# Question 23db6

Jun 9, 2016

If this is a kinetics problem, you need to know the order of the reaction, the integrated rate law, and the rate constant at 420 °C

Since you don’t state the specific problem, let's arbitrarily assume that the reaction is first order, the initial concentration of ${\text{SO"_2"Cl}}_{2}$ is 0.0225 mol/L and that the rate constant is 2.90 × 10^"-4"color(white)(l) "s"^"-1" at 420 °C.

Whenever a question asks, "How much is left after an amount of time?", that is a clue for you to use an integrated rate law.

The integrated rate law for a first order reaction is

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \ln \left({A}_{0} / {A}_{t}\right) = k t \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$\text{A"_0 = "concentration at time 0}$
$\text{A"_t = "concentration at time} \textcolor{w h i t e}{l} t$
$k = \text{rate constant}$
$t = \text{time}$

t = 16.2 color(red)(cancel(color(black)("h"))) × (60 color(red)(cancel(color(black)("min"))))/(1 color(red)(cancel(color(black)("h")))) × "60 s"/(1 color(red)(cancel(color(black)("min")))) = "58 320 s"

Then,

ln("0.0225 mol/L"/"A"_t) = 2.90 × 10^"-4" color(red)(cancel(color(black)("s"^"-1"))) × "58 320" color(red)(cancel(color(black)("s"))) = 16.91

"0.0225 mol/L"/"A"_t= e^16.91 = 2.21× 10^7#

$\text{A"_t = "0.0225 mol/L"/(2.21× 10^7) = 1.02 × 10^"-9" color(white)(l)"mol/L}$