolytic cell and immersed in a solution of a particular metal cation. When the electrolytic cell is operated, a thin coating of the metal cation is reduced to the elemental metal on the surface of the object; the thickness of the coating can be as little as a few micrometers (10−6 m). Jewelry, eating utensils, electrical contacts, and car parts like bumpers are common items that are electroplated. Gold, silver, nickel, copper, and chromium are common metals used in electroplating.
Key Takeaways
• Electrolysis is a forced non-spontaneous redox occurence by the introduction of electricity into a cell from an outside source.
• Electrolysis is used to isolate elements and electroplate objects. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/14%3A_Oxidation_and_Reduction/14.05%3A_Electrolysis.txt |
14.2: Oxidation-Reduction Reactions
1. Is this reaction a redox reaction? Explain your answer.2K(s) + Br2(ℓ) → 2KBr(s)
2. Is this reaction a redox reaction? Explain your answer. 2NaCl(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbCl2(s)
3. Which substance loses electrons and which substance gains electrons in this reaction? 2Mg(s) + O2(g) → 2MgO
4. Which substance loses electrons and which substance gains electrons in this reaction? 16Fe(s) + 3S8(s) → 8Fe2S3(s)
5. Which substance is oxidized and which substance is reduced in this reaction? 2Li(s) + O2(g) → Li2O2(s)
6. Which substance is oxidized and which substance is reduced in this reaction? 2Fe(s) + 3I2(s) → 2FeI3(s)
7. What are two different definitions of oxidation?
1. What are two different definitions of reduction?
2. Assign oxidation numbers to the atoms in each substance.
1. P4
2. SO3
3. SO32
4. Ca3(PO3)2
3. Assign oxidation numbers to the atoms in each substance.
1. PCl5
2. (NH4)2Se
3. Ag
4. Li2O2
4. Assign oxidation numbers to the atoms in each substance.
1. NO
2. NO2
3. CrCl2
4. CrCl3
5. Assign oxidation numbers to the atoms in each substance.
1. NaH
2. N2O3
3. NO2
4. CuNO3
6. Assign oxidation numbers to the atoms in each substance.
1. CH2O
2. NH3
3. Rb2SO4
4. Zn(C2H3O2)2
7. Assign oxidation numbers to the atoms in each substance.
1. C6H6
2. B(OH)3
3. Li2S
4. Au
8. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2NO + Cl2 → 2NOCl
9. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.Sr + SO3 → SrSO3
10. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2KrF2 + 2H2O → 2Kr + 4HF + O2
11. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.SO3 + SCl2 → SOCl2 + SO2
12. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2Rb + MgCl2 → 2RbCl + Mg
13. Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.2C8H18 + 25O2 → 16CO2 + 18H2O
Answers
1. yes because oxidation numbers are changing
2.
3. lose: Mg; gain: O
4.
5. oxidized: Li; reduced: O
6.
1. increase in oxidation number; loss of electrons
2.
1. P: 0
2. S: +6; O: −2
3. S: +4; O: −2
4. Ca: +2; P: +3; O: −2
3.
1. N: +2; O: −2
2. N: +4; O: −2
3. Cr: +2; Cl: −1
4. Cr: +3; Cl: −1
4.
1. C: 0; H: +1; O: −2
2. N: −3; H: +1
3. Rb: +1; S: +6; O: −2
4. Zn: +2; C: 0; H: +1; O: −2
5.
6. oxidized: N; reduced: Cl
7.
8. oxidized: O; reduced: Kr
9.
10. oxidized: Rb; reduced: Mg
14.3: Balancing Redox Reactions
1. Balance these redox reactions by inspection.
1. Na + F2 → NaF
2. Al2O3 + H2 → Al + H2O
2. Balance these redox reactions by inspection.
1. Fe2S3 + O2 → Fe2O3 + S
2. Cu2O + H2 → Cu + H2O
3. Balance these redox reactions by inspection.
1. CH4 + O2 → CO2 + H2O
2. P2O5 + Cl2 → PCl3 + O2
4. Balance these redox reactions by inspection.
1. PbCl2 + FeCl3 → PbCl4 + FeCl2
2. SO2 + F2 → SF4 + OF2
5. Balance these redox reactions by the half reaction method.
1. Ca + H+ → Ca2+ + H2
2. Sn2+ → Sn + Sn4+ (Hint: both half reactions will start with the same reactant.)
6. Balance these redox reactions by the half reaction method.
1. Fe3+ + Sn2+ → Fe + Sn4+
2. Pb2+ → Pb + Pb4+ (Hint: both half reactions will start with the same reactant.)
7. Balance these redox reactions by the half reaction method.
1. Na + Hg2Cl2 → NaCl + Hg
2. Al2O3 + C → Al + CO2
8. Balance these redox reactions by the half reaction method.
1. Br + I2 → I + Br2
2. CrCl3 + F2 → CrF3 + Cl2
9. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. Cu + NO3 → Cu2+ + NO2
2. Fe + MnO4 → Fe3+ + Mn
10. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. CrO3 + Ni2+ → Cr3+ + Ni3+
2. OsO4 + C2H4 → Os + CO2
11. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. ClO + Ti2+ → Ti4+ + Cl
2. BrO3 + Ag → Ag+ + BrO2
12. Balance these redox reactions that occur in aqueous solution. Use whatever water-derived species is necessary; there may be more than one correct balanced equation.
1. H2O2 + NO → N2O3 + H2O
2. VO2+ + NO → V3+ + NO2
13. Explain why this chemical equation is not balanced and balance it if it can be balanced: Cr2+ + Cl2 → Cr3+ + 2Cl
14. Explain why this equation is not balanced and balance it if it can be balanced: O2 + 2H2O + Br2 → 4OH + 2Br
Answers
1. 2Na + F2 → 2NaF
2. Al2O3 + 3H2 → 2Al + 3H2O
1.
1. CH4 + 2O2 → CO2 + 2H2O
2. 2P2O5 + 6Cl2 → 4PCl3 + 5O2
2.
1. Ca + 2H+ → Ca2+ + H2
2. 2Sn2+ → Sn + Sn4+
3.
1. 2Na + Hg2Cl2 → 2NaCl + 2Hg
2. 2Al2O3 + 3C → 4Al + 3CO2
4.
1. 4H+ + Cu + 2NO3 → Cu2+ + 2NO2 + 2H2O in acidic solution; 2H2O + Cu + 2NO3 → Cu2+ + 2NO2 + 4OH in basic solution
2. 24H+ + 3MnO4 + 7Fe → 7Fe3+ + 3Mn + 12H2O in acidic solution; 12H2O + 3MnO4 + 7Fe → 7Fe3+ + 3Mn + 24OH in basic solution
5.
1. 2H+ + ClO + Ti2+ → Cl + H2O + Ti4+ in acidic solution; H2O + ClO + Ti2+ → Cl + Ti4+ + 2OH in basic solution
2. 2H+ + BrO3 + Ag → BrO2 + H2O + Ag+ in acidic solution; H2O + BrO3 + Ag → BrO2 + Ag+ + 2OH in basic solution
6.
