Section 4: Physical Chemistry - IGCSE Chemistry - Shawon Notes

Chemistry > Section 4: Physical Chemistry


a) Acid, Alkalis and Salt

4.1 describe the use of the indicators litmus, phenolphthalein and methyl orange to distinguish between acidic and alkaline solutions

Types of indicators:

  • Litmus paper
  • Methyl orange
  • Phenolphthalein

Litmus paper in acidic solutions turns red and in alkaline solutions it turns blue. Methyl orange turns red in acidic solutions and yellow in alkaline solutions.

Phenolphthalein turns colorless in acidic solutions and bright pink in alkaline solutions.

4.2 understand how the pH scale, from 0-14, can be used to classify solutions as strongly acidic, weakly acidic, neutral, weakly alkaline or strongly alkaline

pH scale contains 14 colors. From 0-6 all are acid. From 8-14 all are alkaline. 7 means neutral. During acid, the lower the number, the higher the strength of acid. During alkaline the higher the number, the higher the strength of the alkaline.

4.3 describe the use of universal indicator to measure the approximate pH value of a solution

Universal indicator is a mixture of several different indicators. Unlike litmus, universal indicator can show us exactly how strongly acidic or alkaline a solution is. This is measured using the pH scale. The pH scale runs from pH 0 to pH 14.

Universal indicator has many different colour changes, from red for strong acids to dark purple for strong bases. In the middle, neutral pH 7 is indicated by green.

4.4 define acids as sources of hydrogen ions, H+, and alkalis as sources of hydroxide ions, OH-

Acid produces hydrogen ions, H+ when it is dissolved in water. Alkalis produce hydroxide ions, OH- when dissolved in water.

4.5 predict the products of reactions between dilute hydrochloric, nitric and sulfuric acids; and metals, metal oxides and metal carbonates (excluding the reactions between nitric acid and metals)

Metals:

  • Hydrochloric acid + metal ==> metal chloride salt + hydrogen
  • Nitric acid + metal ==> Not needed
  • Sulphuric acid + metal ==> metal sulphate + hydrogen

Metal oxides:

  • Hydrochloric acid + metal oxide==> metal chloride salt + water
  • Nitric acid + metal oxide ==> metal nitrate salt + water
  • Sulphuric acid + metal oxide ==> metal sulphate + water

Metal carbonates:

  • Hydrochloric acid + metal carbonate ==> metal chloride salt + water + carbon dioxide
  • Nitric acid + metal carbonate ==> metal nitrate salt + water + carbon dioxide
  • Sulphuric acid + metal carbonate ==> metal sulphate + water + carbon dioxide

4.6 understand the general rules for predicting the solubility of salts in water:

  • all common sodium, potassium and ammonium salts are soluble
  • all nitrates are soluble
  • common chlorides are soluble, except silver chloride
  • common sulfates are soluble, except those of barium and calcium
  • common carbonates are insoluble, except those of sodium, potassium and ammonium

4.7 describe experiments to prepare soluble salts from acids

These all involve reacting a solid with an acid:

  • acid + metal (but only for moderately reactive metals from magnesium to iron in reactivity series)
  • acid + metal oxide or hydroxide
  • acid + carbonate

Making soluble crystals of magnesium sulphate:

1. Add enough magnesium to dilute sulphuric acid, so that there is no acid left and the solution stops bubbling Mg(s) + H2SO4(aq) ==> MgSO4 + H2O

2. Filter the unused magnesium

3. Heat the solution until it forms crystal when it is cooled.

4. Leave the solution to dry up and soluble colourless crystals of magnesium sulphate is formed.

4.8 describe experiments to prepare insoluble salts using precipitation reactions

Preparation of an insoluble salt:

1. Add the sodium salt solution of the anion to the nitrate salt solution of the cation until no more precipitate forms.

2. Filter to collect the residue

3. Wash the residue with cold water

4. Leave residue to dry on filter paper dry in a warm oven

Example: Describe how to prepare a dry solid sample of silver chloride, AgCl, a salt which is insoluble in water.

Sodium chloride + silver nitrate ==> silver chloride + sodium nitrate NaCl (aq) + AgNO3 (aq) ==> AgCl (s) + NaNO 3 (aq)

Silver nitrate solution contains silver ions and nitrate ions in solution. The positive and negative ions are attracted to each other, but the attraction aren't strong enough for them to stick together. Similarly sodium chloride solution contains sodium ions and chloride ions - again, the attractions, aren't strong enough for them to stick together.

