Chapter 5. Reactions

Understanding Chemical Reactions

Chemical Equations

Chemical equation tells how chemical transformation occurs in a reaction. Reactant is separated from the Product by an arrow, as shown below.

For example, production of water from its elements. H2 is the elemental form of hydrogen and O2 is the elemental form of oxygen.

However, there are mismatch in the number of oxygen in the reactant and product.

We have missing oxygen on the product side. So, we multiply 2 for water.

Now, we have 4 hydrogens on the right-hand side, while the left-hand side only has two hyrdogens. So, we multiply H2 by 2 to obtain.

Now, the number of atoms for different elements are the same on the both sides of the equation. It is said to be balanced .

Example:Balance the following chemical equations.

a) SiO2 + HF → SiF4 + H2O
b) CH4 + Cl2 → CH2Cl2 + HCl
c) NO2 + O2 + H2O → HNO3

a)

SiO2 + HF → SiF4 + H2O
First step is perhaps to make the number of oxygen equal to both sides by multiplying 2 on water.
SiO2 + HF → SiF4 + 2H2O

Now, I can see that there are four F on the right-hand side, so multiply 4 to HF.

SiO2 + 4HF → SiF4 + 2H2O
It appears now that the equation is balanced.

b) Simlar line of thought,

CH4 + 2Cl2 → CH2Cl2 + 2HCl

c) This NO2, exhausted out of cars, reacts with oxygen and water causes the acid rain (rain of nitric acid!!). Yikes!

NO2 + O2 + H2O → HNO3

N = 1, O = 5, H = 2 → N =1, O = 3, H = 1

Let's start with multiplying 2 for water

NO2 + O2 + 2H2O → HNO3

N=1, O=6, H=4 → N=1, O=3, H=1

We need four HNO3,

NO2 + O2 + 2H2O → 4HNO3
N=1, O=6, H=4 → N=4, O=12, H=4

Then, make 4 NO2

4NO2 + O2 + 2H2O → 4HNO3
N=4, O=12, H=4 → N=4, O=12, H=4

Reaction Types

We can classify reactions into four categories:
  • Synthesis Reaction
  • Decomposition Reaction
  • Single Replacement Reaction
  • Double Replacement Reaction

    Synthesis Reaction

    A + B → AB
    This type of reaction combines two reactants, A and B, into a single product, AB. One example is:
    CO2 + H2O → H2CO3
    Carbon dioxide dissolved in water becomes carbonic acid, which is responsible for the sour taste off carbonated water (like seltzer water).

    Decomposition Reaction

    AB → A + B

    Single Replacement Reaction

    A + BC → AC + B

    Double Replacement Reaction

    AB + CD → CB + AD

    Reaction Involving Water

    The functional groups that are required to undergo reactions that involves water are: alkenes, alcohols, carboxylic acids, and esters. In all cases, we use H+ (proton = meaning we have acid) as a catalyst for the reaction. Catalyst is a facillitator of reaction but not get consumed by the reaction; just a helper to do the job, and leave when it is done.

    Mechanistically, the reactions are initiated by a catalytic proton, attracted to the partial negative, δ-, charge of oxygen on oxygen containing functional groups or a double bond of alkene. Subsequently, water molecule is attracted to the positively charged protonated functional group.

    Alkanes are not reactive toward water.

    Hydrolysis

    Ester is hydrolyzed to produce a carboxylic acid and an alcohol.

    Hydration

    Water can be added to alkene to produce an alcohol.

    Dehydration

    Alcohol can be decomposed to alkene.

    Example:Fill the blank in the following reactions.

    a)

    b)

    c)

    + Answer to hydrolysis example 1
    + Answer to hydrolysis example 2
    + Answer to dehydration example 1

    Oxidation-Reduction (Red-Ox)

    Reduction = in a chemical process, the compound gains a number of electrons

    Oxidation = in a chemical process, the compound loses a number of electrons

    3Cu2+(aq) + 2Al(s) → 3Cu(s) + 2Al3+(aq)
    In the above equation, each Cu2+ ion gains 2 e- to become Cu metal (solid). So, Cu is reduced in the process. each Al metal loses 3 e- to becom aqueous Al3+ ion. Thus, aluminum metal is oxidized in the reaction.