7. The charges are not properly balanced. The correct balanced equation is 2Cr2+ + Cl2 → 2Cr3+ + 2Cl.
14.4: Applications of Redox Reactions - Voltaic Cells
1. Draw the voltaic cell represented by this reaction and label the cathode, the anode, the salt bridge, the oxidation half cell, the reduction half cell, the positive electrode, and the negative electrode. Use Fig. 14.4.1 as a guide. Zn + 2Ag+ → Zn2+ + 2Ag
2. Draw the voltaic cell represented by this reaction and label the cathode, the anode, the salt bridge, the oxidation half cell, the reduction half cell, the positive electrode, and the negative electrode. Use Fig. 14.4.1 as a guide. 3Mg + 2Cr3+ → 3Mg2+ + 2Cr
3. What is the voltage of this half reaction? 2F → F2 + 2e
4. What is the voltage of this half reaction? Na → Na+ + e
5. What is the voltage of the voltaic cell in Exercise 1? Consult Table 14.4.1.
6. What is the voltage of the voltaic cell in Exercise 2? Consult Table 14.4.1.
7. Balance this redox reaction and determine its voltage. Is it spontaneous? Li+ + Al → Li + Al3+
8. Balance this redox reaction and determine its voltage. Is it spontaneous? Pb2+ + Ni → Pb + Ni2+
9. Balance this redox reaction and determine its voltage. Is it spontaneous? Cu2+ + Ag + Cl → Cu + AgCl
10. Balance this redox reaction and determine its voltage. Is it spontaneous? Mn2+ + Br2 → MnO4 + Br
11. Which reaction represents the cathode reaction in Exercise 7? The anode reaction?
12. Which reaction represents the cathode reaction in Exercise 8? The anode reaction?
13. Which reaction represents the cathode reaction in Exercise 9? The anode reaction?
14. Which reaction represents the cathode reaction in Exercise 10? The anode reaction?
15. A voltaic cell is based on this reaction: Ni + 2Au+ → Ni2+ + 2Au; If the voltage of the cell is 0.33 V, what is the standard reduction potential of the Au+ + e → Au half reaction?
16. A voltaic cell is based on this reaction: 3Pb + 2V3+ → 3Pb2+ + 2V; If the voltage of the cell is −0.72 V, what is the standard reduction potential of the V3+ + 3e → V half reaction?
17. What species is being oxidized and what species is being reduced in a dry cell?
18. What species is being oxidized and what species is being reduced in an alkaline battery?
19. What species is being oxidized and what species is being reduced in a silver oxide button battery?
20. What species is being oxidized and what species is being reduced in a lead storage battery?
21. Based on the data in Table 14.4.1, what is the highest voltage battery you can construct?
22. Based on the data in Table 14.4.1, what is the lowest voltage battery you can construct? (This may be more challenging to answer than Exercise 21.)
Answers
1.
2. −2.87 V
3.
4. 1.56 V
5.
6. 3Li+ + Al → 3Li + Al3+; −1.39 V; not spontaneous
7.
8. Cu2+ + 2Ag + 2Cl → Cu + 2AgCl; 0.12 V; spontaneous
9.
10. cathode reaction: Li+ + e → Li; anode reaction: Al → Al3+ + 3e
11.
12. cathode reaction: Cu2+ + 2e → Cu; anode reaction: Ag + Cl → AgCl + e
13.
14. 0.08 V
15.
16. oxidized: Zn; reduced: Mn
17.
18. oxidized: Zn; reduced: Ag
19.
20. 5.92 V from the reaction of F2 and Li
14.5: Electrolysis
1. Define electrolytic cell.
2. How does the operation of an electrolytic cell differ from a voltaic cell?
3. List at least three elements that are produced by electrolysis.
4. Write the half reactions for the electrolysis of the elements listed in Exercise 3.
5. Based on Table 14.4.1, what voltage must be applied to an electrolytic cell to electroplate copper from Cu2+?
6. Based on Table 14.4.1, what voltage must be applied to an electrolytic cell to electroplate aluminum from Al3+?
Answers
1. an electrochemical cell in which charge is forced through and a nonspontaneous reaction occurs
2.
3. any three of the following: Al, K, Li, Na, Cl2, or Mg
4.
5. 0.34 V
Additional Exercises
1. Oxidation was once defined as chemically adding oxygen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of oxidation: 2Mg + O2 → 2MgO
2. Reduction was once defined as chemically adding hydrogen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of reduction: C2H2 + 2H2 → C2H6
3. Assign oxidation numbers to the atoms in each substance.
1. Kr (krypton)
2. krypton tetrafluoride (KrF4)
3. dioxygen difluoride (O2F2)
4. Assign oxidation numbers to the atoms in each substance.
1. lithium hydride (LiH)
2. potassium peroxide (K2O2)
3. potassium fluoride (KF)
5. N atoms can have a wide range of oxidation numbers. Assign oxidation numbers for the N atom in each compound, all of which are known compounds.
1. N2O5
2. N2O4
3. NO2
4. NO
5. N2H4
6. NH3
6. Cr atoms can have a wide range of oxidation numbers. Assign oxidation numbers for the Cr atom in each compound, all of which are known compounds.
1. Na2CrO4
2. Na2Cr2O7
3. CrF5
4. CrCl3
5. CrCl2
7. Balance this redox reaction by inspection: S8 + O2 → SO2
8. Balance this redox reaction by inspection: C18H38 + O2 → CO2 + H2O
9. Balance this redox reaction by the half reaction method by assuming an acidic solution: Cr2O72 + Fe → Cr3+ + Fe3+
10. Balance the redox reaction in Exercise 9 by the half reaction method by assuming a basic solution.
11. The uranyl ion (UO22+) is a fairly stable ion of uranium that requires strong reducers to reduce the oxidation number of uranium further. Balance this redox reaction using the half reaction method by assuming an acidic solution.UO22+ + HN3 → U + N2
12. Balance the redox reaction in Exercise 11 by the half reaction method by assuming a basic solution.
13. Zinc metal can be dissolved by acid, which contains H+ ions. Demonstrate that this is consistent with the fact that this reaction has a spontaneous voltage: Zn + 2H+ → Zn2+ + H2
14. Copper metal cannot be dissolved by acid, which contains H+ ions. Demonstrate that this is consistent with the fact that this reaction has a nonspontaneous voltage: Cu + 2H+ → Cu2+ + H2
15. A disproportionation reaction occurs when a single reactant is both oxidized and reduced. Balance and determine the voltage of this disproportionation reaction. Use the data in Table 14.4.1 - Standard Reduction Potentials of Half Reactions: Cr2+ → Cr + Cr3+
16. A disproportionation reaction occurs when a single reactant is both oxidized and reduced. Balance and determine the voltage of this disproportionation reaction. Use the data in Table 14.4.1 - Standard Reduction Potentials of Half Reactions: Fe2+ → Fe + Fe3+
17. What would be overall reaction for a fuel cell that uses CH4 as the fuel?
18. What would be overall reaction for a fuel cell that uses gasoline (general formula C8H18) as the fuel?
19. When NaCl undergoes electrolysis, sodium appears at the cathode. Is the definition of cathode the same for an electrolytic cell as it is for a voltaic cell?
