Question

The rate constant for this first‑order reaction is 0.0350 s − 1 0.0350 s−1 at 400 ∘ C. 400 ∘C. A ⟶ products A⟶products After how many seconds will 18.1 % 18.1% of the reactant remain?

Answer 1

Answer:

After 48.84 s, 18.1% of the reactant will remain.

Explanation:

Given that:

The rate constant, k = 0.0350 s⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

18.1 % of the reactant remain which means that 0.181 of $[A_0]$ is remained. So,

$\frac {[A_t]}{[A_0]}$ = 0.181

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.181=e^{-0.0350\times t}$

t = 48.84 s

After 48.84 s, 18.1% of the reactant will remain.