CHEMICAL REACTIONS BASIC CONCEPTS: REACTANTS AND PRODUCTS

CHEMICAL REACTIONS

CONTENTS:

1. Basic concept (Activation energy) and
2. Introduction to collision theory.
3. Factors affecting rate of chemical reactions.
4. Types of chemical reactions – Endothermic and Exothermic.

 

PERIOD 1: BASIC CONCEPTS: REACTANTS AND PRODUCTS

Chemical reactions can be represented simply on paper by chemical equations involving only formula and symbols e.g.

4NH_3(g) + O_2(g) 4NO_(g) + 6H₂O_(g)

Reactants                         Products

Reaction time: The total time taken for a particular reaction to take place.

Reaction rate: is the number of moles of reactions converted or product formed per unit time. Unit is moldm^-3s^-1

In all chemical reactions, existing bonds in the reactants particles must be broken first before new bonds can be formed to form products. The breakings of bonds require energy. An initial energy input is required to activate the reactant particles. This energy is the activation energy of the reaction. The activation energy of a reaction is equivalent to that energy barrier that must be overcome before bonds are broken to enable the reaction to occur. Hence, activation energy can be defined as the minimum amount of energy required for a reaction can occur.  When this activation is acquired by the reactant particles, they form complex particles of high energy content. This complex particle is known as the activated complex. Therefore, activated complex is an unstable molecule with high energy contents which gives the product It is unstable because of its high energy content and so will readily decompose to give the products or the reactants, depending on the nature if the reaction.

EVALUATION

Define activation energy.

 

 

 

PERIOD 2:   INTRODUCTION TO COLLISION THEORY

The collision theory was developed from the kinetic theory of gases to account for the influence of concentration and temperature on reaction rates. The theory is based on the following postulates:

1. Reactions occur as a result of the collision of reactant particles.
2. A reaction results only if collision attains certain minimum energy. This minimum energy is called activation energy.

iii. Collision will not give rise to a reaction unless the colliding particles are correctly oriented to one another.

1. The rate of reaction is proportional to the number of effective collisions. A collision is said to be effective if the energy of the colliding molecules is greater than or equal to the activation and the molecules are correctly oriented to one another.

The collision theory assumes that for a chemical reaction to occur there must be collisions between reactant particles.

EVALUATION

Explain the collision theory.

PERIOD 3: FACTORS AFFECTING THE RATE OF CHEMICAL REACTION

The rate at which reaction takes place will be affected by the following factors:

1. Nature of reactant: The chemical nature of the reactants taking part in the reaction influences the rate of reaction. For instance if iron, zinc and gold metals are placed in different beaker continually Hydro chloric acid, there will be rapid evolution in the beaker containing the acid and zinc metal wherein the beaker containing Iron metal and the acid there will be slower evolution of hydrogen gas and in the beaker containing gold metal and the acid solution there will be no reaction.
2. Effect of concentration / pressure (gases) of reactants: If reactant particles are crowded in a particular place, their frequency of collision will be faster than if the particles are far from one another. Thus, the more the concentration of reactant particles the higher the rate of reaction.

Pressure affects the concentration of gaseous reactant, the higher the pressure of gaseous reactants, the higher the frequency of collision of the particles the higher the rate of reaction.

3. Effect of surface area of the reactants: the more exposed the area of contact of reacting particles to each other the faster the rate of reaction. For solid reactants the exposed surface area must be increased by subdividing or breaking the solid into smaller pieces.

Where the reactants are gases, liquids or solids dissolved in solution, the thoroughness of mixing is of vital importance so as to ensure maximum contact between the reactants.

4. Effect of temperature of reaction mixture: The higher the temperature the higher the rate of reaction of most reactions. When the temperature of a chemical reaction is increased, the number of particles with energies equal to or greater than the activation energy increases. Also the kinetic energy of the reactant particles increase which implies increase in frequency of effective collision and increase in the rate of reaction.
5. Effect of the presence of light: Reactions whose rates are affected by light are called photochemical reactions. They are thus able to overcome the activation energy barrier and react rapidly by a chain reaction.

Examples: i. The decomposition of hydrogen peroxide

1. The reactions between methane and chlorine Photosynthesis

iii. The conversion of silver halides to grey metallic silver

1. The halogenations of alkanes
2. Effect of catalyst: catalyst will alter the rate of a chemical reaction but itself does not undergo any permanent change at the end of the reaction.