20. When NaCl undergoes electrolysis, chlorine appears at the anode. Is the definition of anode the same for an electrolytic cell as it is for a voltaic cell?
21. An award is being plated with pure gold before it is presented to a recipient. If the area of the award is 55.0 cm2 and will be plated with 3.00 µm of Au, what mass of Au will be plated on the award? The density of Au is 19.3 g/cm3.
22. The unit of electrical charge is called the coulomb (C). It takes 96,500 coulombs of charge to reduce 27.0 g of Al from Al3+ to Al metal. At 1,040 cm3, how many coulombs of charge were needed to reduce the aluminum in the cap of the Washington monument, assuming the cap is pure Al? The density of Al is 2.70 g/cm3.
Answer
As oxygen is added to magnesium, it is being oxidized. In modern terms, the Mg atoms are losing electrons and being oxidized, while the electrons are going to the O atoms.
1. Kr: 0
2. Kr: +4; F: −1
3. O: +1; F: −1
1. +5
2. +4
3. +4
4. +2
5. −2
6. −3
S8 + 8O2 → 8SO2
14H+ + Cr2O72 + 2Fe → 2Cr3+ + 7H2O + 2Fe3+
6HN3 + UO22+ → U + 2H2O + 9N2 + 2H+
The voltage of the reaction is +0.76 V, which implies a spontaneous reaction.
3Cr2+ → Cr + 2Cr3+; −0.50 V
CH4 + 2O2 → CO2 + 2H2O
yes because reduction occurs at the cathode
0.318 g | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/14%3A_Oxidation_and_Reduction/14.E%3A_Oxidation-Reduction_Reaction_%28Exercises%29.txt |
Most chemists pay little attention to the nucleus of an atom except to consider the number of protons it contains, because that determines an element's identity. However, in nuclear chemistry, the composition of the nucleus and the changes that occur there are very important. Applications of nuclear chemistry may be more widespread than you realize. Many people are aware of nuclear power plants and nuclear bombs, but nuclear chemistry also has applications ranging from smoke detectors to medicine, from the sterilization of food to the analysis of ancient artifacts. In this chapter, we will examine some of the basic concepts of nuclear chemistry and some of the nuclear reactions that are important in our everyday lives.
• 15.1: Prelude to Nuclear Chemistry
Many people think of nuclear chemistry in connection with the nuclear power industry and atomic bombs, but do not realize that most smoke detectors rely on nuclear chemistry and save countless lives every year. The applications of nuclear chemistry may be more widespread than you think.
• 15.2: Radioactivity
The major types of radioactivity include alpha particles, beta particles, and gamma rays. Fission is a type of radioactivity in which large nuclei spontaneously break apart into smaller nuclei.
• 15.3: Half-Life
Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively. The amount of material left over after a certain number of half-lives can be easily calculated.
• 15.4: Units of Radioactivity
Radioactivity can be expressed in a variety of units, including rems, rads, and curies.
• 15.5: Uses of Radioactive Isotopes
Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and preservation of food.
• 15.6: Nuclear Energy
Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy.
• 15.E: Nuclear Chemistry (Exercises)
These are exercises and select solutions to accompany Chapter 15 of the "Beginning Chemistry" Textmap formulated around the Ball et al. textbook.
Thumbnail: Part of carbon–nitrogen–oxygen (CNO) reaction chain diagram, made just to be illustrative for nuclear reactions in general. (CC BY-SA 3.0; Michalsmid via Wikipedia).
15: Nuclear Chemistry
Most of us have at least one device in our homes that guards our safety and, at the same time, depends on radioactivity to operate properly. This device is a smoke detector.
A typical smoke detector contains an electric circuit that includes two metal plates about 1 cm apart. A battery in the circuit creates a voltage between the plates. Next to the plates is a small disk containing a tiny amount (∼0.0002 g) of the radioactive element americium. The radioactivity of americium ionizes the air between the plates, causing a tiny current to constantly flow between them. (This constant drain on the battery explains why the batteries in smoke detectors should be replaced regularly, whether the alarm has been triggered or not.) When particles of smoke from a fire enter the smoke detector, they interfere with the ions between the metal plates, interrupting the flow of current. When the current drops beneath a set value, another circuit triggers a loud alarm, warning of the possible presence of fire.
Although radioactive, the americium in a smoke detector is embedded in plastic and is not harmful unless the plastic package is taken apart, which is unlikely. Nuclear chemistry is often thought of only in connection with the nuclear power industry and atomic bombs. Although many people have an unfounded fear of radioactivity, smoke detectors save thousands of lives every year. The applications of nuclear chemistry may be more widespread than you think. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.01%3A_Prelude_to_Nuclear_Chemistry.txt |
Learning Objective
• Define and give examples of the major types of radioactivity.
We saw in Chapter 3 that atoms are composed of subatomic particles—protons, neutrons, and electrons. Protons and neutrons are located in the nucleus and provide most of the mass of an atom, while electrons circle the nucleus in shells and subshells and account for an atom's size.
In Chapter 2, we also introduced the notation for succinctly representing an isotope of a particular atom:
$_{6}^{12}\textrm{C}\nonumber$
The element in this example, represented by the symbol C, is carbon. Its atomic number, 6, is the subscript next to the symbol and is the number of protons in the atom. The mass number, the superscript next to the symbol, is the sum of the number of protons and neutrons in the nucleus of this particular isotope. In this case, the mass number is 12, which means that the number of neutrons in the atom is 12 − 6 = 6 (that is, the mass number of the atom minus the number of protons in the nucleus equals the number of neurons). Occasionally, the atomic number is omitted in this notation because the symbol of the element itself conveys its characteristic atomic number. The two isotopes of hydrogen—2H and 3H—are given their own names and symbols: deuterium (D) and tritium (T), respectively.
Atomic theory in the nineteenth century presumed that nuclei had fixed compositions. But in 1896, the French scientist Henri Becquerel found that a uranium compound placed near a photographic plate made an image on the plate, even if the compound was wrapped in black cloth. He reasoned that the uranium compound was emitting some kind of radiation that passed through the cloth to expose the photographic plate. Further investigations showed that the radiation was a combination of particles and electromagnetic rays, with its ultimate source being the atomic nucleus. These emanations were ultimately called, collectively, radioactivity.
There are three main forms of radioactive emissions. The first is called an alpha particle, which is symbolized by the Greek letter α. An alpha particle is composed of 2 protons and 2 neutrons and is the same as a helium nucleus. We often use:
$_{2}^{4}\textrm{He}\nonumber$
to represent an alpha particle. It has a 2+ charge. When a radioactive atom emits an alpha particle, the original atom's atomic number decreases by 2 (because of the loss of 2 protons), and its mass number decreases by 4 (because of the loss of 4 nuclear particles). We can represent the emission of an alpha particle with a chemical equation. For example, the alpha-particle emission of uranium-235 is as follows:
$_{92}^{235}\textrm{U}\rightarrow \; _{2}^{4}\textrm{He}+\: _{90}^{231}\textrm{Th}\nonumber$
Rather than calling this equation a chemical equation, we call it a nuclear equation to emphasize that the change occurs in an atomic nucleus. How do we know that a product of this reaction is $\ce{^{90}_{231}T}$? We use the law of conservation of matter, which says that matter cannot be created or destroyed. This means that we must have the same number of protons and neutrons on both sides of the nuclear equation. If our uranium nucleus loses 2 protons, there are 90 protons remaining, identifying the element as thorium. Moreover, if we lose four nuclear particles of the original 235, there are 231 remaining. So we use subtraction to identify the isotope of the Th atom—in this case,
$_{90}^{231}\textrm{Th}\nonumber$
Chemists often use the names parent isotope and daughter isotope to represent the original atom and the product other than the alpha particle. In the previous example, $_{92}^{235}\textrm{U}\nonumber$ is the parent isotope, and $_{90}^{231}\textrm{Th}\nonumber$ is the daughter isotope. When one element changes into another in this manner, it undergoes radioactive decay.
