Question 19M.2.HL.TZ2.2
| Date | May 2019 | Marks available | [Maximum mark: 17] | Reference code | 19M.2.HL.TZ2.2 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Annotate, Deduce, Determine, Explain, Outline, Sketch | Question number | 2 | Adapted from | N/A |
The thermal decomposition of dinitrogen monoxide occurs according to the equation:
2N2O (g) → 2N2 (g) + O2 (g)
The reaction can be followed by measuring the change in total pressure, at constant temperature, with time.
The x-axis and y-axis are shown with arbitrary units.
This decomposition obeys the rate expression:
= k[N2O]
Explain why, as the reaction proceeds, the pressure increases by the amount shown.
[2]
increase in the amount/number of moles/molecules «of gas» [✔]
from 2 to 3/by 50 % [✔]
Students were able in general to relate more moles of gas to increase in pressure.
Outline, in terms of collision theory, how a decrease in pressure would affect the rate of reaction.
[2]
«rate of reaction decreases»
concentration/number of molecules in a given volume decreases
OR
more space between molecules [✔]
collision rate/frequency decreases
OR
fewer collisions per unit time [✔]
Note: Do not accept just “larger space/volume” for M1.
Few students were able to relate the effect of reduced pressure at constant volume with a decrease in concentration of gas molecules and mostly did not even refer to this, but rather concentrated on lower rate of reaction and frequency of collisions. Many candidates lost a mark by failing to explain rate as collisions per unit time, frequency, etc.
Deduce how the rate of reaction at t = 2 would compare to the initial rate.
[1]
half «of the initial rate» [✔]
Note: Accept “lower/slower «than initial rate»”.
Though the differential equation was considered to be misleading by teachers, most candidates attempted to answer this question, and more than half did so correctly, considering they had the graph to visualize the gradient.
It has been suggested that the reaction occurs as a two-step process:
Step 1: N2O (g) → N2 (g) + O (g)
Step 2: N2O (g) + O (g) → N2 (g) + O2 (g)
Explain how this could support the observed rate expression.
[2]
1 slower than 2
OR
1 rate determinant step/RDS [✔]
1 is unimolecular/involves just one molecule so it must be first order
OR
if 1 faster/2 RDS, second order in N2O
OR
if 1 faster/2 RDS, first order in O [✔]
Most students were able to identity step 1 as the RDS/slow but few mentioned unimolecularity or referred vaguely to NO2 as the only reagent (which was obvious) and got only 1 mark.
The experiment is repeated using the same amount of dinitrogen monoxide in the same apparatus, but at a lower temperature.
Sketch, on the axes in question 2, the graph that you would expect.
[2]
smaller initial gradient [✔]
initial pressure is lower AND final pressure of gas lower «by similar factor» [✔]
Many students drew a lower initial gradient, but most did not reflect the effect of lower temperature on pressure at constant volume and started and finished the curve at the same pressure as the original one.
The experiment gave an error in the rate because the pressure gauge was inaccurate.
Outline whether repeating the experiment, using the same apparatus, and averaging the results would reduce the error.
[1]
no AND it is a systematic error/not a random error
OR
no AND «a similar magnitude» error would occur every time [✔]
Almost all candidates identified the inaccurate pressure gauge as a systematic error, thus relating accuracy to this type of error.
The graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.
The rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.
Annotate and use the graph to outline why a catalyst has this effect.
[2]
catalysed and uncatalysed Ea marked on graph AND with the catalysed being at lower energy [✔]
«for catalysed reaction» greater proportion of/more molecules have E ≥ Ea / E > Ea
OR
«for catalysed reaction» greater area under curve to the right of the Ea [✔]
Note: Accept “more molecules have the activation energy”.
The graph was generally well done, but in quite a few cases, candidates did not mention that increase of rate in the catalyzed reaction was due to E (particles) > Ea or did so too vaguely.
Determine the standard entropy change, in J K−1, for the decomposition of dinitrogen monoxide.
2N2O (g) → 2N2 (g) + O2 (g)
[2]
ΔSθ = 2(Sθ(N2)) + Sθ(O2) – 2(Sθ(N2O))
OR
ΔSθ = 2 × 193 «J mol-1 K-1» + 205 «J mol-1 K-1» – 2 × 220 «J mol-1 K-1» [✔]
«ΔSθ = +»151 «J K-1» [✔]
Note: Award [2] for correct final answer.
Candidates were able to calculate the ΔS of the reaction, though in some cases they failed to multiply by the number of moles.
Dinitrogen monoxide has a positive standard enthalpy of formation, ΔHfθ.
Deduce, giving reasons, whether altering the temperature would change the spontaneity of the decomposition reaction.
[3]
exothermic decomposition
OR
ΔH(decomposition) < 0 [✔]
TΔSθ > ΔHθ
OR
ΔGθ «= ΔHθ – TΔSθ» < 0 «at all temperatures» [✔]
reaction spontaneous at all temperatures [✔]
Though the question asked for decomposition (in bold), most candidates ignored this and worked on the basis of a the ΔH of formation. However, many did write a sound explanation for that situation. On the other hand, in quite a number of cases, they did not state the sign of the ΔH (probably taking it for granted) nor explicitly relate ΔG and spontaneity, which left the examiner with no possibility of evaluating their reasoning.