Kinetics, Energetics, Equilibria and Acid-Base Equilibria Worksheet Answer

Question 1: Kinetics

(a)(i) Collision Theory and Reaction Rate

Collision theory states that a reaction could only be affected when the particles undergoing the reaction collide with enough energy and orient them in the right way. Activation energy (Ea) is the least energy that is needed to allow a successful collision (Gültepe, 2021). Boltzmann distribution tells how at a certain temperature there is only a small proportion of particles above which their energy exceeds Ea. As the temperature is raised, area above Ea also increases, and this implies that a larger number of particles access enough energy to react. Thus, when you raise temperature, the chance of a successful collision considerably rises, therefore, causing the reaction speed to rise.

(a)(ii) Graph of [OH⁻] vs Time

Graph of [OH⁻] vs Time

(a)(iii) Order with respect to OH⁻

The reaction with regard to hydroxide ions is first order. This deduction is due to the graph-shape presented in Figure 1, which indicates OH 179 concentration on time. The graph shows a gentle curve that is downward as opposed to a linear decline (Milanović et al., 2023). In chemical kinetics a linear relationship between concentration and time would correspond to a zero-order reaction, in which the concentration of the reactant does not affect the reaction rate. The curve-like nature of this observation, however, implies that the rate of reaction is reduced as the concentration of the hydroxide ions is reduced.

To strengthen this one may plot natural log [OH^-] versus time. When this plot is a straight line whose slope is negative, it will confirm that the reaction is first-order. The reason is that in first-order reaction rate is proportional to the concentration of the reactant and the integrated rate law has the form ln[OH^- ] = -kt + ln[OH^-] 0, which is the equation of a straight line..

(a)(iv) Mechanism Deduction

This reaction is an SN1 because [OH^-] is the only significant dependence at the rate so the mechanism is two steps: first a chloroalkane forms a carbocation, then it is attacked by OH^- quickly. This corresponds to the first-order rate of kinetics observed.

(b)(i) SO₂Cl₂ Decomposition Graph

(b)(ii) Order and Rate Equation

The particular graph of [SO₂Cl₂] versus time specifically gives an overall straight line. This confirms that the overall decomposition of SO₂Cl₂. These mainly follow first-order type kinetics. Therefore, the rate of the equation is given below.

rate = k[SO₂Cl₂]

(b)(iii) Rate and Rate Constant (k)

At [SO₂Cl₂] is equals to 0.200 mol/dm³, as well as with the particular slope of the [SO₂Cl₂] versus time graph which approximated as –2.5 × 10⁻⁴ s⁻¹. In which the rate is mainly given below.

rate = 2.5 × 10⁻⁴ × 0.200 = 5.0 × 10⁻⁵ mol dm⁻³ s⁻¹

(b)(iv) Two-Step Mechanism

The overall mechanism likely or totally involves a very slow type of first step: SO₂Cl₂ → SO₂ + Cl∙ (rate-determining). It is mainly followed by a fast step which is Cl∙ + SO₂Cl₂ → products.

Question 2: Energetics

(a)(i) Born–Haber Diagram for Ag₂O

The final BornHaber cycle diagram specifically consists of following type species 2Ag(g) above 2Ag(s) representing overall atomisation, O⁻(g) between the O(g) as well as O²⁻(g) for the first electron affinity, as well as another O⁻(g) above which is an intermediate before mainly forming O²⁻(g). These overall additions mainly depend on the changes of enthalpy shown along with completing the overall energy pathway precisely.

(a)(ii) Electron Affinity of Oxygen

The particularly first affinity of electron of oxygen is very negative because overall energy is specifically released when oxygen develops one specific electron that forming O⁻. The affinity of second electron is very positive as the energy is mainly required to mainly add an specific electron towards the already negative O⁻ ion. This specific repulsion between like charges creates the overall process endothermic. These mainly require an input of energy.