Example $1$
Write the nuclear equation that represents the radioactive decay of radon-222 by alpha particle emission and identify the daughter isotope.
Solution
Radon has an atomic number of 86, so the parent isotope is represented as $_{86}^{222}\textrm{Rn} \nonumber \nonumber$.
We represent the alpha particle as
$_{2}^{4}\textrm{He} \nonumber \nonumber$
Use subtraction (222 − 4 = 218 and 86 − 2 = 84) to identify the daughter isotope as polonium:
$_{86}^{222}\textrm{Rn}\rightarrow \; _{2}^{4}\textrm{He}+\: _{84}^{218}\textrm{Th} \nonumber \nonumber$
Exercise $2$
Write the nuclear equation that represents radioactive decay of polonium-208 by alpha particle emission and identify the daughter isotope.
Answer
$_{80}^{208}\textrm{Po}\rightarrow \; _{2}^{4}\textrm{He}+\: _{82}^{204}\textrm{Pb} \nonumber \nonumber$
$_{82}^{204}\textrm{Pb} \nonumber \nonumber$
The second major type of radioactive emission is called a beta particle, symbolized by the Greek letter β. A beta particle is an electron ejected from the nucleus (not from the shells of electrons about the nucleus) and has a 1− charge. We can also represent a beta particle as e10. The net effect of beta particle emission on a nucleus is that a neutron is converted to a proton. The overall mass number stays the same, but because the number of protons increases by one, the atomic number goes up by one. Carbon-14 decays by emitting a beta particle:
$_{6}^{14}\textrm{C}\rightarrow \; _{7}^{14}\textrm{N}+\: _{-1}^{0}\textrm{e}\nonumber$
Again, the sum of the atomic numbers is the same on both sides of the equation, as is the sum of the mass numbers. (Note that the electron is assigned an "atomic number" of –1, equal to its charge.)
The third major type of radioactive emission is not a particle but rather a very energetic form of electromagnetic radiation called gamma rays, symbolized by the Greek letter γ. Gamma rays themselves do not carry an overall electrical charge, but they may knock electrons out of atoms in a sample of matter and make it electrically charged (for which gamma rays are termed ionizing radiation). For example, in the radioactive decay of radon-222, both alpha and gamma radiation are emitted, with the latter having an energy of 8.2 × 10−14 J per nucleus decayed:
$_{86}^{222}\textrm{Rn}\rightarrow \;_{84}^{218}\textrm{Po}+_{2}^{4}\textrm{He}+\gamma\nonumber$
This may not seem like much energy, but if 1 mol of Rn atoms were to decay, the gamma ray energy would be 4.9 × 107 kJ!
Example $3$
Write the nuclear equation that represents the radioactive decay of boron-12 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle.
Solution
The parent isotope is B-12,
$_{5}^{12}\textrm{B}\nonumber$
while one of the products is B-12,
$_{-1}^{0}\textrm{e}\nonumber$
So that the mass and atomic numbers have the same value on both sides, the mass number of the daughter isotope must be 12, and its atomic number must be 6. The element having an atomic number of 6 is carbon. Thus the complete nuclear equation is as follows:
$_{5}^{12}\textrm{B}\rightarrow \;_{6}^{12}\textrm{C}+_{-1}^{0}\textrm{e}+\gamma\nonumber$
The daughter isotope is carbon-12.
Exercise $4$
Answer
$_{43}^{133}\textrm{Tc}\rightarrow \;_{44}^{133}\textrm{Ru}+_{-1}^{0}\textrm{e}+\gamma\nonumber$
Alpha, beta, and gamma emissions have different abilities to penetrate matter. The relatively large alpha particle is easily stopped by matter (although it may impart a significant amount of energy to the matter it contacts). Beta particles penetrate slightly into matter, perhaps a few centimeters at most. Gamma rays can penetrate deeply into matter and can impart a large amount of energy into the surrounding matter. Table $1$ summarizes the properties of the three main types of radioactive emissions.
Table $1$ The Three Main Forms of Radioactive Emissions
Characteristic Alpha Particles Beta Particles Gamma Rays
symbols α, H24 βe10$e−10$ γ
identity helium nucleus electron electromagnetic radiation
charge 2+ 1− none
mass number 4 0 0
penetrating power minimal (will not penetrate skin) short (will penetrate skin and some tissues slightly) deep (will penetrate tissues deeply)
Occasionally, an atomic nucleus breaks apart into smaller pieces in a radioactive process called spontaneous fission (or fission). Typically, the daughter isotopes produced by fission are a varied mix of products, rather than a specific isotope as with alpha and beta particle emission. Often, fission produces excess neutrons that will sometimes be captured by other nuclei, possibly inducing additional radioactive events. Uranium-235 undergoes spontaneous fission to a small extent. One typical reaction is
$_{92}^{235}\textrm{Tc}\rightarrow \;_{56}^{139}\textrm{Ba}+\: _{36}^{94}\textrm{Kr}+2\: _{0}^{1}\textrm{n}\nonumber$
where $\ce{_0^1n}$ is a neutron. As with any nuclear process, the sums of the atomic numbers and mass numbers must be the same on both sides of the equation. Spontaneous fission is found only in large nuclei. The smallest nucleus that exhibits spontaneous fission is lead-208. (Fission is the radioactive process used in nuclear power plants and one type of nuclear bomb.)
Key Takeaways
• The major types of radioactivity include alpha particles, beta particles, and gamma rays.
• Fission is a type of radioactivity in which large nuclei spontaneously break apart into smaller nuclei. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.02%3A_Radioactivity.txt |
Learning Objectives
• Define half-life.
• Determine the amount of radioactive substance remaining after a given number of half-lives.
Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is half life, which is the amount of time it takes for one-half of a radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
Consider the following example. Suppose we have 100.0 g of tritium (a radioactive isotope of hydrogen). It has a half-life of 12.3 y. After 12.3 y, half of the sample will have decayed from hydrogen-3 to helium-3 by emitting a beta particle, so that only 50.0 g of the original tritium remains. After another 12.3 y—making a total of 24.6 y—another half of the remaining tritium will have decayed, leaving 25.0 g of tritium. After another 12.3 y—now a total of 36.9 y—another half of the remaining tritium will have decayed, leaving 12.5 g. This sequence of events is illustrated in Figure $1$ - Radioactive Decay.
