Download CHEMICAL REACTIONS Topic 5: Preliminary course Chemistry

CHEMICAL REACTIONS
Topic 5: Preliminary course
Key Points

Mass is conserved

Atoms are conserved
Phosphate
PO4
3-
Carbonate
CO3
2-
Sulfate
SO4
2-
Nitrate
NO3
1-
Hydroxide
OH
1-
Ammonium
NH4
1+

Key Terms
Syllabus
What is Gay-Lussac’s Law?
Radicals
Name
1.
2.
3.
Symbol
Valency
Synthesis: Cu + S –> CuS
Decomposition:
ZnCO3 –> ZnO + CO2
Combustion: Ethanol
Prac-work
Series of chemical reactions
8.
C2H5OH + O2 –> 2CO2 +
9.
3H2O
4.
7.
Displacement: Iron wool +
copper sulphate –> copper +
Iron sulphate
Neutralisation: Acid + Base +
Universal indicator –> H2O +
NaCl. Hydrogen swaps.
Precipitation: Silver Nitrate +
NaCl –> AgCl + NaNO3. Silver
chloride = photographic.
Corrosion: Fe +O2 –> Fe2O3.


Mg is more reactive to Fe and
so is sacrificed. It looses
New substance formed
Energy change (up or down),
5.
6.
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electrons and becomes ions
floating in the solution. Stops
Fe from losing electrons.
Exothermic: KMnO3 + C3H8O3
(glycerol) –> + energy
Endothermic: Sodium
Bicarbonate + Citric acid (with
water) –> absorbs heat.
Reacting Gas Volumes
Nitric Oxide + Oxygen –> Nitrogen
dioxide (brown gas, soluble in water).
NO2 dissolves in water, leaving
unreacted NO.
How do we tell of a chemical
reaction has occurred?

sound, light, temp.
Colour change


(change in state)
change in physical/chemical
Molar Weight


properties of product compares
to reactants
different volume (gasses)
difficult to reverse.
is the same as the atomic weight.
Eg. 1 mol of Sulphur weighs 32.1
grams. (Mr = 32.1)
Law of conservation of mass
Total number of atoms in reactants =
The weight of the substance in grams
Amount of substance (n)
Amount of substance =
mass of substance(m)
molar mass (Mr or FM)
total number of atoms in products.
(Mass (and atoms) are conserved)
n =
Gay-Lussac’s law of combining gas
volumes (1808)
The law of combing gas volumes
states that the ratio of the volume of
gasses involved in a chemical reaction
are expressed by small, whole
numbers.
2H2 (20ml) + O2 (10ml) –> 2H20
(at 125C)
m
M
n = amount of substance
m = mass of compound
M = Molar mass
Mole Theory and Chemical
Reactions
Balanced chemical equations show the
number of moles of REACTANTS and
Avogadro’s Law states that equal
PRODUCTS needed and produced.
This is particularly useful to chemists
as they can calculate the masses of
reactants and products in the reaction
without performing the experiment.
volumes of gasses, at the same
temperature and pressure, contain
equal numbers of molecules.
Example: Have 160g Methane. How
much O2 is needed to react with this.
Ratio: 2:1:2
Avogadro’s Law
METHANE + OXYGEN –> CARBON DIOXIDE + WATER
1.
Write a balanced chemical
equation
CH4 + 2O2 –> CO2 + 2H2O
1.
Write amount of Moles
Moles 1
:2
:1
:2
1.
Work out weight
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Methane = CH4.
Formula mass = 12 + 4 = 16.
1 mol = 16 g
10 mol = 160 g
10 mol CH4 needs 20 mol O2
1 mol O2 = 32 g
20 mol (2 x 10) = 640 g
20 mol H2O = 2 + 16 = 18 x 2 =
How much H2O and CO2 are
360 g
produced from the reaction?
160 g + 640 g –> ? + ?
Moles:10
20
10
20
10 mol CO2 = 12 + 2(32)= 440g
1.
Write balanced Chemical
Equation
2H2 + O2 –> 2H2O
1.
In terms of Moles
2 mol : 1 mol –>
2 mol
1.
1.
2.
3.
4.
5.
In terms of weights
2H2 = 4, O2 = 32, 2H2O = 36
4g
: 32 g –>
36 g
General Equation
Chemical Equation
Moles
Relative weights (F.M.)
Weights
Example 4: What mass of calcium
metal would be needed to react with
Example 3: What mass H2O would be
produced from the complete
combustion of 20 grams H2 gas.
(20/4 = 5. Therefore all x 5)
20 g : 160 g –>
180 g
Example 3: What mass of magnesium
chloride would be produced from the
reaction of 2.4 g of magnesium with
excess hydrochloric acid.
conditions.
25 C
1 atm
298 K
101.3 kPa
Molar Volume (Vm) - Volume of one
mole of gas under specified conditions.
S.T.P Vm = 22.4
L/mol OC
S.L.C. Vm = 24.5
L/mol 298K
an excess of sulphuric acid to produce
0.5 mol calcium sulphate.
n =
Example 4: What masses of CO2,
NaSO4 and H2O would be produced
from the reaction of 50 g of NaCO3
with excess sulphuric acid?
Molar Volume
S.T.P.: Standard Temperature and
Pressure.
0 C
1 atm
273 K
101.3 kPa
S.L.C.: Standard Laboratory
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V
Vm
n = number of moles
V = Volume of gas
Vm = Molar Volume
Example: No. of moles of gas in 100L
at SLC.
n =
V
100
, n =
= 4.08 mol
Vm
24.5
Example 2: Volume of CO2 produced
when 24.47 L of methane combusted
at SLC.1
Example 3: 2.4 g of a compound of C,
H and O give on combustion 3.52 g of
CO2 and 1.44 g H2O. The relative
molar mass of the compound was 60.
1
24.47 L CO2
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a)
What volume of CO2 is
produced?
b)
What are the EF and MF of the
compound
General Equation
4.
HYDROCARBON + OXYGEN –> WATER +
c)
compound
4.
5.
CARBON DIOXIDE
1.
2.
3.


