Chemical Equations

Introduction:

Changes:

Changes are classified into two types, they are Physical change and Chemical change.

Physical Change:

• Physical change is the change in which there is no formation of a new substance.
• Physical changes are Temporary changes and they can be reversed.

Example:1. Melting of wax

               2. Ice melting and water vapour formation. 

               Eg: On heating, Ice -----> water ------> water vapour

Chemical Change:

• A chemical change is a process in which a new substance is formed with new properties that are totally different from the actual substance.
• It is an irreversible and permanent change.

Example: Burning of crackers, the process of digestion, ripening of fruits, etc.

Observations regarding Chemical Change:

Following are some of the observations from which it can be concluded that a chemical change or reaction has taken place:

1. Change in colour
2. Change in physical state
3. Liberation of gas
4. Formation of precipitate
5. Change in temperature

Activity - 1: Reaction of Quick Lime (Calcium Oxide) with water (Change in Temperature)

Aim: To demonstrate the reaction of quick lime (Calcium Oxide) with water.

Materials Required: Beaker, Litmus paper, quick lime, water.

Procedure:

1.    Take a beaker and add 1gm of Calicium Oxide (Quick Lime) into it.

2.    Then add 10ml of water to the same water

3.    Now touch the beaker with your finger, it observes to be hot.

4.    The hotness of the beaker is due to the reaction of Calcium Oxide(quick lime) with water,         heat energy is released. Hence it is an exothermic reaction.

        CaO_(s) + H₂O_(aq) → Ca(OH)_2(aq) + Heat

5.    Calcium Oxide dissolves in water and produces a colourless solution. 

6.    A red litmus paper turns to a blue colour when it is dipped in water. This clearly indicates that the solution is basic in nature.

Result: In this activity, the reaction is exothermic and the solution formed is basic in nature.

Activity 1 Based Questions:

Q1: What do you notice?

A:    It is noticed that the beaker is hot when we touch it and the reason behind that is the action of calcium oxide (quick lime) with water, heat is emitted.

Q2: What is the nature of the solution?

A:    As red litmus paper turns out to be blue colour when dipped in the solution, it indicates that the solution is basic in nature.

Activity - 2: Formation of Precipitate

Aim: To show the formation of precipitation in some chemical reactions.

Materials Required: Two beakers, sodium sulphate (Na₂SO₄) solution, barium chloride (BaCl₂) solution and water.

Procedure:

    1. Take a beaker and add 100ml of water into it. Then dissolve a small amount of sodium sulphate (Na₂SO₄) into it.

   2. Take another beaker and add 100ml of water into it. Then dissolve a small amount of barium chloride (BaCl₂) into it.

     3. On observing the two beakers, both the solutions present in the beakers are colourless.

     4. Now add Na₂SO₄ solution to BaCl₂ solution.

     5. Then, the formation of a white coloured precipitate (BaSO₄) is formed.

         Na₂SO_4(aq)+BaCl_2(aq) → BaSO_4(s)↓+ 2NaCl_(aq)

Result: From this activity, it is concluded that some chemical reactions are characterized by the formation of a precipitate. This reaction is a double displacement reaction. 

Activity 2 Based Questions:

Q1:    What are the colours of the above solutions?

A:       Both the solutions are colourless.

Q2:    Can you name the solutions obtained by adding Na₂SO₄ solution to BaCl₂ solution?

A:       Barium Sulphate(BaSO₄)and Sodium Chloride(NaCl).

Q3:    Do you observe any change in mixing these solutions?

A:       Yes, a white coloured precipitate of Barium Sulphate is formed.

Activity - 3: Formation Hydrogen Gas

Aim: To show the formation of hydrogen gas by the action of dilute HCl on zinc.

Materials Required: Conical flask, dilute HCl, zinc granules, matchbox.

Procedure:

    1. Take some zinc granules in a conical flask and add 5ml of dilute HCl into it.

    2.  It can be observed that the bubbles formed when zinc granules were added to dilute HCl.

    3. Now place the burning match-stick near the mouth of the conical flask. The match-stick turns off with a pop-up sound which indicates the evolution of hydrogen gas.

    4. As heat is evolved during the reaction, the conical flask is turned out to be hot.

        Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)↑

Result: From this activity, we can conclude that the reaction is a single displacement reaction because zinc displaces hydrogen to produce zinc chloride and hydrogen gas.