(a)(iii) Atomisation of Ag

Atomisation of silver mainly includes converting overall solid Ag throughout gaseous Ag atoms. For the equation of Ag₂O, 2Ag(s) → 2Ag(g), the energy of atomisation is the overall energy that is required to break the overall metallic type bonds (Marcandalli, Villalba, & Koper, 2021). Utilizing Hess’s Law as well as values from the cycle of Born–Haber, the atomisation of enthalpy can be calculated by the help of rearranging the formation of the equation along with isolating the term of atomisation.

(b)(i) Born–Haber Cycle for CuO

The cycle of Born–Haber for CuO specifically includes changing Cu(s) as well as O₂(g) throughout gaseous ions then compacting or combining them to form CuO(s) (Boulebd, 2023). Steps include atomisation of Cu, first and second energies of ionisation, O₂ bond dissociation, oxygen electron affinity, as well as lattice energy (Di Noja, Garrido, Gualandi, & Ragazzon, 2023). Each and every step mainly corresponds towards a specific change of enthalpy. These mainly create a pathway of complete thermo chemicals.

(b)(ii) Lattice Energy of CuO

The lattice CuO energy is specifically calculated utilizing Hess’s Law. Add each and every change of enthalpy for atomisation, Cu ionisation, oxygen bond dissociation, as well as oxygen’s electron affinity. Subtract the overall CuO formation enthalpy (Cooke et al., 2024). The overall result is mainly the lattice energy, which particularly reflects the overall strength of electrostatic forces within the ionic solid type of structure.

(c)(i) Feasibility of Ammonia Formation

Utilizing ΔG = ΔH – TΔS, they mainly calculate ΔG at the value of 298 K for the specific type of ammonia formation. With the value of ΔH equals –92 kJ/mol and ΔS equals –200 J/mol·K, ΔG = –32.4 kJ/mol (Weeraratna et al., 2023). The value of ΔG is negative, the overall reaction is thermodynamically very feasible under the standard type conditions. These indicate automatic or impulsive at room temperature.

(c)(ii) High Temperature Explanation

At very high temperatures, the term of TΔS mainly dominates within the equation of Gibbs free energy. Since the change in entropy or ΔS for the formation of ammonia which is negative. This overall increasing temperature creates –TΔS more very positive. This can mainly cause ΔG towards becoming positive. These make the overall reaction totally non-spontaneous. Thus, the overall procedure becomes very less favourable at higher temperatures.

(c)(iii) Enthalpy via Hess’s Law

To particularly find the change of enthalpy for mainly producing 2-methylpropene from the help of 2-methylpropan-1-ol that specifically applies Hess’s Law: ΔH = ΔHf(products) – ΔHf(reactants). Utilizing provided values, ΔH is equal –20 – (–326) = +306 kJ per mol (Zhou et al., 2025). This specifically indicates the overall reaction is endothermic. These require overall input of heat for the conversion to efficiently proceed under standard type conditions.

Question 3: Acid-Base Equilibria

Part (a) - Bronsted-Lowry Theory of Acids and Bases

According to Bronsted-Lowry theory if the acids are taken as the substance that donating the protons i.e. (H+) ions whereas if the bases are used as the substances then it will accept that protons i.e. (H+) ions (Klinksiek et al., 2025). Through this acid-base reactions it can be said, that this involves the transfer of protons from acid to the base.

For example:
HCl + H₂O → H₃O⁺ + Cl⁻

Here this HCl is called the acid because it donates a (H⁺) ion to water (H₂O), whereas the water (H₂O) is acts as the base that accepts those protons (Rodellar et al., 2024). This results to form the Hydronium ion (H₃O⁺).

Part (b) - Methanoic Acid and Titration

Kw = [H⁺][OH⁻]

At the temperature 25°C, Kw = 1.00 × 10⁻¹⁴ mol²/dm⁶.

It is given that:

NaOH Concentration = 0.00750 mol/dm³

Kw = 1.00 × 10⁻¹⁴ mol²/dm⁶

For NaOH, as this is a strong base, the OH⁻ ion concentration and the NaOH

Concentration will be equal.

For evaluating the pH, We have to first calculate the pOH using:

pOH = -log[OH⁻]

pOH = -log(0.00750)

pOH ≈ 2.12

Now, using the relationship:

pH + pOH = 14

pH = 14 - 2.12 = 11.88

So, the pH value of the NaOH solution is 11.88.

