The decomposition of gaseous acetaldehyde at T(K) follows second order kinetics. The half-life of this reaction is 400 s when initial pressure is 250 torr. What will be the rate constant (in torr) and half-life (in s) respectively, if the initial pressure of acetaldehyde is 200 torr at the same temperature?

Question

The decomposition of gaseous acetaldehyde at T(K) follows second order kinetics. The half-life of this reaction is 400 s when initial pressure is 250 torr. What will be the rate constant (in torr) and half-life (in s) respectively, if the initial pressure of acetaldehyde is 200 torr at the same temperature? (A) 105 and 500  (B) 10-5 and 400  (C) 10-4 and 400  (D) 10-5 and 500

Tardigrade Admin · Accepted Answer

For 2nd order reaction, k = (1/t1/2p°) where P° is initial pressure so k = (1/400 × 250) = 10-5 torr-1 sec -1 when P ° = 200 torr

$1 0^{5}$ and $500$

$1 0^{- 5}$ and $400$

$1 0^{- 4}$ and $400$

$1 0^{- 5}$ and $500$

For $2^{n d}$ order reaction, $k = \frac{1}{t _{1/2} p ^{\circ}}$
where $P^{\circ}$ is initial pressure
so $k = \frac{1}{400 \times 250} = 1 0^{- 5} t or r^{- 1} se c^{- 1}$
when $P^{\circ} = 200 t orr$

105 and 500

10−5 and 400

10−4 and 400

10−5 and 500

For 2nd order reaction, k=t1/2​p∘1​ where P∘ is initial pressure so k=400×2501​=10−5torr−1sec−1 when P∘=200torr

$1 0^{5}$ and $500$

$1 0^{- 5}$ and $400$

$1 0^{- 4}$ and $400$

$1 0^{- 5}$ and $500$

For $2^{n d}$ order reaction, $k = \frac{1}{t _{1/2} p ^{\circ}}$
where $P^{\circ}$ is initial pressure
so $k = \frac{1}{400 \times 250} = 1 0^{- 5} t or r^{- 1} se c^{- 1}$
when $P^{\circ} = 200 t orr$