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## Q. 13.4

The First-Order Integrated Rate Law: Determining the Concentration of a Reactant at a Given Time
In Example 13.3, you determined that the decomposition of $SO_{2}Cl_{2}$ (under the given reaction conditions) is first order and has a rate constant of +2.90×10$^{-4} s^{-1}$. If the reaction is carried out at the same temperature and the initial concentration of  $SO_{2}Cl_{2}$ is 0.0225 M, what is the$SO_{2}Cl_{2}$ concentration after 865 s?

 SORT You are given the rate constant of a first-order reac-tion and the initial concentration of the reactant. You are asked to find the concentration at 865 seconds. GIVEN k = +2.90 ×10$^{-4} s^{-1}$ [ $SO_{2}Cl_{2}]_{0}$ = 0.0225 M FIND [ $SO_{2}Cl_{2}$] at t = 865 s STRATEGIZE Use the first-order integrated rate law to deter mine the $SO_{2}Cl_{2}$ concentration at t = 865 s EQUATION ln[A]$_{t} = -kt + ln[A]_{0}$ SOLVE Substitute the rate constant, the initial concentra-tion, and the time into the integrated rate law. Solve the integrated rate law for the concentration of $[SO_{2}Cl_{2}]_{t}$.

## Verified Solution

ln $[SO_{2}Cl_{2}]_{t} = -kt + ln [SO_{2}Cl_{2}]_{0}$
ln $[SO_{2}Cl_{2}]_{t} = -(2.90\times 10^{-4}\cancel{s^{-1}})865 \cancel{s}$ + ln (0.0225)
ln $[SO_{2}Cl_{2}]_{t}$ = -0.251 – 3.79
$[SO_{2}Cl_{2}]_{t} = e^{-4.04}$
= 0.0175 M

CHECK The concentration is smaller than the original concentration as expected. If the concentration were larger than the ini-tial concentration, this would indicate a mistake in the signs of one of the quantities on the right-hand side of the equation