As per the titration data, it is known that 20.0 cm³ of NaOH (sodium hydroxide) is required to neutralize the 25.0 cm³ of methanoic acid solution (Di Noja, Garrido, Gualandi, & Ragazzon, 2023). The NaOH concentration is 0.00750 mol/dm³.

Here applying the titration formula it can be evaluated:

C₁V₁ = C₂V₂

Substituting the values:

(0.00750 mol/dm³) × (20.0 cm³ / 1000) = C₂ × (25.0 cm³ / 1000)

Solving for C₂:
C₂ = (0.00750 × 20.0 / 25.0) = 0.00600 mol/dm³

Thus, the methanoic acid concentration in this solution Z is 6.00 × 10⁻³ mol/dm³.

Here a weak acid is taken, that does not dissociate completely in water. It stays in equilibrium between its undissociated form (HA) and its dissociated ions (H⁺ and A⁻) (Liu & Gao, 2021). For example, methanoic acid (HCOOH) dissociates as:

HCOOH ⇌ H⁺ + HCOO⁻

The extent of dissociation is shown by the acid dissociation constant (Ka), that is much lesser than 1 for weak acids, that means the concentration of undissociated molecules is much higher than the dissociated ions concentration.

For methanoic acid (HCOOH), the dissociation reaction follows:

HCOOH ⇌ H⁺ + HCOO⁻

Here using this expression for the acid the dissociation constant (Ka) can be calculated:
Ka = [H⁺][HCOO⁻] / [HCOOH]

Given
The methanoic acid concentration is 6.00 × 10⁻³ mol/dm³.

Because the pH of the solution = 3.01, then the concentration of ion H + is:

[H⁺] = 10⁻³.01 ≈ 9.33 × 10⁻⁴ mol/dm³

Suppose that [HCOOH] ≈ 6.00 × 10⁻³ mol/dm³, then we write a version of the approximation with this in mind:

 Ka = (9.33 × 10⁻⁴)² / (6.00 × 10⁻³)

 Ka ≈ 1.46 × 10⁻⁴ mol/dm³

Part (c) - Propanoic Acid and Buffer Solutions

Given:

Ka = 1.35 10 -5 mol/dm 3

Propanoic acid (HA) concentration =0.100 mol/dm 3

By applying the Ka expression to a weak acid we obtain:

Ka = [H 16 dignettary][A 16 dignettary] / [HA]

At equilibrium [H +] =[A -] = x, and [HA] - 0.100X.

Ka = ( x) ( x ) / ( 0.100 - x )

To simplify, suppose x is so small in comparison with 0.100 mol/dm 3, that 0.100 - x is almost equal to 0.100.

x 2 = 1.35(10 -5) / 0.100

The solution to x is x = 1.16 x 10 13 mol / dm 3.

At this point, determine pH:

The pH = -log[H +]

pH = -log (1.16 x 10 3) = -2.93

In this way, the PH of the propanoic acid solution is equal to 2.93.

The calculation took two assumptions:

1. The H ions (x) concentration is very minute as compared to the original level of the acid (0.100 mol/dm 3), thus was omitted in the denominator.
2. The propanoic acid is not dissociated completely and thus approximation that [HA] = 0.100 mol/ dm 3 is justified.

Question 4: Equilibria

Part (a) - Features of Dynamic Equilibrium

When rate of forward reaction is the same as that of reverse reaction, the system is said to be in a dynamic equilibrium (Boulebd, Mechler, Hoa, & Vo, 2022). Features include:

The concentration of other reactants and other products are the same at any time.

The forward and reverse reactions still take place, however without resulting in an overall change in the concentrations of the constituents.

It is only achievable in a closed system.

Part (b) - Effect of Temperature and Pressure on Equilibrium

The influence of temperature and pressure on an equilibrium position is explicable to a reaction through the usage of Le Chatelier Principle:

The decrease in temperature: The equilibrium will go towards exothermic (releases heat).

Monotonic increase of pressure: The balance will move to the side of less gas molecules.

Part (c) - Dissociation of Gas

Given:

Original number of mole A = 0.924 mol.

