The MCAT Chemistry Book - Aryangat A. 2012

General Chemistry
Chemical Reactions

A. INTRODUCTION

Chemical reaction is a process at the molecular or ionic level by which one or more types of substances are transformed into one or more new types of substances by different modes of combination. In this chapter, we will explore the different types of chemical reactions including oxidation-reduction reactions. We will also learn how to balance equations.

B. CHEMICAL REACTIONS

A chemical reaction can be represented by a chemical equation. In a chemical equation representing an irreversible reaction, the substances that react (reactants) are written on the left side, while the resulting substances (products) are written on the right side of an arrow.

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In the reaction shown above, two molecules of hydrogen react with one molecule of oxygen, forming two molecules of water. Balancing of equations will be covered a little later in this chapter.

C. TYPES OF CHEMICAL REACTIONS

There are five types of chemical reactions.

1) Combination reaction

2) Combustion reaction

3) Decomposition reaction

4) Displacement reaction or single-replacement reaction

5) Metathesis reaction or double-replacement reaction

Combination Reaction

A reaction involving the formation of a compound from two or more substances is called a combination reaction.

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Combustion Reaction

A combustion reaction involves the reaction of substances with oxygen, and it is usually accompanied by the release of large amounts of heat. Combustion reactions are thus highly exothermic.

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Decomposition Reaction

A decomposition reaction is a process in which one compound decomposes or splits to form two or more simpler compounds and/or elements.

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Displacement Reaction (single-replacement reaction)

In a single-replacement reaction, an element reacts with a compound, and results in the displacement of an element or group from the compound. An example of a single-replacement reaction is shown.

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Metathesis Reaction (double-replacement reaction)

A metathesis (double-replacement) reaction involves the exchange of two groups or two ions among the reactants. Remember that in a single-replacement reaction, there is only one group or ion being switched. A metathesis reaction can often result in an insoluble product from soluble reactants, and the insoluble compound formed is called a precipitate.

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Note that this reaction involves the formation of a precipitate of AgCl.

With respect to the types of reactions, your objective should be to understand the basis behind the categorization.

D. BALANCING SIMPLE EQUATIONS

A chemical equation is said to be balanced if all the atoms present in the reactants appear in the same numbers among the products. Here is an example.

Example 2-1

Balance the following equation.

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Solution:

Start by balancing the oxygen atoms. There are two oxygen atoms on the reactant side and three oxygen atoms on the product side. To balance this, put 3 as the coefficient of oxygen on the reactant side. When we write ’3 O2,’ that means we have 6 oxygen atoms on the reactant side. To make the same number of oxygen atoms on the product side, let’s put 2 as the coefficient of Fe2O3. Now the oxygen atoms seem to be balanced.

Let’s take a look at Fe. Since the coefficient of Fe2O3 is 2, we have 4 atoms of Fe on the product side. We can balance this by writing 4 as the coefficient of Fe on the reactant side. So the balanced equation is as follows:

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Problem 2-1

Balance the following equations:

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Answers:

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E. OXIDATION NUMBER

Electrons are exchanged during oxidation-reduction reactions. The behavior of atoms or ions in terms of the number of electrons transferred is expressed as the oxidation state (oxidation number). We can define oxidation number as the charge of an atom or ion, based on a set of standard rules. If the given species is an ion containing a single atom, then its oxidation state is its charge itself. Let’s analyze this by looking at a few examples.

In NaCl, the oxidation state of sodium is +1 and the oxidation state of chlorine is —1. Generally, the elements at the top right corner of the periodic table are assigned negative oxidation numbers. Some of the elements on the right side of the periodic table can have positive or negative oxidation numbers depending upon the atom to which the given element is bonded to. The elements in the middle and the left portions of the periodic table have almost exclusively positive oxidation numbers.

Table 1-1

General guidelines for assigning oxidation numbers

1. The elemental natural state oxidation number of any atom is zero. For example, the oxidation number of oxygen atom in O2 is zero.

2. The sum of the oxidation numbers of the atoms in a compound should be zero.

3. The sum of the oxidation numbers of the atoms in an ionic species (a species with a net charge) should equal the net charge of the ionic species.

