Multiple Choice
In the reaction 3 BrO⁻(aq) → BrO₃⁻(aq) + 2 Br⁻(aq), a plot of 1/[BrO⁻] vs. time is linear. What does this indicate about the order of the reaction with respect to BrO⁻?
15. Chemical Kinetics
Integrated Rate Law
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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?
A
0.025 s⁻¹
B
0.045 s⁻¹
C
0.034 s⁻¹
D
0.012 s⁻¹
Verified step by step guidance1
Identify that the reaction is first-order, which means the rate law can be expressed as: \( \text{Rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.
Use the integrated rate law for a first-order reaction: \( \ln([A]_t) = -kt + \ln([A]_0) \), where \( [A]_t \) is the concentration at time \( t \), \( [A]_0 \) is the initial concentration, and \( k \) is the rate constant.
Substitute the given concentrations and times into the integrated rate law equation. For the first time point (25.0 s), \( [A]_t = 7.30 \times 10^{-2} \text{ M} \) and for the second time point (95.0 s), \( [A]_t = 9.50 \times 10^{-3} \text{ M} \).
Rearrange the integrated rate law equation to solve for the rate constant \( k \): \( k = \frac{\ln([A]_0/[A]_t)}{t} \). Use the two time points to calculate \( k \) by substituting the values: \( k = \frac{\ln(7.30 \times 10^{-2} / 9.50 \times 10^{-3})}{95.0 \text{ s} - 25.0 \text{ s}} \).
Calculate the natural logarithm and the difference in time to find the value of \( k \). This will give you the rate constant for the reaction in units of \( \text{s}^{-1} \).
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