docid
stringlengths 25
45
| url
stringlengths 14
987
| title
stringlengths 0
45k
| headings
stringlengths 1
259k
| segment
stringlengths 2
10k
| start_char
int64 0
9.96k
| end_char
int64 2
10k
|
|---|---|---|---|---|---|---|
msmarco_v2.1_doc_00_10634855#15_15413642
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
carbon tetrachloride
silicon dioxide
trisilicon tetranitride
Solution
The name carbon tetrachloride implies one carbon atom and four chlorine atoms, so the formula is CCl 4. The name silicon dioxide implies one silicon atom and two oxygen atoms, so the formula is SiO 2. 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 Si 3 N 4. Test Yourself
Give the formula for each molecule. disulfur difluoride
iodine pentabromide
Answers
S 2 F 2
IBr 5
Some simple molecules have common names that we use as part of the formal system of chemical nomenclature. For example, H 2 O is given the name water, not dihydrogen monoxide. NH 3 is called ammonia, while CH 4 is called methane. We will occasionally see other molecules that have common names; we will point them out as they occur.
| 7,299 | 8,162 |
msmarco_v2.1_doc_00_10634855#16_15414959
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
disulfur difluoride
iodine pentabromide
Answers
S 2 F 2
IBr 5
Some simple molecules have common names that we use as part of the formal system of chemical nomenclature. For example, H 2 O is given the name water, not dihydrogen monoxide. NH 3 is called ammonia, while CH 4 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. Exercises
Which of these formulas represent molecules?
| 7,767 | 8,477 |
msmarco_v2.1_doc_00_10634855#17_15416120
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
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. Exercises
Which of these formulas represent molecules? State how many atoms are in each molecule. Fe
PCl 3
P 4
Ar
Which of these formulas represent molecules? State how many atoms are in each molecule. I 2
He
H 2 O
Al
What is the difference between CO and Co? What is the difference between H 2 O and H 2 O 2 (hydrogen peroxide)?
| 8,162 | 8,752 |
msmarco_v2.1_doc_00_10634855#18_15417164
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
State how many atoms are in each molecule. Fe
PCl 3
P 4
Ar
Which of these formulas represent molecules? State how many atoms are in each molecule. I 2
He
H 2 O
Al
What is the difference between CO and Co? What is the difference between H 2 O and H 2 O 2 (hydrogen peroxide)? Give the proper formula for each diatomic element. In 1986, when Halley’s comet last passed the earth, astronomers detected the presence of S 2 in their telescopes. Why is sulfur not considered a diatomic element? What is the stem of fluorine used in molecule names? CF 4 is one example.
| 8,478 | 9,040 |
msmarco_v2.1_doc_00_10634855#19_15418184
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
Give the proper formula for each diatomic element. In 1986, when Halley’s comet last passed the earth, astronomers detected the presence of S 2 in their telescopes. Why is sulfur not considered a diatomic element? What is the stem of fluorine used in molecule names? CF 4 is one example. What is the stem of selenium used in molecule names? SiSe 2 is an example. Give the proper name for each molecule. PF 3
TeCl 2
N 2 O 3
Give the proper name for each molecule. NO
CS 2
As 2 O 3
Give the proper name for each molecule.
| 8,752 | 9,272 |
msmarco_v2.1_doc_00_10634855#20_15419159
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
What is the stem of selenium used in molecule names? SiSe 2 is an example. Give the proper name for each molecule. PF 3
TeCl 2
N 2 O 3
Give the proper name for each molecule. NO
CS 2
As 2 O 3
Give the proper name for each molecule. XeF 2
O 2 F 2
SF 6
Give the proper name for each molecule. P 4 O 10
B 2 O 3
P 2 S 3
Give the proper name for each molecule. N 2 O
N 2 O 4
N 2 O 5
Give the proper name for each molecule. SeO 2
Cl 2 O
XeF 6
Give the proper formula for each name. dinitrogen pentoxide
tetraboron tricarbide
phosphorus pentachloride
Give the proper formula for each name.
| 9,040 | 9,623 |
msmarco_v2.1_doc_00_10634855#21_15420207
|
http://2012books.lardbucket.org/books/beginning-chemistry/s07-02-molecules-and-chemical-nomencl.html
|
Molecules and Chemical Nomenclature
|
3.2
Molecules and Chemical Nomenclature
3.2 Molecules and Chemical Nomenclature
Learning Objectives
Example 3
Example 4
Key Takeaways
Exercises
Answers
|
XeF 2
O 2 F 2
SF 6
Give the proper name for each molecule. P 4 O 10
B 2 O 3
P 2 S 3
Give the proper name for each molecule. N 2 O
N 2 O 4
N 2 O 5
Give the proper name for each molecule. SeO 2
Cl 2 O
XeF 6
Give the proper formula for each name. dinitrogen pentoxide
tetraboron tricarbide
phosphorus pentachloride
Give the proper formula for each name. nitrogen triiodide
diarsenic trisulfide
iodine trichloride
Give the proper formula for each name. dioxygen dichloride
dinitrogen trisulfide
xenon tetrafluoride
Give the proper formula for each name. chlorine dioxide
selenium dibromide
dinitrogen trioxide
Give the proper formula for each name. iodine trifluoride
xenon trioxide
disulfur decafluoride
Give the proper formula fo
| 9,272 | 10,000 |
msmarco_v2.1_doc_00_10646020#0_15421407
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
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 to 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
A chemical reaction in which one element is substituted for another element in a compound. 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. For example,
2HCl (aq) + Zn (s) → ZnCl2(aq) + H2(g)
is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element—hydrogen—is formed.
| 0 | 1,173 |
msmarco_v2.1_doc_00_10646020#1_15423241
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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
A chemical reaction in which one element is substituted for another element in a compound. 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. For example,
2HCl (aq) + Zn (s) → ZnCl2(aq) + H2(g)
is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element—hydrogen—is formed. Another example of a single-replacement reaction is
2NaCl (aq) + F2(g) → 2NaF (s) + Cl2(g)
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 (see Figure 4.1 "Halogens on the Periodic Table" ).
| 561 | 1,793 |
msmarco_v2.1_doc_00_10646020#2_15425142
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Another example of a single-replacement reaction is
2NaCl (aq) + F2(g) → 2NaF (s) + Cl2(g)
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 (see Figure 4.1 "Halogens on the Periodic Table" ). 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
CaI2(s) + Cl2(g) → CaCl2(s) + I2(s)
will occur, but the reaction
CaF2(s) + Br2(ℓ) → CaBr2(s) + F2(g)
will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry. Figure 4.1 Halogens on the Periodic Table
The halogens are the elements in the next-to-last column on the periodic table. Example 2
Will a single-replacement reaction occur?
| 1,174 | 2,371 |
msmarco_v2.1_doc_00_10646020#3_15427012
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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
CaI2(s) + Cl2(g) → CaCl2(s) + I2(s)
will occur, but the reaction
CaF2(s) + Br2(ℓ) → CaBr2(s) + F2(g)
will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry. Figure 4.1 Halogens on the Periodic Table
The halogens are the elements in the next-to-last column on the periodic table. Example 2
Will a single-replacement reaction occur? If so, identify the products. MgCl 2 + I 2 → ? CaBr 2 + F 2 → ? Solution
Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur. Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF 2 and Br 2.
| 1,794 | 2,702 |
msmarco_v2.1_doc_00_10646020#4_15428595
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
If so, identify the products. MgCl 2 + I 2 → ? CaBr 2 + F 2 → ? Solution
Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur. Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF 2 and Br 2. Test Yourself
Will a single-replacement reaction occur? If so, identify the products. FeI 2 + Cl 2 → ? Answer
Yes; FeCl 2 and I 2
Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds—simply use their relative positions on the periodic table.
| 2,372 | 2,986 |
msmarco_v2.1_doc_00_10646020#5_15429876
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Test Yourself
Will a single-replacement reaction occur? If so, identify the products. FeI 2 + Cl 2 → ? Answer
Yes; FeCl 2 and I 2
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
A list of elements that will replace elements below them in single-replacement reactions. does the same thing the periodic table does for halogens:
| 2,702 | 3,415 |
msmarco_v2.1_doc_00_10646020#6_15431245
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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
A list of elements that will replace elements below them in single-replacement reactions. 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
H 2
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.
| 2,987 | 3,951 |
msmarco_v2.1_doc_00_10646020#7_15432876
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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
H 2
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 3
Use the activity series to predict the products, if any, of each equation. FeCl 2 + Zn → ? HNO 3 + Au → ? Solution
Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl 2 and Fe.
| 3,416 | 4,233 |
msmarco_v2.1_doc_00_10646020#8_15434371
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Example 3
Use the activity series to predict the products, if any, of each equation. FeCl 2 + Zn → ? HNO 3 + Au → ? Solution
Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl 2 and Fe. 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. Test Yourself
Use the activity series to predict the products, if any, of this equation. AlPO 4 + Mg → ?
