UPSC Combined Geo-Scientist, Mains · 2016 · Mains Paper II
Q1(i). Rate-law determination
Question
Derive the rate law for A+B→ products from the data below. Concentrations are in mol dm⁻³ and rates in mol dm⁻³ s⁻¹.tabularccc[A]&[B]&Rate 0.006&0.001&0.012 0.006&0.002&0.024 0.002&0.0015&0.002 0.004&0.0015&0.008tabular
Answer in brief
r=k[A]²[B], where k=3.33×10⁵ dm⁶ mol⁻² s⁻¹.
Key points
- Write r=k[A]m[B]ⁿ. Comparing the first two experiments, doubling [B] at fixed [A] doubles the rate, so n=1.
- Comparing the last two experiments, doubling [A] at fixed [B] quadruples the rate, so m=2. The overall order is three.
- From the first row, k=0.012/[(0.006)²(0.001)]=3.33×10⁵. The other rows give the same value.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversRate-law determinationChemical kinetics
All parts of Question 1
- 1(a) Virial equation5 marks
- 1(b) Inclined capillary5 marks
- 1(c) Entropy and diffraction5 marks
- 1(d) Equipartition in hydrogen5 marks
- 1(e) Expansion cooling of helium5 marks
- 1(f) Temperature dependence of equilibrium5 marks
- 1(g) Ammonium chloride pH5 marks
- 1(h) Reverse osmosis5 marks
- 1(i) Rate-law determination5 marks
- 1(j) Monolayer surface area5 marks
- 1(k) Electrolysis of copper sulfate5 marks
- 1(l) Ionization of helium ion5 marks
- 1(m) Photochemical quantum yield5 marks
- 1(n) Momentum in a box5 marks
- 1(o) Formaldehyde electronic transitions5 marks
- 1(p) NMR distinction of ethers5 marks