Q5(c). Half-life of two parallel first-order reactions
Show that half-life period (t_(1/2)) of the following parallel reaction is (ln(2))/(k₁+k₂).
A k₁ B, A k₂ C (Both reactions are elementary).
Answer in brief
A disappears by both paths at once: -d[A]/dt=(k₁+k₂)[A], a first-order decay with the rate constant k₁+k₂; so [A]=[A]₀e^(-(k₁+k₂)t) and t_(1/2)=2/(k₁+k₂).
Key points
- (d[B])/(dt)=k₁[A], (d[C])/(dt)=k₂[A]
- Loss of A by both paths: -(d[A])/(dt)=k₁[A]+k₂[A]=(k₁+k₂)[A]
- ∫_([A]₀)^([A])(d[A])/([A])=-(k₁+k₂)∫₀^tdt, ln([A]₀)/([A])=(k₁+k₂) t
- at t=t_(1/2), [A]=[A]₀/2: 2=(k₁+k₂) t_(1/2), t_(1/2)=(2)/(k₁+k₂)=(0.693)/(k₁+k₂)
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
- Rate laws (unimolecular elementary steps): (d[B])/(dt)=k₁[A], (d[C])/(dt)=k₂[A]
- Loss of A by both paths: -(d[A])/(dt)=k₁[A]+k₂[A]=(k₁+k₂)[A]
- Integrate from [A]₀ at t=0: ∫_([A]₀)^([A])(d[A])/([A])=-(k₁+k₂)∫₀^tdt, ln([A]₀)/([A])=(k₁+k₂) t
- Half-life: at t=t_(1/2), [A]=[A]₀/2: 2=(k₁+k₂) t_(1/2), t_(1/2)=(2)/(k₁+k₂)=(0.693)/(k₁+k₂)
- Independent of [A]₀, as for any first-order reaction; it is shorter than the half-life of either path alone, because A is consumed by both.
- Products: dividing the two rate laws, d[B]/d[C]=k₁/k₂, so [B]/[C]=k₁/k₂ at every instant: [B]=(k₁)/(k₁+k₂)[A]₀(1-e^(-(k₁+k₂)t)). The measured k₁+k₂ and the product ratio give k₁ and k₂ separately.
Application. the nitration of toluene to o-, m- and p-nitrotoluene, and the decay of a nucleus by two branches (for example ⁴⁰K to ⁴⁰Ca and ⁴⁰Ar), follow this scheme: one overall half-life, with the branching ratio fixed by k₁/k₂.
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All parts of Question 5
- 5(a) Degeneracy of the hydrogenic level with E=-R_H/3610 marks
- 5(b) van der Waals constants of ethylene from T_c and p_c10 marks
- 5(c) Half-life of two parallel first-order reactions10 marks
- 5(d) Signs of Δ H, Δ S and Δ G for spontaneous adsorption10 marks
- 5(e) Decay routes of an electronically excited molecule: Jablonski diagram10 marks