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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 reaction \(A \longrightarrow 2 B+C\), the following data are obtained for \([\mathrm{A}]\) as a function of time: \(t=0 \mathrm{min}\) \([\mathrm{A}]=0.80 \mathrm{M} ; 8 \mathrm{min}, 0.60 \mathrm{M} ; 24 \mathrm{min}, 0.35 \mathrm{M} ; 40 \mathrm{min}\) \(0.20 \mathrm{M}\) (a) By suitable means, establish the order of the reaction. (b) What is the value of the rate constant, \(k ?\) (c) Calculate the rate of formation of \(\mathrm{B}\) at \(t=30 \mathrm{min}\).

In your own words, define or explain the following terms or symbols: (a) \([\mathrm{A}]_{0} ;\) (b) \(\dot{k} ;\) (c) \(t_{1 / 2} ;\) (d) zeroorder reaction; (e) catalyst.

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?

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ Which of these sets of data corresponds to a (a) zero-order, (b) first-order, (c) second-order reaction?

The half-life for the first-order decomposition of nitramide, \(\mathrm{NH}_{2} \mathrm{NO}_{2}(\mathrm{aq}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1),\) is \(123 \min\) at \(15^{\circ} \mathrm{C} .\) If \(165 \mathrm{mL}\) of a \(0.105 \mathrm{M} \mathrm{NH}_{2} \mathrm{NO}_{2}\) solution is allowed to decompose, how long must the reaction proceed to yield \(50.0 \mathrm{mL}\) of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) collected over water at \(15^{\circ} \mathrm{C}\) and a barometric pressure of \(756 \mathrm{mm} \mathrm{Hg} ?\) (The vapor pressure of water at \(15^{\circ} \mathrm{C}\) is \(12.8 \mathrm{mmHg} .)\)

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