Chemical Reactions

1. Introduction: Count! (1/2)

In this section, we will learn to count the number of atoms and the electric charge.

Matter is composed, among other things, of molecules, atoms, and ions.

We know more than a hundred atoms.

The most common atoms:
  • Carbon → symbol: C
  • Oxygen → symbol: O
  • Hydrogen → symbol: H
  • Nitrogen → symbol: N

Atom symbols can also have 2 letters, like copper → Cu, the 2nd letter is always lowercase.

Ions come from atoms that have gained or lost one or more electrons.

Since an electron has a negative charge, an ion must have an electric charge, unlike an atom which has a net electric charge of 0.

Examples of ions:
  • copper ion → Cu2+
  • hydroxide ion → OH-

Atoms can bond together through "covalent" bonds to form molecules.

Since atoms are electrically neutral, molecules are too.

Examples of molecules:
  • water → H2O
  • dioxygen → O2

Similarly, an ion formed from a single atom is called a "monatomic ion," such as the sodium ion: Na+ or the chloride ion: Cl-.

If the ion is formed from several atoms, it is called a polyatomic ion, like the hydroxide ion: OH-.

An electron has a negative charge, often noted as "-e."

Note: "an elementary charge" noted as e is equal to 1.6x10-19C, where C is the symbol of the unit of electric charge: Coulomb.
However, this is not the focus of this section 🌝, just remember that the value of the charge does not change.

For example:
  • Cu2+ (copper(II) ion) carries 2 positive charges
  • Cl- (chloride ion) has one negative charge

Introduction: Count! (2/2)

In the following, I will use only empirical formulas as well as a simplified notation like 3O instead of 3 oxygen atoms.

In the empirical formula of propanol C3H8O, we count 3C, 8H, and 1O because:

  • The subscript numbers indicate only the number of atoms of the element immediately to the left of the number.
  • → In CO2(carbon dioxide) "the 2" concerns only O and not C, so there is 1C and 2O.
  • Generally, we do not indicate the quantity 1.
  • → We write CO2 and not C1O2, even though C1O2 is not incorrect.

Let's move on to electric charges with ions.

The electric charge is indicated at the top right of the empirical formula, not complicated 🌝.

  • Iron(II) ion → Fe2+: 1Fe, 2 positive charges
  • Iron(III) ion → Fe3+: 1Fe, 3 positive charges
  • Permanganate ion → MnO2-4: 1Mn, 4O, and 2 negative charges

Next, I propose an activity like:

I have 4MnO2-4, now count the elements and charges.

  • If 1MnO2-4 gives 1Mn, 4O, and 2 negative charges
  • Then 4MnO2-4 gives 4x(1Mn, 4O, and 2 negative charges)
  • So: 4Mn, 16O, and 8 negative charges

The goal of Activity 1 is to learn to count chemical elements and charges

Find the electric charge and the number of elements:
TEST: 0 / 5
⬅️
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48

2. Balancing a Chemical Equation: Introduction

a) The Sandwich Method


Let's say that to make a sandwich (S), I need:
  • 1 bread (B)
  • 2 slices of ham (H)
  • 3 lettuce leaves (L)

If I want to put my super recipe 😎 into an equation, it could look something like this:

  1 bread + 2 slices of ham + 3 lettuce leaves → 1 sandwich


As a good scientist is a lazy scientist, I will rather use symbols in my equation:

   1 B + 2 H + 3 L→ 1 S.

Now, I can make the link between my recipe and a real chemical reaction equation:

  • The ingredients are called Reactants, they are consumed (not necessarily completely) during the reaction.
  • What we get, here the sandwich, is called the product.
  • The reaction arrow → separates the reactants from the products.
  • The numbers are called stoichiometric coefficients, they indicate the proportion of disappearance of reactants and the proportion of product creation.

b) Let's move on to a "real" reaction equation with the combustion of methane (CH4) in oxygen (O2).

During combustion, methane (CH4) and oxygen (O2) are consumed, so they are reactants.

This reaction produces carbon dioxide (CO2) and water (H2O).

So we have: CH4 + O2 → CO2 + H2O

Unfortunately, this is wrong! 😱 Why?

From activity1, we know how to count atoms. We see that there are 4H in the reactants and 2H in the products, which implies that 2H were created from nothing ... That's impossible.

We must respect the conservation of matter: in a chemical reaction, chemical elements and charges are conserved.

We must therefore add stoichiometric coefficients to have the same number of atoms in the reactants as in the products:

So we have: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

(g) specifies the physical state: gas. We can also find (s): solid, (l): liquid, and (aq): dissolved in water

We have the same number of atoms on each side (reactants/products), the reaction equation is balanced


1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
TEST: 0 / 5

⬇️ drag and drop to adjust the stoichiometric coefficients ⬇️

→

Reactants

Balanced?

