Chapter 1: Chemical Reactions and Equations

Concept : Magnesium is quite reactive. Even at room temperature it combines slowly with the oxygen of air, forming a thin, white layer of magnesium oxide on its surface.

Step 1 – Surface reaction at room temperature
$$\mathrm{2Mg\,(s) + O_2\,(g) \rightarrow 2MgO\,(s)}$$

The oxide film $$\mathrm{MgO}$$ is insoluble and sticks tightly to the metal.

Step 2 – Effect of the oxide film during burning

• The adherent layer blocks contact between atmospheric oxygen and the underlying magnesium.
• Because of this barrier the ribbon either fails to ignite or burns very feebly.

Step 3 – Purpose of cleaning
Rubbing with sand-paper removes the oxide coating, exposing a shiny, fresh metallic surface that can react freely with oxygen. The cleaned ribbon therefore ignites at once and burns with an intense white flame.

Conclusion
A magnesium ribbon must be cleaned before burning so that the oxide layer is removed and rapid combustion can occur.

In Activity 1.7 distilled water is made slightly conducting by adding a few drops of dilute $$\mathrm{H_2SO_4}$$ and is then electrolysed between two inert electrodes.

Step 1 – Write the overall decomposition reaction.

$$\mathrm{2H_2O(l) \xrightarrow[\;\;\text{electricity}\;\;]{} 2H_2(g) + O_2(g)}$$

Step 2 – What happens at each electrode.

• Cathode (negative electrode) : $$\mathrm{H^+}$$ ions from water gain electrons and are reduced to hydrogen gas:
  $$\mathrm{4H^+ + 4e^- \rightarrow 2H_2(g)}$$
• Anode (positive electrode) : $$\mathrm{OH^-}$$ ions (or water molecules) lose electrons and are oxidised to oxygen gas. (Note: free $$\mathrm{O^{2-}}$$ ions do not exist in aqueous solution; what is actually discharged at the anode is $$\mathrm{OH^-}$$ / $$\mathrm{H_2O}$$.)
  $$\mathrm{4OH^- \rightarrow O_2(g) + 2H_2O(l) + 4e^-}$$

Step 3 – Stoichiometry.
The balanced overall equation produces 2 mol of $$\mathrm{H_2}$$ for every 1 mol of $$\mathrm{O_2}$$, i.e. mole ratio $$\mathrm{H_2:O_2}=2:1$$.

Step 4 – Mole–volume relationship.
By Avogadro’s law, equal moles of any gases at the same temperature and pressure occupy equal volumes. Therefore a $$2:1$$ mole ratio translates directly to a $$2:1$$ volume ratio.

Step 5 – Observation in the apparatus.
The test tube placed over the cathode collects twice the volume of gas as the test tube over the anode. The gas with the larger volume is hydrogen; the one with the smaller volume is oxygen.

When an iron nail is immersed in an aqueous solution of copper(II) sulphate, a displacement reaction takes place because iron lies above copper in the reactivity series.

Step 1 – Write the chemical equation
$$\mathrm{Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s)}$$

Step 2 – Compare the positions in the reactivity series

• Iron, $$\mathrm{Fe}$$, is more reactive than copper, $$\mathrm{Cu}$$.
• A more reactive metal can displace a less reactive metal from its salt solution.

Step 3 – Explain the observed colour change

• Before the reaction: the solution contains $$\mathrm{Cu^{2+}}$$ ions, producing the characteristic blue colour of $$\mathrm{CuSO_4}$$.
• During the reaction: iron atoms lose two electrons and go into solution as $$\mathrm{Fe^{2+}}$$ ions;
  $$\mathrm{Fe \rightarrow Fe^{2+} + 2e^-}$$
• Simultaneously, the released electrons reduce the $$\mathrm{Cu^{2+}}$$ ions to copper metal, which deposits as a reddish-brown layer on the nail;
  $$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$$
• After the reaction: the solution now mainly contains $$\mathrm{Fe^{2+}}$$ ions. An aqueous solution of $$\mathrm{FeSO_4}$$ is pale green, so the blue colour gradually fades and changes to light green.

Hence, the colour change occurs because blue $$\mathrm{Cu^{2+}}$$ ions are replaced by pale-green $$\mathrm{Fe^{2+}}$$ ions as iron displaces copper from copper sulphate.

