The decomposition of gaseous acetaldehyde at T(K) follows second order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in $\mathrm{Torr}^{-1} \mathrm{s}^{-1}$ ) and half-life (in s ) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?
(a)$10^{5}$ and 500
(b)$10^{-5}$ and 400
(c)$10^{-4}$ and 400
(d)$10^{-5}$ and 500
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Second order: t₁/₂ = 1/(k p₀) ⇒ k = 1/(400 × 250) = 1×10⁻⁵ Torr⁻¹ s⁻¹. At 200 Torr, t₁/₂ = 1/(10⁻⁵ × 200) = 500 s.