WEEK 9
GAY LUSSAC’S LAW: It states that when gasses react, they do so in volumes and these volumes are in simple ratio to one another and to volume of the product if gaseous provided the temperature and pressure remain constant.
Gay Lussac observed as follows:
H2 + O2 2H2O
Volume 2 1 2
Ratio 2 : 1 : 2
He noticed that the combining volumes as well as the volumes of the products were related by simple ratios of whole number, provided they are gases.
EXERCISE
- 2O cm3 of CO are sparked with 20 cm3 of Oxygen. If all the volumes of gases are measured at s.t.p, calculate the volume of the residual gases after sparking.
2CO (g) + O2(g) 2CO2(g)
SOLUTION
2CO (g) + O2(g) 2CO2(g)
2CO2 (g) + O2 (g) 2CO2 (g)
Combining volume: · 2 : 1 : 2
Vol. before sparking: 20cm3 20cm3 –
Vol. during sparking: 20cm3 10cm3 20cm3
Vol. after sparking: – 10cm3 20cm3
Residual gases = Unreacted gas + Product formed
10 + 20 = 30cm3
2. 100cm3 of Nitrogen gases mixed together with 150 cm3 of Hydrogen. The mixtures were
made to react at low temperature at which the product cannot dissociate.
I. which of the two gases was in excess and by what volume?
ii. What was the volume of NH3 produced?
Equation for the reaction; N2 + 3H2 2NH3
3. What is the volume of Oxygen required to burn completely 45cm3 of Methane gas (CH4)?
Equation for the reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
SOLUTION BY GAY-LUSSAC’S LAW
1 Volume of CH4 requires 2 volumes of oxygen
i.e. 1cm3 of CH4 requires 2cm3 of oxygen
45cm3 of CH4 requires 2 x 45 = 90 cm3
1
Alternatively:
By mole concept, from the equation, 1 mole of CH4 requires 2 moles of O2,
(1 X 22.4) dm3 of CH4 requires (2x 22.4) dm3 of O2.
22.4 dm3 of CH4 require 44.8 dm3 of O2.
Hence, 45cm3 require 44.8 x 45 = 90 cm3
22. 4
AVOGADRO’S LAW: By an Italian Professor Avogadro (1776 – 1856). It states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules
AVOGARO’S LAW AND GAY LUSSAC’S LAW.
Avogadro’s law is used to convert the volume of gases into the number of molecules contained by those gases i.e.
Equation 2H2 (g) + O2(g) 2H2O(g)
Volume 2 1 2
Gay Lussac 2 : 1 2
Avogadro 2mols 1molecules 2molecules.
Therefore, Gay Lussac’s law can be re-stated that when gasses react, they do so in small whole numbers of molecules of reactant to produce small whole numbers of products.
RELATIVE VAPOUR DENSITY OF GASSES.
It is the number of times a given volume of a gas is as heavy as the same volume of Hydrogen gas at a particular temperature and pressure.
V.D = Mass of a given volume of a gas
Mass of an equal volume of Hydrogen gas
Relationship between V.D and R.M.M.
Applying Avogadro’s law into the formula of VD,
V.D = Mass of a given molecule of gas
Mass of an equal molecule of H2
Where the molecule is 1,
V.D = Mass of 1 molecule of a gas
Mass of 1 molecule of H2
However, H2 is diatomic,
V.D = Mass of 1 molecule of a gas
Mass of 2 atoms of H2
Mass of 1 molecule of a gas
2(mass of 1atom of H2)
2 x V.D = mass of 1 molecule of a gas
Mass of 1 atom of H2
NOTE: R. M. M= mass of 1 molecule of a gas
Mass of 1 atom of H2
Substitute this equation
Hence 2 x V.D = R.M.M
V.D = R.M.M/2
GRAHAM’S LAW OF DIFFUSION OF GASSES
Graham in 1833 discovered that a less dense gas can diffuse faster than a denser gas, so the density of the gas determines the rate of diffusion of a gas.
The law states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density.
i.e. R α
Where R = rate of diffusion
= density (Greek letter rho)
For two gases (say 1 and 2)
R1α and R2α ₂
On multiplication
R1 =
R2 =
Note: The rate of a gas is directly proportional to the square root of its molecular mass.
R1 = M1
R2 = M2 where M = Molecular mass
Note: In calculation, M where R is not given, it can be calculated as follows:
R = Volume
Time
CALCULATIONS
1.400 cm3 of a gas A diffuses through a porous partition in 5 seconds and 200cm3 of a gas B diffuses in 20 seconds under the same condition of temperature and pressure:
- Calculate the rates of diffusion of gases A and B
- Which gas is denser?
SOLUTION
- RA = Vol = 400 = 80 cm3
Time 5
RB =200 = 10 cm3
20. - Gas B is Denser
Time 5
20.2.A gas X diffused through a porous partition at the rate of 2.5 cm3/s . Under the same conditions, hydrogen diffused at the rate of 10cm3 /s . Calculate the RMM of the gas. (H = 1) Ans = 32
APPLICATION OF MOLE CONCEPT IN CHEMICAL REACTION
The mole concept can be applied to the following types of calculates which are based upon a
balanced chemical equation.