Step 1 – Recall the definition
A double-displacement (or metathesis) reaction is one in which two ionic compounds exchange their positive and negative ions to form two new compounds.

Step 2 – Choose suitable reactants
Take aqueous solutions of barium chloride and sodium sulphate. Their ions are:

• $$\mathrm{BaCl_2(aq)}: \;Ba^{2+},\;Cl^-$$
• $$\mathrm{Na_2SO_4(aq)}: \;Na^+,\;SO_4^{2-}$$

Step 3 – Write the skeletal equation
$$\mathrm{BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + NaCl}$$

Step 4 – Exchange the ions
The cation $$\mathrm{Ba^{2+}}$$ pairs with the anion $$\mathrm{SO_4^{2-}}$$ to give $$\mathrm{BaSO_4}$$, while $$\mathrm{Na^+}$$ pairs with $$\mathrm{Cl^-}$$ to give $$\mathrm{NaCl}$$.

Step 5 – Balance the equation

1. Count atoms in the skeletal form:
   Ba:1  Cl:2  Na:2  S:1  O:4
2. Product side currently has Ba:1  Cl:1  Na:1  S:1  O:4
3. To equalise Na and Cl, put coefficient 2 in front of $$\mathrm{NaCl}$$:
   $$\mathrm{BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2\,NaCl}$$
4. Now each element has identical counts on both sides, so the equation is balanced.

Step 6 – State the physical changes
$$\mathrm{BaSO_4}$$ is insoluble in water and appears as a white precipitate, confirming that a reaction has occurred.

Final balanced double-displacement reaction
$$\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s)\downarrow + 2\,NaCl(aq)}$$

Step 1 – Write the balanced equation
$$\mathrm{2PbO(s)+C(s)\;\longrightarrow\;2Pb(s)+CO_2(g)}$$

Step 2 – Recall the definitions

• Oxidation: gain of oxygen / loss of electrons.
• Reduction: loss of oxygen / gain of electrons.

Step 3 – Identify the species that gain or lose oxygen

• $$\mathrm{C}$$ gains two atoms of oxygen to give $$\mathrm{CO_2}$$  →  carbon is oxidised.
• $$\mathrm{PbO}$$ loses one atom of oxygen to give $$\mathrm{Pb}$$  →  lead oxide is reduced.

Step 4 – Test each statement

Step 5 – Pick the option that lists all incorrect statements
Incorrect statements: (a) and (b).
Option (i) matches this set.

First, let us recall the characteristic features of the four reaction types mentioned in the options:

• Combination reaction: Two or more reactants combine to give a single product, e.g. $$ ext{A} + ext{B} \rightarrow ext{AB}$$.
• Decomposition reaction: A single compound breaks down into two or more simpler substances, e.g. $$\text{AB} \rightarrow \text{A} + \text{B}$$.
• Double-displacement (double-decomposition) reaction: Ions are exchanged between two compounds to form two new compounds, e.g. $$\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$$.
• Displacement (single-displacement) reaction: A more reactive element displaces a less reactive element from its compound, e.g. $$\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$$.

The given chemical equation is

$$\mathrm{Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe}$$

Let us analyse it step by step.

1. The reactants are $$\mathrm{Fe_2O_3}$$ (iron(III) oxide) and $$\mathrm{Al}$$ (aluminium—an element).
2. The products are $$\mathrm{Al_2O_3}$$ (aluminium oxide) and $$\mathrm{Fe}$$ (iron—an element).
3. Notice that aluminium, which is a more reactive metal than iron (check the reactivity series), takes the place of iron in the oxide. In other words, aluminium removes oxygen from iron(III) oxide, producing aluminium oxide and leaving elemental iron behind.
4. This exactly matches the general pattern of a single displacement reaction: $$\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$$, where
   • $$\text{A}$$ corresponds to $$\mathrm{Al}$$,
   • $$\text{BC}$$ corresponds to $$\mathrm{Fe_2O_3}$$, and
   • the products $$\text{AC}$$ and $$\text{B}$$ correspond to $$\mathrm{Al_2O_3}$$ and $$\mathrm{Fe}$$, respectively.
5. Because only one element (iron) is displaced from its compound by another (aluminium), it is not a double-displacement reaction. Also, two reactants are giving two products, so it cannot be a combination or decomposition reaction.

Therefore, the given reaction is a displacement reaction.

