The half-life for the first-order decomposition of A is \(355 \mathrm{~s}\). How much time must elapse for the concentration of A to decrease to (a) \(\frac{1}{8}[\mathrm{~A}]_{0} ;\) (b) one-fourth of its initial concentration; (c) \(15 \%\) of its initial concentration; (d) one-ninth of its initial concentration?

Short Answer

Expert verified
The time to reach (a) 1/8, (b) 1/4, (c) 15%, and (d) 1/9 of the initial concentration of A is (a) 1065 seconds, (b) 710 seconds, and for both (c) and (d), the times must be calculated using the formula with respective fractional values and a calculator for the logarithmic parts.

Step by step solution

01

Understanding Half-life

The half-life of a first-order reaction is the time it takes for the concentration of a substance to reduce to half of its initial value. The formula to calculate the time for a concentration to reduce to a certain fraction of the initial concentration in a first-order reaction is given by the equation: \( t = t_{1/2} \times \frac{\log(\frac{[A]_0}{[A]})}{\log(2)} \), where \( t \) is the time elapsed, \( t_{1/2} \) is the half-life, \( [A]_0 \) is the initial concentration, and \( [A] \) is the final concentration.
02

Calculating Time for 1/8 Initial Concentration

For the concentration to decrease to \( \frac{1}{8}[A]_0 \), the fraction of the initial concentration is \( \frac{1}{8} \). Substitute this value into the time calculation formula: \( t = 355 \times \frac{\log(\frac{1}{\frac{1}{8}})}{\log(2)} = 355 \times \frac{\log(8)}{\log(2)} \).
03

Solving for the Time to Reach 1/8 Initial Concentration

Calculate the value of \( \frac{\log(8)}{\log(2)} \) which is equivalent to \( \log_2(8) \). Since \( 8 = 2^3 \), \( \log_2(8) = 3 \). Therefore, \( t = 355 \times 3 \).
04

Calculating Time for 1/4 Initial Concentration

For the concentration to decrease to \( \frac{1}{4}[A]_0 \), the fraction is \( \frac{1}{4} \). The calculation is \( t = 355 \times \frac{\log(\frac{1}{\frac{1}{4}})}{\log(2)} = 355 \times \frac{\log(4)}{\log(2)} \).
05

Solving for the Time to Reach 1/4 Initial Concentration

Calculate \( \log_2(4) \) which is 2 since \( 4 = 2^2 \). Then, \( t = 355 \times 2 \).
06

Calculating Time for 15% Initial Concentration

For 15% of the initial concentration, the calculation is \( t = 355 \times \frac{\log(\frac{1}{0.15})}{\log(2)} \).
07

Solving for the Time to Reach 15% Initial Concentration

Calculate the logarithmic part: \( \frac{\log(\frac{1}{0.15})}{\log(2)} \) and multiply the result by 355 to find \( t \).
08

Calculating Time for 1/9 Initial Concentration

For the concentration to decrease to \( \frac{1}{9}[A]_0 \), perform the calculation \( t = 355 \times \frac{\log(\frac{1}{\frac{1}{9}})}{\log(2)} = 355 \times \frac{\log(9)}{\log(2)} \).
09

Solving for the Time to Reach 1/9 Initial Concentration

Finally, calculating \( \log_2(9) \) and multiplying by 355 gives the value for \( t \). Since \( 9 \) is not a power of \( 2 \), you will need to use a calculator for this logarithm part.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life
Imagine you have a magic cube that disappears piece by piece over time. If someone tells you that its half-life is one hour, it means that in one hour, exactly half of your cube will have vanished. In the realm of chemical reactions, half-life is the period it takes for the concentration of a substance involved in a first-order reaction to be reduced by half its initial amount.

Understanding half-life is crucial because it is a constant value for a given reaction at a constant temperature and does not depend on the initial concentration of the reactant. Therefore, half-life provides a handy reference for predicting how long it will take for a substance to reach a certain level of decay, regardless of its starting concentration. This is especially important when dealing with pharmaceuticals, environmental pollutants, or any situation where the decay rate of a substance impacts health and safety.
First-Order Reaction
Let's think of a first-order reaction as a group of marathon runners who all run at different speeds. In this case, the speed of our slowest runner dictates that of the whole group. In chemistry, a first-order reaction is one where the rate of the reaction is directly proportional to just one reactant's concentration.

Like our slowest runner, it's the concentration of this single reactant that sets the pace for the entire reaction. This proportionality to one reactant means that if you were to double the concentration of this reactant, the reaction rate would also double. First-order reactions are common in nature and are often found in processes like radioactive decay and simple enzyme reactions.
Reaction Concentration Decrease
If you have a container of 100 ping pong balls and every minute you take away half of them, how many would you have after one minute? Fifty, right? Now what if you kept going, taking half of whatever's left each minute? Well, this is similar to how the concentration of a reactant decreases over time in many chemical reactions.

In a first-order reaction, the concentration of the reactant falls exponentially, which means that it drops quickly at first, then more and more slowly over time. The reaction 'forgets' its past, in a way, as the rate of decrease is always relative to the current concentration - the fewer reactant molecules there are, the slower they are used up.
Logarithmic Relationships in Kinetics
Mathematics often seems like a secret language, but it helps us unlock the mysteries of the universe. Logarithms are one such magical key. In chemical kinetics, they allow us to relate the concentrations of reactants in a first-order reaction to time in a clear and concise way.

Through logarithms, we can create simple linear relationships from complex exponential decay patterns, enabling us to calculate how concentrations will change over time without having to tally each reactant molecule. It's like having a map that tells you exactly where you'll be and when, on a long winding road trip where distances don't always match the effort it takes to travel them.

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Most popular questions from this chapter

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(3.7 \times 10^{-5} \mathrm{~s}^{-1}\) at \(298 \mathrm{~K}\). (a) What is the half-life (in hours) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(298 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(3.5 \mathrm{~h}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

When the rate of the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow\) \(2 \mathrm{NO}_{2}(\mathrm{~g})\) was studied, the rate was found to double when the \(\mathrm{O}_{2}\) concentration alone was doubled but to quadruple when the NO concentration alone was doubled. Which of the following mechanisms accounts for these observations? Explain your reasoning. (a) Step \(1 \mathrm{NO}+\mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}\) and its reverse (both fast, equilibrium) Step \(2 \mathrm{NO}+\mathrm{NO}_{3} \rightarrow \mathrm{NO}_{2}+\mathrm{NO}_{2}\) (slow) (b) Step \(1 \mathrm{NO}+\mathrm{NO} \rightarrow \mathrm{N}_{2} \mathrm{O}_{2}\) (slow) Step \(2 \mathrm{O}_{2}+\mathrm{N}_{2} \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}\) (fast) Step \(3 \mathrm{~N}_{2} \mathrm{O}_{4} \rightarrow \mathrm{NO}_{2}+\mathrm{NO}_{2}\) (fast)

Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. Step \(1 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{Br}^{-}\) Step \(2 \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}{ }^{+}\) Step \(3 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\)

Derive an expression for the half-life of the reactant A that decays by a third-order reaction with rate constant \(k\).

The decomposition of A has the rate law Rate \(=k[\mathrm{~A}]^{a}\). Show that for this reaction the ratio \(t_{1 / 2} / t_{3 / 4}\), where \(t_{1 / 2}\) is the halflife and \(t_{3 / 4}\) is the time for the concentration of A to decrease to \(\frac{3}{4}\) of its initial concentration, can be written as a function of \(a\) alone and can therefore be used to make a quick assessment of the order of the reaction in A.

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