05. UNIT 5. Chemical reactions

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UNIT 5. CHEMICAL REACTIONS. SUMMARY

KEY IDEAS YOU MUST REMEMBER PHYSICAL CHANGES

Physical changes are changes in which the chemical nature of the compounds is the same at the beginning and at the end of the change. No chemical bonds are broken or formed, so no new substances are formed.

Examples: changes of state, preparing a solution, expansion...

CHEMICAL CHANGES

Chemical changes are changes in which a new substance is produced because of a chemical reaction (a rearrangement of the atoms).

Examples: burning wood, rotting, rusting of iron...

LAW OF CONSERVATION OF MASS. LAVOISIER’S LAW

In any closed system the mass remains constant over time, regardless of any chemical reactions that occur. Mass is neither created nor destroyed in chemical reactions.

A chemical reaction is just a reorganization of the atoms. This is the reason why chemical reactions must be balanced (there must be the same number of atoms of each element on each side of a chemical equation).

BALANCING A CHEMICAL EQUATION

reactants  products

Chemical equations contain information about the state of products and reactants: aqueous solution (dissolved in water, aq), solid (s), liquid (l) or gas (g).

You cannot change subscripts (subíndices) in a chemical formula to balance a chemical equation. You can change only the stoichiometric coefficients (coeficientes estequiométricos). Changing subscripts means you are changing the compounds, so they will have different chemical properties.

Example: Cu (s) + 2 AgNO3 (aq)  Cu(NO3)2 (aq) + 2 Ag (s)

stoichiometric coefficient subscript

COMBUSTION REACTIONS

fuel (organic compound) + O₂ (oxidant)  CO₂ + H₂O combustible + O₂ (comburente)  CO₂ + H₂O Example: C3H8 + 5 O2 3 CO2+ 4 H2O

 Combustion reactions are always exothermic reactions because energy is released in the form of light and heat.

 To balance combustion reactions balance the elements in alphabetical order (C, H, O).

COLLISION THEORY

A chemical reaction is a process in which the reactants (compounds at the beginning) are converted into the products (compounds at the end). They occur when chemical bonds between atoms are broken and new chemical bonds are formed, as a result of collisions between the reactants.

In order to have a chemical reaction, effective chemical collisions are necessary. Two conditions must happen:

- Reactants must collide with a proper orientation. - Collisions must have enough energy.

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CHEMICAL REACTION RATE

In order to increase the rate of a chemical reaction there are some factors that we can modify:

1. Temperature. As temperature rises, there are more effective collisions among the reactants and the reaction rate increases.

2. Concentration of the reactants. When the concentration is higher, the frequency of collisions increases as well.

3. Contact surface. When reactants are in a solid state, the size of the particles matters. When the surface area in contact between the reactants increases, it allows more collisions and the speed is bigger.

4. Use of a catalyst. A catalyst is a substance that speeds up a chemical reaction because it lowers the activation energy (it provides an alternative mechanism for the reaction to take place). It is not a reactant or a product, as it is recovered at the end. Biological catalysts are called enzymes.

ENERGY IN CHEMICAL REACTIONS ( )

Chemical reactions involve an amount of energy associated because some bonds must be broken (energy is needed) and new bonds are created (energy is released). The chemical energy (E_R, also called enthalpy) is the global balance of both.

-  EXOTHERMIC reaction. Energy is released.

-  ENDOTHERMIC reaction. Energy is absorbed.

- E_a: activation energy. It is an energy barrier that must be overcome so the chemical reaction takes place.

ENERGY DIAGRAMS

Endothermic reaction Exothermic reaction

Example: 2 HgO (s)  2 Hg (l) + O₂ (g) H= 181,6 kJ 2 HgO (s) + 181,6 kJ  2 Hg (l) + O₂ Example: C (s) + O₂ (g)  CO₂ (g) H= 393,5 kJ C (s) + O₂ (g)  CO₂ (g) + 393,5 kJ

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PERCENT COMPOSITION (COMPOSICIÓN PORCENTUAL EN MASA)

The percent composition is the percent by mass of each element in a compound.

Example. Calculate the percent composition of H₂CO₃.

Data: Ma H= 1 u; Ma C= 12 u; Ma O= 16 u Mm H2CO3=2·1+1·12+3·16=62 g/mol

The sum of percentages of all the elements in the compound must be 100%.

ATOMIC MASS OF ISOTOPES

The atomic mass of a chemical element is the weighted average mass (media ponderada) of all the isotopes that element has, taking into account their abundance in nature.