Balanced Chemical Equation
Work out moles of each
Weights of each
Recall that solute + solvent –>
solution. (Volume of final
solution)
The concentrations of a solution
is defined as the quantity of
solute in a given volume of
solution (at a fixed temperature
and pressure)
concentration = amount of solute

Concentrations expressed as
mol/L are called molarities (M).

ie. a 1 M solution contains 1 mol
of solvent per litre of solution.
Experiment: exact concentration of
acid. Titration (see book).
Mol Ratio.
Gases
Volume of gas at pressure
n = PV
RT
n = amount, mol
P = Pressure
V = Volume
T = Temperature
R = gas constant (8.314 J/K/mol)
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What are the masses of C, H
and O in 2.4 grams of the
Empirical formula
Using molar mass work out
MF.
Molarity
Volume of solution
c =
n
V
c = concentration
n = no. of moles
V = volume, L
General Equations
Synthesis - Forms more complex
compound
METAL + ACID –> SALT + HYDROGEN
Decomposition - Forms elements or
simpler compounds
ACID + BASE –> SALT + WATER
CARBONATE + ACID –> CO2 + SALT +
WAT
ER
Calculations involving excess or
limiting reagents (make sure can be
done)
Steps
HYDROCARBON + OXYGEN –> CO2 +
WAT
ER
1.
2.
3.
4.
5.






General Equation
Balanced Chemical equation
Work out number of moles
Write all information below each
formula
Do calculations
(amount of mols =
sodium in water
petrol and oxygen
Rust
Temperature of the reactants
(higher temp, higher rate) More
molecules above activation
energy at higher temps (moving
fate so more successful
collisions).
Concentration - More
concentrated - faster reactions,
and more volume of product (1
M –> 6 M).
Surface Area - only the
exposed surface can react.
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concentration x Volume)
(n = c x V)
Rates of Reactions
Reactions vary greatly in rate
Fast
Slow
The rate of a reaction depends on:
Higher SA, faster reaction
Catalyst - A substance which

increases the rate of a chemical
reaction. Works by reacting
with one of the initial
compounds so the chemical
bonds can be broken more
easily (and with less energy).
The catalyst alters the pathway
by which a reaction takes place.
They lower the activation
energy at which a reaction can
take place. (MnO2) catalyst is
not used up
2H2O2 –>MnO2 O2 + 2H2O
Activation Energy - the minimum
energy required to start a chemical
reaction.
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