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Chapter 10: Problem 101

In the first-order reaction, half of the reaction is com pleted in 100 seconds. The time for \(99 \%\) reaction to occur will be (a) \(664.64 \mathrm{~s}\) (b) \(646.6 \mathrm{~s}\) (c) \(660.9 \mathrm{~s}\) (d) \(654.5 \mathrm{~s}\)

Short Answer

Expert verified
The time for 99% reaction to occur is approximately 664.64 seconds (option a).

Step by step solution

01

Determine the Half-Life Formula for a First-Order Reaction

For a first-order reaction, the half-life \( t_{1/2} \) is a constant and is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant.
02

Calculate the Rate Constant

Since half of the reaction is completed in 100 seconds, this means that the half-life \( t_{1/2} \) is 100 seconds. Use the formula from Step 1 to solve for \( k \): \[ 100 = \frac{0.693}{k} \] Solving for \( k \), we get: \[ k = \frac{0.693}{100} = 0.00693 \text{ s}^{-1} \]
03

Use the First-Order Reaction Time Formula

The time required \( t \) for a certain percentage \( \ x \% \) of a first-order reaction can be found using the formula: \[ t = \frac{2.303}{k} \log \left( \frac{100}{100-x} \right) \] We need to determine the time for \( 99\% \) completion.
04

Substitute the Values into the Time Formula

Substitute \( k = 0.00693 \text{ s}^{-1} \) and \( x = 99 \) into the formula: \[ t = \frac{2.303}{0.00693} \log \left( \frac{100}{1} \right) \] \[ t = \frac{2.303}{0.00693} \times 2 \] where \( \log \left( \frac{100}{1} \right) = 2 \).
05

Calculate the Time for 99% Completion

Perform the calculation: \[ t = \frac{2.303 \times 2}{0.00693} \approx 664.64 \text{ s} \]
06

Identify the Correct Answer Choice

The calculated time for \( 99\% \) completion is approximately 664.64 seconds, which corresponds to option (a).

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Key Concepts

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

Rate Constant
The rate constant, often symbolized as \(k\), is a fundamental part of the equation used to determine the rate of a reaction in chemical kinetics. For a first-order reaction, the rate constant is calculated using the equation for half-life:
  • The formula is \(t_{1/2} = \frac{0.693}{k}\).
  • In this specific case, with a half-life of 100 seconds, we derived \(k = 0.00693 \text{s}^{-1}\).
The rate constant is crucial because it helps us predict how long a reaction will take to reach a certain level of completion. A larger rate constant means the reaction proceeds more quickly. The units of \(k\) for a first-order reaction are \(\text{s}^{-1}\), indicating the change with respect to time. This rate constant allows us to link time and concentration changes, offering insights into the speed and efficiency of chemical processes.
Half-Life
Half-life, in the context of reaction kinetics, refers to the time required for half of the reactant concentration to be consumed. This is a defining feature of first-order reactions. When undergoing such reactions:
  • The half-life is independent of the initial concentration of reactants.
  • It remains constant throughout the reaction, reflecting the exponential decay typical of first-order kinetics.
  • Knowing the half-life allows us to predict the time frames for various stages of a reaction.
In our exercise, the half-life was provided as 100 seconds, meaning in this duration, exactly 50% of the reaction completes. This steady time frame makes it straightforward to calculate when significant percentages of the reaction will be achieved.
Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical processes occur and the factors that affect these rates. In first-order reactions:
  • The rate is directly proportional to the concentration of one reactant.
  • It follows an exponential decay, as indicated by the formula \(\frac{d[A]}{dt} = -k[A]\).
  • The formula \(t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \) allows the determination of the time required for a specific completion percentage.
Our exercise demonstrated this by using the kinetics formula to find the time taken for 99% of a reaction to complete. Understanding reaction kinetics is vital for predicting and controlling reaction dynamics, which is particularly useful in industrial and laboratory settings.
Percentage Completion
Percentage completion in a chemical reaction context refers to the proportion of reactants consumed relative to the total amount initially present. It's an essential parameter for assessing reaction progress:
  • It provides insights into how far a reaction has proceeded after a given period.
  • For first-order kinetics, we calculated the time to reach 99% completion using the derived rate constant and concentration ratios.
  • The relevant formula is \(t = \frac{2.303}{k} \log \left( \frac{100}{100-x} \right)\), where \(x\) represents the desired percentage completion.
In practical scenarios, understanding how to calculate percentage completion allows chemists to optimize conditions, ensuring that reactions proceed efficiently and within desired time frames. This metric helps gauge efficiency and effectiveness of chemical reactions in tangible percentages, offering a clear depiction of progress over time.

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Most popular questions from this chapter

The following data are obtained from the decomposition of a gaseous compound Initial pressure in arm \(\begin{array}{lll}1.6 & 0.8 & 0.4\end{array}\) Time for \(50 \%\) reaction in \(\min 80\) \(113 \quad 160\) The order of the reaction is (a) \(0.5\) (b) \(1.0\) (c) \(1.5\) (d) \(2.0\)

\(3 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\) It would be a zero order reaction when (a) the rate of reaction is proportional to square of concentration of \(\mathrm{A}\) (b) the rate of reaction remains same at any concentration of \(\mathrm{A}\) (c) the rate remains unchanged at any concentration of \(\mathrm{B}\) and \(\mathrm{C}\) (d) the rate of reaction doubles if concentration of is increased to double

\(k\) for a zero-order reaction is \(2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). If the concentration of the reactant after \(25 \mathrm{~s}\) is \(0.5 \mathrm{M}\), the initial concentration must have been (a) \(0.5 \mathrm{M}\) (b) \(1.25 \mathrm{M}\) (c) \(12.5 \mathrm{M}\) (d) \(1.0 \mathrm{M}\)

A certain reaction proceeds in a sequence of three elementary steps with the rate constants \(\mathrm{k}_{1}, \mathrm{k}_{2}\) and \(\mathrm{k}_{3} .\) If the observed rate constant of the expressed as \(\mathrm{k}\) (obs) \(=\mathrm{k}(\mathrm{obs})=\left[\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}\right]^{1 / 2} \mathrm{k}_{3}\), the observed energy of activa- tion of the reaction is (a) \(\frac{\mathrm{E}_{3}+\mathrm{E}_{1}}{2}\) (b) \(\frac{1}{2}\left[\frac{E_{1}}{E_{2}}\right]+E_{3}\) (c) \(\mathrm{E}_{3}+\frac{1}{2}\left[\mathrm{E}_{1}-\mathrm{E}_{2}\right]\) (d) \(\mathrm{E}_{3}\left[\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right]^{12}\)

Which of the following is incorrect about order of reaction? (a) it is calculated experimentally (b) it is sum of powers of concentration in rate law expression (c) the order of reaction cannot be fractional (d) there is not necessarily a connection between order and stoichiometry of a reaction.
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