A substance 'A' decomposes in solution following first-order kinetics. Flask 1 contains 11 of \(1 \mathrm{M}\) solution of \(\mathrm{A}^{\prime}\) and flask 2 contains \(100 \mathrm{ml}\) of \(0.6 \mathrm{M}\) solution of 'A'. After \(8.0 \mathrm{~h}\), the concentration of 'A' in flask 1 becomes \(0.25 \mathrm{M}\). In what time, the concentration of 'A' in flask 2 becomes \(0.3 \mathrm{M}\) ? (a) \(8.0 \mathrm{~h}\) (b) \(3.2 \mathrm{~h}\) (c) \(4.0 \mathrm{~h}\) (d) \(9.6 \mathrm{~h}\)

Short Answer

Expert verified
The time when the concentration of 'A' in flask 2 becomes 0.3 M is 4.0 hours, so the correct answer is (c) 4.0 hours.

Step by step solution

01

Determine the Rate Constant for the First-Order Reaction

For a first-order reaction, the rate constant (k) can be calculated using the formula: \( k = \frac{\ln(\frac{[A]_0}{[A]})}{t} \), where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration after time t. Given that the concentration of A in flask 1 decreases from \(1 \text{M}\) to \(0.25 \text{M}\) in 8 hours, we can calculate k.
02

Calculate the Rate Constant

Substitute the given values into the rate constant formula: \( k = \frac{\ln(\frac{1}{0.25})}{8} = \frac{\ln(4)}{8} \). Now calculate k using the natural logarithm of 4 over the time span of 8 hours.
03

Apply the Rate Constant to Flask 2

Now that we have the rate constant k, we can use the same first-order kinetics formula to find the time required for the concentration in flask 2 to decrease from \(0.6 \text{M}\) to \(0.3 \text{M}\). The time (t) needed can be calculated using the rearranged formula: \( t = \frac{\ln(\frac{[A]_0}{[A]})}{k} \). Substitute the values for flask 2 with the previously found k.
04

Calculate the Time for Concentration Change in Flask 2

Substitute the concentration values and k into the time formula to get: \( t = \frac{\ln(\frac{0.6}{0.3})}{k} \). Use the same value of k as calculated from flask 1 to determine the time t.
05

Solve for the Time t

Before plugging in the values, simplify \( \ln(\frac{0.6}{0.3}) \) to \( \ln(2) \) and use the previously calculated value of k to find the time t.

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Key Concepts

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

Rate Constant Calculation
Understanding the rate constant in the context of chemical kinetics is essential for predicting how quickly a reaction will proceed. In a first-order reaction, the rate of the reaction is directly proportional to the concentration of one reactant. This kind of reaction is characterized by the equation:
\( k = \frac{\ln(\frac{[A]_0}{[A]})}{t} \),
where \( k \) is the rate constant, \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \).
To calculate the rate constant, one must take the natural logarithm of the ratio between the initial concentration and the concentration after time \( t \), then divide by \( t \). This value is indicative of the reaction's speed, with a higher \( k \) representing a faster reaction. It's important to remember that the rate constant has different units depending on the order of the reaction, which, in the case of a first-order reaction, is inverse time (e.g., \( s^{-1} \), \( hr^{-1} \)).
Chemical Kinetics
Chemical kinetics deals with the speed or rate at which a chemical reaction occurs and the factors that affect this rate. It's a way to quantify the changes in concentration of reactants or products over time.
For first-order reactions, this rate is directly proportional to the concentration of one reactant. In the classroom or laboratory, students often determine the kinetics of a reaction by measuring the change in concentration over time and can graph these changes to visualize the reaction's progress.
When working with kinetics, specific temperature, physical state, and the presence of a catalyst can all influence the rate of reaction. This makes kinetic studies important not only for academic purposes but also for industrial applications where controlling the rate of a reaction can be crucial to the success of a process.
Concentration-Time Relationship
The concentration-time relationship expresses how the concentration of reactants in a chemical reaction changes as time progresses. In the case of first-order reactions, the relationship between concentration and time is logarithmic, not linear.
An integral part of understanding this relationship is being able to use the rate constant \( k \) to predict how long it will take for a reactant to reach a certain concentration. The formula derived from the integrated first-order rate law is:
\( t = \frac{\ln(\frac{[A]_0}{[A]})}{k} \).
Using this formula involves taking the natural logarithm of the initial to the remaining concentration ratio and dividing it by the rate constant to find the time required for that concentration change. This calculation is crucial in fields ranging from pharmacology, where it predicts how long a drug remains in the bloodstream, to environmental science, for calculating pollutant degradation rates.

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Most popular questions from this chapter

A solution of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) yields by decomposition at \(45^{\circ} \mathrm{C}, 4.8 \mathrm{ml}\) of \(\mathrm{O}_{2}\), 20 min after the start of the experiment and \(9.6 \mathrm{ml}\) of \(\mathrm{O}_{2}\) after a very long time. The decomposition obeys first-order kinetics. What volume of \(\mathrm{O}_{2}\) would have evolved, 40 min after the start? (a) \(7.2 \mathrm{ml}\) (b) \(2.4 \mathrm{ml}\) (c) \(9.6 \mathrm{ml}\) (d) \(6.0 \mathrm{ml}\)

For the reaction: \(\mathrm{A}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}(\mathrm{g})+\frac{1}{2} \mathrm{C}(\mathrm{g})\) pressure of the system increases from 100 to \(120 \mathrm{~mm}\) in 5 min. The average rate of disappearance of \(\mathrm{A}_{2}\) (in \(\mathrm{mm} / \mathrm{min}\) ) in this time interval is (a) 4 (b) 8 (c) 2 (d) 16

For the consecutive first-order reactions: \(\mathrm{A} \stackrel{K_{1}}{\longrightarrow} \mathrm{B} \stackrel{K_{2}}{\longrightarrow} \mathrm{C}\), the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are \(0.2 \mathrm{M}\) and \(0.01 \mathrm{M}\), respectively, at steady state. If \(K_{1}\) is \(2.5 \times 10^{-4} \mathrm{~min}^{-1}\), what is the value of \(K_{2} ?\) (a) \(5.0 \times 10^{-3} \mathrm{~min}^{-1}\) (b) \(2.5 \times 10^{-4} \mathrm{~min}^{-1}\) (c) \(1.25 \times 10^{-5} \mathrm{~min}^{-1}\) (d) \(5.0 \times 10^{-4} \mathrm{~min}^{-1}\)

For a gaseous reaction: \(\mathrm{A}(\mathrm{g}) \rightarrow\) Products (order \(=n\) ), the rate may be expressed as: (i) \(-\frac{\mathrm{d} P_{\mathrm{A}}}{\mathrm{d} t}=K_{1} \cdot P_{\mathrm{A}}^{n}\) (ii) \(-\frac{1}{V} \frac{\mathrm{d} n_{\mathrm{A}}}{\mathrm{d} t}=K_{2} \cdot C_{\mathrm{A}}^{n}\) The rate constants, \(K_{1}\) and \(K_{2}\) are related as \(\left(P_{A}\right.\) and \(C_{A}\) are the partial pressure and molar concentration of \(\mathrm{A}\) at time ' \(t^{\prime}\), respectively) (a) \(K_{1}=K_{2}\) (b) \(K_{2}=K_{1} \cdot(R T)^{n}\) (c) \(K_{2}=K_{1} \cdot(R T)^{1-n}\) (d) \(K_{2}=K_{1} \cdot(R T)^{n-1}\)

A first-order reaction: \(\mathrm{A}(\mathrm{g}) \rightarrow n \mathrm{~B}(\mathrm{~g})\) is started with 'A'. The reaction takes place at constant temperature and pressure. If the initial pressure was \(P_{0}\) and the rate constant of reaction is ' \(K\), then at any time, \(t\), the total pressure of the reaction system will be (a) \(P_{0}\left[n+(1-n) e^{-k t}\right]\) (b) \(P_{0}(1-n) e^{-k t}\) (c) \(P_{0} \cdot n \cdot e^{-k t}\) (d) \(P_{0}\left[n-(1-n) e^{-k t}\right.\)

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