CHEMISTRY 3A PAPER 2 (CHEM09005) 2021
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CHEMISTRY 3A PAPER 2 (CHEM09005)
2021
1. Answer all parts.
(a) A system can adopt one of four different discrete microstates with respective
energies E0 = −5 kJ mol− 1, E1 = −3 kJ mol− 1, E2 = −3 kJ mol− 1 and E3 = −1 kJ mol− 1 .
(i) Compute the value of the partition function at temperature T1 = 100 K. [2]
(ii) Find the value of the temperature for which the population ofthe first excited
energy level equals that of the ground state. [3]
(iii) Compute in kJ K− 1 mol− 1 the value of the Gibbs entropy of the system at the
two temperatures T1 = 100 K and T2 = 105 K. Explain why the entropy is
greater at T2 than T1 . [3]
(b) A monoatomic ideal gas of ⃞ particles of atomic mass ⃞ with a two-fold
degenerate electronic ground state is mixed with a homonuclear diatomic ideal gas of ଶ particles of atomic mass ଶ with bond length ଶ , angular vibrational frequency ଶ , and a single electronic ground state, in a container of volume at a temperature .
(i) Derive an expression for the canonical partition function of the system in terms of de Broglie wavelengths, rotational and vibrational temperatures, under the assumption that equipartition holds. [4]
(ii) Derive an expression for the Helmholtz free energy of the system using
Stirling’s approximation. [2]
(iii) Using the equipartition theorem or otherwise show that the internal energy of
the system is
ൌ ቀ ⃞ ⃞ ଶቁ ⃞ .
(iv) Show that the entropy is
ൌ ⃞ ቈ െ ln ቆ4ൗଶ ቇ⃞ ⃞ଶ ⃞ െ ln ቌ ቍ⃞ .
2. Answer all parts.
A study of the gas-phase isomerisation of cis-but-2-ene to trans-but-2-ene, in a sample of pure cis-but-2-ene, found that the order of the reaction depended on the pressure (concentration) of cis-but-2-ene. At high concentration, first order kinetics were observed, but at low pressure the rate became 2nd order in [cis-but-2-ene].
Write the mechanism of the reaction and explain the processes that are involved. [4]
Show that this mechanism accounts for the experimental observations. [6]
Explain why isomerisation reactions in solution always follow 1st-order
kinetics, regardless of reactant concentration. [2]
The influence of ionic strength on the rate constant for the dimerisation ofa cationic protein was investigated in aqueous solution at 298 K. The data were processed and plotted to give a straight-line graph with a gradient of 9.16 and an intercept of 0.
Explain what was plotted and what can be deduced from the graph. [4]
Explain, in terms of activated complex theory, why the rate of dimerisation of any charged protein would be expected to increase with increasing ionic strength. [4]
The mechanism of the reaction of H2 and Br2 is given below, together with the rate constant of each step, at 500 K and a pressure of 1 atm.
k / dm3mol- 1 s- 1
Br + H2 → HBr + H 3.8 10ଶ (2)
H + Br2 → HBr + Br 9.6 10 ⃞⃞ (3)
H + HBr → H2 + Br 7.2 10ଽ (4)
Br + Br + M → Br2 + M 4.2 10 ⃞ଷ [M] (5)
The rate law according to this mechanism is:
⃞
െdሾHଶ ሿ ଶ ൬ ହ(⃞)⃞ ଶ ሾBrଶ ሿ ଶ(⃞) ሾHଶ ሿ
d ൌ 1 ⃞
(b) Explain why the rate constant of step (2) is much less than that of step (3). [2]
(c) Explain why termination by the recombination of H atoms is not significant and so is not included in the mechanism. [3]
(d) When the initial rate of the reaction was measured, the rate law was found to be:
െdሾHଶ ሿ ⃞
Discuss whether this is consistent with the mechanism. [3]
(e) The chain length, , is the mean number of propagation steps that occur before
termination.
ൌ rate of propagation
rate of termination
Show that, for this reaction, the chain length is given by:
ൌ ଶ ሾHଶ ሿ
ହ ሾBrሿሾMሿ .
[4]
(f) The activation energies of the reactions I + H2 and Br +H2 are 142 kJ mol- 1 and
82 kJ mol- 1 , respectively. Calculate the rate constant for I + H2 at 500 K, stating any assumption that you make. [3]
(g) With reference to your answer to part (f), explain why the reaction H2 + I2 does not
occur by a chain mechanism. [3]
(a) Investigations into phase formation in the Pr2O3-SiO2 phase diagram found the
presence of three binary phases, which were labelled A, B and C. Also, elemental analysis was carried out and the resulting weight % ofPr2O3 for each phase is given in the table below. Determine the chemical formulae for the phases A, B and C.