Metals that are placed above hydrogen in the reactivity series displace hydrogen from dilute acids to form a salt and liberate hydrogen gas.

Iron lies above hydrogen, so it will react with dilute hydrochloric acid.

Write the skeletal equation first:

$$\mathrm{Fe + HCl \rightarrow FeCl_2 + H_2}$$

To balance the chlorine and hydrogen atoms, place the coefficient 2 before $$\mathrm{HCl}:$$

$$\mathrm{Fe + 2\,HCl \rightarrow FeCl_2 + H_2\uparrow}$$

The balanced equation shows that the reaction produces iron(II) chloride $$\mathrm{(FeCl_2)}$$ and hydrogen gas $$\mathrm{(H_2)}.$

Hence, the correct option is (a).

Step 1 ‒ Meaning of a balanced chemical equation

An equation is said to be balanced when the number of atoms of each element present in the reactants side equals the number of atoms of the same element present in the products side.

Example (unbalanced): $$\mathrm{H_2 + O_2 \;\to\; H_2O}$$

• Reactant side: $$\mathrm{H}=2$$ atoms, $$\mathrm{O}=2$$ atoms
• Product side: $$\mathrm{H}=2$$ atoms, $$\mathrm{O}=1$$ atom

Because oxygen is not equal on the two sides, the equation is not balanced.

After balancing: $$\mathrm{2H_2 + O_2 \;\to\; 2H_2O}$$

• Reactant side: $$\mathrm{H}=4$$; $$\mathrm{O}=2$$
• Product side: $$\mathrm{H}=4$$; $$\mathrm{O}=2$$

Now every element shows the same count on both sides, so the equation is balanced.

Step 2 ‒ Why must equations be balanced?

1. Law of conservation of mass. Atoms (and hence mass) can neither be created nor destroyed in an ordinary chemical reaction. Therefore the total mass—and, atom by atom, the total number of each element—must remain unchanged: $$\displaystyle \text{mass}_{\text{reactants}} = \text{mass}_{\text{products}}$$ Balancing an equation enforces this natural law.
2. Correct stoichiometric information. The numerical coefficients obtained while balancing indicate the exact, smallest whole-number mole ratio in which reactants combine and products form. Without balancing, quantitative calculations (e.g. determining how much reactant is needed or how much product will form) would be impossible or wrong.

Hence every chemical equation must be balanced before it can be used reliably in theory or practice.

During a chemical reaction the energy stored in the bonds of the reacting substances (reactants) is rearranged to form the products. The accompanying heat change is expressed by the enthalpy change $$\Delta H$$.

1. Exothermic reactions

• Meaning : Heat is released to the surroundings; therefore $$\Delta H < 0$$.
• Energy profile : (draw reactants at a higher energy level than products; the vertical drop shows “heat released”).
• Illustrative equations
• Combustion of carbon:
  $$\mathrm{C + O_2 \rightarrow CO_2 + 393.5\,kJ}$$
• Slaking of lime:
  $$\mathrm{CaO + H_2O \rightarrow Ca(OH)_2 + 65.2\,kJ}$$

2. Endothermic reactions

• Meaning : Heat is absorbed from the surroundings; therefore $$\Delta H > 0$$.
• Energy profile : (draw reactants lower than products; the vertical rise shows “heat absorbed”).
• Illustrative equations
• Formation of nitric oxide:
  $$\mathrm{N_2 + O_2 + 180.5\,kJ \rightarrow 2NO}$$
• Thermal decomposition of limestone:
  $$\mathrm{CaCO_3 + 178\,kJ \rightarrow CaO + CO_2}$$

Hence : Exothermic reactions liberate heat ($$\Delta H < 0$$) while endothermic reactions absorb heat ($$\Delta H > 0$$).

Step 1 – Recall the definition of an exothermic reaction
A chemical reaction in which energy is released to the surroundings, usually in the form of heat, is called an exothermic reaction. In such reactions the enthalpy (heat content) of the products is less than that of the reactants, so $$\Delta H < 0$$.

