Question
The periodic table is a useful tool in explaining trends of chemical behaviour.
(a) (i) Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
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arrows AND identifies 2s AND 2p sub orbitals
Question
(ii) Sketch the shapes of the occupied orbitals identified in part (a)(i).
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Question
(b) Explain the decrease in first ionization energy from Li to Cs, group 1.
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The first ionization energy is the energy required to remove the outermost electron from an atom in its gaseous state. As we move down Group 1 from lithium (Li) to cesium (Cs), there is a general decrease in the first ionization energy. This trend can be explained by the following factors:
Increase in Atomic Size: As we move down the Group, the number of electron shells or energy levels increases. With each successive element, additional electron shells are added, leading to an increase in atomic size. The outermost electron in cesium is farther from the nucleus compared to the outermost electron in lithium. Therefore, the attraction between the outermost electron and the nucleus decreases, making it easier to remove the outermost electron. This results in a decrease in the first ionization energy from Li to Cs.
Shielding Effect: With the addition of electron shells, there is an increase in the number of inner electrons. These inner electrons repel the outermost electron, effectively shielding it from the attraction of the nucleus. As a result, the effective nuclear charge experienced by the outermost electron decreases as we move down the Group. A weaker effective nuclear charge means that the outermost electron is less tightly held by the nucleus, leading to a decrease in the first ionization energy.
Increase in Electron Repulsion: With the addition of electron shells, there is an increase in electron-electron repulsion within the atom. The outermost electron in cesium experiences greater repulsion from the inner electrons compared to the outermost electron in lithium. This repulsion makes it easier to remove the outermost electron in cesium, contributing to the decrease in the first ionization energy.
In summary, the decrease in first ionization energy from lithium to cesium in Group 1 is primarily due to the increase in atomic size, shielding effect, and increase in electron repulsion as we move down the Group. These factors collectively result in a weaker attraction between the outermost electron and the nucleus, making it easier to remove the outermost electron and leading to a decrease in the first ionization energy.
Question
(c) (i) State the electron domain geometry of the ammonia molecule.
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Tetrahedral
Question
(ii) Deduce the Lewis (electron dot) structure of ammonia and sketch its 3D molecular shape.
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Question
(iii) Explain, with reference to the forces between molecules, why ammonia has a higher boiling point than phosphine \((PH_3)\).
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Ammonia (\(NH_3\)) has a higher boiling point than phosphine (\(PH_3\)) due to differences in the intermolecular forces present in these molecules.
Ammonia:
Ammonia molecules are polar because the nitrogen atom is more electronegative than the hydrogen atoms, resulting in a net dipole moment in the molecule. This polarity allows ammonia molecules to form hydrogen bonds with each other. Hydrogen bonds are strong dipole-dipole interactions that occur between a hydrogen atom bonded to a highly electronegative atom (in this case, nitrogen) and a lone pair of electrons on a nearby electronegative atom (in this case, another nitrogen atom). These hydrogen bonds are relatively strong compared to other intermolecular forces and require more energy to break, leading to a higher boiling point for ammonia.
Phosphine:
Phosphine molecules are nonpolar because the electronegativity difference between phosphorus and hydrogen is smaller compared to nitrogen and hydrogen. Therefore, phosphine molecules do not exhibit significant dipole-dipole interactions or hydrogen bonding. Instead, the intermolecular forces present in phosphine are weak van der Waals forces (specifically, London dispersion forces). London dispersion forces result from temporary fluctuations in the electron distribution within molecules, leading to temporary dipoles. These temporary dipoles induce similar dipoles in neighboring molecules, resulting in weak attractive forces between the molecules. London dispersion forces are generally weaker than hydrogen bonds, leading to a lower boiling point for phosphine compared to ammonia.
In summary, the higher boiling point of ammonia compared to phosphine can be attributed to the presence of stronger hydrogen bonds between ammonia molecules, whereas phosphine molecules exhibit weaker van der Waals forces (London dispersion forces) due to their nonpolar nature.
Question
(d) (i) Ammonia is manufactured by the Haber process.
\(N_2(g)+3H_2(g)\rightleftharpoons 2NH_3 (g)\) \(\Delta H^{\theta}_t = -92.0\) KJ \(mol^{-1}\)
Outline what is meant by dynamic equilibrium.
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Reversible Reaction: The chemical reaction can proceed in both the forward and reverse directions. In the case of the Haber process, the reaction is \(N_2(g)+\) \(3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\), indicating that both the formation of ammonia from nitrogen and hydrogen (forward reaction) and the decomposition of ammonia into nitrogen and hydrogen (reverse reaction) can occur simultaneously.
Constant Concentrations: While the reactions continue to occur in both directions, the concentrations of the reactants and products remain constant over time. This does not mean that the concentrations are equal; rather, it means that they do not change.