Catalysts that speed up the rate of a chemical reaction are called positive catalyst. They lower the activation energies of the reactant particles  by providing alternative pathway so more reactant particles are able to collide effectively to produce more products. E.g. Manganese(IV) oxide, MnO₂ catalysts is used in the production of oxygen thermal decomposition of KClO₃

EVALUATION

List and explain 5 factors that affect the rate of chemical reaction.

PERIOD 3: TYPES OF CHEMICAL REACTIONS: ENDOTHERMIC AND EXOTHERMIC REACTIONS

An endothermic reaction is one during which heat is absorbed from the surrounding. Most decomposition processes are endothermic reaction.  Dissolution of ammonium chloride, ammonium tetraoxosulphate(vi) etc are endothermic process. In endothermic reaction, the enthalpy of product(s)  is greater than that of reactant(s). change in enthalpy is equal to the sum of heat of product minus the sum of heat of reactant i.e △H =      –   . Where    is sum of heat of products and  is sum of heat of reactants. Enthalpy change in this regard is positive. i.e △H = +ve

E

Examples of Endothermic reaction

• N_2(g) + O_2(g) 2NO_(g)    ∆H= +180.6KJmol ^-1

Exothermic reaction is one during which heat is released to the surrounding. Examples of exothermic reaction is combustion  reaction. Dissolution of H₂SO₄,  NaOH,  KOH, neutralization  and AlCl₃ are exothermic process. In exothermic reaction, the enthalpy of product(s)  is less than that of reactant(s). change in enthalpy is equal to the sum of heat of product minus the sum of heat of reactant i.e △H =      –   . Where   is sum of heat of products and  is sumof heat of reactants. Enthalpy change in this regard is negative. i.e △H = -ve

• Example of exothermic reaction.

C_(s) + O_2(g)                 CO_2(g)

∆H = – 408KJmol ^–1

Note: The negative and the positive signs indicated in the above figures indicate exothermic and endothermic reactions respectively

Heat of formation: Heat of formatiom of a compound is the heat change that occurs when one mole of a compound is formed from its elements in their standard states. The standard condition are 25^Oc and 1 atm. The heat of formation of water is -286kJmol^-1. The equation is given below:  H_2(g)  +  ½O_2(g0  →  H₂O_(l)      ΔH_f = -286kJmol^-1

Heat of combustion: This is the heat evolved when one mole of a substance is completely burnt in oxygen under state condition. The heat of combustion of carbon is -393kJmol^-1

Example:

Calculate the heat of reaction in the equation below given that the heat of combustion of carbon and hydrogen are -393kJmol^-1 and -286kJmol^-1 respectively and heat of formation of butane is -125kJmol^-1.

C₄H_10(l)  +   O_2(g)    4CO₂  +  5H₂O_(l)

Solution

△H =      –

=    [-393+(-286)]  –   (-125+0)

=    -2877kJmol^-1

NOTE: The enthalpy of formation of an element is zero

EVALUATION

1. Differentiate between exothermic and endothermic reaction.

GENERAL EVALUATION

OBJECTIVE TEST:

1. Endothermic reaction may be defined as except

• Reaction in which heat is absorbed
• Reaction in which heat is released’
• Reaction in which H is positive
• Reactant than energy of the reactant.

2. Rate of chemical reaction depends on the following factors except.

• Rate at which gas is evolved
• Rate at which products are formed
• Rate at which colour of reactions change.
• Rate at which the reactants diminish.

3. The unit of rate of chemical reaction is (a) Moldm^-3S^-1 (b) Mol^-1S^-1 (c) Mol^-1 (d) Smol^-1
4. The rate of a reaction is proportional to the number of effective collisions occurring per second between the reactants. This statement is associated with the.

• Kinetic theory (b) rate law (c) atomic theory (d) collision theory

5. Which of the following statement is not true about the reaction between dilute hydrochloric acid and marble chips calcium trioxocarbonate(IV)?

• It is faster after three seconds than it is after ten seconds.
• It slows down with time.
• It eventually stops.
• It proceeds at a constant rate

 

 

ESSAY QUESTIONS

1. Use the arguments of the collision theory to explain how the following factors affect the rate of a chemical reaction. (i) concentration (ii) surface area
2. (a) State in three short sentences the main ideas of the collision theory.
3. State how the rate of a chemical reaction is affected by the following factors.