    In this situation, we call Cu2+ ion to be oxidizing agent, and Al to be reducing agent.

    Combustion as a Red-Ox Rxn

    Combustion is a chemical process that add oxygen to carbons of hydrocarbon compounds to produce carbon dioxide and water.
    CH4(aq) + 2O2 → CO2 + 2H2O
    In this process, as you can see, methane (CH4) is oxidized to become carbon dioxide, and oxygen (O2) is reduced to become water.

    Oxidizing Alcohols and Aldehyde

    If the OH is at the terminal carbon, the reaction produces aldehyde.

    If the OH is in the middle, the reaction produces ketone.

    Aldehyde can be converted to carboxylic acid.

    You will see more in Chapter 9!

    Hydrogenation

    Platinum metal, Pt, is used as catalyst to hydrogenate alkene, aldehyde and ketone.

    Mole and Mass Relations

    In this section we examine the relationship between the masses of molecules involved in a reaction as well as the relationship between the number of moles of molecules in the reaction.

    We have looked at molar mass and the mole concept for elements in Chapter 2. We also know how to calculate molar mass of molecule according to Chapter 3. We are now going to relate different species in terms of quantity, either mass or moles, that are involved in the reaction.

    Given the balanced chemical equation:

    2H2 + O2 → 2H2O
    2 molecules 1 molecule 2 molecules Microscopic View
    2 mol 1 mol 2 mol Macroscopic View

    Microscopic view == molecular picture

    Macroscopic view == quantities that can be measured (like g and mL)

    Example:Let's start with simple example. Given that

    2H2 + O2 → 2H2O
    If you start with 10 mol of H2, how many mol of H2O can be produced?

    Since 2 mol H2 yields with 2 mol H2O, it stands to reason that 10 mol H2 produce 10 mol H2O. Let's put this into dimensional analysis. ?mo l H 2 O =10mo l H 2 is how you start.

    Following the rule of dimensional analysis to cancel the available unit with the denominator of the conversion factor placed inside of the parenthesis, we use the balanced chemical equation as our conversion factor, and that is 2 molH2 = 2 molH2O. Then, ?mo l H 2 O =10mo l H 2 ( 2mo l H 2 O 2mo l H 2 )=10mo l H 2 O

    Example:Now, a simlar question, but adding complexity. Given that

    2H2 + O2 → 2H2O
    If you start with 10 mol of H2, how many g of H2O can be produced?

    Again, we start with 2 mol H2, but this time we are seeking the gram quantity of H2O. So we ask: ? g H 2 O =10mo l H 2 Since we started with 10 molH2, the first parenthesis is the same as the example above. However, we now seek gH2O, so we need a way to convert gH2O into molH2O, and the g to mol (or mol to g) conversion is accomplished by using the molar mass. The molar mass of water is given as 2 X (1.008 g/mol) + (16.00 g/mol) = 18.016 g/mol. Using this molar mass of water, we get, ? g H 2 O =10mo l H 2 ( 2mo l H 2 O 2mo l H 2 )( 18.0 g H 2 O 1mo l H 2 O )=180 g H 2 O

    Example:If we can convert mol to g, then we can do g to g conversion! Given again that

    2H2 + O2 → 2H2O
    If you start with 20.16 g of H2, how many g of H2O can be produced?

    Much the same way as before, but this time, we start with 20.16 g of H2, ? g H 2 O =20.16 g H 2 This time, you need to cancel the gH2 in the denominator. When you see a gram of something, you should always think about molar mass first. The molar mass of H2 is 2 X (1.008 g/mol) = 2.016 g/mol. Then, ? g H 2 O =20.16 g H 2 ( 1mo l H 2 2.016 g H 2 )( 2mo l H 2 O 2mo l H 2 )( 18.0 g H 2 O 1mo l H 2 O )=180 g H 2 O

    Given a balanced chemical equation, you can relate product mass from the reactant mass! Try the following questions!

