JUPEB CHEMISTRY THEORY QUESTIONS AND ANSWERS BELOW
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JUPEB
CHM001
General chemistry
No2
ai) The periodic law states that
similar properties recur periodically when elements are arranged according to increasing atomic number
aii)Two atomic properties are atomic number and atomic mass.
The atomic number increases from left to right and from top to bottom in the periodic table. Atomic mass also increases from left to right and from bottom to top.
JUPEB 2023 CHEMISTRY ANSWERS
(4a)
The standard enthalpy of combustion is the change in enthalpy that occurs when one mole of a is completely burned in excess oxygen under standard conditions, which include a temperature of 298 K and a pressure of 1 atmosphere. It is usually measured in units of kilojoules per mole (kJ/mol) and represents the amount of heat released during the combustion reaction. The standard enthalpy of combustion is a useful parameter for calculating the energy content of fuels and other organic compounds.
(4bi)
(i)To calculate the mass of butane used, we need to use the heat released by burning it the heat capacity of water. From the problem, the heat produced by burning butane can be used to raise the temperature of water by 14.3°C, which is equivalent to:
q = (230 g) × (4.2 J/g°C) × (14.3°C) = 57.4 J
Since butane is the only source of heat, the heat released by burning it must be equal to the heat absorbed by water:
q = (mass of butane) × (ΔHcomb)
where ΔHcomb is the standard enthalpy of combustion of butane. Assuming no heat losses, we can use the ideal gas law to find the number of moles of butane:
PV = nRT
n = (PV)/(RT) = (98 kPa × 0.2 L) / [(8.31 J/mol K) × (25 + 273) K] = 0.0073 mol
From the balanced chemical equation for the combustion of butane:
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
we see that 2 moles of butane produce 8 moles of CO2 and 10 moles of water. Therefore, 0.0073 moles of butane produce:
(10/2) × 0.0073 = 0.0365 moles of water
The mass of this amount of water is:
(0.0365 mol) × (18.015 g/mol) = 0.657 g
Therefore, the mass of butane used is:
q / ΔHcomb = (0.657 g) / (0.0073 mol × ΔHcomb)
Substituting the given values and solving for the mass of butane, we get:
m = q / (ΔHcomb) = (13757.4 J) / (0.0073 mol × (-2878 kJ/mol)) = 2.889 g
Therefore, the mass of butane used is 2.889 g.
(4bii) To determine the amount of heat released, we just need to use the specific heat capacity of water:
q = (230 g) × (4.2 J/g°C) × (14.3°C) = 13757.4 J
Therefore, the amount of heat released is 13757.4 J.
(4biii)To calculate the standard enthalpy of combustion of butane, we can use the mass of butane and the amount of heat released:
ΔHcomb = q / (m × n)
where n is the number of moles of butane:
n = m / M = 2.889 g / 58.12 g/mol = 0.0497 mol
Substituting the given values, we get:
ΔHcomb = (13757.4 J) / (2.889 g × 0.0497 mol) = -2851.8 kJ/mol
Therefore, the standard enthalpy of combustion of butane is -2851.8 kJ/mol.
JUPEB CHEMISTRY
NUMBER EIGHT
(8)
(i) Hybridization is the process by which atomic orbitals mix to form new hybrid orbitals in a molecule. It involves combining the orbitals of different electron shells to create a set of hybrid orbitals that are used in bonding.
(ii) The name of the compound CH₂=CHCH₂CH₂COOH is pent-2-enoic acid. The first and last carbon atoms in the compound have sp3 hybridization. In sp3 hybridization one s orbital and three p orbitals combine to form four sp3 hybrid orbitals. These hybrid orbitals are arranged in a tetrahedral shape around the central atom.
(8bi)
Aldehydes and ketones exhibit functional group isomerism as they have different functional groups attached to the same carbon chain.
(8bii)
(i) When butanal is reacted with Na₂Cr₂O/H₂SO₄ it undergoes oxidation to form butanoic acid. When 2-butanone is reacted with Na₂Cr₂O/H₂SO₄ it does not undergo oxidation as ketones are relatively resistant to oxidation. Therefore there is no reaction.
(ii) When butanal is reacted with NaBH₄/H₃O it undergoes reduction to form butanol. When 2-butanone is reacted with NaBH₄/H₃O it undergoes reduction to form 2-butanol.
(8ci)
Polymerization is the process by which small molecules called monomers react with each other to form a larger chain-like molecule called a polymer. This reaction occurs through the repetitive bonding of monomers resulting in the formation of a long macromolecular chain.
(8cii)
(i) In condensation polymerization the reaction process involves the elimination of a small molecule such as water or an alcohol as a byproduct. This occurs during the formation of each covalent bond between monomers. While addition polymerization involves the direct addition of monomers without the elimination of any byproducts.
(ii) Condensation polymerization involves monomers with different functional groups such as a carboxyl group and an amino group which can react with each other to form a covalent bond and eliminate a byproduct. While addition polymerization involves monomers with the same functional group such as a double bond which opens up during the reaction to form a new bond with another monomer.
(8ciii)
(a) In a condensation polymerization reaction two different monomers combine to form a polymer while eliminating a small molecule such as water. The reaction can be represented by the following equation:
Monomer A + Monomer B —-> Polymer + Small molecule
(b) In an addition polymerization reaction monomers containing double bonds react to form a polymer by the addition of monomer units. The reaction can be represented by the following equation:
Monomer + Monomer —> Polymer
(8d)
A chemical test that can be used to distinguish between a reducing sugar and a non-reducing sugar is the Benedict’s test. To perform the test start by preparing a solution of Benedict’s reagent which is a mixture of sodium citrate sodium carbonate and copper sulfate. Then take a sample of the sugar solution and add it to a test tube.
For a reducing sugar such as glucose or fructose the test tube will contain both a reducing sugar and Benedict’s reagent. Place the test tube in a water bath and heat it to around 80-90°C for a few minutes. If the sugar is a reducing sugar it will react with the copper sulfate in Benedict’s reagent causing a color change from blue to green yellow orange or even brick-red precipitate depending on the concentration of the reducing sugar present.
For a non-reducing sugar such as sucrose additional steps are needed. Start by adding a few drops of dilute hydrochloric acid to the sample in the test tube and heating it in a water bath for a couple of minutes. This step helps to hydrolyze the sucrose into its component reducing sugars. After hydrolysis neutralize the solution by adding sodium carbonate to the test tube until effervescence stops. Then proceed with adding Benedict’s reagent and heating the mixture. If the sugar was originally a non-reducing sugar the color change will be observed due to the presence of the reducing sugars formed from hydrolysis.
CHEMISTRY OBJECTIVE
1. A
2. A
3. A
4. A
5. D
6. A.
7. A
8. B
9. A
10. B
11. A
12. C
13. B
14. B
15. A
16. C
17.
18.
19. D
20.
21. B
22. A
23. C
24. D
25.
26. A
27. A
28. A
29. B
30. C
31.
32. A
33. D
34. A
35.
36. B
37. B
38. C
39. C
40. C
41.
42. C
43. B
44. C
45. D
46.
47.
48.
49. B
50.