| 3,951 | 4,485 |
msmarco_v2.1_doc_00_10646020#9_15435565
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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. Test Yourself
Use the activity series to predict the products, if any, of this equation. AlPO 4 + Mg → ? Answer
Mg 3 (PO 4) 2 and Al
A double-replacement reaction
A chemical reaction in which parts of two ionic compounds are exchanged. 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
CuCl2(aq) + 2AgNO3(aq) → Cu (NO3)2(aq) + 2AgCl (s)
There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (
| 4,233 | 5,036 |
msmarco_v2.1_doc_00_10646020#10_15437026
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Answer
Mg 3 (PO 4) 2 and Al
A double-replacement reaction
A chemical reaction in which parts of two ionic compounds are exchanged. 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
CuCl2(aq) + 2AgNO3(aq) → Cu (NO3)2(aq) + 2AgCl (s)
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 4
Predict the products of this double-replacement equation: BaCl 2 + Na 2 SO 4 → ?
| 4,485 | 5,378 |
msmarco_v2.1_doc_00_10646020#11_15438579
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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 4
Predict the products of this double-replacement equation: BaCl 2 + Na 2 SO 4 → ? Solution
Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO 4 and NaCl. Test Yourself
Predict the products of this double-replacement equation: KBr + AgNO 3 → ? Answer
KNO 3 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:
| 5,036 | 5,844 |
msmarco_v2.1_doc_00_10646020#12_15440046
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Solution
Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO 4 and NaCl. Test Yourself
Predict the products of this double-replacement equation: KBr + AgNO 3 → ? Answer
KNO 3 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
A chemical reaction in which two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve. 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
A solid that falls out of solution in a precipitation reaction. .
| 5,378 | 6,261 |
msmarco_v2.1_doc_00_10646020#13_15441583
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
the precipitation reaction. A precipitation reaction
A chemical reaction in which two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve. 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
A solid that falls out of solution in a precipitation reaction. . 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
General statements that predict which ionic compounds dissolve and which do not. , which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). Table 4.1 "Some Useful Solubility Rules" lists some general solubility rules.
| 5,845 | 6,806 |
msmarco_v2.1_doc_00_10646020#14_15443193
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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
General statements that predict which ionic compounds dissolve and which do not. , which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). Table 4.1 "Some Useful Solubility Rules" 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 in Table 4.1 "Some Useful Solubility Rules". If a compound is soluble, we use the (aq) label with it, indicating 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 4.1 Some Useful Solubility Rules
These compounds generally dissolve in water (are soluble):
| 6,262 | 7,324 |
msmarco_v2.1_doc_00_10646020#15_15444905
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in Table 4.1 "Some Useful Solubility Rules". If a compound is soluble, we use the (aq) label with it, indicating 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 4.1 Some Useful Solubility Rules
These compounds generally dissolve in water (are soluble): Exceptions: All compounds of Li +, Na +, K +, Rb +, Cs +, and NH 4+
None
All compounds of NO 3− and C 2 H 3 O 2−
None
Compounds of Cl −, Br −, I −
Ag +, Hg 22+, Pb 2+
Compounds of SO 42
Hg 22+, Pb 2+, Sr 2+, Ba 2+
These compounds generally do not dissolve in water (are insoluble): Exceptions: Compounds of CO 32− and PO 43−
Compounds of Li +, Na +, K +, Rb +, Cs +, and NH 4+
Compounds of OH −
Compounds of Li +, Na +, K +, Rb +, Cs +, NH 4+, Sr 2+, and Ba 2+
For example, consider the possible double-replacement reaction between Na 2 SO 4 and SrCl 2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag +, Hg 22+, and Pb 2+, which are not being considered here.
| 6,807 | 8,065 |
msmarco_v2.1_doc_00_10646020#16_15446862
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Exceptions: All compounds of Li +, Na +, K +, Rb +, Cs +, and NH 4+
None
All compounds of NO 3− and C 2 H 3 O 2−
None
Compounds of Cl −, Br −, I −
Ag +, Hg 22+, Pb 2+
Compounds of SO 42
Hg 22+, Pb 2+, Sr 2+, Ba 2+
These compounds generally do not dissolve in water (are insoluble): Exceptions: Compounds of CO 32− and PO 43−
Compounds of Li +, Na +, K +, Rb +, Cs +, and NH 4+
Compounds of OH −
Compounds of Li +, Na +, K +, Rb +, Cs +, NH 4+, Sr 2+, and Ba 2+
For example, consider the possible double-replacement reaction between Na 2 SO 4 and SrCl 2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag +, Hg 22+, and Pb 2+, which are not being considered here. Therefore, Na 2 SO 4 and SrCl 2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO 4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO 4? Compounds of the sulfate ion are generally soluble, but Sr 2+ is an exception:
| 7,324 | 8,350 |
msmarco_v2.1_doc_00_10646020#17_15448583
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Therefore, Na 2 SO 4 and SrCl 2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO 4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO 4? Compounds of the sulfate ion are generally soluble, but Sr 2+ 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
Na2SO4(aq) + SrCl2(aq) → 2NaCl (aq) + SrSO4(s)
You would expect to see a visual change corresponding to SrSO 4 precipitating out of solution ( Figure 4.2 "Double-Replacement Reactions" ). Figure 4.2 Double-Replacement Reactions
Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution. Source: Photo courtesy of Choij, http://commons.wikimedia.org/wiki/File:Copper_solution.jpg.
| 8,066 | 8,911 |
msmarco_v2.1_doc_00_10646020#18_15450087
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be
Na2SO4(aq) + SrCl2(aq) → 2NaCl (aq) + SrSO4(s)
You would expect to see a visual change corresponding to SrSO 4 precipitating out of solution ( Figure 4.2 "Double-Replacement Reactions" ). Figure 4.2 Double-Replacement Reactions
Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution. Source: Photo courtesy of Choij, http://commons.wikimedia.org/wiki/File:Copper_solution.jpg. Example 5
Will a double-replacement reaction occur? If so, identify the products. Ca (NO 3) 2 + KBr → ? NaOH + FeCl 2 → ? Solution
According to the solubility rules, both Ca (NO 3) 2 and KBr are soluble.
| 8,351 | 9,115 |
msmarco_v2.1_doc_00_10646020#19_15451522
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Example 5
Will a double-replacement reaction occur? If so, identify the products. Ca (NO 3) 2 + KBr → ? NaOH + FeCl 2 → ? Solution
According to the solubility rules, both Ca (NO 3) 2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr 2 and KNO 3. 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. According to the solubility rules, both NaOH and FeCl 2 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.
| 8,911 | 9,625 |
msmarco_v2.1_doc_00_10646020#20_15452896
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr 2 and KNO 3. 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. According to the solubility rules, both NaOH and FeCl 2 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
2NaOH (aq) + FeCl2(aq) → 2NaCl (aq) + Fe (OH)2(s)
Test Yourself
Will a double-replacement equation occur? If so, identify the products. Sr (NO 3) 2 + KCl → ?
| 9,116 | 9,980 |
msmarco_v2.1_doc_00_10646020#21_15454422
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-02-types-of-chemical-reactions-si.html
|
Types of Chemical Reactions: Single- and Double-Displacement Reactions
|
4.2
Types of Chemical Reactions: Single- and Double-Displacement Reactions
4.2 Types of Chemical Reactions: Single- and Double-Displacement Reactions
Learning Objectives
Example 2
Activity Series for Cation Replacement in Single-Replacement Reactions
Example 3
Example 4
Example 5
Key Takeaways
Exercises
Answers
|
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
2NaOH (aq) + FeCl2(aq) → 2NaCl (aq) + Fe (OH)2(s)
Test Yourself
Will a double-replacement equation occur? If so, identify the products. Sr (NO 3) 2 + KCl → ? Answer
No reaction;
| 9,626 | 10,000 |
msmarco_v2.1_doc_00_10661183#0_15455455
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
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 dissolving 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.
| 0 | 826 |
msmarco_v2.1_doc_00_10661183#1_15456783
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
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 dissolving 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: NaCl (s) → H 2 O Na + (aq) + Cl − (aq)
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 4.3 "Ionic Solutions" ). This process is called dissociation
The process of an ionic compound separating into ions when it dissolves. ;
| 335 | 1,289 |
msmarco_v2.1_doc_00_10661183#2_15458258
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
We can use a chemical equation to represent this process—for example, with NaCl: NaCl (s) → H 2 O Na + (aq) + Cl − (aq)
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 4.3 "Ionic Solutions" ). This process is called dissociation
The process of an ionic compound separating into ions when it dissolves. ; we say that the ions dissociate. Figure 4.3 Ionic Solutions
When an ionic compound dissociates in water, water molecules surround each ion and separate it from the rest of the solid. Each ion goes its own way in solution. 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.
| 827 | 1,697 |
msmarco_v2.1_doc_00_10661183#3_15459656
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
we say that the ions dissociate. Figure 4.3 Ionic Solutions
When an ionic compound dissociates in water, water molecules surround each ion and separate it from the rest of the solid. Each ion goes its own way in solution. 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 CaCl 2 dissolves, the one Ca 2+ ion and the two Cl − ions separate from each other: CaCl 2 (s) → H 2 O Ca 2 + (aq) + Cl − (aq) + Cl − (aq) or CaCl 2 (s) → H 2 O Ca 2 + (aq) + 2Cl − (aq)
That is, the two chloride ions go off on their own.
| 1,290 | 2,254 |
msmarco_v2.1_doc_00_10661183#4_15461161
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
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 CaCl 2 dissolves, the one Ca 2+ ion and the two Cl − ions separate from each other: CaCl 2 (s) → H 2 O Ca 2 + (aq) + Cl − (aq) + Cl − (aq) or CaCl 2 (s) → H 2 O Ca 2 + (aq) + 2Cl − (aq)
That is, the two chloride ions go off on their own. They do not remain as Cl 2 (that would be elemental chlorine; these are chloride ions); they do not stick together to make Cl 2− or Cl 22−. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved. Example 6
Write the chemical equation that represents the dissociation of each ionic compound.
| 1,698 | 2,614 |
msmarco_v2.1_doc_00_10661183#5_15462622
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
They do not remain as Cl 2 (that would be elemental chlorine; these are chloride ions); they do not stick together to make Cl 2− or Cl 22−. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved. Example 6
Write the chemical equation that represents the dissociation of each ionic compound. KBr
Na 2 SO 4
Solution
KBr (s) → K + (aq) + Br − (aq)
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)
Test Yourself
Write the chemical equation that represents the dissociation of (NH 4) 2 S.
