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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper I

Q6(a)(ii). Alkaline hydrolysis of ethyl acetate is second order

  • 10 marks
  • Answered
Question

In the hydrolysis of ethyl acetate using equal concentrations of ester and NaOH, the following results were obtained:
1.2tabular|l|c|c|c|c|c| t (minutes) & 0 & 5 & 15 & 25 & 35 HCl (ml) & 16.0 & 10.24 & 6.13 & 4.32 & 3.41tabular
Show that the reaction is of second order.

Answer in brief

k=1/t·(V₀-Vₜ)/(V₀Vₜ) is constant (0.0070, 0.0067, 0.0068, 0.0066; mean 6.8×10⁻³ mL⁻¹ min⁻¹), and 1/V against t is a straight line: the reaction is second order.

Key points

  • a sample of the mixture is run into acid (or titrated with HCl); the titre V is proportional to the OH⁻ not yet used, and with equal starting concentrations [OH⁻]=[ester]. So V₀∝ a and Vₜ∝(a-x).
  • (1)/(a-x)-1/a=kt, in titres k'=1/t((1)/(Vₜ)-(1)/(V₀))=1/t·(V₀-Vₜ)/(V₀Vₜ)
  • The second-order constant is steady within experimental scatter: k'=6.8×10⁻³ mL⁻¹ min⁻¹ (mean)
  • The first-order “constant” falls steadily from 0.089 to 0.044 min⁻¹, so the reaction is not first order.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. What is titrated: a sample of the mixture is run into acid (or titrated with HCl); the titre V is proportional to the OH⁻ not yet used, and with equal starting concentrations [OH⁻]=[ester]. So V₀∝ a and Vₜ∝(a-x).
  2. Second-order law for equal concentrations: (1)/(a-x)-1/a=kt, in titres k'=1/t((1)/(Vₜ)-(1)/(V₀))=1/t·(V₀-Vₜ)/(V₀Vₜ)
t / minVₜ / mL1/Vₜ / mL⁻¹k'=(V₀-Vₜ)/(t V₀Vₜ) / mL⁻¹ min⁻¹first order: 1tln(V₀)/(Vₜ) / min⁻¹
016.00.0625——
510.240.09775.76/(5×16.0×10.24)=0.007030.0893
156.130.16319.87/(15×16.0×6.13)=0.006710.0640
254.320.231511.68/(25×16.0×4.32)=0.006760.0524
353.410.293312.59/(35×16.0×3.41)=0.006590.0442
  1. The second-order constant is steady within experimental scatter: k'=6.8×10⁻³ mL⁻¹ min⁻¹ (mean)
  2. The first-order “constant” falls steadily from 0.089 to 0.044 min⁻¹, so the reaction is not first order.
  3. Graph: 1/Vₜ against t is a straight line (correlation 0.9998), slope 6.6×10⁻³ mL⁻¹ min⁻¹, intercept ≈1/V₀. Hence the reaction is second order: first order in ester and first order in OH⁻.
  4. t_(1/2)=1/k'V₀≈9.5 min: the titre falls to 8.0 mL near 9–10 min, as the data show between 5 and 15 min. To express k in dm³ mol⁻¹ min⁻¹, multiply k' by the volume of sample and divide by the molarity of the HCl (neither is given).

Application. saponification of esters (soap making from fats with NaOH) follows this second-order rate law; following it by conductance or titration is a standard laboratory method for second-order kinetics.

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Model answer to Q6(a)(ii), 2018 Paper I: solution page 36
Solution page 36 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 6