When you mix the two solutions, the various ions meet each other. When silver ions meet chloride ions, the attractions are so strong that the ions clump together and form a solid. The sodium and nitrate ions remain in the solution.

Ag+(aq) + Cl- (aq) ==> AgCl(s)

Example: Making pure barium sulfate

Ba2+(aq) + SO42-(aq) ==> BaSO4(s)

Barium chloride and dilute sulphuric acid are mixed together. Hydrogen ions and chloride ions are spectator ions and aren't involved in the reaction at all. Barium and sulphate ions attract together to form white precipitate of barium sulphate.

The mixture is filtered to get the precipitate. The solid barium sulphate is impure because of the presence of the spectator ions and any excess barium chloride solution or sulphuric acid. It is washed with pure water while it is still on the filter paper and then left to dry.

4.9 describe experiments to carry out acid-alkali titrations.

Titration:

i. fill the acid up to the mark in the burette

ii. pipette 25.0cm3 sodium hydroxide into a conical flask

iii. add a few drops of methyl orange indicator

iv. add acid from the burette drop wise with swirling of flask

v. stop when colour change is permanent (turns pink red)

vi. note burette readings

vii. repeat until concordant results are obtained (results are within 0.1 of each other)

viii. take average of results

Uses:

If they ask you how to prepare a soluble salt using an acid and an alkali, titration must be used. You first carry out a normal titration, and find out the exact amount of acid needed to neutralise the alkali. Then you repeat it without an indicator so that the salt is not contaminated with its colour. You remove the salt from the neutralised solution by evaporation, then you dry it.

b) Energetics

4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermic

A reaction that gives out heat is said to be exothermic.

A reaction that absorbs energy is said to be endothermic.

4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation in which heat energy changes can be calculated from measured temperature changes

Measuring energy changes involving solution

Measuring the heal evolved when magnesium reacts with a acid

When magnesium reacts with dilute sulfuric acid, the mixture gets very warm. The reaction is:

Mg(s) + H2SO4(aq) ==> MgSO4(aq) + H2(g)

50 cm3 of dilute sulfuric acid is run into a polystyrene cup using a pipette or burette and the temperature of the acid is measured. A small amount of magnesium powder is placed in a weighing bottle, and the mass of the bottle plus magnesium is recorded.

The magnesium is then tipped into the acid, and the maximum temperature reached is measured on the thermometer.

The mass of the empty weighing bottle is found and then the experiment is repeated to check the reliability of the result.

Volume of acid used = 50.0 cm3

Experiment 1

Experiment 2

Mass of weighing bottle + Mg(g)

10.810

10.800

Mass of weighing bottle afterwards (g)

10.687

10.685

Mass of Mg used (g)

0.123

0.115

Initial temperature (oC)

17.4

17.3

Maximum temperature (oC)

27.5

26.7

Temperature rise (oC)

10.1

9.4

If you do a reaction using a known mass of solution and measure the temperature rise, the amount of heat given out during the reaction is given by:

Heat given out = mass x specific heat x temperature rise

The specific heat is the amount of heat needed to raise the temperature of 1 gram of a substance by 1oC.

For water, the value is 4.18 J g-1C-1 (joules per gram per degree Celsius)

You can normally assume that dilute solutions have the same specific heat and density as water. You can also assume that negligibly small amounts of heat are used to warm up the cup and the thermometer.

In this case, we will take the mass of the solution as 50g. The mass of the magnesium is so small that it can be ignored. There are other major sources of error in the experiment which will make much more difference to the results.

Calculation:

Heat evolved when 0.123 g Mg reacts = 50 x 4.18 x 10.1 J = 2111 J

Energy changes in reactions are always quoted in kJ. Diving by 1000 gives 2.111 kJ evolved when 0.123g of Mg react.

Now we need to calculate how much heat is evolved when 24.3 g of Mg react. 24.3 is an accurate value for the relative atomic mass of the magnesium.

