claureka
contestada

The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the concentration of SOCl2 to drop from 0.36 M to 0.045 M?

Respuesta :

4.1 h = 14760 s 

t 1/2 = ln 2 / k 

k = rate reaction = 4.97 x 10^-5 

ln 0.045 / 0.36 = - 4.97 x 10^-5 t 

2.08 = 4.97 x 10^-5 t 

t = 41839.9 s = 11 h 37 min 19 s

The formation of the product by the splitting of the reactant is called a decomposition reaction. It will take 11 h 37 min 19 s for the concentration to drop.

What is a first-order reaction?

In the first-order reactions, the rate of the reaction is in linear relation to the concentration of a single reactant.

Given,

Half life of the reaction = 14760 sec

Reaction rate (k) = [tex]4.97 \times 10^{-5}[/tex]

Initial concentration = 0.36 M

Final concentration = 0.045 M

Time (t) is calculated as:

[tex]\begin{aligned} \rm t \dfrac {1}{2} &= \dfrac{0.693}{\rm k}\\\\2.08 &= 4.97 \times 10^{-5} \;\rm \times t\\\\&= 41839.9 \;\rm s\end{aligned}[/tex]

Therefore, it will take 11 h 37 min 19s for the concentration to drop from 0.36 M to 0.045 M.

Learn more about the first-order reaction here:

https://brainly.com/question/15494549