Answer
(NH 4) 2 S (s) → 2NH 4+ (aq) + S 2− (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
A chemical equation in which the dissolved ionic compounds are written as separated ions. 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.
| 2,255 | 3,467 |
msmarco_v2.1_doc_00_10661183#6_15464389
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
KBr
Na 2 SO 4
Solution
KBr (s) → K + (aq) + Br − (aq)
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)
Test Yourself
Write the chemical equation that represents the dissociation of (NH 4) 2 S.
Answer
(NH 4) 2 S (s) → 2NH 4+ (aq) + S 2− (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
A chemical equation in which the dissolved ionic compounds are written as separated ions. 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 AgNO 3 (aq) in a double-replacement reaction to precipitate AgCl (s) and form NaNO 3 (aq), the complete ionic equation includes NaCl, AgNO 3, and NaNO 3 written as separated ions: Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → AgCl (s) + Na+(aq) + NO3−(aq)
This is more representative of what is occurring in the solution. Example 7
Write the complete ionic equation for each chemical reaction. KBr (aq) + AgC 2 H 3 O 2 (aq) → KC 2 H 3 O 2 (aq) + AgBr (s)
MgSO 4 (aq) + Ba (NO 3) 2 (aq) → Mg (NO 3) 2 (aq) + BaSO 4 (s)
Solution
For any ionic compound that is aqueous, we will write the compound as separated ions. The complete ionic equation is
K+(aq) + Br−(aq) + Ag+(aq) + C2H3O2−(aq) → K+(aq) + C2H3O2−(aq) + AgBr (s)
The complete ionic equation is
Mg2+(aq) + SO42−(aq) + Ba2+(aq) + 2NO3−(aq) → Mg2+(aq) + 2NO3−(aq) + BaSO4(s)
Test Yourself
Write the complete ionic equation for
CaCl2(aq) + Pb (NO3)2(aq) → Ca (NO3)2(aq) + PbCl2(s)
Answer
Ca 2+ (aq) + 2Cl − (aq) + Pb 2+ (aq) + 2NO 3− (aq) → Ca 2+ (aq) + 2NO 3− (aq) + PbCl 2 (s)
You may notice that in a complete ionic equation, some ions do not change their chemical form;
| 2,614 | 4,628 |
msmarco_v2.1_doc_00_10661183#7_15467055
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
For example, when NaCl (aq) reacts with AgNO 3 (aq) in a double-replacement reaction to precipitate AgCl (s) and form NaNO 3 (aq), the complete ionic equation includes NaCl, AgNO 3, and NaNO 3 written as separated ions: Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → AgCl (s) + Na+(aq) + NO3−(aq)
This is more representative of what is occurring in the solution. Example 7
Write the complete ionic equation for each chemical reaction. KBr (aq) + AgC 2 H 3 O 2 (aq) → KC 2 H 3 O 2 (aq) + AgBr (s)
MgSO 4 (aq) + Ba (NO 3) 2 (aq) → Mg (NO 3) 2 (aq) + BaSO 4 (s)
Solution
For any ionic compound that is aqueous, we will write the compound as separated ions. The complete ionic equation is
K+(aq) + Br−(aq) + Ag+(aq) + C2H3O2−(aq) → K+(aq) + C2H3O2−(aq) + AgBr (s)
The complete ionic equation is
Mg2+(aq) + SO42−(aq) + Ba2+(aq) + 2NO3−(aq) → Mg2+(aq) + 2NO3−(aq) + BaSO4(s)
Test Yourself
Write the complete ionic equation for
CaCl2(aq) + Pb (NO3)2(aq) → Ca (NO3)2(aq) + PbCl2(s)
Answer
Ca 2+ (aq) + 2Cl − (aq) + Pb 2+ (aq) + 2NO 3− (aq) → Ca 2+ (aq) + 2NO 3− (aq) + PbCl 2 (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 NO 3− (aq) ions stay as Na + (aq) ions and NO 3− (aq) ions. These two ions are examples of spectator ions
An ion that does nothing in the overall course of a chemical reaction. , ions that do nothing in the overall course of a chemical reaction. They are present, but they do not participate in the overall chemistry.
| 3,468 | 5,192 |
msmarco_v2.1_doc_00_10661183#8_15469429
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
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 NO 3− (aq) ions stay as Na + (aq) ions and NO 3− (aq) ions. These two ions are examples of spectator ions
An ion that does nothing in the overall course of a chemical reaction. , 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: Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq) → AgCl (s) + Na + (aq) + NO 3 − (aq)
What remains when the spectator ions are removed is called the net ionic equation
A chemical equation with the spectator ions removed. , 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 don’t experience any net chemical change, so they are not written in a net ionic equation. Example 8
Write the net ionic equation for each chemical reaction.
| 4,629 | 5,909 |
msmarco_v2.1_doc_00_10661183#9_15471289
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
It is common to cancel spectator ions (something also done with algebraic quantities) on the opposite sides of a chemical equation: Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq) → AgCl (s) + Na + (aq) + NO 3 − (aq)
What remains when the spectator ions are removed is called the net ionic equation
A chemical equation with the spectator ions removed. , 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 don’t experience any net chemical change, so they are not written in a net ionic equation. Example 8
Write the net ionic equation for each chemical reaction. K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2− (aq) → K + (aq) + C 2 H 3 O 2− (aq) + AgBr (s)
Mg 2+ (aq) + SO 42− (aq) + Ba 2+ (aq) + 2NO 3− (aq) → Mg 2+ (aq) + 2NO 3− (aq) + BaSO 4 (s)
Solution
In the first equation, the K + (aq) and C 2 H 3 O 2− (aq) ions are spectator ions, so they are canceled: K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2 − (aq) → K + (aq) + C 2 H 3 O 2 − (aq) + AgBr (s)
The net ionic equation is
Br−(aq) + Ag+(aq) → AgBr (s)
In the second equation, the Mg 2+ (aq) and NO 3− (aq) ions are spectator ions, so they are canceled: Mg 2 + (aq) + SO 4 2 − (aq) + Ba 2 + (aq) + 2NO 3 − (aq) → Mg 2 + (aq) + 2NO 3 − (aq) + BaSO 4 (s)
The net ionic equation is
SO42−(aq) + Ba2+(aq) → BaSO4(s)
Test Yourself
Write the net ionic equation for
CaCl2(aq) + Pb (NO3)2(aq) → Ca (NO3)2(aq) + PbCl2(s)
Answer
Pb 2+ (aq) + 2Cl − (aq) → PbCl 2 (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.
| 5,193 | 7,055 |
msmarco_v2.1_doc_00_10661183#10_15473833
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2− (aq) → K + (aq) + C 2 H 3 O 2− (aq) + AgBr (s)
Mg 2+ (aq) + SO 42− (aq) + Ba 2+ (aq) + 2NO 3− (aq) → Mg 2+ (aq) + 2NO 3− (aq) + BaSO 4 (s)
Solution
In the first equation, the K + (aq) and C 2 H 3 O 2− (aq) ions are spectator ions, so they are canceled: K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2 − (aq) → K + (aq) + C 2 H 3 O 2 − (aq) + AgBr (s)
The net ionic equation is
Br−(aq) + Ag+(aq) → AgBr (s)
In the second equation, the Mg 2+ (aq) and NO 3− (aq) ions are spectator ions, so they are canceled: Mg 2 + (aq) + SO 4 2 − (aq) + Ba 2 + (aq) + 2NO 3 − (aq) → Mg 2 + (aq) + 2NO 3 − (aq) + BaSO 4 (s)
The net ionic equation is
SO42−(aq) + Ba2+(aq) → BaSO4(s)
Test Yourself
Write the net ionic equation for
CaCl2(aq) + Pb (NO3)2(aq) → Ca (NO3)2(aq) + PbCl2(s)
Answer
Pb 2+ (aq) + 2Cl − (aq) → PbCl 2 (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 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 NaCl soluble but 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 (CaCO 3 ).
| 5,909 | 7,606 |
msmarco_v2.1_doc_00_10661183#11_15476173
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
For most ionic compounds, there is also a limit to the amount of compound 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 NaCl soluble but 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 (CaCO 3 ). However, CaCO 3 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, CaCO 3 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. Most homes in the United States have a tank-type water heater like this one.
| 7,056 | 8,258 |
msmarco_v2.1_doc_00_10661183#12_15477879
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
However, CaCO 3 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, CaCO 3 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. Most homes in the United States have a tank-type water heater like this one. © Thinkstock
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!
| 7,607 | 8,667 |
msmarco_v2.1_doc_00_10661183#13_15479449
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
© Thinkstock
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! The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. © Thinkstock
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.