If: 0.123 g Mg produce

2.111 kJ

Then: 24.3g Mg produce

24.3

2.111 = 417

0.123

The amount of heat given out by the reaction is therefore:

Mg(s) + H2SO4(aq) ==> MgSO4(aq) + H2(g)

ΔH = -417 kJ mol-1

Measuring the heat evolved in burning liquids

Volume of water

= 100 cm3

Mass of burner + ethanol before experiment

= 37.355 g

Mass of burner + ethanol after experiment

= 36.575 g

Original temperature of water

= 21.5 oC

Final temperature of water

= 62.8 oC

Mass of ethanol burnt

= 0.780 g

Water temperature increase

= 41.3 oC

Mass of water being heated

= 100 g

Heat gained by water

= 100 x 4.18 x 41.3 J

= 17260 J

= 17.26kJ

Burning 0.780g of ethanol produces 17.26 kJ

Amount of heat produced when 1g of ethanol burns = 17.26 / 0.780 = 22.1

The amount of heat produced from 1 mole of ethanol = 17.26 / 0.780 x 46 kJ = 1020kJ

4.12 calculate molar enthalpy change from heat energy change

4.13 understand the use of ΔH to represent enthalpy change for exothermic and endothermic reaction

ΔH is the symbol that represents the amount of energy lost or gained in a reaction.

+ΔH is endothermic (because it gains heat)
- Δ H is exothermic (because it looses heat)

4.14 represent exothermic and endothermic reactions on a simple energy level diagram

Endothermic reaction Exothermic reaction

4.15 understand that the breaking of bonds is endothermic and that the making of bonds is exothermic

During chemical reactions, bonds in the reactants have to be broken, and new ones formed to make the products. Breaking bonds needs energy and energy is released when new bonds are made.

4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction.

Bond Energies

Bonds

Energies

mol

H-H

= 436

C-C

= 348

C-H

= 413

C-O

= 443

O = O

=466

O = H

= 463

H-Cl

= 431

N=_N

= 946

N = H

= 488

Cl - Cl

=242

H2 + O2 ==> 2H2O

Bond Breaking:

When the bonds in the hydrogen molecule and oxygen molecule are breaking, they have 2 - free hydrogen atom. This requires energy to be absorbed. That means the process is endothermic.

Energy required to break 1 mol of H-H bonds = +436kJ

Δ H = 436 kJ

Energy required to break 2 mol of H-H bonds = 2 x 436 kJ = +872kJ

Δ H = +872 kJ

Energy required to break 1 mol of O=O bonds = +496 kJ

Total energy = 872 + 496 = +1368kJ

Bond making:

Energy released on forming 2 mol of O-H bonds = -468kJ x 2 = -926 kJ

Therefore, energy released on forming 2 mol of H2O molecules = 2 x -926 = -1852 kJ

The overall enthalpy change for the reaction = (+1368) + (-1852) = -484 kJ

c) Rates of Reaction

4.17 describe experiments to investigate the effects of changes in surface area of a solid, concentration of solutions, temperature and the use of a catalyst on the rate of a reaction

Experiment: How temperature affects the speed of Reaction

Sodium thiosulphate reacts with hydrochloric acid to form yellow deposit of sulfur.

Na2S2O3 (aq) +2HCl (aq) ==> 2NaCl (aq) + SO2 (g) + H2O (l) +S (s)

1. Measure out 10 cm3 of the sodium thiosulphate solution (40 g dm-3) into a conical flask and add 40 cm3 of water.

2. If necessary, very gently warm the solution to a temperature of approximately 30 oC.

3. Add 5 cm3 of 2 mol dm-3 hydrochloric acid and start the stopclock.

4. Swirl the mixture and place over the cross marked on the white sheet of paper.

5. Look down vertically onto the cross, and time its

6. Repeat the experiment at about 40 oC, 50 oC, 60 oC, 70 oC by heating the thiosulphate solution before you add the acid. It is advisable to heat the solution to a little above these temperatures, to allow for the temperature to fall before you actually add the acid and start the clock.

Experiment: To study the effect of concentration on the speed of the reaction between magnesium and hydrochloric acid

1. Two beakers are taken: Beaker A: 50 cm3 dilute (1 mol dm3) hydrochloric acid and 5cm clean magnesium ribbon. Beaker B: 25 cm3 dilute (1 mol dm3) hydrochloric acid + 25 cm3 distilled water and 5cm clean magnesium ribbon.