4. The oxidation number of a given ion containing a single atom is its charge itself.

Oxidation numbers of some common elements

1. The common oxidation number of Group IA metals is +1. E.g., lithium, sodium, potassium.

2. The common oxidation number of Group IIA metals is +2. E.g., beryllium, magnesium, calcium.

3. The common oxidation number of Group IIIA is +3. E.g., aluminum, boron.

4. The common oxidation number of Group IVA is +4. +2 is also seen in some compounds such as CO.

5. The common oxidation numbers of Group V A are +5 and —3.

6. The common oxidation number of Group VIA is —2.

7. The common oxidation number of Group VIIA is —1.

8. The common oxidation number of H is +1. In some metal hydrides, hydrogen shows an oxidation number of —1.

The above list of common oxidation numbers is not comprehensive. Nevertheless, it gives you a basic and essential picture about assigning oxidation numbers in common compounds and ionic species. Most elements can have multiple oxidation states, depending on the element or ionic species they are bonded to. You have to always follow the general guidelines in Table 1-1, and check whether the items listed are satisfied.

Now that we have learned the theory of assigning oxidation numbers, let’s do an example to see how it works.

Example 2-2

What is the oxidation number of sulfur in sulfuric acid?

Solution:

Sulfuric acid is H2SO4. The oxidation number of hydrogen is +1. But we have two hydrogens which add up to a charge of +2. Since the total charge of this molecule should be zero, we can say that the charge of sulfate ion is —2. We also know that the oxidation number of oxygen is —2. But there are four oxygens in a sulfate ion. So the charge adds to —8. Now let’s solve this algebraically.

ONSulfur - Oxidation number of sulfur

ONOxygen - Oxidation number of oxygen

ONSulfur + 4 (ONOxygen) = —2

ONSulfur + 4 (—2) = —2

ONSulfur = +6

So the oxidation number of sulfur in sulfuric acid is +6.

Problem 2-2

Calculate the oxidation state of the element indicated in each of the following problems.

(a) What is the oxidation state of hydrogen in MgH2?

(b) What is the oxidation state of chlorine in ClO3?

(c) What is the oxidation state of oxygen in Na2O2?

(d) What is the oxidation state of nitrogen in NH3?

(e) What is the oxidation state of oxygen in O2?

(f) What is the oxidation state of bromine in HBrO2?

(g) What is the oxidation state of manganese in KMnO4?

Answers:

(a) —1

(b) +5

(c) —1

(d) —3

(e) 0

(f) +3

(g) +7

F. OXIDATION-REDUCTION REACTIONS

Oxidation-reduction (redox) reactions involve the transfer of electrons from one compound or species to another. In this section, we will discuss oxidation-reduction reactions and learn how to balance them.

Oxidation is the process by which an atom or species loses its electrons. In reduction, an atom or species gains electrons. Let’s first consider oxidation and reduction separately.

Consider the conversion of iron from its neutral elemental state to its ionic form.

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Notice that in the process iron lost electrons. The process is oxidation.

Consider another example. An example involving the conversion of bromine to its ionic form.

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Notice that in the process bromine gained electrons. The process is reduction.

Now let’s go one step forward. Consider the next reaction.

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This reaction is a typical example of an oxidation-reduction reaction. The oxidation number of iron on the reactant side is 0. The oxidation number of bromine on the reactant side is also 0. What are the oxidation numbers of iron and bromine in FeBr3? Well, we know that bromine has an oxidation number of —1. So the oxidation number of iron in FeBr3 is +3. Thus iron is oxidized and bromine is reduced. The species that gets oxidized is called the reducing agent. The species that gets reduced is called the oxidizing agent. In this reaction, evidently iron acts as the reducing agent, and bromine acts as the oxidizing agent.

Oxidation results in an increase in the oxidation number. In the process of oxidation, electrons are lost.

Reduction results in a decrease in the oxidation number. In the process, electrons are gained.

Balancing Oxidation-Reduction Reactions

Balancing of an oxidation-reduction reaction is a little more complex than balancing a simple reaction. The main rule that you have to follow when balancing oxidation-reduction reactions is that the absolute value of the increase in oxidation number of all the atoms that are oxidized should equal the absolute value of the decrease in oxidation number of all the atoms that are reduced. Balancing oxidation-reduction reaction is sometimes time-consuming and quite often frustrating. We will look at two methods of balancing oxidation-reduction reactions.

Method A

1. Write the unbalanced equation.

2. Find the oxidation numbers of the atoms that undergo change in oxidation states and write on top of each atom the corresponding oxidation number.

3. By this process, we will be able to see which atoms are getting oxidized and reduced.

4. Compare and indicate the change in oxidation numbers from the reactant side and the product side, and write down the change in the oxidation numbers.

5. Make the necessary changes by writing coefficients that will equalize the changes in the oxidation numbers. In other words, the net decrease in the oxidation numbers should equal the net increase in the oxidation numbers. Add water if necessary.

6. Do a final check on whether all the atoms and charges balance out.

Method B

1. Write the unbalanced equation.

2. Separate the two half-reactions, write them out, and balance any of the atoms. From this point onward, we balance the reactions separately. Do the obvious or the simple balancing by inspection if possible.

