Multiple Choice
The reactant concentration in a first-order reaction was 7.30 × 10^(-2) M after 25.0 s and 9.50 × 10^(-3) M after 95.0 s. What is the rate constant for this reaction?
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15. Chemical Kinetics
Integrated Rate Law
Multiple Choice
For the reaction A → B, the rate constant is 0.0837 M–1•sec–1. How long would it take for [A] to decrease by 85%?
A
0.0211 s
B
1.94 s
C
179 s
D
0.677 s
Verified step by step guidance1
Identify the order of the reaction based on the units of the rate constant. Since the rate constant has units of M\(\textsuperscript{–1}\)•s\(\textsuperscript{–1}\), this indicates a second-order reaction.
Write the integrated rate law for a second-order reaction: \(\frac{1}{[A]} = \frac{1}{[A]_0} + k t\), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.
Express the final concentration \([A]\) in terms of the initial concentration \([A]_0\) and the percentage decrease. Since \([A]\) decreases by 85%, \([A] = 0.15 [A]_0\).
Substitute \([A] = 0.15 [A]_0\) into the integrated rate law and solve for time \(t\): \(\frac{1}{0.15 [A]_0} = \frac{1}{[A]_0} + k t\).
Rearrange the equation to isolate \(t\): \(t = \frac{1}{k} \left( \frac{1}{0.15 [A]_0} - \frac{1}{[A]_0} \right)\), then plug in the known values of \(k\) and \([A]_0\) (if given) to calculate the time.
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