• Concentration (b) surface area

4. (i) What is a catalyst?

• Write a relevant equation to show the industrial process in which V₂O₅ was used as a catalysts.

5. Explain briefly how temperature affects the rate of a chemical reaction.

 

 

 

TOPIC: CHEMICAL REACTIONS

CONTENTS:

1. CHEMICAL EQUILIBRIUM:
2. CHARACTERISTICS OF A CHEMICAL SYSTEM AT EQUILIBRIUM,
3. LE CHATELIER’S PRINCIPLE AND FACTORS AFFECTING EQUILIBRIUM OF CHEMICAL REACTION
4. EQUILIBRIUM CONSTANT, KC

PERIOD 1:   CHEMICAL EQUILIBRIUM

Studies have shown that some reactions are such that all the reactants are never completely converted into products. In such reactions, the products are converted into the reactant as they are been formed.

A reaction is said to be reversible, if it can be made to proceed forward and backward, under a given set of conditions.

For a chemical reaction involving a gas to be reversible, it must be carried out in a closed vessel, i.e. an isolated or closed system. Two opposing arrows in a chemical equation represent a reversible reaction:

1. a) H₂O_(l) H₂O_(g)
2. b) N_2(g) + 3H_2(g) 2NH_3(g)
3. c) CaCO_3(s) CaO_(s)  + CO_2(g)

Such reactions usually have relatively low activation energies for the reverse reactions.

A reaction in which the products cannot readily be combined to give the reactants is said to be irreversible. In such a reaction, a single arrow is used pointing to the product(s), e.g. combustion reactions

1. a) C_(s) + O_2(g) CO_2(g)
2. b) 2Mg_(s) + O_2(g) 2MgO_(s)

Such reactions usually have relatively high activation energies for the reverse reaction.

 

REACTIONS AT EQUILIBRIUM

Dynamic Equilibrium

When the rates of forward and backward reactions in a reversible reaction are equal, the reaction is said to be at equilibrium.  At equilibrium, both the forward and backward reactions are still going on, but because there rates are the same, the concentrations of the reactants and products do no longer change with time.  Hence, chemical equilibrium is dynamic and not static.

How to Identify a Reaction at Equilibrium

At equilibrium, certain observable properties become constant depending on the type of reaction. Such properties include;

• Concentrations of the reactants and products;
• Pressure of gases;
• Density or intensity of colour of a solution;
• Temperature of the system.

EVALUATION

1. When is a reaction said to be reversible?
2. Describe a natural process that is reversible.

 

PERIOD 2: CHARACTERISTICS PROPERTIES OF A SYSTEM AT EQUILIBRIUM

1. Equilibrium is dynamic; that is,

• The reactants and products are present at equilibrium, i.e. equilibrium can be achieved from either direction:

H_2(g) + I_2(g)   2HI_(g)

2HI_(g)          H_2(g) + I_2(g) 

2. The concentrations of the reactants and the products, pressure, density or intensity of colour, and the temperature of the system are constant- depending on the type of the reaction.
3. The rate of forward and backward reactions are equal.
4. The equilibrium position is not affected by the presence of the catalyst, it only quickens the rate at which equilibrium is achieved.
5. At equilibrium, the free energy change, ∆G, of the reaction is zero.
6. A system at equilibrium will resist a change.

 

Information Obtained from an Equilibrium Equation

Consider a typical reversible reaction:

N_2(g) + 3H_2(g)    2NH_3(g)        ∆H  = – 92kJ

The following information can be obtained:

1. The two opposing reactions are occurring at the same time – as indicated by the opposing double arrow.
2. The two opposing reactions are at equilibrium.
3. The amounts, in moles, of the reactants and products at equilibrium are proportional to the stoichiometry in the balanced equation.
4. The forward reaction is exothermic (∆H is negative; hence, the backward (or reverse) reaction will be endothermic (∆H is positive), i.e.

2NH_3(g)     N_2(g) + 3H_2(g)     ∆H  = + 92kJ

EVALUATION

1. What is meant by an equilibrium system?
2. Give three properties of a chemical system at equilibrium.

 

LE CHATELIER’S PRINCIPLE

Le Chatelier’s principle stares that if an external constraint such as a change to temperature, pressure or concentration is imposed on a chemical system in equilibrium, the equilibrium will shift so as to annul or neutralize the constraint.

