CHEMISTRY REVISION 2 FOR TEST 2

COLLISION THEORY AND RATE OF REACTION

When substances are mixed together a chemical reaction sometimes occurs. Chemical
reactions always involve the collision of particles. For any particles to react they must (a) collide
with sufficient energy to break old bonds and form new ones and (b) collide with correct
orientation to allow particles to rearrange and make new compounds.

Reaction rates can be increased by increasing the concentration of solutions (more solution
particles, leads to higher rates of collision), breaking up solid substances into smaller pieces to
increase surface area, and heating up the substances to provide more energy when they
collide. A catalyst can also be added. A catalyst speeds up the rate of reaction by lowering the
amount of energy needed to form new products. Some catalysts provide a surface on which the
two reactants can meet in the correct orientation, allowing products to be formed.

BALANCING CHEMICAL EQUATIONS

The way in which particles collide to produce a reaction and then new substances is
summarised in a chemical equation. To write a correct chemical equation you must:

1. Write correct formula for each substance involved. Once the correct formula is written it
must not be changed.

2. Balance the equation to make sure it obeys the Law of Conservation of Mass. To
balance an equation a COEFFICIENT (number) is written in front of each formula to
make the number of atoms of each atom on the reactants side the same as the number
of atoms of that element on the product side. REMEMBER: DO NOT CHANGE THE
FORMULA, ONLY CHANGE THE COEFFICIENT. If the coefficient is 1, no number is
written.

Methane (CH4) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). Write
a balanced equation.

CH4 + O2  CO2 + H2 O

All the formulae are written correctly, so balance numbers by adding coefficients.

CH4 + 2O2  CO2 + 2H2O

The mass of individual atoms is very small and hence it is more convenient to use relative
masses. This means that the mass of each atom is compared to a standard which is an isotope
of carbon, carbon-12. The relative atomic mass (A r) of an element is the average mass of one
atom of it compared to the mass of an atom of carbon-12. The relative formula or molecular
mass (Mr) an element or compound is the mass of one of its molecules or formula compared to
the mass of an atom of carbon-12. The relative molecular mass or formula mass can be
calculated by simply adding the relative masses of all the atoms shown in the formula.

Mr (H2O) = 2 x Ar(H) + 1 x Ar(O) = 2 x 1 + 1 x 16 = 18

The mole (n) is a number that chemists use to count large numbers of particles. 1 mole of a
substance = 6.02 x 1023 particles of that substance.

N
23
n = 6 . 02×10 where n = number of moles, N = number of particles.

23
EXAMPLE ONE: How many moles of carbon dioxide molecules are there in 3.60 x 10
molecules of this substance?

N 3 . 60×10 23
23 23
n = 6 . 02×10 = 6 . 02×10 = 5.98 mol of CO2 molecules

EXAMPLE TWO: How many atoms of sodium are there in 4.0 moles of this substance?

N = n x 6.02 x 1023 = 4 x 6.02 x 1023 = 2.41 x 1024 atoms of sodium

In 3 moles of NH3 there are 3 moles of Nitrogen atoms and 9 moles of Hydrogen atoms.

The mass of one mole of a substance is called its molar mass (M). This mass is simply the
relative atomic mass (Ar) or relative molecular or formula mass (Mr) expressed in grams.

The molar mass of water is: M = 18 grams

The number of moles of a substance in a given mass of that substance can be calculated using
the following relationship.

m
n= M where n = number of moles, m = mass of substance in grams, M = molar mass of
substance in grams
 CHEMISTRY REVISION 2 FOR TEST 2

EXAMPLE THREE: Calculate the number of moles of sulfuric acid in 240 g of this substance.

M (H2SO4) = 98.1 g
m 240
n= M = 98.1 = 2.45 mol

EXAMPLE FOUR: Calculate the mass of 5.0 moles of glucose (C 6H12O6)

m = n x M = 5 x 180 = 900 g

Chemical reactions can be represented by chemical equations if all the substances involved are
known. The equation will tell us in what proportion (mole ratio) the reactants and products are
involved.

Hydrogen gas + oxygen gas  water

In calculations involving chemical equations a set formula is followed.

Step 1: Write a balanced equation.

Step 2: Identify the known and unknown quantities.

Step 3: Convert the known quantity to number of moles.

Step 4: Find the number of moles of the unknown quantity using the mole ratios of
the unknown over the known.
Step 5: Convert the number of moles of the unknown to the value required.

EXAMPLE FIVE: 16 g of oxygen gas reacts with excess hydrogen gas to produce water. What
mass of water is produced?

