Q4(b)(ii). Rate coefficients of two parallel first-order reactions

Given: C_(B,0)=0.5, C_B=0.05 and C_(D')=0.1 kmol m⁻³ at t=40 min; no D or D' at the start.

Method, and why: both steps are elementary and unimolecular, so each is first order in B:

-(dC_B)/(dt)=(k₁+k₂) C_B, (dC_D)/(dt)=k₁C_B, dC_(D')dt=k₂C_B

Dividing the last two, dC_D/dC_(D')=k₁/k₂. Both products start at zero, so C_D/C_(D')=k₁/k₂ at every instant.

1. Total rate constant from the decay of B: k₁+k₂=1/tlnC_(B,0)C_B=(ln(0.5/0.05))/(40 min)=(2.303)/(40)=0.0576min⁻¹ (9.59×10⁻⁴s⁻¹)
2. Mass balance: B lost =0.5-0.05=0.45 kmol m⁻³, so C_D=0.45-0.10=0.35 kmol m⁻³.
3. Ratio: (k₁)/(k₂)=C_DC_(D')=(0.35)/(0.10)=3.5
4. k₁=0.0576×(0.35)/(0.45), k₁=0.0448min⁻¹=7.46×10⁻⁴s⁻¹
5. k₂=0.0576×(0.10)/(0.45), k₂=0.0128min⁻¹=2.13×10⁻⁴s⁻¹

Reading the answer: k₁+k₂=0.0448+0.0128=0.0576 min⁻¹, as required. B has a half-life of 2/(k₁+k₂)=12.0 min, and the measured drop to one-tenth in 40 min is 3.3 half-lives. Of all B that reacts, 77.8 % gives D and 22.2 % gives D', so at the end C_D=0.389 and C_(D')=0.111 kmol m⁻³. The product ratio is fixed by the rate constants, not by the stabilities of D and D' (kinetic control).

Application. the gas-phase decomposition of acetic acid at 1189 K runs by two parallel first-order paths, to CH₄ + CO₂ (k₁=3.74 s⁻¹) and to ketene CH₂CO + H₂O (k₂=4.65 s⁻¹), so the largest possible yield of ketene is k₂/(k₁+k₂)=55.4 %.