Consider the ionisation of a monoprotic acid (HX) in water and in the gas phase at 298 K.

HX(aq)  ⇌  H⁺(aq) + X^-(aq)       ΔG°_water = +15.40 kJ mol⁻¹

HX(g)    ⇌  H⁺(g)   + X^-(g)        ΔG°_gas = +1050.50 kJ mol⁻¹

Please give your numerical answers to 2 decimal places.

1. Calculate the pK_a of HX in water. [1 mark]
   pK_a =

2. The degree of ionisation refers to the percentage of acid that exists in the ionised form. Calculate the degree of ionisation of a 0.8 mol L-1 aqueous solution of HX. [1 mark]
   % ionisation = 

3. A thermodynamic cycle can be constructed to relate the gas phase and aqueous phase Gibbs free energies of ionisation as shown below.

   

   According to this cycle, ΔG

   ^o_water can be expressed as some linear combination of ΔG^o_gas and the hydration free energies (ΔG°_hyd) of the reactant and products. [1 mark]
   The correct expression for ΔG°_water 

   is

   ΔG°_gas - ΔG°_hyd(H⁺) - ΔG°_hyd(X^-) - ΔG°_hyd(HX)

   ΔG°_gas + ΔG°_hyd(H⁺) + ΔG°_hyd(X^-) + ΔG°_hyd(HX)

   ΔG°_gas + ΔG°_hyd(H⁺) + ΔG°_hyd(X^-) - ΔG°_hyd(HX)

   ΔG°_gas - ΔG°_hyd(H⁺) - ΔG°_hyd(X^-) + ΔG°_hyd(HX)

4. Using the relationship in part (b), and given that the ΔG°_hyd(X^-) and ΔG°_hyd(HX) are -400 and -20 kJ mol^-1 respectively, determine the value of ΔG°_hyd(H⁺) [1 mark]
   kJ mol^-1.

5. The reason why HX is a stronger acid in water than in the gas phase is because [1 mark]
   the entropy of the reaction is very positive in water

   the ionic products have large negative hydration free energies

   the acid is more stable in the gas phase

   The H-X bond is weaker in water

1. 5.0 g of NaOH was added to 1.0 L of 1.0 M HNO₂ (K_a = 4.0 × 10^–4). How much more NaOH is needed to be added to achieve a pH of 3.00? [3 marks]
   g

2. A 500 mL sample of the buffer created above was used to host a chemical experiment that had to be kept within a pH range of 2.9–3.2 to be considered ‘valid’. Over the course of the experiment, a total of 45 mL of a 0.1 M HCl solution was added to the buffer solution. What would the pH of the solution be at the end of the experiment? [2 marks]

3. Would the experiment be considered ‘valid’ at the end of the experiment? [1 mark]
   yes

   no

   not enough information to judge

Complete the table below by selecting the correct response from each of the drop-down boxes to identify the size of the marked angle and the hybridisation of the central atom for the hypothetical molecule given in each row. [1 mark each]

Molecule	Size of marked angle	Hybridisation of central atom
	180° slightly less than 180° slightly greater than 180° slightly less than 120° slightly greater than 120° 120°	sp sp² sp³ sp⁴ unhybridised
	109.5° slightly less than 109.5° slightly greater than 109.5° slightly less than 120° slightly greater than 120° 120°	sp sp² sp³ sp⁴ unhybridised
	109.5° slightly less than 109.5° slightly greater than 109.5° slightly less than 120° slightly greater than 120° 120°	sp sp² sp³ sp⁴ unhybridised

 The partition coefficient (logP) is a measure of the equilibrium distribution of a solute between two immiscible liquids, usually water and 1-octanol.

When phenol is behaving as the solute with water and 1-octanaol as the solvents its log P value is 30.2. What does this value say about the relative solubility of a phenol molecule in each solvent liquids? Explain your answer with reference to theory covered in the course regarding miscibility and intermolecular forces. [4 marks]

 Note: in the formula above the equilibrium concentration of a solute S in solvent x is denoted as [S]_x ( e.g. [S]_aq denotes what concentration of a solute is dissolved in the water layer)

The equilibrium constant in terms of concentrations (K_c) for this reaction:

N₂(g)  + 3H₂(g) ⇌ 2NH₃(g)

equals 4.6 × 10^-3. If, at equilibrium, the concentration of H₂ = 5.9 mol/L and the concentration of NH₃ = 2.2 mol/L, what is the equilibrium concentration of N₂(g)?