2. The time taken for each piece of magnesium ribbon to dissolve is recorded.

Beaker

A

B

Time taken for magnesium to dissolve

39

78

(g)

In the above experiment, the acid in beaker A is twice as concentrated as the acid in beaker B. the time taken for magnesium to react completely in beaker A is shorter. Thus, we can conclude that a reaction proceeds faster when the concentration of a reactant is increased.

Experiment: To study the effect of particle size on the speed of reaction. 1. A conical flask is set up and attached to a gas syringe(to measure the amount of CO2 produced). Marbles chips is placed in hydrochloric acid in the flask. The volume of gas produced is recorded at one-minute intervals for investigation 1. 2. The experiment is repeated for investigation II, with marble chips that have been crushed into much smaller pieces. 3. Result will show small chips will form carbon dioxide faster than larger ones

4.18 describe the effects of changes in surface area of a solid, concentration of solutions, pressure of gases, temperature and the use of a catalyst on the rate of a reaction

Temperature: The higher the temperature, the faster the rate of reaction.

Concentration: The more concentrated the substance used, the faster the speed of a chemical reaction.

Pressure: The speed of reactions which involve gases is faster at higher pressures.

Particle Size: Smaller particles like powders have a much greater surface area than larger lumps or crystals. Within a greater surface area, the other reactant can attack it more easily and thereby increase the speed of reaction.

4.19 understand the term activation energy and represent it on a reaction profile

Activation energy is the minimum energy that molecules must possess during their collisions in order for a chemical reaction to occur.

4.20 explain the effects of changes in surface area of a solid, concentration of solutions, pressure of gases and temperature on the rate of a reaction in terms of particle collision theory

Temperature: When the temperature is raised, the reactant particles have a greater heat energy, causing them to move about more and with a greater kinetic energy. They, therefore, stand a better chance of colliding into another reactant molecule with sufficient energy to convert into product molecules.

Concentration: When concentrated substance is used, there is a greater likelihood that reacting molecules will collide with one another with sufficient energy to form products.

Pressure: In higher pressure the concentration within a certain volume increases, therefore more collision occurs.

Particles Size: Smaller pieces have large surface area and therefore more collisions.

4.21 explain that a catalyst speeds up a reaction by providing an alternative pathway with lower activation energy.

A catalyst is a substance which increases the rate of chemical reaction, without itself being chemically changed at the end of the reaction.

Catalysts work by proving a more direct route from reactants to products. If we plot energy against time for a reaction, there is an energy barrier over which the reactants have to pass. The height of this barrier above the energy of the reactants is called the activation energy. What catalysts do, is to lower this activation energy, so as to allow the reactants to change into products more quickly.

d) Equilibria

4.22 understand that some reactions are reversible and are indicated by the symbol in equations

Reaction which can be reversed is called reversible reaction. It is indicated by the symbol ⇌.

4.23 describe reversible reactions such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chloride

Heating copper(II) sulphate crystals

If you heat blue copper(II) sulphate gently, the blue crystals turn to a white powder and water is driven off. Heating causes the crystals to lose their water of crystallisation and white anhydrous copper(II) sulphate is formed.

CuSO4 . 5H2O(s) ==> CuSO4(s) + 5H2O(l)

Now if you add water to the white solid, it will turn blue and will get warm.

CuSO4(s) + 5H2O(l) ==> CuSO4 . 5H2O(s)

Heating ammonium chloride

If you heat ammonium chloride, the white crystals disappear from the bottom of the tube and reappear further up. Heating ammonium chloride splits it into the colourless gases ammonia and hydrogen chloride.

NH4Cl(s) ==> NH3(g) + HCl(g)

These gases recombine when the condition are changed from hot to cool.

NH3(g) +HCl(g) ==> NH4Cl(s)

4.24 understand the concept of dynamic equilibrium

Dynamic equilibrium is when a reversible reaction is happening both ways at the same time, at the same rate.

4.25 predict the effects of changing the pressure and temperature on the equilibrium position in reversible reactions.

Le Chatelier’s Principle

When a reversible reaction is in equilibrium and you make a change, it will do what it can to oppose that change.

The effect of pressure:

If you increase the pressure, the equilibrium will move to reduce it again by producing fewer molecules. If you lower the pressure, the equilibrium will move to increase it again by producing more molecules.

The effect of temperature:

If you decrease the temperature, the equilibrium will move to produce more heat to counteract the change you have made.

If you increase the temperature, the equilibrium will move to lower it again by favoring the endothermic change.