We can determine the amount of a radioactive isotope remaining after a given number half-lives by using the following expression:
$\text{amount remaining} = \text{initial amount} \times \left ( \frac{1}{2} \right )^{n}\nonumber$
where $n$ is the number of half-lives. This expression works even if the number of half-lives is not a whole number.
Example $1$
The half-life of fluorine-20 is 11.0 s. If a sample initially contains 5.00 g of fluorine-20, how much remains after 44.0 s?
Solution
If we compare the time that has passed to the isotope's half-life, we note that 44.0 s is exactly 4 half-lives, so we use the previous expression n = 4. Substituting and solving results in the following:
\begin{align*} \text{amount remaining} &= 5.00\,g \times \left ( \frac{1}{2} \right )^{4} \[4pt] & =\: 5.00\,g\times \left ( \frac{1}{16} \right ) \[4pt] &= 0.313\,g \end{align*}\nonumber
Less than one-third of a gram of fluorine-20 remains.
Exercise $1$
The half-life of titanium-44 is 60.0 y. A sample of titanium contains 0.600 g of titanium-44. How much remains after 240.0 y?
Answer
0.0375 g
Half-lives of isotopes range from fractions of a microsecond to billions of years. Table $1$ - Half-Lives of Various Isotopes, lists the half-lives of some isotopes.
Table $1$ Half-Lives of Various Isotopes
Isotope Half-Life
3H 12.3 y
14C 5730 y
40K 1.26 × 109 y
51Cr 27.70 d
90Sr 29.1 y
131I 8.04 d
222Rn 3.823 d
235U 7.04 × 108 y
238U 4.47 × 109 y
241Am 432.7 y
248Bk 23.7 h
260Sg 4 ms
Chemistry is Everywhere: Radioactive Elements in the Body
You may not think of yourself as radioactive, but you are. A small portion of certain elements in the human body are radioactive and constantly undergo decay. The following table summarizes radioactivity in the normal human body.
Table with four columns and eight rows. The first column on the left is labeled as Radioactive Isotope, underneath in the rows has radioactive isotopes. The second column is labeled as Half-Life (y), underneath in the rows has half-lives. The third column is labeled as Isotope Mass in the Body (g), underneath in the rows has isotope masses. The fourth column on right is labeled as Activity in the Body (decays/s), underneath in the rows has the activities in the body as the decays.
Radioactive Isotope Half-Life (y) Isotope Mass in the Body (g) Activity in the Body (decays/s)
40K 1.26 × 109 0.0164 4,340
14C 5,730 1.6 × 10−8 3,080
87Rb 4.9 × 1010 0.19 600
210Pb 22.3 5.4 × 10−10 15
3H 12.3 2 × 10−14 7
238U 4.47 × 109 1 × 10−4 5
228Ra 5.76 4.6 × 10−14 5
226Ra 1,620 3.6 × 10−11 3
The average human body experiences about 8,000 radioactive decays.
Most of the radioactivity in the human body comes from potassium-40 and carbon-14. Potassium and carbon are two elements that we absolutely cannot live without, so unless we can remove all the radioactive isotopes of these elements, there is no way to escape at least some radioactivity. There is debate about which radioactive element is more problematic. There is more potassium-40 in the body than carbon-14, and it has a much longer half-life. Potassium-40 also decays with about 10 times more energy than carbon-14, making each decay potentially more problematic. However, carbon is the element that makes up the backbone of most living molecules, making carbon-14 more likely to be present around important molecules, such as proteins and DNA molecules. Most experts agree that while it is foolhardy to expect absolutely no exposure to radioactivity, we can and should minimize exposure to excess radioactivity.
What if the elapsed time is not an exact number of half-lives? We can still calculate the amount of material we have left, but the equation is more complicated. The equation is
$\text{amount remaining} = (\text{amount initial}) \times e^{-0.693t/t_{1/2}}\nonumber$
where e is the base of natural logarithms (2.71828182…), t is the elapsed time, and t1/2 is the half-life of the radioactive isotope. The variables t and t1/2 should have the same units of time, and you may need to make sure you know how to evaluate natural-logarithm powers on your calculator (for many calculators, there is an "inverse logarithm" function that you can use; consult your instructor if you are not sure how to use your calculator). Although this is a more complicated formula, the length of time t need not be an exact multiple of half-lives.
Example $2$
The half-life of fluorine-20 is 11.0 s. If a sample initially contains 5.00 g of fluorine-20, how much remains after 60.0 s?
Solution
Although similar to Example 3, the amount of time is not an exact multiple of a half-life. Here we identify the initial amount as 5.00 g, t = 60.0 s, and t1/2 = 11.0 s. Substituting into the equation:
amount remaining = (5.00 g) × e−(0.693)(60.0 s)/11.0 s
Evaluating the exponent (and noting that the s units cancel), we get
amount remaining = (5.00 g) × e−3.78
Solving, the amount remaining is 0.114 g. (You may want to verify this answer to confirm that you are using your calculator properly.)
Exercise $2$
The half-life of titanium-44 is 60.0 y. A sample of titanium contains 0.600 g of titanium-44. How much remains after 100.0 y?
Answer
0.189 g
Key Takeaways
• Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively.
• The amount of material left over after a certain number of half-lives can be easily calculated. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.03%3A_Half-Life.txt |
Learning Objective
• Express amounts of radioactivity in a variety of units.
In Section 15.2, we used mass to indicate the amount of radioactive substance present. This is only one of several units used to express amounts of radiation. Some units describe the number of radioactive events occurring per unit time, while others express the amount of a person's exposure to radiation.
Perhaps the direct way of reporting radioactivity is the number of radioactive decays per second. One decay per second is called one becquerel (Bq). Even in a small mass of radioactive material, however, there are thousands upon thousands of decays or disintegrations per second. The unit curie (Ci), now defined as 3.7 × 1010 decays/s, was originally defined as the number of decays per second in 1 g of radium. Many radioactive samples have activities that are on the order of microcuries (µCi) or more. Both the becquerel and the curie can be used in place of grams to describe quantities of radioactive material. As an example, the amount of americium in an average smoke detector has an activity of 0.9 µCi. (The curie is named after Polish scientist Marie Curie, who performed some of the initial investigations into radioactive phenomena in the early 1900s; the becquerel is named after Henri Becquerel, who discovered radioactivity in 1896.)
Example $1$
A sample of radium has an activity of 16.0 mCi (millicuries). If the half-life of radium is 1,600 y, how long before the sample's activity is 1.0 mCi?
Solution
The following table shows the activity of the radium sample over multiple half-lives:
Solutions to Example 15.4.1
Time in Years Activity
0 16.0 mCi
1,600 8.0 mCi
3,200 4.0 mCi
4,800 2.0 mCi
6,400 1.0 mCi
Over a period of 4 half-lives, the activity of the radium will be halved four times, at which point its activity will be 1.0 mCi. Thus it takes 4 half-lives, or 4 × 1,600 y = 6,400 y, for the activity to decrease to 1.0 mCi.
Exercise $1$
A sample of radon has an activity of 60,000 Bq. If the half-life of radon is 15 h, how long before the sample's activity is 3,750 Bq?
Answer
60 h
Example $2$
A sample of radium has an activity of 16.0 mCi. If the half-life of radium is 1,600 y, how long before the sample's activity is 5.6 mCi?