2H2 + O2  2H2O
known unknown

m 16
n (O2) = M = 32 = 0.5 mol

unknown mole ratio 2

m (H2O) = n x M = 1 x 18 = 18 g
QUESTIONS

1. Determine the relative molecular mass for CO2.

44 g

2. For each of the following sets of molecules determine which is the heaviest: Cl 2, Br2, I2.

I2

3. Calculate the number of silver atoms in 0.25 moles of silver (Ag).

1.505 x 1023

4. Determine the molar mass of ethanoic acid (CH 3COOH).

60 g

5. Calculate the mass of 3 moles of calcium carbonate (CaCO 3).

300.3 g

6. Determine the molar mass of a substance if 2 moles of the substance has a mass of 196g.

98 g

7. Determine the number of moles of sodium in 80 g of sodium carbonate (Na 2CO3).

0.75 mol
0.75 * 2 = 1.5 mol of sodium

8. When magnesium burns in air it forms magnesium oxide as shown in the equation below.
Determine the number of moles of oxygen gas consumed in producing 5 mole of magnesium
oxide.

2Mg + O2  2MgO

X/5 = 1/2
X = 2.5 mol of oxygen gas
 CHEMISTRY REVISION 2 FOR TEST 2

9. Determine the mass of sodium chloride required to react with 2 mole of silver nitrate
according to the following equation.

AgNO3 + NaCl  NaNO3 + AgCl

x/2 = 1/1
x = 2 mol of sodium chloride
117 g of sodium chloride

10. Sodium hydrogencarbonate decomposes on heating to form sodium carbonate, carbon

2 NaHCO3  Na2CO3 + CO2 + H2O

23.8 mol of sodium hydrogencarbonate

x/23.8 = ½
x = 11.9 mol of carbon dioxide
523 g of carbon dioxide

11. Balance the following equations. For the word equations write the correct formula and then
balance them.

Cr(OH)3 + H2SO4  Cr2(SO4)3 + H2O

2 Cr(OH)3 + 3 H2SO4  Cr2(SO4)3 + 6 H2O

HCl + Fe(OH)3  FeCl3 + H2O

3 HCl + Fe(OH)3  FeCl3 + 3 H2O

N2 + H2  NH3

N2 + 3 H2  2 NH3

Nitric acid + barium oxide  barium nitrate + water

2 HNO3 + BaO  Ba(NO3)2 + H2O

Zinc + hydrochloric acid  zinc chloride + hydrogen gas

Zn + 2 HCl  ZnCl2 + H2
12. Write the correct formula for the following:

(a) Dinitrogen tetraoxide: N2O4 (b) Iron III Carbonate: Fe2(CO3)3

(c) Sodium: Na (d) Nitric Acid: HNO3

(e) Sulfuric Acid: H2SO4 (f) Hydrochloric Acid: HCl

(g) Sulfur dioxide: SO2 (h) Potassium sulfide: K2S

(i) Sodium nitrite: NaNO2 (j) Carbon tetrafluoride: CFl4

13. Propane is commonly used for gas barbeques. A particular cylinder contains 4.5 kg of
propane. If the gas is pure propane (C 3H8), determine the mass of carbon dioxide and water
produced from its complete combustion.

C3H8 + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

102.2 mol of propane

x/102.2 = 4/1
x = 409.1 mol of water
7363 g of water

14. A 9.63 mole sample of aluminium is dissolved in excess sulfuric acid. Determine the mole of
sulfuric acid consumed and the moles of aluminium sulfate produced.

2 Al (s) + 3 H2SO4 (aq)  Al2(SO4)3 + 3 H2 (g)

x/9.63 = 3/2
14.445 mol of sulfuric acid

x/9.63 = ½
4.815 mol of aluminium sulfate

15. Are products formed every time molecules of reactants collide?

No, products only form when they collide with sufficient amount of energy and with correct
orientation

16. Why does increasing the surface area increase the rate of reaction?

Increasing the surface area exposes more reactant particles to the reaction which would
increase the frequency of particle collisions which in all increases the rate of reaction
 CHEMISTRY REVISION 2 FOR TEST 2

17. Why does diluting a solution decrease the rate of reaction?

Diluting a solution decreases the amount of reactant particles within the reaction which would
decrease the frequency of particle collisions which would decrease the rate of reaction

18. Why does a reaction occur faster when the reactants are stirred together?

Agitation allows more of the particles to be exposed throughout the reaction which would
increase the frequency of particle collisions which would increase the rate of reaction

19. How does collision theory explain the dramatic increase in the rate of a reaction as the
reactants are heated?

As the particles are heated, they possess more energy and they move more faster. With
reaction theory, the more energy a particle has and the faster frequency it collides the faster the
rate of reaction, heating the particles gives them more energy and allows them to move faster
which would increase the frequency at which particles collide at

20. Explain the two ways in which catalysts can work.

Catalysts lowers the activation energy needed for a reaction to occur and they also help
orientate the particles in order to achieve a successful reaction.