FACTORS AFFECTING EQUILIBIUM OF A CHEMICAL REACTION

1. EFFECT OF A CHANGE IN TEMPERATURE e.g. For the following systems at equilibrium

• 2SO_2(g) + O_2(g) ⇌2SO_3(g)H = – 395.7 KJmol ^–1

The forward equation is exothermic i.e. it involves increase in temperature whereas the backward equation is endothermic or it involves decrease in temperature.

If the temperature of the system is increased:

Increase in the temperature of the system will shift the equilibrium position o backward reaction, that is, reactant formation is favoured.

Increasing the temperature of the system will shift the equilibrium position to the left, favouring the backward reaction i.e. reactants formation (i.e. SO₂ and O₂).

If the temperature of the system is decreased:

Decrease in the temperature of the system will cause the equilibrium position to shift to the right favouring the forward reaction, that is, product formation.

• N2(g) + O₂⇌ 2NO(g) ΔH   =   +90.4KJmol ^–1

When the system is at equilibrium, the forward reaction is endothermic and the backward reaction is exothermic.

If the temperature of the system is increased:

Increase in the temperature of the system will cause the equilibrium position to shift to the right favouring forward reaction. i.e. product formation.

If the temperature of the system is decreased:

Decrease in the temperature of the system will cause the equilibrium position to shift to the left favouring the backward reaction.

2. EFFECT OF A CHANGE IN PRESSURE:

For a change in pressure to affect a chemical system in equilibrium.

• One of the reactants in the or products in the reversible reaction must be gaseous.
• The total number of gaseous molecules on the left side of the equation must be different from the total number of moles of gaseous molecules on the right side.

For the following systems at equilibrium

• 3H₂(g) + N₂(g) ⇌       2NH₃(g)

The forward reaction is from high pressure to low pressure and the backward reaction is from low pressure to high pressure.

Increase in pressure of the system:

When the pressure of the system is increased the equilibrium position will shift to the right favouring the forward reaction that is, the product formation.

Decrease in the pressure of the system:

When the pressure of the system is decreased the equilibrium position will shift to the left favouring the backward reaction that is reactant formation.

• N₂O₄(g) ⇌       2NO₂(g)

When the system is at equilibrium, the forward reaction involves an increase in the pressure of the system and the backward reaction involved a decrease in the pressure of the system.

If the pressure of the system is decreased:

A decrease in the pressure of the system will cause the equilibrium position to shift to the right favouring the forward reaction that is, product formation.

If the pressure of the system is increased:

An increase in the pressure of the system will cause the equilibrium position to shift to the left favouring the backward reaction that is, reactant formation

• H₂(g) + I₂(g) ⇌      2HI(g)

A change in pressure will not affect this system at equilibrium because the number of moles of the reactants is the same as the number of moles of the product.

• 3Fe(s) + 4H₂O(g) ⇌     Fe₃O₄(s) +4H₂(g)

A change in pressure will not affect this system at equilibrium because the number of moles of gaseous reactant and product is the same.

3. EFFECT OF A CHANGE IN CONCENTRAION

E.g. 3Fe(s) + 4H₂O(g)         ⇌      Fe₃O4(s) + 4H₂(g)

If the concentration of any of the reactants is increased the equilibrium position will shift to the right but if it is decreased it will shift to the left.

Also removal of any of the reactant or product will also cause the equilibrium position to shift. E.g. removal of hydrogen gas from the system will cause the equilibrium position to shift to the right.

4.EFFECT OF A CATALYST

Addition of a catalyst to a system in equilibrium will not affect the equilibrium position. Instead, addition of catalysts will only make equilibrium state to be reached attained faster.

EQUILIBRIUM CONSTANT, K_C

In a reversible reaction, there is a fixed relationship, at constant temperature, between the concentrations in moldm^-3 of the products and the reactants. This relationship is accounted for in the study carried out by Guldberg and Waage. This law, known as the Law of Mass Action states that at constant temperature, the rate of reaction is proportional to the active masses (concentration raise to the power of its coefficient) of each of the reactants.

EXPRESSION OF EQUILIBRIUM CONSTANT, K_C

In a reaction: aA + bB       ⇌      cC +dD

The equilibrium constant K_C is expressed as follows:

K_C   =     . In the cause of a chemical reaction, it the concentration of gases and aqueous species the normally change. The concentrations of solid and pure liquid are always constant, hence, thay cannot appear in equilibrium constant expression.