Equal Rates: At dynamic equilibrium, the rates of the forward and reverse reactions are equal. This means that the rate of formation of products from reactants (forward reaction) is equal to the rate of formation of reactants from products (reverse reaction).
No Net Change: Despite the ongoing forward and reverse reactions, there is no net change in the concentrations of reactants and products. The concentrations of reactants and products reach a steady state where the forward and reverse reactions balance each other out.
Macroscopic Properties: While there is no net change in the concentrations of reactants and products, macroscopic properties such as pressure, temperature, and color may still change over time as the system adjusts to reach dynamic equilibrium.
In summary, dynamic equilibrium describes a situation in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, leading to a steady state where the concentrations of reactants and products remain constant over time.
Question
(ii) Deduce the \(K_c\) expression for the reaction in part (d)(i).
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The equilibrium constant (\(K_c\)) expression for a chemical reaction in terms of concentrations is determined by the stoichiometry of the balanced equation. In the given reaction:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
The expression for \(K_c\) is given by the ratio of the concentration of the products raised to their respective stoichiometric coefficients to the concentration of the reactants raised to their respective stoichiometric coefficients.
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
In this expression:
\([NH_3]\) represents the concentration of ammonia (NH₃) in mol/L,
\([N_2]\) represents the concentration of nitrogen (N₂) in mol/L,
\([H_2]\) represents the concentration of hydrogen (H₂) in mol/L.
The stoichiometric coefficients from the balanced equation (2 for NH₃, 1 for N₂, and 3 for H₂) are used as the exponents in the \(K_c\) expression. This expression represents the equilibrium constant in terms of concentrations for the given reaction.
Question
(iii) The Haber process requires a catalyst. State how a catalyst functions.
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A catalyst functions by providing an alternative reaction pathway that has a lower activation energy compared to the uncatalyzed reaction. This lower activation energy allows the reaction to proceed at a faster rate, facilitating the conversion of reactants into products. The catalyst itself remains unchanged at the end of the reaction and can be reused multiple times. In essence, a catalyst speeds up the reaction by lowering the energy barrier required for the reaction to occur without being consumed in the process.
Question
(iv) Sketch a Maxwell–Boltzmann distribution curve showing the activation energies with and without a catalyst.
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A Maxwell-Boltzmann distribution curve depicts the distribution of molecular speeds in a gas or the distribution of kinetic energies of particles in a system. When a catalyst is present in a reaction, it provides an alternative reaction pathway with a lower activation energy compared to the uncatalyzed reaction. As a result, the Maxwell-Boltzmann distribution curve for the reaction with a catalyst will show a greater number of molecules with energies equal to or greater than the lower activation energy provided by the catalyst, compared to the curve for the reaction without a catalyst.
Here’s a sketch of the Maxwell-Boltzmann distribution curve showing the activation energies with and without a catalyst:
In this sketch:
- The curve on the right represents the Maxwell-Boltzmann distribution for the reaction with a catalyst, showing a greater number of molecules with energies equal to or greater than the lower activation energy provided by the catalyst.
- The curve on the left represents the Maxwell-Boltzmann distribution for the reaction without a catalyst, showing fewer molecules with energies equal to or greater than the higher activation energy required for the uncatalyzed reaction.
The presence of a catalyst shifts the Maxwell-Boltzmann distribution curve to the right, indicating that a greater proportion of molecules have sufficient energy to overcome the lower activation energy barrier and participate in the reaction.
Question
(v) Suggest how the progress of the reaction could be monitored.
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he progress of the Haber process reaction, or any chemical reaction, can be monitored using various techniques. Some common methods include:
Spectroscopic Techniques: Spectroscopic methods such as UV-Vis spectroscopy, infrared spectroscopy (IR), or Raman spectroscopy can be used to monitor changes in the concentration of reactants or products by measuring absorbance or emission of light at specific wavelengths.
Chromatography: Techniques like gas chromatography (GC) or high-performance liquid chromatography (HPLC) can be employed to separate and quantify reactants and products as they elute from the chromatographic column.
Mass Spectrometry: Mass spectrometry (MS) can be used to identify and quantify reactants and products by measuring their mass-to-charge ratio.
Titration: Titration techniques can be used to monitor changes in the concentration of reactants or products by adding a titrant of known concentration to the reaction mixture until the reaction reaches completion.
Pressure Monitoring: Since the Haber process involves gases as reactants and products, the progress of the reaction can also be monitored by measuring changes in pressure using a pressure gauge or manometer.
Temperature Monitoring: Monitoring changes in temperature can also provide information about the progress of the reaction, as some reactions are exothermic or endothermic and may generate or absorb heat.
Online Monitoring: Online monitoring techniques, such as online sensors or automated sampling systems, can continuously monitor the reaction progress in real-time and provide data for analysis.
Overall, the choice of monitoring technique depends on factors such as the nature of the reaction, the reactants and products involved, and the desired level of sensitivity and accuracy.