Ethanol is commonly used as a solvent in organic synthesis and for the preparation of a range of commercial products, including hand sanitiser. The evaporation of one mole of ethanol at 298 K has a Gibbs free energy change of 5.40 kJ mol⁻¹.

CH₃CH₂OH(l)  ⇌  CH₃CH₂OH(g)       ΔG°_{298 K} = +5.40 kJ mol⁻¹

1. Predict the sign for ΔH° and ΔS° for the evaporation of 1 mole of ethanol. [0.5 mark each]

   • ΔH° will be
     negative

     positive

     equal to zero
     .
   • ΔS° will be
     negative

     positive

     equal to zero
     .

2. Predict if the evaporation of ethanol under standard conditions at 298 K will be always spontaneous, never spontaneous or only sometimes spontaneous. [1 mark]
   The evaporation of ethanol will be
   always spontaneous

   never spontaneous

   only sometimes spontaneous
   .

The rate equation for a given reaction with reactants A and B was found to be

r = k[A][B]

1. The reaction is 1st order with respect to reactant A

2. The reaction is 0 order with respect to reactant B

3. The reaction is 2nd order overall

4. The units for the rate constant are mol^–1 L s^–1

Quantum chemistry calculations can predict the value of the partial charge on an atom when bonded to other atoms, i.e. the excess or deficiency of electrons compared to the number of protons around the atom. A quantum chemistry calculation for SCl₂ (with the sulfur as the central atom) was performed, and the calculation gave partial charges of –0.402 for the sulfur atom and 0.201 for both chlorine atoms.

• Based on these results, sulfur has
  a more negative

  the same

  a more positive
  electronegativity than chlorine. [1 mark]

• If a similar calculation was done for water, the magnitude of the partial charge on oxygen would be 
  more negative than

  the same as

  more positive than
  the partial charge of sulfur in SCl₂. [1 mark]

Identify which of the following spectra would be

seen by the space telescope through path A.

The data in the table below were obtained for the reaction:

The order of the reaction with respect to [A] is ________

A decomposition reaction was determined to be first order.

After 627 seconds only 42.2% of the starting material was remaining.

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 ionic products have large negative hydration free energies

   the entropy of the reaction is very positive in water

   the acid is more stable in the gas phase

   The H-X bond is weaker in water

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

N₂(g)  + 3H₂(g) ⇌ 2NH₃(g)

equals 4.7 × 10^-3. If, at equilibrium, the concentration of H₂ = 5.9 mol/L and the concentration of NH₃ = 4.6 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 319 K:

2NO(g)  +  O₂(g)  ⇌  2NO₂(g)

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