Note: The species must be written in a square bracket as shown above e.g. in the reaction:

• 2SO_2(g) + O_2(g)⇌ 2SO_3(g)

The equilibrium expression

K_C =

• N_2(g) +3H_2(g)⇌ 2NH_3(g)

K_C =

(c).   CaCO_3(s)⇌  CaO_(s)   +   CO_2(g)

K_C =  ]

IMPORTANCE OF K_C

KC determines the yield of chemical reaction at equilibrium. If Kc is greater than one, it more product is formed 1.e forward reaction is favoured. But if Kc is less than one, it less product is formed (poor yield) i.e backward reaction is favoured. Kc is temperature dependent and change with change in temperature.

RELATIONSHIP BETWEEN KC AND △G

Free energy (G) is the energy for doing useful work. Gibb’s free energy cannot be measured directly but the change in free energy is usually measured. Change in free energy determines the spontaneity of a given reaction. For a spontaneous reaction, △G is negative.

△G  =  -RTInKc. Where n = no of mole of specie,  R = gas constant (8.314Jmol^-1K^-1 and T = temperature in Kelvin

EVALUATION:

1. Write the expression for the equilibrium constant K_C for the following

• 2H_2(g) + O_2(g)⇌ 2H₂O_(g)

 

• 2CO_(g) + O_2(g) ⇌2CO_2(g)

GENERAL EVALUATION:

OBJECTIVE TEST:

1. Two boys balanced in a sea-saw game is an example of

(a) Static equilibrium (b) dynamic equilibrium (c) homogenous equilibrium (d) mutual equilibrium

2. In the reaction, 2SO_2(g) + O_2(g)⇌2SO_3(g)

(a) The reaction is in physical equilibrium

(b) Contact process

(c) The reaction is the rate determining step

3. In the decomposition ofCaCO_3(s)⇌CaO(_s) + CO_2(g)

The reaction will attain dynamic equilibrium

• When the reaction is in open system
• When the reaction is heated strongly
• When the reaction is in a closed system
• When the reaction is catalysed

4. Factors affecting equilibrium reaction include the following except.

(a) Pressure (for solid system)

(b) Concentration

(c) Temperature

(d) Pressure (for gaseous system)

5. All except one is not a condition for considering pressure in an equilibrium system.

(a)There must be concentration gradient between the reactants and the products.

(b) The reactants could be gas while the products may be solids

(c) Both products and reactants must be gaseous.

(d) The reaction takes place in a closed system

ESSAY QUESTIONS

1. What is the effect of each of the following on the equilibrium position of the system indicated below?

A_(s) + B_(g)⇌C_(g) Δ H = – xKLmol^-1

• Cooling the system
• Removing
• As soon as it is formed

2. Write the expression of equilibrium constant for

2AB_2(g) + B_2(g)⇌  2AB_3(g)

3. Consider the reaction represented by the equation

N₂O_4(g)⇌ 2NO_2(g)△H = +57.2KJmol^-1

• When is the reaction said to be at equilibrium
• Mention two conditions that can favour the forward reaction.
• Name the principle involved in (b) above
• Addition of catalyst.

 

 

 

 

TOPIC: STOICHIOMETRY

1. S.I. UNIT OF QUANTITIES: LENGTH, MASS VOLUME ETC
2. BASIC CONCEPTS: MOLE, MOLAR QUANTITIES, MOLALITY, STANDARD TEMPERATURE AND PRESSURE (S.T.P.)
3. CALCULATIONS INVOLVING MASS AND VOLUME
4. MORE CALCULATIONS ON MASS AND VOLUME RELATIONSHIP.

 

PERIOD 1:  S.I. UNIT OF QUANTITIES: LENGTH, MASS VOLUME ETC

S.I. UNITS OF QUANTITIES.