Phase |
Weight % Pr2O3 |
A |
84.6 |
B |
78.5 |
C |
73.3 |
[3]
(b) The three binary phases melt congruently. Phase B was found to have a lower limit
of stability, forming above 1650 °C. The system has a region of liquid immiscibility. From the information below, construct a temperature versus composition (as mol%) phase diagram for the Pr2O3-SiO2 system. Focus on the temperature range above 1500 °C for this phase diagram. Label the phases present in each region of the phase diagram.
Melting temperature of Pr2O3(s) |
2200 °C |
Melting temperature of SiO2(s) |
1750 °C |
Congruent melting of A(s) |
1900 °C |
Lower limit of stability of B(s) |
1650 °C |
Congruent melting of B(s) |
1950 °C |
Congruent melting of C(s) |
1800 °C |
Eutectic composition I |
40 mol% SiO2 , 1800 °C |
Eutectic composition II |
55 mol% SiO2 , 1850 °C |
Eutectic composition III |
65 mol% SiO2 , 1750 °C |
Eutectic composition IV |
70 mol% SiO2 , 1550 °C |
Eutectic composition V |
98 mol% SiO2 , 1700 °C |
Peritectic composition |
75 mol% SiO2 , 1700 °C |
Upper critical temperature |
90 mol% SiO2 , 2200 °C |
Liquid immiscibility boundary I |
80 mol% SiO2 , 2000 °C |
Liquid immiscibility boundary II |
85 mol% SiO2 , 2100 °C |
Liquid immiscibility boundary III |
95 mol% SiO2, 2000 °C |
[10]
(c) 45 mol% Pr2O3 and 55 mol% SiO2 were ground together and heated to 2000 °C. The mixture was cooled to 1500 °C.
(i) Draw the cooling curve, labelling any discontinuities. [4]
(ii) Determine what phases co-exist at 1600 °C and determine the proportion of
each phase. [3]
5. Answer all parts.
(a) The rotational-vibrational spectrum of a diatomic molecule in the gas phase is
shown below. The reduced mass of the molecule is 0.9953 g mol− 1 .
(i) Calculate the rotational constant, B, and the bond length of the molecule. Assume that B is the same in both the v = 0 and v = 1 vibrational levels. [3]
(ii) Estimate the temperature of the gas based on the data below. [2]
(b) The fundamental and first overtone bands for N−O were reported as 1876.2 cm− 1
and 3724.6 cm− 1 .
(i) Use this data to estimate the equilibrium vibrational frequency, ωe, and the anharmonicity constant, ωexe . [3]
(ii) Calculate the force constant. [3]
(c) Fluorescence data from a molecule are shown below.
(i) From the time-resolved emission data in Figure (a), it is possible to find the emission lifetime, τ0, of the excited state. Calculate τ0 based on the linear fit to the natural logarithm of the emission intensity I(t). The noise in the measurement becomes dominant for times longer than ca. 500 ns. [5]
.(ii) In another experiment, the change in emission intensity, IQ, as a function of
concentration, [Q], of a quencher was measured (Figure (b)). I0 is the maximum intensity measured without any quencher. Use the data in the graph and your answer from part (i) to calculate the quenching rate constant, kQ . [4]
6. Answer all parts.
The following text appeared in a recent paper which described the synthesis and characterisation of the complex [RuH2(NHC)4]. The structure of the N-heterocyclic carbene ligand NHC is shown below.
“Diffraction data were collected at 150 K on an Agilent Supernova diffractometer using X-rays of wavelength 1.54184 Å . C20H34N8Ru, space group P4/nnc, a = b = 10. 16000(8) Å, c = 11.36236(15) Å, α = β = γ = 90 o . The structure was solved by charge flipping (SUPERFLIP) and completed and refined by iterative cycles of Fourier syntheses and least-squares refinement (SHELXL). Final R factor was 1.86% [for 580 data which had F > 4(F)], for 38 variables. H-atoms were located in Fourier maps, and anisotropic displacement parameters were refined for all non-H atoms. The final difference map extremes were +0.29 and –0.31 e Å–3 .”
A plot illustrating the structure of the complex is shown below. The Ru, N and H atoms have been labelled, all other atoms are carbon. The H atoms of the NHC ligand are omitted for clarity.
(b) Comment on the quality of this crystal structure determination. [1]
(c) The H atoms were located in a Fourier synthesis. Why are H atoms difficult to locate in crystal structures such as this one? [2]
2022-04-22