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For the reaction 2N2O5(g) → 4NO2(g) + O2(g) with a rate constant k = 2.8×10⁻³ sec⁻¹, if the initial concentration of [N2O5] is 1.58 mol/L, what is the concentration of N2O5 after 300 seconds?
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15. Chemical Kinetics
Integrated Rate Law
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The decomposition of HI to H2 and I2 is a second-order reaction with a rate constant of 3.8 × 10⁻⁵ M⁻¹·s⁻¹ at a certain temperature. If the initial concentration of HI is 0.554 M, calculate the amount of time (in days) it will take to consume 72.4% of the initial concentration.
A
3.7 days
B
4.5 days
C
1.5 days
D
2.3 days
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Identify the type of reaction: The problem states that the decomposition of HI is a second-order reaction. This means the rate law is given by \( \text{Rate} = k[\text{HI}]^2 \), where \( k \) is the rate constant.
Use the integrated rate law for a second-order reaction: The integrated rate law for a second-order reaction is \( \frac{1}{[A]} = \frac{1}{[A]_0} + kt \), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant.
Calculate the concentration of HI after 72.4% has been consumed: If 72.4% of the initial concentration is consumed, then 27.6% remains. Calculate \([\text{HI}]\) at time \( t \) using \([\text{HI}] = 0.276 \times [\text{HI}]_0\).
Substitute the known values into the integrated rate law: Use \([\text{HI}]_0 = 0.554 \text{ M}\), \([\text{HI}] = 0.276 \times 0.554 \text{ M}\), and \( k = 3.8 \times 10^{-5} \text{ M}^{-1}\cdot\text{s}^{-1}\) in the equation \( \frac{1}{[\text{HI}]} = \frac{1}{[\text{HI}]_0} + kt \).
Solve for \( t \) and convert to days: Rearrange the equation to solve for \( t \) in seconds, then convert the time from seconds to days by dividing by the number of seconds in a day (86400 seconds/day).
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