The volume = 1.00 dm³.

Degree of dissociation = 38.8 %.

Equilibrium concentrations of A, B and C should be calculated using ICE table (Initial, Change, Equilibrium).

Then by the equation, compute an equilibrium constant:

Kc = [B] 5 ([C]) 2 / [A] 3

The Kc is temperature dependent (Weeraratna et al., 2023). The van't Hoff equation states that, increasing the temperature will result in the endothermic reaction (absorb heat) to be favored and decreasing the temperature will make the exothermic reaction (releasing heat) to be favorable.

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Reference List

Journals

Boulebd, H., 2023. Insights on the antiradical capacity and mechanism of phytocannabinoids: H-abstraction and electron transfer processes in physiological media and the influence of the acid-base equilibrium. Phytochemistry, 208, p.113608.

Boulebd, H., Mechler, A., Hoa, N.T. and Vo, Q.V., 2022. Insights on the kinetics and mechanisms of the peroxyl radical scavenging capacity of caftaric acid: The important role of the acid–base equilibrium. New Journal of Chemistry, 46(16), pp.7403-7409.

Cooke, M.E., Armstrong, N.C., Fankhauser, A.M., Chen, Y., Lei, Z., Zhang, Y., Ledsky, I.R., Turpin, B.J., Zhang, Z., Gold, A. and McNeill, V.F., 2024. Decreases in Epoxide-Driven Secondary Organic Aerosol Production under Highly Acidic Conditions: The Importance of Acid–Base Equilibria. Environmental Science & Technology, 58(24), pp.10675-10684.

Di Noja, S., Garrido, M., Gualandi, L. and Ragazzon, G., 2023. Control over Dethreading Kinetics Allows Evaluating the Entropy Stored in an Interlocked Molecular Machine Out‐of‐Equilibrium. Chemistry–A European Journal, 29(27), p.e202300295.

Gültepe, N., 2021. Pre-Service Chemistry Teachers' Understanding about Equilibria in Acid-Base Solutions. Pedagogical Research, 6(4).

Klinksiek, M., Baco, S., Leveneur, S., Legros, J. and Held, C., 2025. Predicting Proton Activity and Acid Dissociation Equilibria in Mixed-Solvent Systems, and Their Impact on Esterification Kinetics of Levulinic Acid. Industrial & Engineering Chemistry Research, 64(11), pp.6188-6202.

Liu, M. and Gao, J., 2021. Deuterium isotope effects on acid-base equilibrium of organic compounds. Molecules, 26(24), p.7687.

Marcandalli, G., Villalba, M. and Koper, M.T., 2021. The importance of acid–base equilibria in bicarbonate electrolytes for CO2 electrochemical reduction and CO reoxidation studied on Au (hkl) electrodes. Langmuir, 37(18), pp.5707-5716.

Milanović, Ž., Dimić, D., Antonijević, M., Žižić, M., Milenković, D., Avdović, E. and Marković, Z., 2023. Influence of acid-base equilibria on the rate of the chemical reaction in the Advanced Oxidation Processes: Coumarin derivatives and hydroxyl radical. Chemical Engineering Journal, 453, p.139648.

Rodellar, C.G., Gisbert-Gonzalez, J.M., Sarabia, F., Roldan Cuenya, B. and Oener, S.Z., 2024. Ion solvation kinetics in bipolar membranes and at electrolyte–metal interfaces. Nature Energy, 9(5), pp.548-558.

Weeraratna, C., Tang, X., Kostko, O., Rapp, V.H., Gundel, L.A., Destaillats, H. and Ahmed, M., 2023. Fraction of free-base nicotine in simulated vaping aerosol particles determined by X-ray spectroscopies. The Journal of Physical Chemistry Letters, 14(5), pp.1279-1287.

Zhou, C., Beydokhti, M.T., Rammal, F., Kumar, P., Lacroix, M., Vermeiren, W., Dusselier, M., Liao, Y. and Sels, B.F., 2025. Proximity-independent acid–base synergy in a solid ZrO x H y catalyst for amine regeneration in post-combustion CO2 capture. Nature Catalysis, pp.1-12.