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In the reaction \(A \longrightarrow\) products, 4.40 min after the reac- tion is started, \([\mathrm{A}]=0.588 \mathrm{M}\). The rate of reaction at this point is rate \(=-\Delta[\mathrm{A}] / \Delta t=2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}.\) Assume that this rate remains constant for a short period of time. (a) What is \([\mathrm{A}] 5.00\) min after the reaction is started? (b) At what time after the reaction is started will \([\mathrm{A}]=0.565 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The concentration of A 5.00 min after the reaction started is approximately 0.575 M. (b) The concentration of A will be 0.565 M approximately 5.45 min after the reaction has started.

Step by step solution

01

Using Given Reaction Rate

The given rate of reaction is \(-\Delta[\mathrm{A}] / \Delta t = 2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}\). Assuming this rate to be constant, it can be applied to find out the changes in concentration over given periods of time.
02

(a) Find Concentration of A After 5 min

The time difference between 5.00 min and 4.40 min (when the concentration was last known) is 0.60 min. As the rate is given in M min^-1, we multiply this time period by the rate: \( \Delta[\mathrm{A}] = (2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1})(0.60 \mathrm{min}) = 0.0132 \mathrm{M}\). As this is the amount of A used up (hence negative), the concentration after 5.00 min will be \(0.588 \mathrm{M} - 0.0132 \mathrm{M}\).
03

(b) Find Time When Concentration of A is 0.565 M

Now we need to find out how much time it takes for the concentration to decrease from 0.588 M to 0.565 M. The difference in concentration is \( \Delta[\mathrm{A}] = 0.588 \mathrm{M} - 0.565 \mathrm{M} = 0.023 \mathrm{M}\). The time this takes at the given rate can be found with: \( \Delta t = \Delta[\mathrm{A}] / rate = 0.023 \mathrm{M} / 2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}\). Add this time duration to the initial 4.40 min to find the time after the reaction begins.

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Key Concepts

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

Rate of Reaction
A key concept in reaction kinetics is the rate of reaction. This refers to how quickly reactants are turned into products. For the reaction given, the rate of reaction is represented by the equation: \[ ext{rate} = -\frac{\Delta [A]}{\Delta t} = 2.2 \times 10^{-2} \text{ M min}^{-1} \]This equation tells you the change in concentration (\([A]\)) over time (\(\Delta t\)). Think of it as the speedometer of a chemical reaction. When you see more negative changes, the reaction proceeds faster. Understanding the rate helps predict how long a reaction might take under constant conditions. This is particularly useful when setting up experiments or industrial processes, where time can be a critical factor.

Sometimes the rate isn't constant and changes as reactants are consumed. However, here, it's assumed to be constant for simplicity. This allows for straightforward calculations in a short time window.
Concentration Changes
Concentration changes are at the heart of understanding reaction dynamics. In our exercise, you start with a concentration of \(0.588\, \text{M}\) for \([A]\) at 4.40 minutes. To find the concentration at any moment, you can use:\[ [A]_{\text{final}} = [A]_{\text{initial}} - \Delta [A] \]where \(\Delta [A]\) can be computed using time intervals and reaction rate. It's key to track these changes to know how much reactant is left and how much product is formed.

For example, in part (a), we calculated \(\Delta [A]\) by multiplying the rate (\(2.2 \times 10^{-2}\,\text{M min}^{-1}\)) by the time period (\(0.60\, \text{min}\)). Hence, the concentration change was \(0.0132 \text{ M}\). Deducting this from the initial concentration gives you the current state of the reaction.
Rate Constant
The rate constant is a fundamental part of understanding chemical kinetics. It is usually denoted by \(k\) and is a proportionality factor in the rate equation. Unlike the reaction rate, the rate constant is not dependent on concentration but rather on temperature and the presence of a catalyst. The general form of the rate equation is:\[ ext{Rate} = k[A]^n \]In this particular exercise, the specific rate constant (denoted implicitly through the given rate, assuming first-order kinetics) helps calculate how quickly concentrations change over a set period.

It is important to distinguish the rate constant from the rate of reaction. Equations involving the rate constant can provide deep insights into the reactions' sensitivity to various factors: a powerful tool when predicting outcomes in multi-step reactions.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the speed at which chemical reactions occur. It not only helps predict reaction times but also reveals the pathways of the reactions. Through understanding kinetics, we gain insights into different types of reactions and mechanisms. This knowledge is crucial for developing new products, optimizing industrial processes, and even understanding natural phenomena.

In this exercise, kinetics comes into play to determine crucial reaction parameters such as rate, concentration change, and how these evolve over time. The approach, involving constant reaction rates, is a simplification often utilized to provide foundational understanding. Typically, thermodynamics complements kinetics to paint a complete picture of any reaction.

Knowing kinetic principles is indispensable for chemists and engineers in crafting routes for synthetic schemes or scaling up processes from lab to industrial scale.

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Most popular questions from this chapter

For the reversible reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D},\) the enthalpy change of the forward reaction is \(+21 \mathrm{kJ} / \mathrm{mol}\) The activation energy of the forward reaction is \(84 \mathrm{kJ} / \mathrm{mol}.\) (a) What is the activation energy of the reverse reaction? (b) In the manner of Figure 14-10, sketch the reaction profile of this reaction.

In the first-order decomposition of substance A the following concentrations are found at the indicated times: \(t=0 \mathrm{s},[\mathrm{A}]=0.88 \mathrm{M} ; t=50 \mathrm{s},[\mathrm{A}]=0.62 \mathrm{M} ; t=100 \mathrm{s},[\mathrm{A}]=0.44 \mathrm{M} ; t=150 \mathrm{s},[\mathrm{A}]=0.31 \mathrm{M}.\) Calculate the instantaneous rate of decomposition at \(t=100 \mathrm{s}.\)

The reaction \(A \longrightarrow\) products is first order in A. (a) If \(1.60 \mathrm{g} \mathrm{A}\) is allowed to decompose for 38 min, the mass of A remaining undecomposed is found to be 0.40 g. What is the half-life, \(t_{1 / 2}\), of this reaction? (b) Starting with \(1.60 \mathrm{g} \mathrm{A},\) what is the mass of \(\mathrm{A}\) remaining undecomposed after \(1.00 \mathrm{h} ?\)

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.

A first-order reaction, \(\mathrm{A} \longrightarrow\) products, has a halflife of \(75 \mathrm{s},\) from which we can draw two conclusions. Which of the following are those two (a) the reaction goes to completion in 150 s; (b) the quantity of \(A\) remaining after 150 s is half of what remains after 75 s; (c) the same quantity of A is consumed for every 75 s of the reaction; (d) one- quarter of the original quantity of A is consumed in the first 37.5 s of the reaction; (e) twice as much A is consumed in 75 s when the initial amount of \(\mathrm{A}\) is doubled; (f) the amount of \(\mathrm{A}\) consumed in 150 s is twice as much as is consumed in 75 s.

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