3. Balance the oxygen atoms by adding water on the appropriate side of the half-reaction.

4. Balance the hydrogen atoms by adding H+ on the appropriate side of the half-reaction.

5. Add sufficient number of electrons so that the charges are balanced.

6. Once the half-reactions are balanced, combine the half-reactions and cancel out any common terms that appear on both sides of the equation to get the refined and balanced oxidation-reduction equation.

Example 2-3

Balance the following oxidation-reduction reaction.

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Method A

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Method B

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Half-reaction I

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Half-reaction II

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Before we combine the equations, the electrons need to be balanced out. So multiplying the balanced half-reaction # (I) by 4, we get

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Combining the two half reactions

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G. CALCULATIONS INVOLVING CHEMICAL REACTIONS

In this section, we will look at some chemical equations and find what we can do with the information represented in a chemical equation. The equation shown below represents the reaction of methane and oxygen to form carbon dioxide and water.

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From this balanced equation we can infer many things. Let’s consider a few.

1) One molecule of methane reacts with 2 molecules of oxygen to form 1 molecule of carbon dioxide and 2 molecules of water.

2) We can also say that 1 mole of methane reacts with 2 moles of oxygen to form 1 mole of carbon dioxide and 2 moles of water.

3) Since one mole contains Avogadro number of molecules, we can say that 6.023 x 1023 molecules of methane reacts with 1.2046 x 1024 (= 2 x 6.023 x 1023) molecules of oxygen to form 6.023 x 1023 molecules of carbon dioxide and 1.2046 x 1024 molecules of water.

4) We can confidently say that 16 g of methane reacts with 64 g of oxygen to form 44 g of carbon dioxide and 36 g of water.

Example 2-4

Calculate the number of moles of water produced when 5.25 moles of methane undergo the reaction depicted below. (Assume there is plenty of oxygen for the reaction)

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Solution:

From the equation, it is clear that for every mole of methane, 2 moles of water are formed. So without any elaborate calculations, you should be able to come up with the correct answer. It is very much like a ratio problem. Since there are 5.25 moles of methane, 10.5 moles of water will be formed.

Number of moles of water formed = 5.25 x 2 = 10.5 moles

Example 2-5

110 g of CO2 were formed as a result of the reaction shown below. How many grams of oxygen must have reacted to form that much carbon dioxide?

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Solution:

According to the equation, 2 moles of oxygen result in 1 mole of carbon dioxide. For example, if 2 moles of carbon dioxide were formed, 4 moles of oxygen must have reacted. Here, the amount of carbon dioxide formed is given in terms of grams. So the first step is to convert the grams to moles.

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Hence, 5 moles of oxygen must have reacted to form 2.5 moles of carbon dioxide. But the question asks for this quantity in grams. So the final step is to convert moles to grams.

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H. THE CONCEPT OF LIMITING REAGENT

So far we have been considering reactions in which all the reactants exist in adequate quantities. In this section, we will consider what happens when the amount of one of the reactants available is less than the amount required to complete the reaction. When such a condition exists, we call that reactant or reagent the limiting reagent.

We will further explore this scenario through the following examples.

Example 2-6

A reaction mixture contains 60.75 g magnesium and 146 g hydrogen chloride. Predict the limiting reagent if the reaction occurs as shown below.

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Solution:

First, we have to convert the grams of the substances to moles. Then make the comparison to see which one is the limiting reagent. By now, you should be comfortable with the conversion of moles to grams and vice versa. The number of moles of magnesium present is 2.5 moles. According to the equation, 1 mole of magnesium reacts with 2 moles of hydrogen chloride. For magnesium to completely react, there should be at least 5 moles of hydrogen chloride present. If you calculate the number of moles of hydrogen chloride present, you will get 4 moles. This amount of hydrogen chloride is not enough to completely react with the amount of magnesium present. So the limiting reagent is hydrogen chloride.

I. PERCENT YIELD

If we know the chemical equation and the amounts of reactants, we can calculate the theoretical yield of that reaction. But in reality, the yield depends on many other factors also. Most of the time in synthesis reactions, even in your own lab experiments, you probably noticed that the actual yield is lower than the theoretical yield. The percent yield denotes the amount of actual yield in terms of the theoretical yield. The formula to find the percent yield is given below:

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Example 2-7

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A student conducted the above reaction in a lab as a part of her research assignment. She used 138 g of sodium nitrite, with excess of hydrogen chloride. What is the percent yield of HNO2, if the actual yield of HNO2 was 61.1 g?

