The first order rate constant for a unimolecular gas phase reaction A → products that follows Lindemann mechanism is $2.0 \mathrm{s}^{-1}$ at $\mathrm{p}_{\mathrm{A}}=1 \mathrm{atm}$ and $4.0 \mathrm{s}^{-1}$ at $\mathrm{p}_{\mathrm{A}}=2 \mathrm{atm}$. The rate constant for the activation step is
(a)$1.0 \mathrm{atm}^{-1} \mathrm{s}^{-1}$
(b)$2.0 \mathrm{atm}^{-1} \mathrm{s}^{-1}$
(c)$4.0 \mathrm{atm}^{-1} \mathrm{s}^{-1}$
(d)$8.0 \mathrm{atm}^{-1} \mathrm{s}^{-1}$
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
k_uni doubles when p doubles, so we are in the low-pressure limit k_uni = k₁p: k₁ = 2.0 atm⁻¹ s⁻¹.