Solution
In this case we do not have an exact number of half-lives, so we need to use the more complicated equation (Section 15.2) and solve for time. If the initial amount is represented by 16.0 mCi and the final amount is 5.6 mCi, we have
$5.6\, mCi = (16.0\, mCi)e^{−0.693t/(1,600\, y)} \nonumber \nonumber$
To solve, we divide both sides of the equation by 16.0 mCi to cancel the millicurie units:
$\frac{5.6}{16.0}= e^{-0.693t/(1,600y)}\nonumber \nonumber$
By taking the natural logarithm of both sides; the natural logarithm cancels the exponential function. The natural logarithm of 5.6/16.0 is −1.050. So
$−1.050 = \dfrac{−0.692t}{1,600\, y}\nonumber \nonumber$
The negative sign cancels, and we solve for $t$. Thus
$t = 2,420\, y\nonumber \nonumber$
It makes sense that the time is greater than one half-life (1,600 y) because we have less than one-half of the original activity left.
Exercise $2$
A sample of radon has an activity of 60,000 Bq. If the half-life of radon is 15 h, how long before the sample's activity is 10,000 Bq?
Answer
38.8 h
Other measures of radioactivity are based on the effects it has on living tissue. Radioactivity can transfer energy to tissues in two ways: through the kinetic energy of the particles hitting the tissue, and through the electromagnetic energy of the gamma rays being absorbed by the tissue. Either way, like the thermal energy from boiling water, the transferred energy can damage the tissue.
The rad (an acronym for radiation absorbed dose) is a unit equivalent to 1 g of tissue absorbing 0.01 J:
1 rad = 0.01 J/g
Another unit of radiation absorption is the gray (Gy):
1 Gy = 100 rad
The rad is more common. To get an idea of the amount of energy this represents, consider that the absorption of 1 rad by 70,000 g of water (approximately the same mass as a 150 lb person) would increase the temperature of the water by only 0.002°C. This may not seem like a lot, but it is enough energy to break about 1 × 1021 molecular C–C bonds in a person's body. That amount of damage would not be desirable.
Predicting the effects of radiation is complicated by the fact that different types of emissions affect various tissues differently. To quantify these effects, the unit rem (an acronym for röntgen equivalent man) is defined
asrem = rad × factor
where factor is a number greater than or equal to 1, that takes into account the type of radioactive emission and sometimes the type of tissue being exposed. For beta particles, the factor equals 1. For alpha particles striking most tissues, the factor is 10, but for eye tissue the factor is 30. Most radioactive emissions that people are exposed to are on the order of a few dozen millirems (mrem) or less; a medical X-ray is about 20 mrem. A sievert (Sv) is a related unit and is defined as 100 rem.
What is a person's annual exposure to radioactivity and radiation? Table $1$, lists the sources and annual amounts of radiation exposure. It may surprise you to learn that fully 82% of the radioactivity and radiation exposure we receive is from natural sources—sources we cannot avoid. Fully 10% of the exposure comes from our own bodies, largely from carbon-14 and potassium-40.
Table $1$: Average Annual Radiation Exposure (Approximate)
Source Amount (mrem)
radon gas 200
medical sources 53
radioactive atoms in the body naturally 39
terrestrial sources 28
cosmic sources 28*
consumer products 10
nuclear energy 0.05
Total 358
*Flying from New York City to San Francisco adds 5 mrem to your overall radiation exposure because the plane flies above much of the atmosphere, which protects us from cosmic radiation.
The actual effects of radioactivity and radiation exposure on a person's health depend on the type of radioactivity, the length of exposure, and the tissues exposed. Table $2$ - Effects of Short-Term Exposure to Radioactivity and Radiation, lists the potential threats to health at various amounts of exposure over short periods of time (hours or days).
Table $2$: Effects of Short-Term Exposure to Radioactivity and Radiation
Exposure (rem) Effect
1 (over a full year) no detectable effect
∼20 increased risk of some cancers
∼100 damage to bone marrow and other tissues; possible internal bleeding; decrease in white blood cell count
200–300 visible "burns" in skin, nausea, vomiting, fatigue
>300 loss of white blood cells; hair loss
∼600 death
One of the simplest ways of detecting radioactivity is by using a piece of photographic film embedded in a badge or a pen. On a regular basis, the film is developed and checked for exposure. Comparing the exposure level of the film with a set of standard exposures indicates the amount of radiation a person was exposed to.
Another means of detecting radioactivity is an electrical device called a Geiger counter (Figure $1$). It contains a gas-filled chamber with a thin membrane on one end that allows radiation emitted from radioactive nuclei to enter the chamber and knock electrons off atoms of gas (usually argon). The presence of electrons and positively charged ions causes a small current, which is detected by the Geiger counter and converted to a signal on a meter.
Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter.
Key Takeaway
• Radioactivity can be expressed in a variety of units, including rems, rads, and curies. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.04%3A_Units_of_Radioactivity.txt |
Learning Objective
• Learn some applications of radioactivity.
Radioactive isotopes have a variety of applications. Generally, however, they are useful because either we can detect their radioactivity or we can use the energy they release.
Radioactive isotopes are effective tracers because their radioactivity is easy to detect. A tracer is a substance that can be used to follow the pathway of that substance through some structure. For instance, leaks in underground water pipes can be discovered by running some tritium-containing water through the pipes and then using a Geiger counter to locate any radioactive tritium subsequently present in the ground around the pipes. (Recall that tritium is a radioactive isotope of hydrogen.)
Tracers can also be used to follow the steps of a complex chemical reaction. After incorporating radioactive atoms into reactant molecules, scientists can track where the atoms go by following their radioactivity. One excellent example of this is the use of carbon-14 to determine the steps involved in photosynthesis in plants. We know these steps because researchers followed the progress of carbon-14 throughout the process.
Radioactive Dating
Radioactive isotopes are useful for establishing the ages of various objects. The half-life of radioactive isotopes is unaffected by any environmental factors, so the isotope acts like an internal clock. For example, if a rock is analyzed and is found to contain a certain amount of uranium-235 and a certain amount of its daughter isotope, we can conclude that a certain fraction of the original uranium-235 has radioactively decayed. If half of the uranium has decayed, then the rock has an age of one half-life of uranium-235, or about $4.5\times 10^{9}y$. Many analyses like this, using a wide variety of isotopes, have indicated that age of the earth itself is over $4\times 10^{9}y$.
In another interesting example of radioactive dating, hydrogen-3 dating has been used to verify the stated vintages of some old fine wines.
One isotope, carbon-14, is particularly useful in determining the age of once-living artifacts. A tiny amount of carbon-14 is produced naturally in the upper reaches of the atmosphere, and living things incorporate some of it into their tissues, building up to a constant, albeit very low, level. Once a living thing dies, it no longer acquires carbon-14; as time passes the carbon-14 that was in the tissues decays. (The half-life of carbon-14 is 5,370 y.) If a once-living artifact is discovered and analyzed many years after its death and the remaining carbon-14 is compared to the known constant level, an approximate age of the artifact can be determined. Using such methods, scientists determined that the age of the Shroud of Turin (Figure $1$ purported by some to be the burial cloth of Jesus Christ and composed of flax fibers, a type of plant) is about 600–700 y, not 2,000 y as claimed by some. Scientists were also able to use radiocarbon dating to show that the age of a mummified body found in the ice of the Alps was 5,300 y.