Products

3. Interpreting a Reaction Equation: Introduction

a) The Mole

A drop of water contains billions and billions... of water molecules.

To express the amount of substance in chemistry, we use a unit that contains a lot of elements:

  • In chemistry, we count elements in moles (abbreviated as mol).
  • 1 mole contains 6.02x1023 entities.
  • The amount of substance is expressed in moles with the symbol n.

Example: n(H2O) = 1 mole corresponds to 6.02x1023 water molecules.


b) Interpreting a Reaction Equation

Consider the balanced reaction equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g).

This equation states:

  • 1 CH4 and 2 O2 disappear.
  • 1 CO2 and 2 H2O are formed.

What is important to understand is that this is a proportion.

If 100 CH4 molecules disappear:

  • 1x100 CH4 and 2x100 O2 disappear.
  • → 100 CH4 and 200 O2 disappear.
  • 1x100 CO2 and 2x100 H2O are formed.
  • → 100 CO2 and 200 H2O are formed.

If 5 moles of CH4 disappear:

  • 5 moles of CH4 and 10 moles of O2 disappear.
  • 5 moles of CO2 and 10 moles of H2O are formed.

In Activity 3, you will check if you are able to interpret a reaction equation.

Test : 0 / 5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
➡️

4. Final state of a complete reaction: introduction

a) We limit ourselves to complete and rapid reactions

A complete reaction is a reaction that ends when at least one of the reactants is depleted.

In this case, these reactants are called limiting reactants. At the end of the reaction, their amount of substance is equal to 0 mol.

The end of the reaction is called the final state. If the amounts of substance of all the reactants in the final state are 0, we say that the reactants were introduced in stoichiometric proportions.

Notes:

  • If the reaction is not complete, it is limited. In this case, the products recombine to form reactants, and the final state is an equilibrium state where the amounts of substances stabilize.
  • In the reaction equation, the reaction arrow → should only be used for complete reactions.

b) The progress (X) of a reaction

The progress (denoted X) of a reaction corresponds to a product that has a stoichiometric coefficient of 1, X is expressed in moles.

It is important to understand that this is a fictitious quantity, to be seen as a calculation tool to study the evolution of amounts of substances.

Let’s take the example of methane combustion again: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

  • In the initial state, X = 0; in the final state, X = Xmax
  • If X = 1 mol: CH4 loses 1 mol and O2 loses 2 mol, CO2 gains 1 mol and 2H2O gains 2 mol, according to the stoichiometric coefficients of the equation.
  • If X = 3 mol: CH4 loses 3 mol and O2 loses 6 mol, CO2 gains 3 mol and 2H2O gains 6 mol, etc.

c) Finding the maximum progress (Xmax) and the limiting reactant(s)

The initial amount of substance is denoted ni(...) and the final amount of substance is denoted nf(...).

The higher the stoichiometric coefficient, the faster the reactant is consumed. To find the limiting reactant, you need to:

  • Calculate for all reactants: ni(r)/c: the initial amount of the reactant divided by its stoichiometric coefficient.
  • The smallest value corresponds to both Xmax and the limiting reactant. (If the values are equal → stoichiometric proportion)

For the example of methane combustion: you need to calculate ni(CH4)/1 and ni(O2)/2, then choose the smallest value.

If we obtain, for example, 0.4 mol and 0.5 mol, Xmax = 0.4 mol and CH4 is the limiting reactant.


d) Final state

Let’s take the example of methane combustion again: 1CH4(g) + 2O2(g) → 1CO2(g) + 2H2O(g) (the stoichiometric coefficient 1 is written only to illustrate the following)

At the end of the reaction for the reactants:

  • CH4 loses 1 x Xmax = Xmax, remaining at the end of the reaction: nf(CH4) = ni(CH4) - Xmax
  • O2 loses 2 x Xmax = 2Xmax, remaining at the end of the reaction: nf(O2) = ni(O2) - 2Xmax

At the end of the reaction for the products:

  • CO2 gains 1 x Xmax = Xmax, at the end of the reaction: nf(CO2) = ni(CO2) + Xmax
  • H2O gains 2 x Xmax = 2Xmax, at the end of the reaction: nf(H2O) = ni(H2O) + 2Xmax

In activity 4, find Xmax, the limiting reactant(s), and the final state. It’s your turn! 🌝

SuperCalculator 😁
÷ =
÷ =
🔼 drag / drop 🔼
SuperCalculator 😁
-
 (
x
) 
=
x =
🔼 drag / drop 🔼