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The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

Short Answer

Expert verified
The percentage of a sample of A that remains unreacted 900 s after a reaction has been started is 10%. The rate of the reaction when [A]=0.50 M is 0.001925 M s^{-1}.

Step by step solution

01

Calculate the Rate Constant from the Half-life

The relationship between the half-life (\(t_{1/2}\)) and the rate constant (k) for a first-order reaction is given by the equation \(t_{1/2} = \frac{0.693}{k}\). We can rearrange this equation to solve for k: \(k = \frac{0.693}{t_{1/2}}\). Substituting the given half-life of 180 s into this equation gives: \(k = \frac{0.693}{180 s} = 0.00385 s^{-1}\).
02

Determine the Percentage of A that Remains Unreacted After 900 s

The amount of reactant A remaining after a certain time (t) in a first-order reaction can be found using the equation \(A_t = A_0 \times e^{-kt}\), where \(A_t\) is the amount of A at time t, \(A_0\) is the initial amount of A, k is the rate constant and t is the time. However, we want to find the percentage of A that remains, so we can normalize this equation to \( \frac{A_t}{A_0} = e^{-kt}\). Substituting the calculated k = 0.00385 s^{-1} and t = 900 s into this equation gives: \( \frac{A_t}{A_0} = e^{-0.00385 s^{-1} \times 900 s}\). This equation gives us approximately 0.10, or 10%. So, 10% of the sample of A remains unreacted after 900 s.
03

Calculate the Rate of Reaction when [A]=0.50 M

The rate of a first-order reaction can be found using the equation \( \text{rate} = k[A] \), where k is the rate constant, and [A] is the concentration of reactant A. Substituting the calculated k = 0.00385 s^{-1} and [A] = 0.50 M into this equation gives: \(\text{rate} = 0.00385 s^{-1} \times 0.50 M = 0.001925 M s^{-1}\). So, the rate of the reaction when [A] = 0.50 M is 0.001925 M s^{-1}.

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Key Concepts

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

reaction kinetics
Reaction kinetics is the study of how fast or slow a reaction occurs. It provides vital insights into the rate at which reactants convert into products. For first-order reactions, the rate is directly proportional to the concentration of the reactant. This means if we were to double the concentration of the reactant, the reaction rate would also double.

In the context of first-order reactions, the rate at which the reaction occurs can be expressed by the formula: \[\text{rate} = k[A]\]where \(k\) is the rate constant, and \([A]\) signifies the concentration of the reactant A. This equation indicates that the speed of the reaction depends on both the rate constant and the concentration of A. Observing how these two factors interact helps scientists predict and control the reaction better.

Reaction kinetics helps chemists and chemical engineers optimize conditions for industrial reactions, ensuring that they run efficiently and safely. Understanding the rate also helps predict how long a reaction will take and how changing conditions will impact the overall process.
half-life calculations
Half-life is a concept often used with reactions, especially radioactive decay, but it is also crucial in other first-order reactions. It describes the time taken for half of the initial amount of reactant to be converted into products.

For first-order reactions, the half-life \(t_{1/2}\) has a unique relationship with the rate constant \(k\):\[t_{1/2} = \frac{0.693}{k}\]This formula is pivotal because it shows that the half-life is constant and does not depend on the initial concentration of the reactant. In our exercise, the half-life is given as 180 seconds, leading us to calculate \(k\) as 0.00385 s\(^{-1}\).

By knowing the half-life and the rate constant, we can predict how much of a reactant remains after any given time. For instance, after multiple half-lives, the quantity of the reactant continues to halve until it becomes negligible. Understanding half-life helps in various applications like pharmacology, where it determines how often a drug should be administered to maintain its efficacy in the human body.
rate constant
The rate constant \(k\) is a fundamental component of reaction kinetics. It is a measure of the reaction speed under specific conditions. The rate constant varies depending on factors like temperature, presence of a catalyst, and the nature of the reactant and product.

In our problem, after calculating from the given half-life, the rate constant was found to be 0.00385 s\(^{-1}\). This value signifies how quickly the reaction progresses in this particular setup.

The rate constant has units that depend on the reaction order. For first-order reactions, it's usually in s\(^{-1}\), indicating the change in concentration over time. The importance of \(k\) is underscored by its ability to help chemists understand and tweak reaction conditions to achieve desired outcomes more efficiently.
  • Ensures that reactions run smoothly in industrial applications.
  • Helps predict reaction behavior under altered conditions.
  • Crucial for kinetic modeling and simulation tasks in various chemical processes.
Understanding \(k\) and its implications allows scientists to delve deeper into the molecular world and extract critical information about reaction behaviors.

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Most popular questions from this chapter

One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

In the reaction \(A \longrightarrow\) products, \([A]\) is found to be \(0.485 \mathrm{M}\) at \(t=71.5 \mathrm{s}\) and \(0.474 \mathrm{M}\) at \(t=82.4 \mathrm{s} .\) What is the average rate of the reaction during this time interval?

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

Certain gas-phase reactions on a heterogeneous catalyst are first order at low gas pressures and zero order at high pressures. Can you suggest a reason for this?

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