S.I.  unit means international system [used to describe units of measurements from French ‘system international’] S.I. units

For example

QUANTITY	SYMBOL	UNITS
Length	L	m, cm
Mass	M	g or kg
Molar mass	M	gmol-1
Volume	V	cm3, dm3
Amount	N	mol
Molar concentration	L	mol.dm-3
Mass concentration	P	gdm-3
Avogadro’s law	L	mol-1
Avogadro’s number	N	——
N.B molar mass	M	g.mol-1

 

Relationship between quantities

(a) M =   i.e molar mass =

Molar volume  =

(b) Molar volume =  =

(c) Avogadro’s constant = L =  ==

TYPES OF MASS AND VOLUME PROBLEMS

There are two types of mass and volume relationship problems namely:

1. Mass- mass problem:

In this type of problem, the reacting mass of substance and its molar mass is usually given and we may be required to determine the mass of another substance produced and its molar mass also given. The reversed form of this is when the mass of another substance product produced is given and one is asked to find the mass of the substance which produced it.

2. Mass- gas volume problem:

In this case, you may be given a certain reacting mass of a reactant and required to find the volume of another substance produced at S.T.P. or under any other conditions. The reverse of this type is when you are given the volume of a certain gas liberated at S.T.P or other conditions and the gas produced.

Note: This method is related to simple proportion as in mathematics.

3. Gas volume – gas volume problem

In this class of reactions, only the volumes of the reactants and products are involved in the calculations.

4. Mass – Liquid volume problem

In this type, the mass I given and the volume of liquid has to be determined.

5. Liquid volume – volume problem

In this group of calculations, one volume is given while another is to be found.

EVALUATION

1. What is a mole of a chemical substance?
2. Mention two elementary units of a chemical substance.
3. What mass, in grams, of aluminium contains the same number of atoms as in 12.0 g of carbon? (C = 12.0, Al = 27.0).

PERIOD 2:BASIC MOLE CONCEPT

Definition: A mole is the amount of a chemical substance that contains as many elementary particles as there atoms in 12.0 g of C-12 isotope. It means that a mole of an element are collections of its atoms such that the total mass in grams, is numerically equal to its relative atomic mass. That is: if 1 mole of carbon atoms weighs 12.0 g, then, 1 mole of oxygen atoms will weigh 16.0 g; 1 mole of sodium atoms will weigh 23.0 g, e.t.c.  The number of elementary units or particles in 1 mole of a chemical substance is constant, and it is called Avogadro’s constant, L. It is numerically equal to 6.02 x 10²³=. Hence:

• 1 mole of carbon atoms weighs 12.0 g and contains 6.02 x 10²³
• 1 mole of sodium atoms weighs 23.0 g and contains 6.02 x 10²³
• 1 mole of sulphur atoms weighs 32.0 g and contains 6.02 x 10^{23 ­}
• The molar mass of every element contains the number of atoms.
• The molar mass of every compound contains the number of molecule
• One faraday of electricity (9630 coulombs) is said to have flowed past a point in a circuit when this number of electrons have flowed through.
• The molar volume of every gas, 22.4 dm³ at s.t.p contains this umber of molecules. One mole of very particle, atom, molecule, ion or electron contains the Avogadro’s number of particles.

The formula of a compound tells us the elements present in it. It also tells us the elements present in it. It also tells us the amounts of the different elements present in it. These amounts are usually expressed as a mole ratio of the different elements in the compounds.

For example:

Formula

Elementpresent           Sodium              sulphur     oxygen  

Number of Moles             2                          1              4

Mole ratio                          2                           1             4

A balanced equation of a chemical reaction tells us the relationship of the amounts of the reactants to one another and for the products is known as of stoichiometry reactions.

MOLE RATIO AND MASS RELATIONSHIP

Again, a mole is the amount of substance which contains and many elementary particles as there are carbon atoms in 0.012kg of carbon-12.

The mole is the amount of substance which contains as many formula units as their atoms in 12gram of carbon-12.

The numerical coefficients of a balanced equation represent the numbers of moles of reactants and products. From coefficient, we get the mole ratio of the reactants and product in a reaction. For example

Equation 1

Number of moles                    1                 2                   1

Mole ratio                                1                 2                    1

Therefore, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reaction

Equation 2.

Mg_(s) + 2HCl_(aq)    MgCl_2(s) + H_2(g)

Number of moles      1              2              1               1

Mole ratio                 1    :         2             1    :            1

Molar mass              24g          36.5g        95g            2g

Reacting mass          24g         (36.5 ×2)     95g           2g

From the equation 2, we see that

Amount × molar mass = reacting mass

Amount =

Molality– molality is defined as the amount of substance (in mole) of solute divided by the mass in kilogram of the solvent

Molality =

Molality should not be confused with molarity as they are both measures of concentration of solutions but Molarity is the ratio of moles to the volume while molality is the ratio of the moles to the mass of the solution. Molality is the number of solute per kilogram of solvent. It is important that the mass of solvent issued and not the mass of the solutions.