| 8,258 | 9,104 |
msmarco_v2.1_doc_00_10661183#14_15480811
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. © Thinkstock
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. Exercises
Write a chemical equation that represents NaBr (s) dissociating in water. Write a chemical equation that represents SrCl 2 (s) dissociating in water. Write a chemical equation that represents (NH 4) 3 PO 4 (s) dissociating in water. Write a chemical equation that represents Fe (C 2 H 3 O 2) 3 (s) dissociating in water. Write the complete ionic equation for the reaction of FeCl 2 (aq) and AgNO 3 (aq).
| 8,667 | 9,518 |
msmarco_v2.1_doc_00_10661183#15_15482173
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
Exercises
Write a chemical equation that represents NaBr (s) dissociating in water. Write a chemical equation that represents SrCl 2 (s) dissociating in water. Write a chemical equation that represents (NH 4) 3 PO 4 (s) dissociating in water. Write a chemical equation that represents Fe (C 2 H 3 O 2) 3 (s) dissociating in water. Write the complete ionic equation for the reaction of FeCl 2 (aq) and AgNO 3 (aq). You may have to consult the solubility rules. Write the complete ionic equation for the reaction of BaCl 2 (aq) and Na 2 SO 4 (aq). You may have to consult the solubility rules. Write the complete ionic equation for the reaction of KCl (aq) and NaC 2 H 3 O 2 (aq). You may have to consult the solubility rules.
| 9,104 | 9,829 |
msmarco_v2.1_doc_00_10661183#16_15483402
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-03-ionic-equations-a-closer-look.html
|
Ionic Equations: A Closer Look
|
4.3
Ionic Equations: A Closer Look
4.3 Ionic Equations: A Closer Look
Learning Objectives
Example 6
Example 7
Example 8
Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds
Key Takeaways
Exercises
Answers
|
You may have to consult the solubility rules. Write the complete ionic equation for the reaction of BaCl 2 (aq) and Na 2 SO 4 (aq). You may have to consult the solubility rules. Write the complete ionic equation for the reaction of KCl (aq) and NaC 2 H 3 O 2 (aq). You may have to consult the solubility rules. Write the complete ionic equation for the reaction of Fe 2 (SO 4) 3 (aq) and Sr (NO 3) 2 (aq). You may have to consult the solubility rules. Write the net ionic equation
| 9,519 | 9,999 |
msmarco_v2.1_doc_00_10673298#0_15484386
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Identify an acid and a base. Identify a neutralization reaction and predict its products. In Chapter 3 "Atoms, Molecules, and Ions", Section 3.5 "Acids", 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
A compound that increases the amount of H + ions in an aqueous solution. 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
A compound that increases the amount of OH − ions in an aqueous solution. is a compound that increases the amount of hydroxide ion (OH −) in an aqueous solution.
| 0 | 865 |
msmarco_v2.1_doc_00_10673298#1_15485673
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
an acid
A compound that increases the amount of H + ions in an aqueous solution. 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
A compound that increases the amount of OH − ions in an aqueous solution. 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.
| 440 | 1,299 |
msmarco_v2.1_doc_00_10673298#2_15486950
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
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
H 3 O + ( aq), a water molecule with an extra hydrogen ion attached to it. as H 3 O +, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H + and H 3 O + are treated equivalently. The reaction of an acid and a base is called a neutralization reaction
The reaction of an acid with a base to produce water and a salt.
| 866 | 1,919 |
msmarco_v2.1_doc_00_10673298#3_15488411
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
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
H 3 O + ( aq), a water molecule with an extra hydrogen ion attached to it. as H 3 O +, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H + and H 3 O + are treated equivalently. The reaction of an acid and a base is called a neutralization reaction
The reaction of an acid with a base to produce water and a salt. . 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
acid + base → water + salt
where the term salt
Any ionic compound that is formed from a reaction between an acid and a base. 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.)
| 1,300 | 2,457 |
msmarco_v2.1_doc_00_10673298#4_15489996
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
. 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
acid + base → water + salt
where the term salt
Any ionic compound that is formed from a reaction between an acid and a base. 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
HCl (aq) + KOH (aq) → H2O (ℓ) + KCl (aq)
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 H 2 O are required to balance the chemical equation: 2HCl (aq) + Mg (OH)2(aq) → 2H2O (ℓ) + MgCl2(aq)
Here, the salt is MgCl 2. ( This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid.)
| 1,919 | 3,058 |
msmarco_v2.1_doc_00_10673298#5_15491594
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
For example, the balanced chemical equation for the reaction between HCl (aq) and KOH (aq) is
HCl (aq) + KOH (aq) → H2O (ℓ) + KCl (aq)
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 H 2 O are required to balance the chemical equation: 2HCl (aq) + Mg (OH)2(aq) → 2H2O (ℓ) + MgCl2(aq)
Here, the salt is MgCl 2. ( This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid.) Example 11
Write the neutralization reactions between each acid and base. HNO 3 (aq) and Ba (OH) 2 (aq)
H 3 PO 4 (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. The expected products are water and barium nitrate, so the initial chemical reaction is
HNO3(aq) + Ba (OH)2(aq) → H2O (ℓ) + Ba (NO3)2(aq)
To balance the equation, we need to realize that there will be two H 2 O molecules, so two HNO 3 molecules are required: 2HNO3(aq) + Ba (OH)2(aq) → 2H2O (ℓ) + Ba (NO3)2(aq)
This chemical equation is now balanced.
| 2,458 | 3,695 |
msmarco_v2.1_doc_00_10673298#6_15493299
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Example 11
Write the neutralization reactions between each acid and base. HNO 3 (aq) and Ba (OH) 2 (aq)
H 3 PO 4 (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. The expected products are water and barium nitrate, so the initial chemical reaction is
HNO3(aq) + Ba (OH)2(aq) → H2O (ℓ) + Ba (NO3)2(aq)
To balance the equation, we need to realize that there will be two H 2 O molecules, so two HNO 3 molecules are required: 2HNO3(aq) + Ba (OH)2(aq) → 2H2O (ℓ) + Ba (NO3)2(aq)
This chemical equation is now balanced. The expected products are water and calcium phosphate, so the initial chemical equation is
H3PO4(aq) + Ca (OH)2(aq) → H2O (ℓ) + Ca3(PO4)2(s)
According to the solubility rules, Ca 3 (PO 4) 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: 2H3PO4(aq) + 3Ca (OH)2(aq) → 6H2O (ℓ) + Ca3(PO4)2(s)
This chemical equation is now balanced. Test Yourself
Write the neutralization reaction between H 2 SO 4 (aq) and Sr (OH) 2 (aq).
| 3,058 | 4,249 |
msmarco_v2.1_doc_00_10673298#7_15494947
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
The expected products are water and calcium phosphate, so the initial chemical equation is
H3PO4(aq) + Ca (OH)2(aq) → H2O (ℓ) + Ca3(PO4)2(s)
According to the solubility rules, Ca 3 (PO 4) 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: 2H3PO4(aq) + 3Ca (OH)2(aq) → 6H2O (ℓ) + Ca3(PO4)2(s)
This chemical equation is now balanced. Test Yourself
Write the neutralization reaction between H 2 SO 4 (aq) and Sr (OH) 2 (aq). Answer
H 2 SO 4 (aq) + Sr (OH) 2 (aq) → 2H 2 O (ℓ) + SrSO 4 (aq)
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
3HCl (aq) + Fe (OH)3(s) → 3H2O (ℓ) + FeCl3(aq)
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.
| 3,695 | 5,110 |
msmarco_v2.1_doc_00_10673298#8_15496822
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Answer
H 2 SO 4 (aq) + Sr (OH) 2 (aq) → 2H 2 O (ℓ) + SrSO 4 (aq)
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
3HCl (aq) + Fe (OH)3(s) → 3H2O (ℓ) + FeCl3(aq)
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),
HCl (aq) + NaOH (aq) → H2O (ℓ) + NaCl (aq)
the complete ionic reaction is
H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → H2O (ℓ) + Na+(aq) + Cl−(aq)
The Na + (aq) and Cl − (aq) ions are spectator ions, so we can remove them to have
H+(aq) + OH−(aq) → H2O (ℓ)
as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H 3 O + (aq), we would write it as
H3O+(aq) + OH−(aq) → 2H2O (ℓ)
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: 2H+(aq) + 2Cl−(aq) + Cr (OH)2(s) → 2H2O (ℓ) + Cr2+(aq) + 2Cl−(aq)
The chloride ions are the only spectator ions here, so the net ionic equation is
2H+(aq) + Cr (OH)2(s) → 2H2O (ℓ) + Cr2+(aq)
Example 12
Oxalic acid, H 2 C 2 O 4 (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? (
| 4,249 | 6,217 |
msmarco_v2.1_doc_00_10673298#9_15499338
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
For example, in the reaction of HCl (aq) and NaOH (aq),
HCl (aq) + NaOH (aq) → H2O (ℓ) + NaCl (aq)
the complete ionic reaction is
H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → H2O (ℓ) + Na+(aq) + Cl−(aq)
The Na + (aq) and Cl − (aq) ions are spectator ions, so we can remove them to have
H+(aq) + OH−(aq) → H2O (ℓ)
as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H 3 O + (aq), we would write it as
H3O+(aq) + OH−(aq) → 2H2O (ℓ)
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: 2H+(aq) + 2Cl−(aq) + Cr (OH)2(s) → 2H2O (ℓ) + Cr2+(aq) + 2Cl−(aq)
The chloride ions are the only spectator ions here, so the net ionic equation is
2H+(aq) + Cr (OH)2(s) → 2H2O (ℓ) + Cr2+(aq)
Example 12
Oxalic acid, H 2 C 2 O 4 (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, C 2 O 42− .) 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. Test Yourself
What is the net ionic equation between HNO 3 (aq) and Ti (OH) 4 (s)? Answer
4H + (aq) + Ti (OH) 4 (s) → 4H 2 O (ℓ) + Ti 4+ (aq)
Key Takeaways
The Arrhenius definition of an acid is a substance that increases the amount of H + in an aqueous solution.