OUT OF SYLLABUS of CHEM1832

Calculate the value of K_p for the reaction below given equilibrium concentrations of [NO] = 0.0080 M, [O₂] = 0.014 M and [NO₂] = 0.012 M at 299 K:

2NO(g)  +  O₂(g)  ⇌  2NO₂(g)

Your answer must be given to the correct number of significant figures to receive full marks.

Consider the ionisation of a monoprotic acid (HX) in water and in the gas phase at 298 K.

HX(aq)  ⇌  H⁺(aq) + X^-(aq)       ΔG°_water = +15.40 kJ mol⁻¹

HX(g)    ⇌  H⁺(g)   + X^-(g)        ΔG°_gas = +1050.50 kJ mol⁻¹

Please give your numerical answers to 2 decimal places.

1. Calculate the pK_a of HX in water. [1 mark]
   pK_a =

2. The degree of ionisation refers to the percentage of acid that exists in the ionised form. Calculate the degree of ionisation of a 0.8 mol L-1 aqueous solution of HX. [1 mark]
   % ionisation = 

3. A thermodynamic cycle can be constructed to relate the gas phase and aqueous phase Gibbs free energies of ionisation as shown below.

   

   According to this cycle, ΔG

   ^o_water can be expressed as some linear combination of ΔG^o_gas and the hydration free energies (ΔG°_hyd) of the reactant and products. [1 mark]
   The correct expression for ΔG°_water 

   is

   ΔG°_gas - ΔG°_hyd(H⁺) - ΔG°_hyd(X^-) + ΔG°_hyd(HX)

   ΔG°_gas + ΔG°_hyd(H⁺) + ΔG°_hyd(X^-) - ΔG°_hyd(HX)

   ΔG°_gas - ΔG°_hyd(H⁺) - ΔG°_hyd(X^-) - ΔG°_hyd(HX)

   ΔG°_gas + ΔG°_hyd(H⁺) + ΔG°_hyd(X^-) + ΔG°_hyd(HX)

4. Using the relationship in part (b), and given that the ΔG°_hyd(X^-) and ΔG°_hyd(HX) are -400 and -20 kJ mol^-1 respectively, determine the value of ΔG°_hyd(H⁺) [1 mark]
   kJ mol^-1.

5. The reason why HX is a stronger acid in water than in the gas phase is because [1 mark]
   the acid is more stable in the gas phase

   the ionic products have large negative hydration free energies

   The H-X bond is weaker in water

   the entropy of the reaction is very positive in water

Dibromoethane and potassium iodide react in methanol (the solvent) according to the equation:

C₂H₄Br₂ + 3KI → C₂H₄ + 2KBr + KI₃

The initial rate of this reaction was determined in a series of experiments at 65 °C, giving the data below:

a) Determine the order of the reaction with respect to C₂H₄Br₂

b) Determine the order of the reaction with respect to KI

c) Determine the value of the rate constant and select the appropriate units for it.

1. 5.0 g of NaOH was added to 1.0 L of 1.0 M HNO₂ (K_a = 4.0 × 10^–4). How much more NaOH is needed to be added to achieve a pH of 3.00? [3 marks]
   g

2. A 500 mL sample of the buffer created above was used to host a chemical experiment that had to be kept within a pH range of 2.9–3.2 to be considered ‘valid’. Over the course of the experiment, a total of 45 mL of a 0.1 M HCl solution was added to the buffer solution. What would the pH of the solution be at the end of the experiment? [2 marks]

3. Would the experiment be considered ‘valid’ at the end of the experiment? [1 mark]
   yes

   no

   not enough information to judge

The following table contains thermodynamic data that may be helpful to answer the following TWO questions. (Note this is not the true values for these parameters.)

1. Consider the dissociation of O₃(g) to O₂(g) + O(g) in the absence of light. The standard entropy of the reaction at 298 K is J K^–1 mol^–1 while the standard Gibbs free energy at 298 K is kJ mol^–1. [2 marks]

2. This reaction
   would

   would not
   be spontaneous in the absence of light. The spontaneity of the reaction 
   would

   would not
   increase with temperature. [2 marks]