WORKED EXAMPLE

1. A 4 g of sugar cube (sucrose) – C₁₂H₂₂O₁₁ is dissolved in a 350mol teacup of 8⁰OC water. What is the molality of the sugar solution?

(Given- density of water at 80⁰ =0.975gmol^-1)

Solution

Molality is the number of moles of solute per kilogram per solvent.

Step 1: Determine the number of moles of sucrose in 4g

Molar mass =  12(12)   +    1(22)  +  16(11)

=    144  +   22   +  176

=  342gmol^-1

Divide this amount into the size of the sample.

=    0.0117mol

Step 2 Determine mass of solvent in kg

Density =  ;                       mass    =    density    ×   volume

Mass   =     0.975gmol ×350mol

Mass  =   341.25g

Change gramme to kilogramme = 0.341kg

Step 3: Determine the molarity of the sugar solution

Molality = =

Molality = 0.034mol

EVALUATION

1. How many molecules are there in 14g of nitrogen at s.t.p.? [N=14. Avogadro’s number = 6.02 ×10²³]
2. What amount, in mole, of copper is deposited when 13.0g of zinc reacts with excess copper(II) tetraoxosulphate(VI) solution according to the following reaction?

Zn_(s)  +  CuSO_4(aq)            ZnSO_4(aq)  +     Cu_(s)    [Cu=63.5, Zn =65 ]

3. Calculate the volume of chloride at s.t.p. that would be required to react completely with 3.70g of dry slaked lime according to the following equation: Ca(OH)_2(g) + Cl_2(g) →       CaCl₂  + H₂O  [H=1; O=16; Ca=40; 1mole of gas occupies 22.4dm3 at s.t.p.]

 

PERIOD 3: CALCULATION INVOLVING MASS AND VOLUME

Example 1: mass- mass problem

1. What mass of copper is deposited when 6.5g of granulated zinc reacts with excess copper(II) tetraoxosulphate(vi) solution according to the following equation

Zn(s) + CuSO₄(aq)           ZnSO₄ + Cu(s)

Cu = 64, Zn =65

Given: molar mass of Zn =65gmol-, molar

Mass of Cu = 64gmol-1

Reacting mass of Zn=6.5g

Zn + CuSO₄ ZnSO₄ + Cu

From the above, 65g of Zn deposited 64g of Cu.

1g of Zn will deposit bof cu

65g of Zn will deposit g of cu

= 6.4g

2. What volume of carbon(IV) oxide is produced at s.t.p. when 2.5g of calcium trioxocarbonate (iv) reacts with excess acid according to the following equation

CaCO₃ +2HCl              CaCl₂ + H₂O + CO₂

[CaCO_{3 =} 100, molar volume of gas at s.t.p= 22.4dm₃]

 

Solution

CaCO₃ +2HCl               CaCl₂ + H₂O + CO₂

100g                                                    22.4dm₃

From the equation, 100g of CaCO₃ produces 22.4 dm₃ ofCO₂

1g of CaCo₃ will produce  of CO₂

                                        = 0.56dm³

 

EVALUATION

1. CaO_(s) + H₂O_(l) Ca(OH)_2(s).

From the equation above, calculate the mass in grams of calcium hydroxide produced by 5.6g of calcium oxide.   [Ca = 40, O = 16, H = 1].

2. What volume in dm³ will 0.5g of H₂ occupy at s.t.p? [H = 1, 1 mole of gas at s.t.p = 22.4dm³].
3. Consider the reaction represented by the following equation: 2NaCl_(s) + H₂SO_4(aq) → Na₂SO_4(aq) + 2HCl_(g). Calculate the volume of HCl gas that can be obtained at s.t.p. from 5.85 g of sodium chloride. [ Na = 23.0, Cl = 35.5, Molar volume of gas s.t.p. = 22.4 dm³]

 

PERIOD 4: MORE CALCULATIONS ON MASS- VOLUME

Phosphorous burns in the limited supply of air to form phosphorous(iii) oxide, P₄O₆ and in unlimited supply of air to form phosphorous(v) oxide P₄O₁₀.If 3.3g of phosphorus(III) oxide is formed

(a) what mass oxygen from air is used?