Calculation involving mass-mass relationship
Exercise 1: Calculate the mass of calcium oxide formed when 1.5g of calcium is completely burnt in oxygen
(Ca =40, 0 = 16)
2Ca(s) + 02g 2Ca 0(s) answer = 2.1g of Ca0(s)
EXERCISE 2
Calculate the mass of oxygen gas formed when 10g of potassium trioxonitrate (v) (potassium nitrate) is heated strongly.
Equation for the reaction
2KN03(s) 2KN02(s) + 02(g)
Answer = 1.58g of 02(g)
EXERCISE 3
When 1.4g of impure calcium trioxocarbonate (calcium carbonate) reacts with hydrochloric acid, 0.01mole of carbon (iv) oxide (carbon dioxide) gas was evolved calculate
i. Percentage purity
ii. Percentage impurity of calcium carbonate (Ca =40, C=12, 0=16, H=1)
SOLUTION:
CaC03(s) + 2HCl (aq)CaCl2 (aq) + H2 (aq) + C02 (g)
1 mole C02 = 1 mole CaC03
1 mole = (40 + 12 + 16 x 3) CaC03
1 mole C02 = 100 CaC03
:.0.01 mole C02 = 100 x 0.01 CaC03
Mas of pure = 1g CaC03
Total mass of CaC03 = 1.4g
Mas of pure CaC03 = 1.04g
Mass of impure CaC03
= 1.4 – 1.0 = 0.4g
i. Percentage purity
= mass of prime x 100
Total mass of CaC03 1
= 1.0 x 100
1.4
= 71.4%
EXERCISE 2
Calculate the volume of ammonia gas formed at s.t.p and r.t.p when 0.01g of hydrogen reacts
with nitrogen gas
(N14, H=1 M.V = 22.4drm3 at s.t.p and 24dm3 at r.t.p )
SOLUTION:
Equation for the reaction
N2 (g) + 3H2 (g) ______2NH3 (g)
Answer:
0.075dm3 NH3 at s.t.p
8dm3 NH3 at r.t.p
EXERCISE 3
Calculate the volume of oxygen gas at s.t.p and r.t.p needed to burn 1:20g of magnesium,
according to the equation below
2mg(s) + 02(g) _________2mg0(s)
(Mg = 24, 0 = 16, Mr. = 22.4dm3 at s.t.p, 24dm3 at r.t.p)
Answer:
= 0.56dm3 02 at s.t.p
= 0.56dm3 02 at r.t.p
EXERCISE 4:
The complete combustion of methane in oxygen is represented by the equation
CH4(g) + 02(g)) _____ C02 (g) + 2H20(g)
If 1000cm3 of methane was completely combusted in oxygen at s.t.p
Calculate
i. The mole of methane combusted
ii. The volume of oxygen used for the combustion
iii. The mass of carbon (iv) C02 produced
- The number of water molecule produced.
SOLUTION:i. Answer = 0.04 mole of CH4
ii. Answer = 2000cm3 of 02
iii. Answer = 1.96g of C02
iv. Answer = 5.38 x 1022 molecule of water
GAS VOLUME – GAS VOLUME RELATIONSHIP
EXERCISE 1:
40cm3 of nitrogen gas (N2) reacts with 60cm3 of hydrogen gas to form ammonia gas. Calculate
the volume of unused and volume of ammonia gas formed at the same temperature and pressure
equation for the reaction
N2 (g) + 3H2 (g) ________2NH3 (g)
IV 3vols 2vols
40cm3 60cm3 —
3vols 3H2 (g) = 60cm3
1 vol = 60/3 = 20cm3
:. 40cm3 60
1 x 20cm3 3 x 20 = 0cm3
20cm3 60cm3 ___________2 x 2 = 40cm3
:.Volume of unused gas = nitrogen
40cm3n 20cm3 = 20cm3
EXERCISE 2
100cm3 of sulphur (iv) oxide (sulphur dioxide-so2) gas reacts with 80cm3 of oxygen gas to produce sulphur (vi) oxide (sulphur trioxide). Calculate the volume of the resulting gas and the volume of unused gas measured at the same temperature and pressure.