Example. Chlorine has two common isotopes: 35Cl and 37Cl. If the abundance of 35Cl (34,97 u) is 75,77% and the abundance of 37Cl (36,96 u) is 24,23%, what is the average atomic mass of chlorine?

Atomic mass (u) Abundance (%)

Chlorine-35 34,97 75,77 Chlorine -37 36,96 24,23

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ATOMIC MASS OF ISOTOPES

Example. Rubidium has two naturally occurring isotopes, 85Rb (relative mass 84,91 amu) and 87Rb (relative mass 86,92 amu). If rubidium has an average atomic mass of 85,47 amu, what is the abundance of each isotope (in percent)?

Example. Method 1. You call x the abundance of one isotope and (100x) is the abundance of the other isotope. Those abundances are percentages (porcentajes).

Rubidium-85 84,91 x Rubidium-87 86,92 (100x) Average mass Rb = 85,47 u 

Rubidium-85 84,91 x= 72,14

Rubidium-87 86,92 10072,14 = 27,86 SOLUTION: 72,14% 85Rb and 27,86% 87Rb

Example. Method 2. You call y the abundance of one isotope and (1y) is the abundance of the other isotope. Those abundances are decimals (tantos por uno).

Atomic mass (u) Abundance (decimals)

Rubidium-85 84,91 y Rubidium-87 86,92 (1y) Average mass Rb = 85,47 u

Do not forget this is a decimal, so to get the percentage you must multiply by 100.

Atomic mass (u) Abundance (decimals)

Rubidium-85 84,91 y = 0,7214

Rubidium-87 86,92 10,7214 = 0,2786 SOLUTION: 72,14% 85Rb and 27,86% 87Rb

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SOLUTIONS (solute + solvent) DISOLUCIONES (soluto + disolvente)

A solution is a homogeneous mixture composed of two or more substances. The solute is dissolved in the solvent (there is always more solvent than solute).

- The solute is present in the smaller amount. - The solvent is present in the larger amount.

CONCENTRATION OF A SOLUTE

It expresses the amount of solute present in a certain amount of solution.

1. MOLARITY (molaridad)

2. MASS PERCENTAGE (%) (porcentaje en masa)

3. VOLUME PERCENTAGE (%) (porcentaje en volumen)

4. MASS PER VOLUME

DENSITY OF A SUBSTANCE

IDEAL GAS LAW

 P = pressure (atm)  V = volume (L)  n = number of moles  T = temperature (K)  R = 0,082 atm·L/(mol·K) (data)

- 1 atm = 760 mm Hg; 1 atm = 1,01·105 Pa; T (K) = T (ºC) + 273

- 1 mol of any ideal gas ALWAYS occupies 22,4 L in normal conditions. An ideal gas is a gas that obeys these assumptions:

- There are not intermolecular forces between the gas molecules (gas molecules do not repel or attract each other).

- The volume occupied by the gas molecules is negligible compared to the volume of the container (we can consider the gas molecules have no volume because there is a lot of empty space between the molecules, the real volume).

There are not gases that are exactly ideal, but it is a useful approximation for many situations (at room temperature and pressures near atmospheric pressure many gases behave as ideal).

Normal conditions (STP, Standard Temperature and Pressure):

- T = 0ºC

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CHEMICAL CALCULATIONS

 NA = Avogadro’s number = 6,02·1023 (data). It is the number of units (depending on the

nature of the substance, it can be atoms, molecules…) in 1 mol of any substance.

 The molar mass (Mm) is the mass of 1 mol of a substance, measured in g/mol. It has the same value than the molecular mass, measured in amu (atomic mass unit, u).

STEPS IN SOLVING STOICHIOMETRIC PROBLEMS

Notice that the stoichiometric proportions can only be used when: - Given moles and asked to find moles.

- Given molecules and asked to find molecules.

- Given volume and asked to find volume (only with gases at the same conditions of pressure and temperature).

The stoichiometric coefficients in a balanced chemical equation relate the number of moles (or molecules or litres in gases) that reacts.

1. Balance the chemical equation and pay attention to the compounds the problem relates.

2. Convert the data of the compound you are given to moles (unless it is already in moles).

3. Go to the balanced chemical equation and, using the stoichiometric proportion (in mol), change the amount of moles of the given substance to moles of the desired substance.

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ACIDS AND BASES. Arrhenius’ theory

Acids Bases

An acid is a substance that produces H+ (hydrogen ions, protons) when it is dissolved in water.

Example:

A base is a substance that produces OH (hydroxide ions) when it is dissolved in water. Example:

An important exception which cannot be explained with Arrhenius’s theory is NH3,

which behaves as a base.