Activity 3 Based Questions:

Q1: What changes do you notice?

A:    Hydrogen gas formation.

Q2: What happens to the burning match stick?

A:   Burning match-stick puts off with a pop-up sound when hydrogen gas is liberated.

Q3: Is there any change in temperature?

A:    Yes, the temperature increases.

Chemical Equation:

Def: A chemical equation is the symbolic representation of reactants and products involved in the reaction.

• Reactants: The substances which are involved in the chemical reaction are known as Reactants.
• Products: The new substance(s) which forms after a chemical reaction, known as Products.
• If the number of reactants and products are more than one in a chemical reaction, then they are indicated by the '+' sign between them.
• In any chemical equation, reactants are written on the left-hand side and products are written on the right-hand side.

Representation / Writing a chemical equation:

• The chemical equations can be represented or written in two forms namely Word Equation and Symbolic/formula equation. 
• The symbolic equation is easier to write than the words equation as word representation is much lengthy and time is taken.
• A compound is represented by using its chemical formulae that include symbols of the elements and it also uses subscript which indicates the number of atoms present in the compound.

         Example: Calcium oxide reacts with water and produces calcium hydroxide. The chemical equation for the mentioned reaction can be represented in the following forms:

• Word Equation: calcium oxide + water -------> calcium hydroxide

• Symbolic / formula equation: CaO + H₂O → Ca(OH)₂

Think and Discuss Question:

Q1: You have brushed the wall with an aqueous suspension of Ca(OH)₂. After two days the wall turned to white colour.

    (i) What are the steps involved in the whitewashing of the wall?

    (ii) Write the balanced chemical reactions using the appropriate symbols.

A: (i) Aqueous suspension of slaked lime (Ca(OH)₂)is not white. After applying on the wall, it reacts with carbon dioxide (CO₂) gas present in the air and forms calcium carbonate (CaCO3), water (H₂O). The resultant water gets evaporated and calcium carbonate which is quite white in colour gets stuck to the wall.

    (ii) Balanced chemical reaction:

         Ca(OH)_2(aq) + CO_2(g) = CaCO_3(s) + H₂O_(l)

        

Balancing Chemical Equations:

A chemical equation needs to be balanced to follow the law of conservation of mass. According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products formed should be equal to the mass of reactants consumed. This means atoms are neither be created nor destroyed during the chemical reaction.

Unbalanced/Skeleton Chemical Equation:

A chemical equation in which the number of atoms of each element on the reactant side is not equal to the number of atoms of the same element on the product side is known as an unbalanced chemical equation.

Example: Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl

In the above example, the number of atoms of Na and Cl is not equal on both sides.

Balanced Chemical Equation:

A chemical equation in which the number of atoms of each element on the reactant side is equal to the number of atoms of the same element on the product side is known as a balanced chemical equation.

Example: 2Mg + O₂ → 2MgO

Explanation of Steps involved in Balancing Chemical Equation through Example:

Hydrogen reacts with Oxygen and forms Water:

Step 1: Write an unbalanced chemical equation by using the chemical formula of all substances.

H₂ + O₂ → H₂O

Step 2: Compare the atoms of each element on both sides.

There exist 2 'O' atoms on LHS but there is only one 'O' atom on RHS. To balance it, add a coefficient of 2 to the reactant (H₂) and to the product water (H₂O). Thus equation can be rewritten as:

2H₂ + O₂ → 2H₂O

Step 3: Divide the coefficients of all the substances with a suitable number to get the lowest ratio of coefficients for reactants and products. But in the above equation, there is no common factor. Hence there is no requirement for the division of coefficients.

Step 4: Verification of balanced chemical equation.

Count the number of atoms of each element on both sides and verify it.

Therefore, 2H₂ + O₂ → 2H₂O is a balanced chemical equation.

Making Chemical Equation More Informative:

The chemical equation can be more informative if it is expressed with the following characteristics:

• Physical state
• Heat changes
• Gas evolution
• Formation of precipitate

Interpreting a Balanced Chemical Equation:

    A chemical equation provides information about the involvement of the number of molecules of reactants and products in a chemical reaction. From this equation, the relative masses of reactants and products can also be obtained, since the molecular masses are expressed in Unified Masses (U).