Equation for the reaction 2S02 (g) + 02(g) __________2S03 (g)
Answer:
Volume of unused 02 = 30cm3
Volume of resulting gas 2S03 = 100cm3
MASS – LIQUID VOLUME CALCULATIONS
1. What volume of 2M HCl will be needed to react complete by with 4.0g of calcium.
SOLUTION:
Equation for the reaction Ca (g) + 2HCl (aq) ______CaCl2 (aq) + H2(g)
1 mole Ca(s) __________ 2 moles HCl(aq)
40g Ca(s)______________ 2 moles HCl(aq)
1g Ca(s _____________ 2 mole HCl
40
4.0g of Ca(s) = 2 x 4 mole HCl
40
= 0.2 mole HCl
Concentration of HCl = 2M
Mole = cone in mol/dm3 x Vol in cm3
1000
0.2 mole = 2 x V
1000
0.2 x 1000 = 2 x V
1000
:. V = 0.2 x 1000
2
V = 1000Cm3
2. Calculate the mass of calcium which will complete by react with 500cm3 of 0.1 MHCl
SOLUTION:
Equation for the reaction
Ca(s) + 2 HCl (aq) _________CaCl2 (aq) + H2 (q)
Mole = cone mol/dm3 x Vol in cm3
1000
= 0.1 x 500
1000
= 0.05 mole
From the equation of reaction Ca(s) +2HCl (aq) _______ CaCl2 (aq) +H2(q)
2 moles of HCl (aq) 1 mole of Ca(s)
2 moles HCl (aq) = 40g of Ca(s)
1 mole HCl (aq) = 40 moles HCl
20
:. 0.05 molHCl = 40 x 0.05 mole C
2
= 1g of Ca
THE MOLE FRACTION AND MOLE PERCENT
DEFINITION
The mole fraction can be defined as the number of moles of a particular substance in a mixture divided by the total number of moles of all the substances present in the mixture.
Example
A mixture of 1 moles of chloroform and 3 moles of ethanol were kept in a measuring cylinder calculate
i. The mole fraction of chloroform and ethanol
ii. The mole percent of chloroform and ethanol.
SOLUTION
Total number of moles of mixture = 1 + 3 = 4 moles
ia. The mole fraction of chloroform = 1
4
b. The mole fraction of ethanol = 3/4
iia. The mole percent of chloroform = ¼ x 100
1
= 25%
b. The mole percent of chloroform = 3 x 100 = 75%
4 1
NOTE:
i. When the masses of the substances in the mixture are given in grammes they are converted to moles by dividing with their relative molecular masses before the mole fractions are calculated.
ii. When the volumes of gasses at stated temperature and pressure are given, they are converted to moles before the mole fractions are calculated.
EXAMPLE 2
46g of ethanol was mixed with 36g of water in a reaction vessel, calculate:
i. The mole fraction of water and ethanol
ii. The mole percent of water and ethanol.
SOLUTION:
Rmm of water (H20)
= (1 x 2) + 16 = 18glmol
Rmm of ethanol (C2H50H)
= (12 x 2 + 1 x 5 + 16 + 1) = 46g/mol
No of mole of H20 molecules
= 36`
18 = 2 moles
No of molecules of C2 H5OH
= 46
46 = 1 moles
Total number of moles present in the mixture = 2 + 1 = 3 moles
ia. Mole fraction of water 2
3
b. Mole fraction of ethanol = 1
3
iia. The mole percent of H20
= 2 x 100
3 1 = 66.7%
b. The mole percentage of C2H50H
= 1 x100
3 1 = 33.3%
NOTE:
That one mole of a gas (the relative formula mass) will always tale up a volume of 24dm3 and 24000cm3 at room temperature and pressure (r.t.p)
This means that 28g of N2 will take up a volume of 24dm3 as will 71g of Cl2 also
R.f.m of N2 = 14 x 2 = 28
R.fm of Cl2 = 35.5 x 2 = 71
Formula
Vol of gas (dm3) = mass of gas (g)
24dm3 Rf.m of gas
= 8
2 = 4 moles
Vol of gas = moles x 24dm3
= 4 x 24 = 96dm3
This calculation shows that 8g of hydrogen will take up a volume of 96dm3
QUESTION
1.What volume is taken up by 10g of Ne?
2. What volume is taken up by 56g of N2?
3. What volume is taken up by 14.2g of Cl2?
4. If 0.1 mole of AgN03 reacts with HCl acid, what mass of AgCl could be produced according to the equation AgN03 (aq) + HCl (aq) _____AgCl(s) + HN03 (aq)
(Ag = 108, Cl = 35.5, H = 1, N = 14, 0 = 16)
Answer = 14.35g of AgCl
5. What mass of zinc metal would be required to react with dilute HCl to produce 0.5dm3 of H2 gas at s.t.p according to the equation below?
Zn(s) + 2HCl (aq) ________ZnCl2 (aq) + H2 (g)
(Zn) = 65, H = 1, Cl = 35.5, G.m.v = 22.4dm3 at s.t.p)
Answer = 1.45g of Zn
6. Find the number of molecules of 02 needed to convert 5.60dm3 of S02 gas measured at s.t.p to form S03
(G.m.v = 22.4dm3, NA= 6.02 x 1023molecules)
Answer =1.505 x 1023 molecules
7. When 10g of Na0H is dissolved in 1000cm3 of water, what will e the molar, concentration of the solution formed?
Answer = 0.25 moles
5. A gaseous mixture consist of 500cm3 of hydrogen 250cm3 of nitrogen and 1000cm3 of oxygen at s.t.p
i. The mole fraction of each component
- The mole percent of the gaseous mixture