Irradiation of Food
The radiation emitted by some radioactive substances can be used to kill microorganisms on a variety of foodstuffs, extending the shelf life of these products. Produce such as tomatoes, mushrooms, sprouts, and berries are irradiated with the emissions from coba . This exposure kills a lot of the bacteria that cause spoilage, so the produce stays fresh longer. Eggs and some meat, such as beef, pork, and poultry, can also be irradiated. Contrary to the belief of some people, irradiation of food does not make the food itself radioactive.
Medical Applications
Radioactive isotopes have numerous medical applications in diagnosing and treating illness and diseases. One example of a diagnostic application is using radioactive iodine-131 to test for thyroid activity (Figure $1$). The thyroid gland in the neck is one of the few places in the body with a significant concentration of iodine. To evaluate thyroid activity, a measured dose of 131I is administered to a patient, and the next day a scanner is used to measure the amount of radioactivity in the thyroid gland. The amount of radioactive iodine that collects there is directly related to the activity of the thyroid, allowing trained physicians to diagnose both hyperthyroidism and hypothyroidism. Iodine-131 has a half-life of only 8 d, so the potential for damage due to exposure is minimal. Technetium-99 can also be used to test thyroid function. Bones, the heart, the brain, the liver, the lungs, and many other organs can be imaged in similar ways by using the appropriate radioactive isotope.
Very little radioactive material is needed in these diagnostic techniques because the radiation emitted is so easy to detect. However, therapeutic applications usually require much larger doses because their purpose is to preferentially kill diseased tissues. For example, if a thyroid tumor were detected, a much larger infusion (thousands of rem, as opposed to a diagnostic dose of less than 40 rem) of iodine-131 could help destroy the tumor cells. Similarly, radioactive strontium is used to not only detect, but also to ease the pain of bone cancers. Table $1$ lists several radioactive isotopes and their medical uses.
Table $1$: Some Radioactive Isotopes with Medical Applications
Isotope Use
32P cancer detection and treatment, especially in eyes and skin
59Fe anemia diagnosis
60Co gamma ray irradiation of tumors
99mTc* brain, thyroid, liver, bone marrow, lung, heart, and intestinal scanning; blood volume determination
131I diagnosis and treatment of thyroid function
133Xe lung imaging
198Au liver disease diagnosis
*The "m" means that it is a metastable form of this isotope of technetium.
In addition to the direct application of radioactive isotopes to diseased tissue, the gamma ray emissions of some isotopes can be directed toward the tissue to be destroyed. Cobalt-60 is a useful isotope for this kind of procedure.
Food and Drink Application: Radioactivity in Wines
Wine lovers put some stock in vintages, or the years in which the wine grapes were grown before they were turned into wine. Wine can differ in quality depending on the vintage. Some wine lovers willingly pay much more for a bottle of wine with a certain vintage. But how does one verify that a bottle of wine was in fact part of a certain vintage? Is the label a fake? Is that stash of wine found in the basement of a French chateau really from the 1940s, or was it made in 2009?
Cesium-137 is a radioactive isotope that has a half-life of 30.1 y. It was introduced into the atmosphere in the 1940s and 1950s by the atmospheric testing of nuclear weapons by several countries after World War II. A significant amount of cesium-137 was released during the Chernobyl nuclear disaster in 1986. As a result of this atmospheric contamination, scientists have precise measurements of the amount of cesium-137 available in the environment since 1950. Some of the isotope of cesium is taken up by living plants, including grape vines. Using known vintages, oenologists (wine scientists) can construct a detailed analysis of the cesium-137 of various wines through the years.
The verification of a wine's vintage requires the measurement of the activity of cesium-137 in the wine. By measuring the current activity of cesium-137 in a sample of wine (the gamma rays from the radioactive decay pass through glass wine bottles easily, so there's no need to open the bottle), comparing it to the known amount of cesium-137 from the vintage, and taking into account the passage of time, researchers can collect evidence for or against a claimed wine vintage.
Before about 1950, the amount of cesium-137 in the environment was negligible, so if a wine dated before 1950 shows any measurable cesium-137 activity, it is almost surely a fake, so do not shell out lots of money for it! It may be a good wine, but it is almost definitely not over 60 years old.
Summary
Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and preservation of food. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.05%3A_Uses_of_Radioactive_Isotopes.txt |
Learning Objectives
• Explain where nuclear energy comes from.
• Describe the difference between fission and fusion.
Nuclear changes occur with a simultaneous release of energy. Where does this energy come from? If we could precisely measure the masses of the reactants and products of a nuclear reaction, we would notice that the amount of mass drops slightly in the conversion from reactants to products. Consider the following nuclear equation, in which the molar mass of each species is indicated to four decimal places:
$_{235.0439}^{235}\textrm{U}\rightarrow _{138.9088}^{139}\textrm{Ba}+_{93.9343}^{94}\textrm{Kr}+_{2\times 1.0087}^{2^{1}}\textrm{n}\nonumber$
If we compare the mass of the reactant (235.0439) to the masses of the products (sum=234.8605) we notice a mass difference of -0.1834 g or -0.0001834 kg. Where did this mass go?
According to Albert Einstein's theory of relativity, energy (E) and mass (m) are related by the following equation:
$E = mc^2\nonumber$
where c is the speed of light, or $c=3.00\times 10^{8}\, m/s\nonumber$
In the course of the chemical reaction for uranium, the mass difference is converted to energy, which is given off by the reaction:
$E=(-0.0001834\, kg)(3.00\times 10^{8}\, m/s)^{2}=-1.65\times 10^{13}J=-1.65\times 10^{10}kJ\nonumber$
(For the units to work out, mass must be expressed in units of kilograms.) That is, 16.5 billion kJ of energy is given off every time 1 mol of uranium-235 undergoes this nuclear reaction. This is an extraordinary amount of energy. Compare it to combustion reactions of hydrocarbons, which give off about 650kJ/mol of energy for every CH2 unit in the hydrocarbon, on the order of hundreds of kilojoules per mole. Nuclear reactions give off billions of kilojoules per mole.
If this energy could be properly harvested, it would be a significant source of energy for our society. Nuclear energy involves the controlled harvesting of energy from fission reactions. The reaction can be controlled because the fission of uranium-235 (and a few other isotopes, such as plutonium-239) can be artificially initiated by injecting a neutron into a uranium nucleus. The overall nuclear equation, with energy included as a product, is then as follows:
$_{}^{235}\textrm{U}\: +\: _{ }^{1}\textrm{n}\rightarrow \: _{ }^{139}\textrm{Ba}\: +\: _{ }^{94}\textrm{Kr}\: +\: 3_{ }^{1}\textrm{n}\nonumber$
Thus by the careful addition of extra neutrons into a sample of uranium, we can control the fission process and obtain energy that can be used for other purposes. (Artificial or induced radioactivity, in which neutrons are injected into a sample of matter that subsequently cause fission, was first demonstrated in 1934 by Irène Joliot-Curie and Frédéric Joliot, the daughter and son-in-law of Marie Curie.)