    Example:If 5.0 g of molten magnesium (Mg(l)) is used in the following reaction, how many g of titanium (Ti) is produced?

    TiCl4 + 2Mg(l) → Ti + 2MgCl2

    Example:If 5.0 g of water is produced in the following reaction, how many g of C2H2 did you start with?

    2C2H2 + 5O2 → 4CO2 + 2H2O

    Example:If 5.0 g of calcium (Ca) is used in the following reaction, how many g of vanadium (V) is produced?

    5Ca + V2O5 → 5CaO + 2V

    Calculating the Yield

    The calculations we did above for the product mass is called theoretical yield. The product that one actually obtains from experiment is called acutal yield. Usually, actual yield is less than theoretical yield, and we often calculate % yield. % yield = m actual m theoretical ×100% where mactual is the actual yield in g, and mtheoretical is the theoretical yield in g. Here is how you might use it.

    Example:What is the % yield of H2O if 20.16g of H2 is reacted with O2, and produced 155 g of water, given the following reaction:

    2H2 + O2 → 2H2O

    We know we started with 20.16 g of H2, and it produced 155 g H2O. It means that 155 g is the actual yield. And then, we need to calculate theoretical yield of H2O using 20.16 g H2. The calculation is done in the previous example: ? g H 2 O =20.16 g H 2 ( 1mo l H 2 2.016 g H 2 )( 2mo l H 2 O 2mo l H 2 )( 18.0 g H 2 O 1mo l H 2 O )=180 g H 2 O Then, the % yield is, % yield = 155g 180g ×100%=86.1%

    Free Energy and Rxn Rate

    Energy = ability to do work.

    Work = moving an object with a mass by a force to some distance away. Heat is also energy. In chemistry, many reactions deal with heat. Only gas reactions involve with mechanical work.

    Heat = Enthalpy = heat energy.

    Free Energy = The energy that makes reaction go spontaneously.

    For example, a battery is connected to a LED light, it turns light immediately. Ice melts at room temperature. Hydrocarbon is combusted (burned) to produce CO2 and water. These are all spontaneous reactions.

    ΔH > 0 is endothermic process
    ΔH < 0 is exothermic    		 ΔG > 0 is nonspontaneous
    ΔG < 0 is spontaneous

    Energy Diagram

    We often portray how reactions proceed by illustrating energy in terms of reaction path, similar to the one shown above explaining heat of endothermic and exothermic processes or the free energies of spontaneous and nonspontaneous reactions, but little more detail.

    The following pictures illustrate energy of a reaction as it proceeds from left (reactant, which is the initial state) to right (product which is the final state). The energies of the reactant and product are at the bottoms of the "U" shapes in the diagram. As you can see, the energy in the middle part is higher than the reactant and product. This is called transition state.

    Endothermic Reaction Exothermic Reaction

    These diagrams tell you is that the reaction proceeds by changing the geometry of the reactant gradually, which makes the enegy goes higher, to become the transition state where the energy of the reaction is highest, then proceed to become the product, where the energy is stable. The energy of the transition state compared to the energy of the reactant is called activation energy is literally the wall that prevents reaction to go quickly. The rate of reaction slows down as baarrier height becomes larger. As I said combustion of hydrocarbon to produce CO2 and water is a spontaneous reaction, but you know just mixing the hydrocarbon with O2 would not combust. This is because of the activation energy of the combustion process is quite high, preventing from sustained combustion. You need a spark of some sort to make the combustion to start. For example, you see the spark on your gas stove or a spark plug in your car to initiate ignition.

    Enzyme is a catalyst, made of protein that facillitate reactions. In terms of energy diagram of a given reaction, the enzyme makes the reaction proceeds at much faster rate by reducing the barrier height. The situation is given below: