## Chapter 13

## 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}. |

## Step-by-Step

## 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