(b) How many molecules of oxygen are contained in this mass?

(c) How many molecules of phosphorous are converted to the oxide?

(d) How many molecules of oxygen are needed to convert this oxide into phosphorous v oxide?

(e) What volume would the oxygen in (d) above occupy at s.t.p.?

Solution:                                                                                                 

Molar mass of P_{4   =} 31 × 4 = 124gmol^-1

O₂ =16 × 2 = 32 g.mol^{– 1}

P₄ O_{6 =}  gmol^–

• 220g of P₄ O₆ used (32 ×3) 96g of oxygen

3.3 of P₄O₆ requires g of oxygen

= 1.44g

(b) 1.44g of oxygen = mole

1 mole contain 6.02 ×  molecules

Number of moles in 1.44g of oxygen =

=2.71 molecules

(c) Mass of phosphorous in 3.3g of P₄O₆  =  (3.3 – 1.44)g

= 1.86g

1.86 g of phosphorous is mole

The number of molecules in (c) above i.e. 1.86g of phosphorus

= = 9.03 molecules

• P₄O₆ +20₂ P₄O₁₀

220 of P₄O₆ require 2 moles of oxygen

3.3g of P₄O₆ will require

Number of molecules of oxygen contained in this

olecule

= 1.81 molecules

• 02 molecules occupy 22.4dm³at s.t.p

1.81  molecule occupy 22.4dm³

= 0.067 dm³or 6.73.49 cm³

 

EVALUATION

1. Calculate the mass of sodium trioxocarbonate(IV) produced by the complete decomposition of 16.8g of sodium hydrogen trioxocarbonate(IV) (H= 1, O= 16, Na= 23, S= 33).
2. On carrying out a chemical analysis discovered that the component of egg shells is CaCO₃^. If 25.0 cm³ of 0.700 moldm^-3 HCl reacted with all the CaCO₃ in a sample of egg shells, calculate the:

(i) mass of CaCO₃ in the egg shells;   (ii) volume of CO₂ evolved at s.t.p.

The equation for the reaction is:

CaCO_3(aq) + 2HCl_(aq)    CaCl_2(aq)    + H₂O_(l)  + CO_2(g)

( H = 1.00; O = 16.0; Ca = 40.0;  1 mole of gas occupies 22.4dm³ at s.t.p.)

GENERAL EVALUATION

OBJECTIVE TEST:

1. What volume of steam is produced when 10g of propyne is burnt in excess oxygen at s.t.p?

(a) 44.8dm (b) 11.2 dm³(c) 22.4 dm³(d) 11.2cm³

2. What mass of anhydrous sodium trioxocarbonate(iv), Na₂CO₃, present in 500cm³ of 0.1 moldm^-3?[ Na = 23, C= 12, O = 16] (a) 10.6g (b) 212g (c) 53 (d) 106g.
3. The mass of potassium hydroxide required to make 300.00cm³ of 0.4moldm^-3 solution is [KOH = 56.0] A.26.88 g B. 13.44 g C. 6.72 g    D. 3.36 g
4. A solution of sodium hydroxide containing 6.0g in 250cm³ of solution has a concentration of A. 0.04 mol. dm³ B. 0.60 mol. dm³ C. 0.96 mol. dm³ D. 0.15 mol. dm³ [molar mass of NaOH = 40g mol.^-1
5. The volume occupied by 17g of H₂S at s.t.p. is [H=1.00, S=32.0, Molar volume =22.4 dm³] A. 11.2 dm³ B.17.0 dm³ C. 34.0 dm³ D.  44.8 dm³

 

 

ESSAY QUESTIONS

1. If 11 g of a gas occupies 5.6 dm³at s.t.p. calculate the vapour density [1 mole of a gas occupies 22.4 dm³at s.t.p]
2. 100 cm³ of a gas at 27 exert a pressure of 750mnHg. Calculate its pressure. If its volume is increased to 250cm³at 127 .
3. What is an S.I unit?

4(a) Write down five examples of quantities with their S.I units.(b) Define mole.

5. What volume will 22.4dm₃ of oxygen measured at 25^oC and and 1.12 × 10⁵ Nm occupy at standard pressure of 1.01 ×10⁵ Nm.
6. What volume of steam is produced when 10g of propyne is burnt in excess?

 

 

 

WEEK 6