| 5,111 | 6,828 |
msmarco_v2.1_doc_00_10673298#10_15501606
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
The anion in oxalic acid is the oxalate ion, C 2 O 42− .) 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. Test Yourself
What is the net ionic equation between HNO 3 (aq) and Ti (OH) 4 (s)? Answer
4H + (aq) + Ti (OH) 4 (s) → 4H 2 O (ℓ) + Ti 4+ (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. Exercises
What is the Arrhenius definition of an acid? What is the Arrhenius definition of a base?
| 6,217 | 7,230 |
msmarco_v2.1_doc_00_10673298#11_15503064
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
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. Exercises
What is the Arrhenius definition of an acid? What is the Arrhenius definition of a base? Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs. HCl and KOH
H 2 SO 4 and KOH
H 3 PO 4 and Ni (OH) 2
Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs. HBr and Fe (OH) 3
HNO 2 and Al (OH) 3
HClO 3 and Mg (OH) 2
Write a balanced chemical equation for each neutralization reaction in Exercise 3.
| 6,828 | 7,634 |
msmarco_v2.1_doc_00_10673298#12_15504289
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs. HCl and KOH
H 2 SO 4 and KOH
H 3 PO 4 and Ni (OH) 2
Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs. HBr and Fe (OH) 3
HNO 2 and Al (OH) 3
HClO 3 and Mg (OH) 2
Write a balanced chemical equation for each neutralization reaction in Exercise 3. Write a balanced chemical equation for each neutralization reaction in Exercise 4. Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels. HI (aq) + KOH (aq) → ? H 2 SO 4 (aq) + Ba (OH) 2 (aq) → ?
| 7,230 | 7,909 |
msmarco_v2.1_doc_00_10673298#13_15505391
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Write a balanced chemical equation for each neutralization reaction in Exercise 4. Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels. HI (aq) + KOH (aq) → ? H 2 SO 4 (aq) + Ba (OH) 2 (aq) → ? Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels. HNO 3 (aq) + Fe (OH) 3 (s) → ? H 3 PO 4 (aq) + CsOH (aq) → ? Write the net ionic equation for each neutralization reaction in Exercise 7.
| 7,634 | 8,181 |
msmarco_v2.1_doc_00_10673298#14_15506365
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels. HNO 3 (aq) + Fe (OH) 3 (s) → ? H 3 PO 4 (aq) + CsOH (aq) → ? Write the net ionic equation for each neutralization reaction in Exercise 7. Write the net ionic equation for each neutralization reaction in Exercise 8. Write the complete and net ionic equations for the neutralization reaction between HClO 3 (aq) and Zn (OH) 2 (s). Assume the salt is soluble. Write the complete and net ionic equations for the neutralization reaction between H 2 C 2 O 4 (s) and Sr (OH) 2 (aq). Assume the salt is insoluble.
| 7,909 | 8,549 |
msmarco_v2.1_doc_00_10673298#15_15507422
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Write the net ionic equation for each neutralization reaction in Exercise 8. Write the complete and net ionic equations for the neutralization reaction between HClO 3 (aq) and Zn (OH) 2 (s). Assume the salt is soluble. Write the complete and net ionic equations for the neutralization reaction between H 2 C 2 O 4 (s) and Sr (OH) 2 (aq). Assume the salt is insoluble. 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 HNO 3 (aq) and RbOH. 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. 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? Write the complete and net ionic equations for the neutralization reaction between HClO 3 (aq) and Zn (OH) 2 (s) using the hydronium ion in place of H +.
| 8,181 | 9,301 |
msmarco_v2.1_doc_00_10673298#16_15508949
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
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 HNO 3 (aq) and RbOH. 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. 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? Write the complete and net ionic equations for the neutralization reaction between HClO 3 (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
An Arrhenius acid increases the amount of H + ions in an aqueous solution. KCl and H 2 O
K 2 SO 4 and H 2 O
Ni 3 (PO 4) 2 and H 2 O
HCl + KOH → KCl + H 2 O
H 2 SO 4 + 2KOH → K 2 SO 4 + 2H 2 O
2H 3 PO 4 + 3Ni (OH) 2 → Ni 3 (PO 4) 2 + 6H 2 O
HI (aq) + KOH (aq) → KCl (aq) + H 2 O (ℓ)
H 2 SO 4 (aq) + Ba (OH) 2 (aq) → BaSO 4 (s) + 2H 2 O (ℓ)
H + (aq) + OH − (aq) → H 2 O (ℓ)
2H + (aq) + SO 42− (aq) + Ba 2+ (aq) + 2OH − (aq) → BaSO 4 (s) + 2H 2 O (ℓ)
Complete ionic equation: 2H+(aq) + 2ClO3−(aq) + Zn2+(aq) + 2OH−(aq) → Zn2+(aq) + 2ClO3−(aq) + 2H2O (ℓ)
Net ionic equation:
| 8,549 | 9,967 |
msmarco_v2.1_doc_00_10673298#17_15510881
|
http://2012books.lardbucket.org/books/beginning-chemistry/s08-05-neutralization-reactions.html
|
Neutralization Reactions
|
4.5
Neutralization Reactions
4.5 Neutralization Reactions
Learning Objectives
Example 11
Example 12
Key Takeaways
Exercises
Answers
|
Assume the salt is soluble. What difference does it make when using the hydronium ion? Answers
An Arrhenius acid increases the amount of H + ions in an aqueous solution. KCl and H 2 O
K 2 SO 4 and H 2 O
Ni 3 (PO 4) 2 and H 2 O
HCl + KOH → KCl + H 2 O
H 2 SO 4 + 2KOH → K 2 SO 4 + 2H 2 O
2H 3 PO 4 + 3Ni (OH) 2 → Ni 3 (PO 4) 2 + 6H 2 O
HI (aq) + KOH (aq) → KCl (aq) + H 2 O (ℓ)
H 2 SO 4 (aq) + Ba (OH) 2 (aq) → BaSO 4 (s) + 2H 2 O (ℓ)
H + (aq) + OH − (aq) → H 2 O (ℓ)
2H + (aq) + SO 42− (aq) + Ba 2+ (aq) + 2OH − (aq) → BaSO 4 (s) + 2H 2 O (ℓ)
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)
Be
| 9,302 | 10,000 |
msmarco_v2.1_doc_00_10684544#0_15512111
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
Formation Reactions
7.6 Formation Reactions
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 now we don’t have 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
A chemical reaction that forms one mole of a substance from its constituent elements in their standard states. are chemical reactions that form one mole of a substance from its constituent elements in their standard states. By standard states we mean as a diatomic molecule if that is how the element exists and the proper phase at normal temperatures (typically room temperature).
| 0 | 1,183 |
msmarco_v2.1_doc_00_10684544#1_15513760
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
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
A chemical reaction that forms one mole of a substance from its constituent elements in their standard states. are chemical reactions that form one mole of a substance from its constituent elements in their standard states. By standard states we mean as a diatomic molecule if that is how the element exists and 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 (CH 4) is
C (s) + 2H2(g) → CH4(g)
The formation reaction for carbon dioxide (CO 2) is
C (s) + O2(g) → CO2(g)
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 H 2 O—
2H2(g) + O2(g) → 2H2O (ℓ)
—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:
| 631 | 1,898 |
msmarco_v2.1_doc_00_10684544#2_15515517
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
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 (CH 4) is
C (s) + 2H2(g) → CH4(g)
The formation reaction for carbon dioxide (CO 2) is
C (s) + O2(g) → CO2(g)
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 H 2 O—
2H2(g) + O2(g) → 2H2O (ℓ)
—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: H2(g) + 1/2O2(g) → H2O (ℓ)
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 11
Which of the following are proper formation reactions? H 2 (g) + Cl 2 (g) → 2HCl (g)
Si (s) + 2F 2 (g) → SiF 4 (g)
CaO (s) + CO 2 → CaCO 3 (s)
Solution
In this reaction, two moles of product are produced, so this is not a proper formation reaction. In this reaction, one mole of a substance is produced from its elements in their standard states, so this is a proper formation reaction.
| 1,184 | 2,579 |
msmarco_v2.1_doc_00_10684544#3_15517434
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
H2(g) + 1/2O2(g) → H2O (ℓ)
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 11
Which of the following are proper formation reactions? H 2 (g) + Cl 2 (g) → 2HCl (g)
Si (s) + 2F 2 (g) → SiF 4 (g)
CaO (s) + CO 2 → CaCO 3 (s)
Solution
In this reaction, two moles of product are produced, so this is not a proper formation reaction. In this reaction, one mole of a substance is produced from its elements in their standard states, so this is a proper formation reaction. 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. Test Yourself
Is this a proper formation reaction? Explain why or why not. 2Fe (s) + 3P (s) + 12O (g) → Fe2(PO4)3(s)
Answer
This is not a proper formation reaction because oxygen is not written as a diatomic molecule.