MEASURING ACIDITY 1. THE pH SCALE

- pH < 7 acidic  [H+] > [OH]

- pH >7  basic  [H+] = [OH] pH: acidity scale between 0 and 14. - pH =7  neutral [H+] < [OH]

The pH scale is used to measure how acidic or basic a solution is. The pH scale is logarithmic: pH= log [H+]

- A pH of 5 means that: 5 = log [H+]  [H+] = 105 M

- When pH decreases from a value of 6 to 3, it doesn’t mean that, for example, [H+] increases the double, but [H+] is 1000 (103) times bigger.

- When pH increases from a value of 2 to 6, it means that [H+] is 10000 (104) times lower. - When pH increases from a value of 3 to 4, it means that [H+] is 10 (101) times lower.

Source: Wikipedia 2. METHODS TO MEASURE pH

- Universal indicator. It is a piece of paper that has been impregnated with a mixture of several indicators. On contact with a solution it will change its colour.

- Indicator. It is a substance that changes its colour depending on the pH because it shows a colour when it is contact with an acid and another colour when it reacts with a base. Example: phenolphthalein (fenolftaleína).

- pH meter. It is a device with a sensor that is submerged in a solution and gives a digital pH reading, which is more precise.

NEUTRALISATION REACTIONS

Acids and bases react with each other in neutralisation reactions: acid + base  salt + water Examples:

H₂SO₄+ 2 NaOH  Na₂SO₄ + 2 H₂O 2 HNO₃ + Ba(OH)₂ Ba(NO₃)₂ + 2 H₂O

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BALANCING CHEMICAL REACTIONS. ALGEBRAIC METHOD

A good way to balance chemical equations is trial and error (método de tanteo). But when it is difficult, it is better to use the algebraic method (método algebraico o de los coeficientes indeterminados).

This reaction cannot be easily balanced, so we are going to use the algebraic method: P2I4 + P4 + H2O  PH4I + H3PO4

What does “balanced” mean? It means that for every element there is the same number of atoms on both sides of the reaction equation. Our reaction has five coefficients a, b, c, d and e (the amount of different compounds that we have).

We need five coefficients, and there are only four equations (one for each element present).

a P₂I₄ + b P₄ + c H₂O  d PH₄I + e H₃PO₄ Setting up equations: [1] P: 2·a + 4·b = d + e [2] I: 4·a = d [3] H: 2·c = 4·d + 3·e [4] O: c = 4·e

Now we can assume that one of the variables equals 1 to calculate values of all others. Take a look to find what variable is the most convenient to assume. Assuming e = 1 and simply substituting calculated values we have:

[1] 2a + 4b = d + 1 [2] 4a = d [3] 2c = 4d + 3 [4] c = 4 [3]  2·4 = 4d +3 d= 5/4 [2]  4a = 5/4  a= 5/16 [1]  2·5/16 + 4·b = 5/4 + 1  b = 13/32

Now we have all the coefficients, but as they are fractions, all we have to do is to find the smallest common denominator to have whole coefficients. In this case it is 32, so if we multiply all numbers by 32 we get:

a = 10 b = 13 c = 128 d = 40 e = 32

At the end you can check if the coefficients are correct.

10 P₂I₄ + 13 P₄ + 128 H₂O 40 PH₄I + 32 H₃PO₄ Other examples with solution:

1) AsCl₃ + NaBH₄ AsH₃ + NaCl + BCl₃ 2) Cu + HNO3 Cu(NO3)2 + NO + H2O

3) KI + KClO₃ + HCl  I₂ + H₂O+ KCl

4) PbS + HNO3  Pb(NO3)2 + NO + S + H2O

5) K2Cr2O7 + H2S + H2SO4  K2SO4 + Cr2(SO4)2 + S + H2O

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SOLUTIONS:

1) 4 AsCl₃ + 3 NaBH₄  4 AsH₃ + 3 NaCl + 3 BCl₃ 2) 3 Cu + 8 HNO₃ 3 Cu(NO₃)₂ + 2 NO + 4 H₂O 3) 6 KI + KClO₃ + 6 HCl  3 I₂ + 3 H₂O + 7 KCl

4) 3 PbS + 8 HNO3  3 Pb(NO3)2 + 2 NO + 3 S + 4 H2O

5) K₂Cr₂O₇+ 3 H₂S +4 H₂SO₄  K₂SO₄ + Cr₂(SO₄)₂ + 3 S + 7 H₂O 6) 15 Pb(N3)2 + 44 Cr(MnO4)2 22 Cr2O3 + 88 MnO2 + 5 Pb3O4 + 90 NO