Step 2 – Write the overall chemical equation for respiration
During respiration the simple sugar glucose, $$\mathrm{C_6H_{12}O_6}$$, reacts with oxygen taken in from the air and is converted to carbon dioxide and water:

$$\mathrm{C_6H_{12}O_6 + 6\,O_2 \;\longrightarrow\; 6\,CO_2 + 6\,H_2O}$$

Step 3 – Show that energy is produced
Measurements in the laboratory give the enthalpy change for this reaction approximately as

$$\Delta H \approx -2800\,\text{kJ\,mol}^{-1}$$

The negative sign means that about $$2800\,\text{kJ}$$ of heat is released when one mole (180 g) of glucose is oxidised. This heat maintains the body temperature of warm-blooded organisms and supplies energy for all life processes such as muscle contraction, protein synthesis, active transport, etc.

Step 4 – Relate the heat release to the definition
Because the oxidation of glucose gives out a large amount of energy to the surroundings and has $$\Delta H<0$$, it satisfies the definition of an exothermic reaction.

Therefore respiration is regarded as an exothermic chemical reaction.

Step 1 ‒ Recall the two reaction types

• Combination reaction: two (or more) simple substances react to give one single product.
• Decomposition reaction: one single compound breaks down to give two (or more) simpler substances.

Step 2 ‒ Explain why they are opposites

In a combination reaction the number of chemical species decreases (many → one), whereas in a decomposition reaction the number of chemical species increases (one → many). Thus the direction of chemical change is exactly reversed, so decomposition reactions are called the opposite of combination reactions.

Step 3 ‒ Illustrate with paired equations

Because the second equation regenerates exactly the reactants consumed in the first, it shows clearly that decomposition is the reverse — the “opposite” — of combination.

In a decomposition reaction a single compound splits into two or more simpler substances. The energy needed for this splitting may be supplied in different forms, for example heat (thermal decomposition), light (photochemical decomposition) or electricity (electrolytic decomposition). Below we write one balanced chemical equation for each case.

1. Decomposition by heat (thermal decomposition)
   When calcium carbonate is strongly heated it decomposes into calcium oxide and carbon dioxide:
   $$\mathrm{CaCO_3\,(s)} \xrightarrow{\;\Delta\;} \mathrm{CaO\,(s)} + \mathrm{CO_2\,(g)}$$
2. Decomposition by light (photolysis)
   White silver chloride turns grey in sunlight because it breaks up into silver metal and chlorine gas:
   $$\mathrm{2AgCl\,(s)} \xrightarrow{\;\text{sunlight}\;} 2\mathrm{Ag\,(s)} + \mathrm{Cl_2\,(g)}$$
3. Decomposition by electricity (electrolysis)
   Passing electric current through water acidified with a small amount of $$\mathrm{H_2SO_4}$$ splits it into hydrogen and oxygen gases:
   $$\mathrm{2H_2O\,(l)} \xrightarrow{\;\text{electricity}\;} 2\mathrm{H_2\,(g)} + \mathrm{O_2\,(g)}$$

Step 1 – Recall the meaning of a chemical reaction
Whenever two or more substances interact to form new substance(s) with different properties, we say a chemical reaction has taken place. The skeletal chemical equation is then balanced to respect the Law of Conservation of Mass.

Step 2 – Define a displacement reaction
In a displacement reaction a more reactive free element pushes (displaces) another element out of a compound.
General form (one element + one compound): $$\mathrm{A + BC \;\rightarrow\; AC + B}$$ where $$\mathrm{A}$$ is more reactive than $$\mathrm{B}$$.

Worked example 1
Replace copper from copper(II) sulphate solution using iron filings.

• Write the reactants: $$\mathrm{Fe}$$ (iron) and $$\mathrm{CuSO_4}$$ (copper sulphate).
• Predict products: iron will take the place of copper, giving $$\mathrm{FeSO_4}$$ and free copper.
• Balanced equation: $$\mathrm{Fe + CuSO_4 \;\rightarrow\; FeSO_4 + Cu}$$.

Worked example 2
Zinc metal put into hydrochloric-acid solution:

• General idea: zinc is more reactive than hydrogen, so it displaces hydrogen from $$\mathrm{HCl}$$.
• Balanced equation: $$\mathrm{Zn + 2HCl \;\rightarrow\; ZnCl_2 + H_2 \uparrow}$$.

Step 3 – Define a double-displacement reaction
In a double displacement reaction (also called a double decomposition reaction) two ionic compounds in solution exchange their positive and negative ions with each other.
General form (compound + compound): $$\mathrm{AB + CD \;\rightarrow\; AD + CB}$$.