| 1,898 | 2,941 |
msmarco_v2.1_doc_00_10684544#4_15518970
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
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. Test Yourself
Is this a proper formation reaction? Explain why or why not. 2Fe (s) + 3P (s) + 12O (g) → Fe2(PO4)3(s)
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 12
Write formation reactions for each of the following. FeO (s)
C 2 H 6 (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. Fe (s) + 1/2O 2 (g) → FeO (s)
2C (s) + 3H 2 (g) → C 2 H 6 (g)
Test Yourself
Write the equation for the formation of CaCO 3 (s). Answer
Ca (s) + C (s) + 3/2O2(g) → CaCO3(s)
The enthalpy change for a formation reaction is called the enthalpy of formation
The enthalpy change for a formation reaction;
| 2,579 | 3,598 |
msmarco_v2.1_doc_00_10684544#5_15520476
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
Given the formula of any substance, you should be able to write the proper formation reaction for that substance. Example 12
Write formation reactions for each of the following. FeO (s)
C 2 H 6 (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. Fe (s) + 1/2O 2 (g) → FeO (s)
2C (s) + 3H 2 (g) → C 2 H 6 (g)
Test Yourself
Write the equation for the formation of CaCO 3 (s). Answer
Ca (s) + C (s) + 3/2O2(g) → CaCO3(s)
The enthalpy change for a formation reaction is called the enthalpy of formation
The enthalpy change for a formation reaction; denoted Δ H f.
and is given the symbol Δ Hf. The subscript f is the clue that the reaction of interest is a formation reaction. Thus, for the formation of FeO (s),
Fe (s) + 1 2 O 2 (g) → FeO (s) Δ H ≡ Δ H f = − 272 kJ/mol
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.
| 2,941 | 4,267 |
msmarco_v2.1_doc_00_10684544#6_15522331
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
denoted Δ H f.
and is given the symbol Δ Hf. The subscript f is the clue that the reaction of interest is a formation reaction. Thus, for the formation of FeO (s),
Fe (s) + 1 2 O 2 (g) → FeO (s) Δ H ≡ Δ H f = − 272 kJ/mol
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 7.2 "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 H 2 O). 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 NO 2 and the formation reaction of N 2 O 4: We must multiply the first reaction by 2 to get the correct overall balanced equation.
| 3,599 | 4,916 |
msmarco_v2.1_doc_00_10684544#7_15524167
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
Table 7.2 "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 H 2 O). 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 NO 2 and the formation reaction of N 2 O 4: 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 Δ Hf values of the formation reactions. Table 7.2 Enthalpies of Formation for Various Substances
Compound
Δ Hf (kJ/mol)
Compound
Δ Hf (kJ/mol)
Compound
Δ Hf (kJ/mol)
Compound
Δ Hf (kJ/mol)
Ag (s)
0
Ca (s)
0
Hg 2 Cl 2 (s)
−265.37
NaHCO 3 (s)
−950.81
AgBr (s)
−100.37
CaCl 2 (s)
−795.80
I 2 (s)
0
NaN 3 (s)
21.71
AgCl (s)
−127.01
CaCO 3 (s, arag)
−1,207.1
K (s)
0
Na 2 CO 3 (s)
−1,130.77
Al (s)
0
CaCO 3 (s, calc)
−1,206.9
KBr (s)
−393.8
Na 2 O (s)
−417.98
Al 2 O 3 (s)
−1,675.7
Cl 2 (g)
0
KCl (s)
−436.5
Na 2 SO 4 (s)
−331.64
Ar (g)
0
Cr (s)
0
KF (s)
−567.3
Ne (g)
0
Au (s)
0
Cr 2 O 3 (s)
−1,134.70
KI (s)
−327.9
Ni (s)
0
BaSO 4 (s)
−1,473.19
Cs (s)
0
Li (s)
0
O 2 (g)
0
Br 2 (ℓ)
0
Cu (s)
0
LiBr (s)
−351.2
O 3 (g)
142.67
C (s, dia)
1.897
F 2 (g)
0
LiCl (s)
−408.27
PH 3 (g)
22.89
C (s, gra)
0
Fe (s)
0
LiF (s)
−616.0
Pb (s)
0
CCl 4 (ℓ)
−128.4
Fe 2 (SO 4) 3 (s)
−2,583.00
LiI (s)
−270.4
PbCl 2 (s)
−359.41
CH 2 O (g)
−115.90
Fe 2 O 3 (s)
−825.5
Mg (s)
0
PbO 2 (s)
−274.47
CH 3 COOH (ℓ)
−483.52
Ga (s)
0
MgO (s)
−601.60
PbSO 4 (s)
−919.97
CH 3 OH (ℓ)
−238.4
HBr (g)
−36.29
NH 3 (g)
−45.94
Pt (s)
0
CH 4 (g)
−74.87
HCl (g)
−92.31
NO (g)
90.29
S (s)
0
CO (g)
−110.5
HF (g)
−273.30
NO 2 (g)
33.10
SO 2 (g)
−296.81
CO 2 (g)
−393.51
HI (g)
26.5
N 2 (g)
0
SO 3 (g)
−395.77
C 2 H 5 OH (ℓ)
−277.0
HNO 2 (g)
−76.73
N 2 O (g)
82.05
SO 3 (ℓ)
−438
C 2 H 6 (g)
−83.8
HNO 3 (g)
−134.31
N 2 O 4 (g)
9.08
Si (s)
0
C 6 H 12 (ℓ)
−157.7
H 2 (g)
0
N 2 O 5 (g)
11.30
U (s)
0
C 6 H 12 O 6 (s)
−1277
H 2 O (g)
−241.8
Na (s)
0
UF 6 (s)
−2,197.0
C 6 H 14 (ℓ)
−198.7
H 2 O (ℓ)
−285.83
NaBr (s)
−361.1
UO 2 (s)
−1,085.0
C 6 H 5 CH 3 (ℓ)
12.0
H 2 O (s)
−292.72
NaCl (s)
−385.9
Xe (g)
0
C 6 H 6 (ℓ)
48.95
He (g)
0
NaF (s)
−576.6
Zn (s)
0
C 10 H 8 (s)
77.0
Hg (ℓ)
0
NaI (s)
−287.8
ZnCl 2 (s)
−415.05
C 12 H 22 O 11 (s)
−2,221.2
Sources: National Institute of Standards and Technology’s Chemistry WebBook, http://webbook.nist.gov/chemistry; D. R. Lide, ed., CRC Handbook of Chemistry and Physics, 89th ed. (
| 4,268 | 6,945 |
msmarco_v2.1_doc_00_10684544#8_15527928
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
We are simply using Hess’s law in combining the Δ Hf values of the formation reactions. Table 7.2 Enthalpies of Formation for Various Substances
Compound
Δ Hf (kJ/mol)
Compound
Δ Hf (kJ/mol)
Compound
Δ Hf (kJ/mol)
Compound
Δ Hf (kJ/mol)
Ag (s)
0
Ca (s)
0
Hg 2 Cl 2 (s)
−265.37
NaHCO 3 (s)
−950.81
AgBr (s)
−100.37
CaCl 2 (s)
−795.80
I 2 (s)
0
NaN 3 (s)
21.71
AgCl (s)
−127.01
CaCO 3 (s, arag)
−1,207.1
K (s)
0
Na 2 CO 3 (s)
−1,130.77
Al (s)
0
CaCO 3 (s, calc)
−1,206.9
KBr (s)
−393.8
Na 2 O (s)
−417.98
Al 2 O 3 (s)
−1,675.7
Cl 2 (g)
0
KCl (s)
−436.5
Na 2 SO 4 (s)
−331.64
Ar (g)
0
Cr (s)
0
KF (s)
−567.3
Ne (g)
0
Au (s)
0
Cr 2 O 3 (s)
−1,134.70
KI (s)
−327.9
Ni (s)
0
BaSO 4 (s)
−1,473.19
Cs (s)
0
Li (s)
0
O 2 (g)
0
Br 2 (ℓ)
0
Cu (s)
0
LiBr (s)
−351.2
O 3 (g)
142.67
C (s, dia)
1.897
F 2 (g)
0
LiCl (s)
−408.27
PH 3 (g)
22.89
C (s, gra)
0
Fe (s)
0
LiF (s)
−616.0
Pb (s)
0
CCl 4 (ℓ)
−128.4
Fe 2 (SO 4) 3 (s)
−2,583.00
LiI (s)
−270.4
PbCl 2 (s)
−359.41
CH 2 O (g)
−115.90
Fe 2 O 3 (s)
−825.5
Mg (s)
0
PbO 2 (s)
−274.47
CH 3 COOH (ℓ)
−483.52
Ga (s)
0
MgO (s)
−601.60
PbSO 4 (s)
−919.97
CH 3 OH (ℓ)
−238.4
HBr (g)
−36.29
NH 3 (g)
−45.94
Pt (s)
0
CH 4 (g)
−74.87
HCl (g)
−92.31
NO (g)
90.29
S (s)
0
CO (g)
−110.5
HF (g)
−273.30
NO 2 (g)
33.10
SO 2 (g)
−296.81
CO 2 (g)
−393.51
HI (g)
26.5
N 2 (g)
0
SO 3 (g)
−395.77
C 2 H 5 OH (ℓ)
−277.0
HNO 2 (g)
−76.73
N 2 O (g)
82.05
SO 3 (ℓ)
−438
C 2 H 6 (g)
−83.8
HNO 3 (g)
−134.31
N 2 O 4 (g)
9.08
Si (s)
0
C 6 H 12 (ℓ)
−157.7
H 2 (g)
0
N 2 O 5 (g)
11.30
U (s)
0
C 6 H 12 O 6 (s)
−1277
H 2 O (g)
−241.8
Na (s)
0
UF 6 (s)
−2,197.0
C 6 H 14 (ℓ)
−198.7
H 2 O (ℓ)
−285.83
NaBr (s)
−361.1
UO 2 (s)
−1,085.0
C 6 H 5 CH 3 (ℓ)
12.0
H 2 O (s)
−292.72
NaCl (s)
−385.9
Xe (g)
0
C 6 H 6 (ℓ)
48.95
He (g)
0
NaF (s)
−576.6
Zn (s)
0
C 10 H 8 (s)
77.0
Hg (ℓ)
0
NaI (s)
−287.8
ZnCl 2 (s)
−415.05
C 12 H 22 O 11 (s)
−2,221.2
Sources: National Institute of Standards and Technology’s Chemistry WebBook, http://webbook.nist.gov/chemistry; 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:
| 4,917 | 7,046 |
msmarco_v2.1_doc_00_10684544#9_15531134
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
Boca Raton, FL: CRC Press, 2008); J. A. Dean, ed., Lange’s Handbook of Chemistry, 14th ed. ( New York: McGraw-Hill, 1992). Example 13
Show that the reaction
Fe2O3(s) + 3SO3(g) → Fe2(SO4)3(s)
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 SO 3 will be multiplied by 3 because there are three moles of SO 3 in the balanced chemical equation.