Plutonium-239 can absorb a neutron and undergo a fission reaction to make an atom of gold-204 and an atom of phosphorus-31. Write the balanced nuclear equation for the process and determine the number of neutrons given off as part of the reaction.
Solution
Using the data given, we can write the following initial equation:
$_{0}^{1}\textrm{n}\: +\: _{94}^{239}\textrm{Pu}\rightarrow \: _{79}^{204}\textrm{Au}\: +\: _{15}^{31}\textrm{P}\: +\: ?_{0}^{1}\textrm{n}\nonumber$
In balanced nuclear equations, the sums of the subscripts on each side of the equation are the same, as are the sums of the superscripts. The subscripts are already balanced:
0 + 94 = 94 and 79 + 15 = 94.
The superscripts on the left equal 240 (1 + 239) but equal 235 (204 + 31) on the right. We need five more mass number units on the right. Five neutrons should be products of the process for the mass numbers to balance. (Because the atomic number of a neutron is zero, including five neutrons on the right does not change the overall sum of the subscripts.) Thus the balanced nuclear equation is as follows:
$_{0}^{1}\textrm{n}\: +\: _{94}^{239}\textrm{Pu}\rightarrow \: _{79}^{204}\textrm{Au}\: +\: _{15}^{31}\textrm{P}\: +\: 5_{0}^{1}\textrm{n}\nonumber$
We predict that the overall process will give off five neutrons.
Exercise $1$
Uranium-238 can absorb a neutron and undergo a fission reaction to produce an atom of cesium-135 and an atom of rubidium-96. Write the balanced nuclear equation for the process and determine the number of neutrons given off as part of the reaction.
Answer
$_{0}^{1}\textrm{n}\: +\: _{92}^{238}\textrm{U}\rightarrow \: _{37}^{96}\textrm{Rb}\: +\: _{55}^{135}\textrm{Cs}\: +\: 8_{0}^{1}\textrm{n}\nonumber$
eight neutrons
One balanced nuclear reaction for the fission of plutonium-239 is as follows:
$\underset{1.0087}{_{0}^{1}\textrm{n}}\: +\: \underset{239.0522}{_{94}^{239}\textrm{Pu}}\rightarrow \: \underset{203.9777}{_{79}^{204}\textrm{Au}}\: +\: \underset{30.9738}{_{15}^{31}\textrm{P}}\: +\: \underset{5\times 1.0087}{5_{0}^{1}\textrm{n}}\nonumber$
The molar mass in grams of each species is given for each particle. What is the energy change of this fission reaction?
Solution
We start by adding the masses of all species on each side of the nuclear equation. Then we determine the difference in mass as the reaction proceeds and convert this to an equivalent amount of energy. The total mass of the reactants is as follows:
1.0087 + 239.0522 = 240.0609 g
The total mass of the products is as follows:
203.9777 + 30.9738 + (5 × 1.0087) = 239.9950 g
The change is mass is determined by subtracting the mass of the reactants from the mass of the products:
change in mass = 239.9950 − 240.0609 = −0.0659 g
This mass change must be converted into kilogram units:
$-0.0659g\times \frac{1kg}{1,000g}=-0.0000659kg\nonumber$
Now we can use Einstein's equation to determine the energy change of the nuclear reaction:
E = (−0.0000659 kg)(3.00 × 108 m/s)2 = −5.93 × 1012 J
Exercise $2$
The nuclear equation for the fission of uranium-238 is as follows:
$\underset{1.0087}{_{0}^{1}\textrm{n}}\: +\: \underset{238.0508}{_{92}^{238}\textrm{U}}\rightarrow \: \underset{95.9342}{_{37}^{96}\textrm{Rb}}\: +\: \underset{134.9060}{_{55}^{135}\textrm{Cs}}\: +\: \underset{8\times 1.0087}{8_{0}^{1}\textrm{n}}\nonumber$
The molar mass in grams of each species is given for each particle. What is the energy change of this fission reaction?
Answer
−1.35 × 1013 J
A nuclear reactor is an apparatus designed to carefully control the progress of a nuclear reaction and extract the resulting energy for useful purposes. Figure $1$ shows a simplified diagram of a nuclear reactor. The energy from the controlled nuclear reaction converts water into high-pressure steam, which is used to run turbines that generate electricity.
Although the fission of large nuclei can produce different products, on average the fission of uranium produces two more free neutrons than were present to begin with. These neutrons can themselves stimulate other uranium nuclei to undergo fission, releasing yet more energy and even more neutrons, which can in turn induce even more uranium fission. A single neutron can thus begin a process that grows exponentially in a phenomenon called a chain reaction.
1 → 2 → 4 → 8 → 16 → 32 → 64 → 128 → 256 → 512 → 1,024 → 2,048 → 4,096 → 8,192 → 16,384 →…
Because energy is produced with each fission event, energy is also produced exponentially and in an uncontrolled fashion. The quick production of energy creates an explosion. This is the mechanism behind the atomic bomb. (The first controlled chain reaction was achieved on December 2, 1942, in an experiment supervised by Enrico Fermi in a laboratory underneath the football stadium at the University of Chicago.)
Although fairly simple in theory, an atomic bomb is difficult to produce, in part because uranium-235, the isotope that undergoes fission, makes up only 0.7% of natural uranium; the rest is mostly uranium-238, which does not undergo fission. (Remember that the radioactive process that a nucleus undergoes is characteristic of the isotope.) To make uranium useful for nuclear reactors, the uranium in uranium-235 must be enriched to about 3%. The enrichment of uranium is a laborious and costly series of physical and chemical separations. To be useful in an atomic bomb, uranium must be enriched to 70% or more. At lesser concentrations, the chain reaction cannot sustain itself, so no explosion is produced.
Fusion is another nuclear process that can be used to produce energy. In this process, smaller nuclei are combined to make larger nuclei, with an accompanying release of energy. One example is hydrogen fusion, which makes helium:
$4^{1}H\rightarrow _{ }^{4}\textrm{He}+2.58\times 10^{12}\: J\nonumber$
Notice that the amount of energy given off per mole of reactant is only one-tenth of the amount given off by the fission of 1 mol of uranium-235. On a mass (per gram) basis, however, hydrogen fusion gives off 10 times more energy than fission does. In addition, the product of fusion is helium gas, not a wide range of isotopes (some of which are also radioactive) produced by fission.
Fusion occurs in nature: the sun and other stars use fusion as their ultimate energy source. Fusion is also the basis of very destructive weapons that have been developed by several countries around the world. However, one current goal is to develop a source of controlled fusion for use as an energy source. The practical problem is that to perform fusion, extremely high pressures and temperatures are necessary. Currently, the only known stable systems undergoing fusion are the interiors of stars. The conditions necessary for fusion can be created using an atomic bomb, but the resulting fusion is uncontrollable (and the basis for another type of bomb, a hydrogen bomb). Currently, researchers are looking for safe, controlled ways for producing useful energy using fusion.
Summary
• Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur.
• In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy. | textbooks/chem/Introductory_Chemistry/Beginning_Chemistry_(Ball)/15%3A_Nuclear_Chemistry/15.06%3A_Nuclear_Energy.txt |
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