Half-life equation for first-order reactions: t1/2 = 0.693/k where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s-1). What is the half-life of a first-order reaction with a rate constant of 1.00times10-4 s-1? Express your answer with the appropriate units. What is the rate constant of a first-order reaction that takes 538seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units. A certain first-order reaction has a rate constant of 1.00times10-3 s-1. How long will it take for the reactant concentration to drop to 1/8 of its initial value? Express your answer with the appropriate units.
The Correct Answer and Explanation is :
Calculations and Answers:
- Half-life of a reaction with ( k = 1.00 \times 10^{-4} \, \text{s}^{-1} ): The formula for the half-life of a first-order reaction is:
[
t_{1/2} = \frac{0.693}{k}
]
Substituting ( k = 1.00 \times 10^{-4} \, \text{s}^{-1} ):
[
t_{1/2} = \frac{0.693}{1.00 \times 10^{-4}} = 6930 \, \text{s}
] Answer: ( t_{1/2} = 6930 \, \text{s} ).
- Rate constant ( k ) for a reaction with ( t_{1/2} = 538 \, \text{s} ): Rearrange the half-life formula to solve for ( k ):
[
k = \frac{0.693}{t_{1/2}}
]
Substituting ( t_{1/2} = 538 \, \text{s} ):
[
k = \frac{0.693}{538} = 1.29 \times 10^{-3} \, \text{s}^{-1}
] Answer: ( k = 1.29 \times 10^{-3} \, \text{s}^{-1} ).
- Time to drop to ( \frac{1}{8} ) of initial concentration with ( k = 1.00 \times 10^{-3} \, \text{s}^{-1} ): For a first-order reaction, the concentration is related to time by:
[
\ln\left(\frac{[A]_0}{[A]}\right) = k t
]
When the concentration drops to ( \frac{1}{8} ), ( \frac{[A]_0}{[A]} = 8 ). Substituting ( \ln(8) = 2.079 ) and ( k = 1.00 \times 10^{-3} \, \text{s}^{-1} ):
[
2.079 = (1.00 \times 10^{-3}) t
]
Solve for ( t ):
[
t = \frac{2.079}{1.00 \times 10^{-3}} = 2079 \, \text{s}
] Answer: ( t = 2079 \, \text{s} ).
Explanation (300 words):
The half-life (( t_{1/2} )) of a first-order reaction is constant and independent of the reactant concentration. It is calculated using the equation ( t_{1/2} = 0.693/k ), where ( k ) is the rate constant. For ( k = 1.00 \times 10^{-4} \, \text{s}^{-1} ), the half-life is 6930 seconds, showing that the reaction takes this long for the reactant concentration to halve.
Conversely, when the half-life is known, the rate constant can be determined by rearranging the equation to ( k = 0.693/t_{1/2} ). For ( t_{1/2} = 538 \, \text{s} ), ( k = 1.29 \times 10^{-3} \, \text{s}^{-1} ), which quantifies the reaction speed.
For the final scenario, determining how long it takes for the concentration to drop to ( \frac{1}{8} ) involves the integrated rate law for first-order reactions: ( \ln([A]_0/[A]) = k t ). A drop to ( \frac{1}{8} ) implies ( [A]_0/[A] = 8 ), and substituting ( k = 1.00 \times 10^{-3} \, \text{s}^{-1} ) gives ( t = 2079 \, \text{s} ). This demonstrates that first-order reactions exhibit exponential decay, and the time required for a specific concentration change depends on ( k ) and the initial concentration ratio. These principles are central to understanding chemical kinetics in both experimental and practical applications.