    The chemical equation also gives information about the molar ratios of reactants and products, if they are expressed in grams. If gases are involved, we can equate their masses to their volumes and calculate the volumes or those gases liberated at a given condition of temperature and pressure using the molar mass and molar volume relationship. We can also get the number of molecules and atoms of different compounds by using molar mass and Avagadro's number.

    From the equation, we can get the below-mentioned factors:

• mass - mass relationship
• mass - volume relationship
• volume - volume relationship
• mass - volume - number of molecules relationship etc.

Example Problems:

Q1: Al_(s) + Fe₂O₃ → Al₂O₃ + Fe

       (Atomic masses of Al=27U, Fe=56U, and O=16U)

        2Al_(s) + Fe₂O₃ → Al₂O₃ + 2Fe

        (2*27)U + (2*56+3*16)U → (2*27+3*16)U + (2*56)U

                    (or)

        54U + 160U → 102U + 112U

                    (or)

        54g +160g → 102g + 112g 

Suppose that you are asked to calculate the amount of aluminium required to get 1120kg of iron by the above reaction.

Sol: According to the balanced chemical equation,

        54g of Al required to get 112g of Fe

        

        i.e., 54g → 112g

        The amount of Al required to get 1120kg of Fe to be 'x' g

        i.e., 'x'g → 1120kg

        here 1120kg = 1120000g

        ⇒ 54g → 112g

            x g → 1120000g

        ⇒ x = 54 * 1120000 / 112

            x = 540000g

            x = 540kg

            Therefore the amount of aluminium required to get 1120kg of iron is 540kg.

Q2: Calculate the volume, mass and number of molecules of hydrogen liberated when 230g of sodium reacts with excess water at STP.

        (Atomic masses of Na=23U, O=16U, and H=1U)

        2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g)↑ is a balanced equation

        (2*23)U + 2(2*1+1*16)U → 2(23+16+1)U+ (2*1)U

        46U + 36U → 80U + 2U     

                        (or)

        46g + 36g → 80g + 2g

Sol: According to the balanced chemical equation,

Mass Calculation:

46g of sodium → 2g of hydrogen

230g of sodium → ?

= 230*2 / 46

= 10g

Therefore, hydrogen gas liberated from the reaction of 230g of sodium with excess water is 10g

Volume Calculation:

At STP, 2g of hydrogen occupies 22.4 litres

10g of hydrogen occupies ?

= 10*22.4 / 2

= 112 litres

Therefore, 10g of hydrogen occupies 112 litres at STP

The number of Molecules Calculation:

2g of hydrogen contains 6.623*10²³  number of molecules

10g of hydrogen contains?

= 10*6.623*10²³ / 2

= 3.01*10²⁴ molecules.

Therefore, 10g of hydrogen contains 3.01*10²⁴ molecules.

Q3. Calculate the volume and number of molecules of CO₂ liberated at STP. If 50g of CaCO₃ is treated with dilute HCl which contains 7.3g of dissolved HCl gas.

CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq) + H₂O_(l) + CO_2(g) is a balanced equation.

According to metric-stoichio equation, 100g of CaCO₃ reacts with 73g of HCl to liberate 44g of CO₂.

Sol: In the given problem, 50g of CaCO₃ is treated with 7.3g of HCl.

But, according to the metric-stoichio equation, 100g of CaCO₃ reacts with 73g of HCl.

As 50g of CaCO₃ is half of the 100g then it should be react with 36.5g of HCl, which is half of 73g, but it react with 7.3g of HCl only.

Hence the product CO₂ obtained depends upon the amount of HCl, which is in the least amount but not on the amount of CaCO₃ which is in excess amount.

Here, the reactant HCl which is in least amount is called limiting reagent, because it limits the amount of product to be formed. 

Mass Calculation:

73g of HCl → 44g of CO₂

7.3g of HCl → ?

= 7.3*44 / 7.3

= 4.4g

Therefore, 7.3g of HCl produce 4.4g of CO₂

Volume Calculation:

44g of CO₂ occupies 22.4 litres at STP

4.4g of CO₂ occupies ?

= 4.4*22.4 / 44

= 2.24 litres.

Therefore, 4.4g CO₂ of occupies 2.24 litres.

The number of Molecules Calculation:

44g of CO₂ contains 6.623*10²³ molecules

4.4g of CO₂ contains ?

= 4.4*6.623*10²³ / 44

= 6.623*10²² mol.

Therefore, 4.4g of CO₂ occupies 6.623*10²² molecules.