Cyanomethane CH3CN, commonly known as acetonitrile, is a toxic volatile liquid, used as solvent to purify steroids and to extract fatty acids from fish oils. Acetonitrile can be synthesised from methyl iso-nitrile by the isomerisation reaction: CH3NC (g) → CH3CN (g) This reaction obeys the first-order rate law: Rate = k [CH3NC], in the presence of Argon. The activation energy for this reaction is 161 kJ mol−1, and the rate constant is 6.6 × 10−4 s−1 at 500. K. Calculate the time (in seconds) required for the concentration of CH3NC to decrease to 95% of its initial value, at 300. K.

Question

Cyanomethane CH3CN, commonly known as acetonitrile, is a toxic volatile liquid, used as solvent to purify steroids and to extract fatty acids from fish oils. Acetonitrile can be synthesised from methyl iso-nitrile by the isomerisation reaction: CH3NC (g) → CH3CN (g) This reaction obeys the first-order rate law: Rate = k [CH3NC], in the presence of Argon. The activation energy for this reaction is 161 kJ mol−1, and the rate constant is 6.6 × 10−4 s−1 at 500. K. Calculate the time (in seconds) required for the concentration of CH3NC to decrease to 95% of its initial value, at 300. K.

BlackTom AI · Accepted Answer

Arrhenius equation relates k at two temperatures: k2 = k1 · exp[−Ea/R · (1/T2 − 1/T1)].

1) Relevant concepts
- First‑order kinetics: [A] = [A]0 e^(−kt), so t = −ln([A]/[A]0)/k.
- Temperature dependence: Arrhenius law k = A e^(−Ea/RT); to change T use the ratio form above.
- Constants: Ea = 161 kJ mol−1 = 161000 J mol−1, R = 8.314 J mol−1 K−1, k1 = 6.6×10−4 s−1 at T1 = 500 K, T2 = 300 K.

2) Step‑by‑step calculation
- Compute the reciprocal temperature difference:
  1/T2 − 1/T1 = 1/300 − 1/500 = 0.0033333333 − 0.002 = 0.0013333333 K−1.
- Compute Ea/R:
  Ea/R = 161000 J mol−1 / 8.314 J mol−1 K−1 = 19367.462 (approximately).
- Multiply to get the exponent:
  Ea/R · (1/T2 − 1/T1) = 19367.462 × 0.0013333333 = 25.823283.
- Thus k at 300 K:
  k300 = 6.6×10−4 × exp(−25.823283) = 6.6×10−4 × 6.0910×10−12 ≈ 4.0190×10−15 s−1.

3) Time for 95% remaining (first order)
- Fraction remaining = 0.95 = e^(−k t) ⇒ t = −ln(0.95)/k300.
- −ln(0.95) = 0.051293.
- t = 0.051293 / 4.0190×10−15 s−1 = 1.2704503425155×10^13 s.

4) Why this is correct
- The Arrhenius relation correctly scales the known k at 500 K to 300 K using the given Ea. The first‑order decay formula then gives the time for a specified fractional decrease.

5) Other options
- No multiple‑choice options provided; alternative approaches would be algebraically equivalent.

Conclusion: The required time is 1.2704503425155×10^13 s, i.e. 12704503425155 seconds, matching the provided answer.

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