DP Chemistry (first assessment 2025)
Question 19M.2.SL.TZ2.2c(ii)
| Date | May 2019 | Marks available | [Maximum mark: 2] | Reference code | 19M.2.SL.TZ2.2c(ii) |
| Level | SL | Paper | 2 | Time zone | TZ2 |
| Command term | Explain | Question number | c(ii) | Adapted from | N/A |
c(ii).
[Maximum mark: 2]
19M.2.SL.TZ2.2c(ii)
(c(ii))
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]
Markscheme
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 [✔]
Examiners report
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.
Syllabus sections
Reactivity 2. How much, how fast and how far? » Reactivity 2.2—How fast? The rate of chemical change » Reactivity 2.2.6—Many reactions occur in a series of elementary steps. The slowest step determines the rate of the reaction. Evaluate proposed reaction mechanisms and recognize reaction intermediates. Distinguish between intermediates and transition states, and recognize both in energy profiles of reactions.