Question

In a first order decomposition, how long does it take (in seconds) for 75% of the compound to decompose given that k is 0.0909 sec-1 (give answer to 1 decimal place)?

Answer 1

Answer:

15.3 s

Explanation:

Let's consider the first-order decomposition of a generic compound A. We can find the concentration at a certain time using the following expression.

$[A]_{t}=[A]_{0}.e^{-k\times t }$

where,

$[A]_{t}$: concentration of A at a time t

$[A]_{0}$: initial concentration of A

k: rate constant

When 75% of A decomposes, the remaining concentration is 25% ($[A]_{t}$ = 0.25 $[A]_{0}$). The required time is:

$[A]_{t}/[A]_{0}=e^{-0.0909s^{-1} \times t }$

$0.25[A]_{0}/[A]_{0}=e^{-0.0909s^{-1} \times t }$

t = 15.3 s