Worked example 3 – precipitation type

• Reactants: aqueous sodium sulphate and aqueous barium chloride.
• Ions exchanged: $$\mathrm{Na^+}$$ exchanges with $$\mathrm{Ba^{2+}}$$, $$\mathrm{SO_4^{2-}}$$ exchanges with $$\mathrm{Cl^-}$$.
• Balanced equation: $$\mathrm{Na_2SO_4(aq) + BaCl_2(aq) \;\rightarrow\; BaSO_4(s) \downarrow + 2NaCl(aq)}$$ (white precipitate of $$\mathrm{BaSO_4}$$).

Worked example 4 – neutralisation type

• Reactants: aqueous hydrochloric acid and aqueous sodium hydroxide.
• Balanced equation: $$\mathrm{HCl(aq) + NaOH(aq) \;\rightarrow\; NaCl(aq) + H_2O(l)}$$.

Step 4 – Tabulate the differences for quick revision

Step 5 – State the final answer
A displacement reaction involves the replacement of one element by a more reactive element, whereas a double displacement reaction involves mutual exchange of ions between two compounds. Representative balanced equations are:

• Displacement: $$\mathrm{Fe + CuSO_4 \;\rightarrow\; FeSO_4 + Cu}$$
• Double displacement: $$\mathrm{Na_2SO_4 + BaCl_2 \;\rightarrow\; BaSO_4 \downarrow + 2NaCl}$$

The question asks for the balanced chemical equation that represents the recovery of silver metal (Ag) from an aqueous solution of silver nitrate (AgNO₃) when copper metal (Cu) is introduced.

Step 1 ─ Identify the type of reaction
Copper is placed above silver in the reactivity series of metals, so metallic copper can displace silver from its salt solution. This is a single-displacement (substitution) reaction.

Step 2 ─ Write the word equation
Copper + Silver nitrate → Copper(II) nitrate + Silver

Step 3 ─ Convert to chemical formulas
Cu + AgNO₃ → Cu(NO₃)₂ + Ag

Step 4 ─ Balance the equation

• Count atoms on both sides. Initially: left – 1 Cu, 1 Ag, 1 NO₃; right – 1 Cu, 1 Ag, 2 NO₃.
• There are two nitrate ions on the right but only one on the left. Multiply AgNO₃ by 2.

$$\mathrm{Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag}$$

Re-check atom balance: Cu (1 = 1), Ag (2 = 2), NO₃ (2 = 2). The equation is now balanced.

Conclusion
The balanced chemical equation for the recovery of silver from silver nitrate solution using copper is:
$$\mathrm{2AgNO_3(aq) + Cu(s) \rightarrow Cu(NO_3)_2(aq) + 2Ag(s)}$$

Definition of a precipitation reaction

A precipitation reaction is a type of chemical reaction that occurs in aqueous solution in which two soluble salts (or a salt and an acid/base) react to form one product that is insoluble in water. The insoluble product separates out of the mixture as a solid called a precipitate. The appearance of this solid (often observed as cloudiness or settled solid) is the visual evidence that a precipitation reaction has taken place.

In symbolic form we can write (aq = aqueous, s = solid):
$$\text{Soluble salt}_1\,(\text{aq}) + \text{Soluble salt}_2\,(\text{aq}) \;\longrightarrow\; \text{Insoluble solid (precipitate)}\,(s) + \text{(possible) soluble salt}\,(\text{aq})$$

Why the precipitate forms

When the ions from the two solutions are brought together, they may combine to give a compound whose solubility in water is very low (its solubility product is exceeded). As soon as the ionic product exceeds this low limit, the excess compound comes out of solution as a solid.

Illustrative examples

1. Reaction of sodium sulphate with barium chloride

   $$\mathrm{Na_2SO_4\,(aq)} + \mathrm{BaCl_2\,(aq)} \;\longrightarrow\; \mathrm{BaSO_4\,(s)} \downarrow + \; 2\,\mathrm{NaCl\,(aq)}$$

   • Both $$\mathrm{Na_2SO_4}$$ and $$\mathrm{BaCl_2}$$ are highly soluble in water.
   • $$\mathrm{BaSO_4}$$ is almost completely insoluble, so it precipitates as a dense white solid.
   • The downward arrow (↓) is used to indicate the precipitate.