| 6,945 | 7,509 |
msmarco_v2.1_doc_00_10684544#10_15532168
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
McGraw-Hill, 1992). Example 13
Show that the reaction
Fe2O3(s) + 3SO3(g) → Fe2(SO4)3(s)
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 SO 3 will be multiplied by 3 because there are three moles of SO 3 in the balanced chemical equation. The formation reactions are as follows: Fe 2 O 3 (s) → 2 Fe (s) + 3 2 O 2 (g) 3 × [ SO 3 (g) → S (s) + 3 2 O 2 (g)]
2Fe (s) + 3S (s) + 6O2(g) → Fe2(SO4)3(s)
When these three equations are combined and simplified, the overall reaction is
Fe2O3(s) + 3SO3(s) → Fe2(SO4)3(s)
Test Yourself
Write the formation reactions that will yield
2SO2(g) + O2(g) → 2SO3(g). Answer
2 × [ SO 2 (g) → S (s) + O 2 (g)] 2 × [ S (s) + 3 2 O 2 (g) → 2SO 3 (g)]
Now that we have established formation reactions as the major type of thermochemical reaction we will be interested in, do we always need to write all the formation reactions when we want to determine the enthalpy change of any random chemical reaction? No. There is an easier way.
| 7,047 | 8,229 |
msmarco_v2.1_doc_00_10684544#11_15533873
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
The formation reactions are as follows: Fe 2 O 3 (s) → 2 Fe (s) + 3 2 O 2 (g) 3 × [ SO 3 (g) → S (s) + 3 2 O 2 (g)]
2Fe (s) + 3S (s) + 6O2(g) → Fe2(SO4)3(s)
When these three equations are combined and simplified, the overall reaction is
Fe2O3(s) + 3SO3(s) → Fe2(SO4)3(s)
Test Yourself
Write the formation reactions that will yield
2SO2(g) + O2(g) → 2SO3(g). Answer
2 × [ SO 2 (g) → S (s) + O 2 (g)] 2 × [ S (s) + 3 2 O 2 (g) → 2SO 3 (g)]
Now that we have established formation reactions as the major type of thermochemical reaction we will be interested in, do we always need to write all the formation reactions when we want to determine the enthalpy change of any random chemical reaction? No. There is an easier way. You may have noticed in all our examples that we change the signs on all the enthalpies of formation of the reactants, and we don’t change the signs 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,
Δ H rxn = ∑ n p Δ H f,p − ∑ n r Δ H f,r
where np and nr are the number of moles of products and reactants, respectively (even if they are just 1 mol), and Δ Hf,p and Δ Hf,r are the enthalpies of formation of the product and reactant species, respectively.
| 7,510 | 9,046 |
msmarco_v2.1_doc_00_10684544#12_15535975
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
You may have noticed in all our examples that we change the signs on all the enthalpies of formation of the reactants, and we don’t change the signs 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,
Δ H rxn = ∑ n p Δ H f,p − ∑ n r Δ H f,r
where np and nr are the number of moles of products and reactants, respectively (even if they are just 1 mol), and Δ Hf,p and Δ Hf,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 14
Use the products-minus-reactants approach to determine the enthalpy of reaction for
2HBr (g) + Cl 2 (g) → 2HCl (g) + Br 2 ( ℓ) Δ H f − 36. 3 0 − 92. 3 0 kJ/mol
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:
| 8,230 | 9,797 |
msmarco_v2.1_doc_00_10684544#13_15538079
|
http://2012books.lardbucket.org/books/beginning-chemistry/s11-06-formation-reactions.html
|
Formation Reactions
|
7.6
Formation Reactions
7.6 Formation Reactions
Learning Objectives
Example 11
Example 12
Example 13
Example 14
Food and Drink App: Calories and Nutrition
Key Takeaways
Exercises
Answers
|
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 14
Use the products-minus-reactants approach to determine the enthalpy of reaction for
2HBr (g) + Cl 2 (g) → 2HCl (g) + Br 2 ( ℓ) Δ H f − 36. 3 0 − 92. 3 0 kJ/mol
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: Δ H rxn = [ ( 2 mol) ( − 92. 3 kJ/ mol) + ( 1 mol) ( 0 kJ/ mol)] − [ ( 2 mol) ( − 36. 3 kJ/ mol) + ( 1 mol) ( 0 kJ/ mol)]
All the mol units cancel. Multiplying and combining all the values, we get
ΔHrxn
| 9,047 | 10,000 |
msmarco_v2.1_doc_00_10701457#0_15539545
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
Violations of the Octet Rule
9.5 Violations of the Octet Rule
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
A molecule with an odd number of electrons in the valence shell of an atom. represent 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 NO, NO 2, and ClO 2.
| 0 | 843 |
msmarco_v2.1_doc_00_10701457#1_15540801
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
Odd-electron molecules
A molecule with an odd number of electrons in the valence shell of an atom. represent 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 NO, NO 2, and ClO 2. The Lewis electron dot diagram for NO is as follows: Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds. Electron-deficient molecules
A molecule with less than eight electrons in the valence shell of an atom. represent the second violation to the octet rule.
| 423 | 1,263 |
msmarco_v2.1_doc_00_10701457#2_15542050
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
The Lewis electron dot diagram for NO is as follows: Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds. Electron-deficient molecules
A molecule with less than eight electrons in the valence shell of an atom. represent 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 B atom. A well-known example is BF 3:
| 844 | 1,669 |
msmarco_v2.1_doc_00_10701457#3_15543283
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
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 B atom. A well-known example is BF 3: 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
A molecule with more than eight electrons in the valence shell of an atom. . 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 PF 5.
| 1,264 | 2,143 |
msmarco_v2.1_doc_00_10701457#4_15544571
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
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
A molecule with more than eight electrons in the valence shell of an atom. . 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 PF 5. The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds: Formally, the P atom has 10 electrons in its valence shell. Example 9
Identify each violation to the octet rule by drawing a Lewis electron dot diagram. ClO
SF 6
Solution
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:
| 1,669 | 2,600 |
msmarco_v2.1_doc_00_10701457#5_15545914
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds: Formally, the P atom has 10 electrons in its valence shell. Example 9
Identify each violation to the octet rule by drawing a Lewis electron dot diagram. ClO
SF 6
Solution
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: In SF 6, the central 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: Test Yourself
Identify the violation to the octet rule in XeF 2 by drawing a Lewis electron dot diagram. Answer
The Xe atom has an expanded valence shell with more than eight electrons around it. Key Takeaway
There are three violations to the octet rule:
| 2,144 | 3,031 |
msmarco_v2.1_doc_00_10701457#6_15547216
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
In SF 6, the central 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: Test Yourself
Identify the violation to the octet rule in XeF 2 by drawing a Lewis electron dot diagram. Answer
The Xe atom has an expanded valence shell with more than eight electrons around it. Key Takeaway
There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules. Exercises
Why can an odd-electron molecule not satisfy the octet rule? Why can an atom in the second row of the periodic table not form expanded valence shell molecules? Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule. NO 2
XeF 4
Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule.
| 2,600 | 3,494 |
msmarco_v2.1_doc_00_10701457#7_15548523
|
http://2012books.lardbucket.org/books/beginning-chemistry/s13-05-violations-of-the-octet-rule.html
|
Violations of the Octet Rule
|
9.5
Violations of the Octet Rule
9.5 Violations of the Octet Rule
Learning Objective
Example 9
Key Takeaway
Exercises
Answers
|
odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules. Exercises
Why can an odd-electron molecule not satisfy the octet rule? Why can an atom in the second row of the periodic table not form expanded valence shell molecules? Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule. NO 2
XeF 4
Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule. BCl 3
ClO 2
Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule. POF 3
ClF 3
Draw an acceptable Lewis electron dot diagram for these molecules that violate the octet rule. SF 4
BeH 2
Answers
There is no way all electrons can be paired if there are an odd number of them.
