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}. |
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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