2. Reaction of lead(II) nitrate with potassium iodide

   $$\mathrm{Pb(NO_3)_2\,(aq)} + 2\,\mathrm{KI\,(aq)} \;\longrightarrow\; \mathrm{PbI_2\,(s)} \downarrow + 2\,\mathrm{KNO_3\,(aq)}$$

   • $$\mathrm{PbI_2}$$ is insoluble and separates as a bright yellow precipitate.
   • $$\mathrm{KNO_3}$$ remains in solution because it is soluble.

Key observations that confirm a precipitation reaction

• Sudden appearance of turbidity (cloudiness) in the reaction mixture.
• Formation of a coloured or white solid that eventually settles at the bottom of the container.
• No gas bubbles or large temperature change by themselves; the main evidence is the solid.

This is what is meant by a precipitation reaction, accompanied by two typical examples that Class 10 students commonly perform in the laboratory.

When the shiny brown metal is heated in air, it reacts with the oxygen present in the atmosphere and a black layer is deposited on its surface.

Step 1 – Identify the brown metal
Among the common metals of Class 10 syllabus, copper is distinctly brown and lustrous. Hence element $$X$$ is copper.

Step 2 – Write the reaction with oxygen
Copper combines with oxygen on heating to give copper(II) oxide, which is black in colour:

$$\mathrm{2Cu\,(s) + O_2\,(g) \;\xrightarrow{\;heat\;} 2CuO\,(s)}$$

Step 3 – Identify the black solid
The product $$\mathrm{CuO}$$ (copper(II) oxide) is a black solid that coats the surface of the metal.

Therefore,

• Element $$X$$  :  Copper (Cu)
• Black compound formed  :  Copper(II) oxide, $$\mathrm{CuO}$$

Iron objects lying in the open are exposed to the gases present in air and also to the water vapour always present in the atmosphere.

When iron $$\mathrm{(Fe)}$$ comes in contact with moist air it undergoes a slow chemical reaction called rusting (a type of corrosion):

$$4\mathrm{Fe}+3\mathrm{O_2}+x\mathrm{H_2O}\;\longrightarrow\;2\mathrm{Fe_2O_3}\cdot x\mathrm{H_2O}$$

The reddish–brown flaky substance $$\mathrm{Fe_2O_3}\cdot x\mathrm{H_2O}$$ is called rust. Rusting weakens the article and finally destroys it.

Paint is a non-reactive, water-repelling layer. When we coat an iron article with paint, the paint layer:

• cuts off direct contact between iron and atmospheric oxygen,
• prevents water vapour from reaching the metal surface.

Because both $$\mathrm{O_2}$$ and $$\mathrm{H_2O}$$ are necessary for the above rusting reaction, blocking either of them effectively stops the reaction. Thus the painted coating protects iron from corrosion and prolongs the life of the article.

Food items such as potato chips, biscuits, fried snacks and other preparations rich in oils or fats can spoil when they stand in contact with the oxygen present in air.

Step 1 – Recall what happens to oils/fats in air
Oils and fats are made up of long-chain hydrocarbons that are easily oxidised. In ordinary air, the reaction proceeds slowly:

$$\mathrm{Oil\;/\;fat + O_2 \longrightarrow\; oxidised\; products}$$

These oxidised products are usually aldehydes, ketones and short-chain organic acids that have an unpleasant smell and taste. The slow oxidation that spoils oils and fats is called rancidity.

Step 2 – How can we slow or stop oxidation?
We need to remove or drastically reduce the amount of the reactant oxygen around the food. One practical way is to fill ("flush") the packet with a gas that is

• chemically inert (does not react with the food), and
• readily available and inexpensive.

Step 3 – Why nitrogen?
Nitrogen, $$\mathrm{N_2}$$, fulfils both conditions:

• It is highly unreactive under ordinary conditions because of its strong $$\mathrm{N \equiv N}$$ triple bond.
• It makes up about 78 % of the atmosphere, so it is easy and cheap to obtain.

When a packet is flushed with nitrogen, the oxygen concentration inside becomes so low that the oxidation of fats slows to a negligible rate, thereby preventing rancidity and extending the shelf life of the food.

Conclusion
Oil- and fat-containing food items are flushed with nitrogen to displace oxygen, thus preventing the oxidation (rancidity) of the fats and keeping the food fresh for a longer time.