Solution:

First, you should find the number of moles of NaNO2. Since she used 138 g, the number of moles of NaNO2 is 2 (Mol.wt of NaNO2 is 69 g/mol). Since the ratio of formation of HNO2 is 1:1 with respect to NaNO2, theoretically 2 moles of HNO2 should be formed. Two moles of HNO2 correspond to 94 g. But actually, only 61.1 g of HNO2 was formed. Now it is just a matter of plug and chug in the percent yield formula.

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In this experiment, the actual yield was not high (i.e., only 65% of the theoretically predicted yield) as expected.

Example 2-8

Match the following reactions with the appropriate type of reaction

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Solution

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CHAPTER 2 PRACTICE QUESTIONS

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1. The reaction shown here can be best classified as a:

A. combustion reaction.

B. combination reaction.

C. single-replacement reaction.

D. double-replacement reaction.

2. All the following reactant-combinations are neutralization reactions, except:

A. HNO2 + NaOH

B. KOH + HCl

C. Al+ + 3OH

D. H3PO4 + NaOH

3. An unbalanced equation is shown below. What is the coefficient of aluminum hydroxide in the final balanced equation?

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A. 1

B. 2

C. 3

D. 7

4. How many grams of sodium chloride are required to synthesize 73 grams of hydrogen chloride, if the reaction involves sodium chloride and sulfuric acid?

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A. 58.5 grams

B. 117 grams

C. 175.5 grams

D. 234 grams

5. In some reactions, you will often encounter ions in aqueous solutions which are not actually involved in the reaction. Such ions are best termed:

A. cations.

B. anions.

C. salt-bridge ions.

D. spectator ions.

6. Some substances can act as a base or an acid. Such substances are called:

A. aliphatic substances.

B. amphibasic substances.

C. lyophilic substances.

D. amphoteric substances.

7. Predict the coefficient of iron in the balanced equation for a reaction in which iron reacts with oxygen to form Fe2O3.

A. 1

B. 2

C. 3

D. 4

8. The number of molecules are the same in which of the following pairs?

A. 32 grams of O2 and 32 grams of SO2

B. 49 grams of NO2 and 40 grams of NaOH

C. 20 grams of HF and 36 grams of H2O

D. 60 grams of C2H6 and 156 grams of C6H6

Questions 9-15 are based on the following passage.

Passage 1

Information regarding the amounts of substances that actually react to form the products are of extreme help. Analyzing reactions and having the balanced equations help chemists to determine the correct and optimum proportions of reactants to be used. There are various other factors besides the amount of reactants, which determine the efficiency of reactions.

Experiment 1

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Student 1 used 40.5 g of aluminum and 80 g of Fe2O3 for Reaction I. A second student conducted the same reaction with a different amount for one of the reactants. The second student used 40.5 g of aluminum, and 90 grams of Fe2O3. A third student also conducted the same reaction with 54 g of aluminum and the same amount of Fe2O3 used by the first student.

Experiment 2

Reaction II involved the production of Al2O3 from aluminum hydroxide This was done by heating aluminum hydroxide.

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Experiment 3

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9. What is the most likely identity of product x?

A. aluminum

B. water

C. hydrogen

D. oxygen

10. In Experiment 3, which of the following is the limiting reagent?

A. Fe2O3 because there is only 1 mole of it

B. CO because of its gas phase

C. Fe because of its insufficiency

D. Cannot be determined without more data

11. Based on the given information, which of the following are true regarding Experiment 1?

I. The second student had the highest yield for aluminum oxide.

II. The third student had a higher yield for aluminum oxide than the first student.

III. The first and the third students had the same yield for aluminum oxide.

A. I only

B. I and II only

C. I and III only

D. II and III only

12. Which of the following changes will further increase the overall yield for the reactions conducted in the Experiment 1?

A. Increasing the amount of aluminum used by Student 1

B. Increasing the amount of aluminum used by Student 2

C. Increasing the amount of Fe2O3 used by Student 2

D. None of the above

13. Roughly 80 g of Fe2O3 was present in Experiment 3, and upon completion of the reaction, it was measured that 22 g of CO2 was formed. If this is true, how much CO must have reacted before reaching completion of the reaction?

A. 14 grams

B. 28 grams

C. 56 grams

D. 84 grams

14. In the previous question, which of the reactants acts as the limiting reagent?

A. Fe2O3

B. CO

C. CO2

D. Cannot be determined

15. If 0.1 mol of CO was reacted with excess of Fe2O3, how many molecules of carbon dioxide will be produced?

A. 6.0 x 1023 molecules

B. 6.0 x 1022 molecules

C. 3.0 x 1023 molecules

D. 18.0 x 1023 molecules