2023 JUPEB Chemistry Theory Question And Answers

JUPEB CHEMISTRY
01-10: CACCDCDECE
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21-30: ECABEBEDCA
31-40: DADACEEBBB
41-50: EECBCBBDAE

COMPLETED.

(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.

(5a)
(i) A ligand is an atom ion or molecule that can donate a pair of electrons to form a chemical bond with a central metal atom or ion. It acts as a Lewis base or electron pair donor in a coordination compound.

(ii) Co-ordination number refers to the number of ligands that are bonded to a central metal atom or ion in a coordination compound. It represents the number of donor atoms surrounding the central atom. The coordination number gives information about the spatial arrangement or geometry of the compound.

(5b)
(i) The coordination number of the complex ion [Cu(NH₃)₄]²⁺ is 4. This is because there are 4 ammonia (NH₃) ligands attached to the central copper (Cu) ion.

(ii) The IUPAC name of the complex ion [Cu(NH₃)₄]²⁺ is tetraamminecopper(II) ion. This name indicates that there are four ammonia ligands attached to a copper ion with a +2 oxidation state.

(iii) The oxidation state of the central copper (Cu) ion is +2. This is because the complex ion has a net charge of +2 (the [Cu(NH₃)₄]²⁺ ion is positively charged) and each ammonia ligand has a neutral charge.

(iv) The shape of the complex ion [Cu(NH₃)₄]²⁺ is tetrahedral. This is because the four ammonia ligands are arranged around the central copper ion in a tetrahedral geometry with bond angles of approximately 109.5 degrees.

(5c)
(i) Iodine is a solid because it has a higher boiling point and a higher melting point compared to bromine and chlorine. The reason for this is the increasing strength of the van der Waals forces between the molecules as you go down Group 17.

(ii) Bromine is a liquid at room temperature because it has a lower boiling point and melting point compared to iodine and chlorine. Bromine molecules have weaker van der Waals forces compared to iodine making it easier to overcome these forces and transition from a solid to a liquid state.

(iii) Chlorine is a gas because it has the lowest boiling point and melting point among the three. The intermolecular forces in chlorine molecules are the weakest resulting in a lower boiling and melting point. This weak force allows the molecules to separate easily and move around as a gas.

(5d)
(i) Atomic size: The alkaline earth metals have smaller atomic radii compared to the alkali metals. This results in stronger metallic bonding between the atoms which leads to increased hardness and higher melting points. The smaller atomic size also allows for more efficient packing of atoms in the solid state increasing the strength of metallic bonding.

(ii) Electronegativity: The alkaline earth metals have higher electronegativities than the alkali metals. This means that the alkaline earth metals have a greater tendency to attract and hold on to electrons in their outermost energy levels. This increased electron density in the metallic bond leads to stronger attraction between the metal cations and the delocalized electrons resulting in stronger metallic bonding and harder metals with higher melting points.

(iii) Number of valence electrons: The alkali metals have one valence electron which is relatively easy to remove. This leads to a low ionization energy and a low melting point. In contrast the alkaline earth metals have two valence electrons which require more energy to remove. Consequently the alkaline earth metals have higher ionization energies and higher melting points.

(iv) Crystal structure: The alkaline earth metals adopt a simple cubic or hexagonal closely-packed crystal structure which allows for stronger intermolecular forces and higher melting points compared to the alkali metals which typically have a body-centered cubic or simple cubic structure.

6a

I) Halogens form interhalogen compounds due to their tendency to gain or share electrons in chemical reactions. When halogens combine with each other, they can form stable compounds because they have similar electronegativities and valence electron configurations.

Ii) Four examples of interhalogen compounds are:
a) Chlorine trifluoride (ClF3)
b) Bromine pentafluoride (BrF5)
c) Iodine monochloride (ICl)
d) Chlorine monofluoride (ClF)

6b
1) The ionization energy of manganese is greater than iron due to electron configuration. Manganese (Mn) has one more electron in the 3d subshell than iron (Fe) in the same period. The added electron-electron repulsion in the 3d subshell makes it harder to remove an electron, leading to higher ionization energy.

2) Compounds of Sc3+ and Zn2+ are mostly white because they have a fully filled or half-filled d-orbital configuration, which doesn’t absorb visible light. In contrast, Sc2+ and Cu2+ have partially filled d-orbitals, allowing them to absorb specific wavelengths and exhibit colored compounds.

3) The slightly higher ionization energy of gallium (Ga) compared to aluminum (Al) can be attributed to the increased effective nuclear charge in gallium due to its smaller atomic size. This stronger attraction between the electrons and nucleus requires more energy to remove an electron from gallium, resulting in a higher ionization energy.