| 3,032 | 3,807 |
msmarco_v2.1_doc_00_10705671#0_15549716
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
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
The process of a solid becoming a liquid. (an older term that you may see sometimes is fusion ).
| 0 | 598 |
msmarco_v2.1_doc_00_10705671#1_15550810
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
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
The process of a solid becoming a liquid. (an older term that you may see sometimes is fusion ). The opposite process, a liquid becoming a solid, is called solidification
The process of a liquid becoming a solid. . For any pure substance, the temperature at which melting occurs—known as the melting point
The characteristic temperature at which a solid becomes a liquid. —is a characteristic of that substance. It requires energy for a solid to melt into a liquid.
| 294 | 967 |
msmarco_v2.1_doc_00_10705671#2_15551985
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
The opposite process, a liquid becoming a solid, is called solidification
The process of a liquid becoming a solid. . For any pure substance, the temperature at which melting occurs—known as the melting point
The characteristic temperature at which a solid becomes a liquid. —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)
The amount of energy needed to change from a solid to a liquid or from a liquid to a solid. of the substance, represented as Δ Hfus. Some Δ Hfus values are listed in Table 10.2 "Enthalpies of Fusion for Various Substances"; it is assumed that these values are for the melting point of the substance.
| 599 | 1,429 |
msmarco_v2.1_doc_00_10705671#3_15553330
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
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)
The amount of energy needed to change from a solid to a liquid or from a liquid to a solid. of the substance, represented as Δ Hfus. Some Δ Hfus values are listed in Table 10.2 "Enthalpies of Fusion for Various Substances"; it is assumed that these values are for the melting point of the substance. Note that the unit of Δ Hfus is kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ Hfus is always tabulated as a positive number. However, it can be used for both the melting and the solidification processes as long as you keep in mind that melting is always endothermic (so Δ H will be positive), while solidification is always exothermic (so Δ H will be negative). Table 10.2 Enthalpies of Fusion for Various Substances
Substance (Melting Point)
Δ Hfus (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 2
What is the energy change when 45.7 g of H 2 O melt at 0°C? Solution
The Δ Hfus of H 2 O is 6.01 kJ/mol.
| 968 | 2,172 |
msmarco_v2.1_doc_00_10705671#4_15555122
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
Note that the unit of Δ Hfus is kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ Hfus is always tabulated as a positive number. However, it can be used for both the melting and the solidification processes as long as you keep in mind that melting is always endothermic (so Δ H will be positive), while solidification is always exothermic (so Δ H will be negative). Table 10.2 Enthalpies of Fusion for Various Substances
Substance (Melting Point)
Δ Hfus (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 2
What is the energy change when 45.7 g of H 2 O melt at 0°C? Solution
The Δ Hfus of H 2 O 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 H 2 O, which is 18.0 g/mol. Then we can use Δ Hfus as a conversion factor. Because the substance is melting, the process is endothermic, so the energy change will have a positive sign. 45.7 g H 2 O × 1 mol H 2 O 18.0 g × 6.01 kJ mol = 15.3 kJ
Without a sign, the number is assumed to be positive. Test Yourself
What is the energy change when 108 g of C 6 H 6 freeze at 5.5°C?
| 1,430 | 2,680 |
msmarco_v2.1_doc_00_10705671#5_15556970
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
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 H 2 O, which is 18.0 g/mol. Then we can use Δ Hfus as a conversion factor. Because the substance is melting, the process is endothermic, so the energy change will have a positive sign. 45.7 g H 2 O × 1 mol H 2 O 18.0 g × 6.01 kJ mol = 15.3 kJ
Without a sign, the number is assumed to be positive. Test Yourself
What is the energy change when 108 g of C 6 H 6 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
A process that does not change the temperature. process because a substance stays at the same temperature. Only when all of a substance is melted does any additional energy go to changing its temperature.
| 2,173 | 3,071 |
msmarco_v2.1_doc_00_10705671#6_15558389
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
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
A process that does not change the temperature. process because a substance stays at the same temperature. Only when all of a substance is melted does any 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.
| 2,680 | 3,526 |
msmarco_v2.1_doc_00_10705671#7_15559733
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
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 10.3 "Liquids and Gravity" ). Figure 10.3 Liquids and Gravity
(a) A liquid fills the bottom of its container as it is drawn downward by gravity and the particles slide over each other. ( b) A liquid floats in a zero-gravity environment. The particles still slide over each other because they are in the liquid phase, but now there is no gravity to pull them down.
| 3,071 | 4,115 |
msmarco_v2.1_doc_00_10705671#8_15561276
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
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 10.3 "Liquids and Gravity" ). Figure 10.3 Liquids and Gravity
(a) A liquid fills the bottom of its container as it is drawn downward by gravity and the particles slide over each other. ( b) A liquid floats in a zero-gravity environment. The particles still slide over each other because they are in the liquid phase, but now there is no gravity to pull them down. Source: Photo on the left © Thinkstock. Photo on the right courtesy of NASA, http://www.nasa.gov/mission_pages/station/multimedia/Exp10_image_009.html. 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.
| 3,527 | 4,471 |
msmarco_v2.1_doc_00_10705671#9_15562725
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
Source: Photo on the left © Thinkstock. Photo on the right courtesy of NASA, http://www.nasa.gov/mission_pages/station/multimedia/Exp10_image_009.html. 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)
The process of a liquid becoming a gas. , while the process of a gas becoming a liquid is called condensation
The process of a gas becoming a liquid. . 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
The characteristic temperature at which a liquid becomes a gas.
| 4,115 | 5,046 |
msmarco_v2.1_doc_00_10705671#10_15564154
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
The process of a liquid becoming a gas is called boiling (or vaporization)
The process of a liquid becoming a gas. , while the process of a gas becoming a liquid is called condensation
The process of a gas becoming a liquid. . 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
The characteristic temperature at which a liquid becomes a gas. , can change with surrounding pressure. Therefore, we define the normal boiling point
The characteristic temperature at which a liquid becomes a gas when the surrounding pressure is exactly 1 atm. 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.
| 4,472 | 5,528 |
msmarco_v2.1_doc_00_10705671#11_15565706
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
, can change with surrounding pressure. Therefore, we define the normal boiling point
The characteristic temperature at which a liquid becomes a gas when the surrounding pressure is exactly 1 atm. 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
The amount of energy needed to change from a liquid to a gas or from a gas to a liquid. (or heat of vaporization), represented as Δ Hvap. Some Δ Hvap values are listed in Table 10.3 "Enthalpies of Vaporization for Various Substances"; 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 Δ Hvap is also kilojoules per mole, so we need to know the quantity of material to know how much energy is involved.
| 5,046 | 6,120 |
msmarco_v2.1_doc_00_10705671#12_15567290
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
The amount of energy required to convert a liquid to a gas is called the enthalpy of vaporization
The amount of energy needed to change from a liquid to a gas or from a gas to a liquid. (or heat of vaporization), represented as Δ Hvap. Some Δ Hvap values are listed in Table 10.3 "Enthalpies of Vaporization for Various Substances"; 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 Δ Hvap is also kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ Hvap 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 Δ H will be positive), while condensation is always exothermic (so Δ H will be negative). Table 10.3 Enthalpies of Vaporization for Various Substances
Substance (Normal Boiling Point)
Δ Hvap (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 3
What is the energy change when 66.7 g of Br 2 (g) condense to a liquid at 59.5°C? Solution
The Δ Hvap of Br 2 is 15.4 kJ/mol. Even though this is a condensation process, we can still use the numerical value of Δ Hvap as long as we realize that we must take energy out, so the Δ H value will be negative.
| 5,529 | 6,934 |
msmarco_v2.1_doc_00_10705671#13_15569285
|
http://2012books.lardbucket.org/books/beginning-chemistry/s14-02-phase-transitions-melting-boil.html
|
Phase Transitions: Melting, Boiling, and Subliming
|
10.2
Phase Transitions: Melting, Boiling, and Subliming
10.2 Phase Transitions: Melting, Boiling, and Subliming
Learning Objectives
Example 2
Example 3
Key Takeaways
Exercises
Answers
|
The Δ Hvap 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 Δ H will be positive), while condensation is always exothermic (so Δ H will be negative). Table 10.3 Enthalpies of Vaporization for Various Substances
Substance (Normal Boiling Point)
Δ Hvap (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 3
What is the energy change when 66.7 g of Br 2 (g) condense to a liquid at 59.5°C? Solution
The Δ Hvap of Br 2 is 15.4 kJ/mol. Even though this is a condensation process, we can still use the numerical value of Δ Hvap as long as we realize that we must take energy out, so the Δ H value will be negative. To determine the magnitude of the energy change, we must first convert the amount of Br 2 to moles. Then we can use Δ Hvap as a conversion factor. 66.7 g Br 2 × 1 mol Br 2 159.8 g × 15.4 kJ mol = 6.43 kJ
Because the process is exothermic, the actual value will be negative: Δ H = −6.43 kJ.
Test Yourself
What is the energy change when 822 g of C 2 H 5 OH (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;
| 6,121 | 7,445 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.