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Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta \mathrm{t}\).)

Short Answer

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Question: Derive the integrated rate law for a zero-order reaction starting from the relation \(-\Delta[\mathrm{A}] = k\Delta t\). Answer: The integrated rate law for a zero-order reaction is \([\mathrm{A}] = [\mathrm{A}]_{0} - kt\), where \([\mathrm{A}]\) represents the concentration of reactant A at any time \(t\), \([\mathrm{A}]_{0}\) is the initial concentration of A, and \(k\) is the rate constant.

Step by step solution

01

Write down the given differential relationship

We are provided with the relation \(-\Delta[\mathrm{A}] = k\Delta t\). We can rewrite this with differentials as: $$-\mathrm{d}\left[\mathrm{A}\right] = k\,\mathrm{d}t.$$
02

Integrate both sides of the equation

Now, we will integrate both sides of the equation. Let's denote the initial concentration of A at time \(t = 0\) as \([\mathrm{A}]_0\) and the concentration at any time \(t\) as \([\mathrm{A}]\): $$\int_{[\mathrm{A}]_{0}}^{[\mathrm{A}]} { -\frac{\mathrm{d}\left[\mathrm{A}\right]}{1} } = \int_{0}^{t}{k\,\mathrm{d}t}.$$
03

Perform the integrations

The integrations on both sides are straightforward since we are integrating constants: $$-[\mathrm{A}] \Big|_{[\mathrm{A}]_{0}}^{[\mathrm{A}]} = kt \Big|_0^t.$$
04

Evaluate the expressions with limits

Now, let's evaluate both expressions using the given limits: $$-([\mathrm{A}]-[\mathrm{A}]_{0}) = kt - 0.$$
05

Solve for the concentration of A

Finally, we need to solve for the concentration of reactant A as a function of time: $$[\mathrm{A}] = [\mathrm{A}]_{0} - kt.$$ Thus, we have derived the integrated rate law for a zero-order reaction: $$[\mathrm{A}] = [\mathrm{A}]_{0} - kt.$$

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Key Concepts

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

Zero-Order Reaction
In chemical kinetics, which is the branch of physical chemistry concerned with understanding the rates of chemical reactions, a zero-order reaction refers to a scenario where the rate of reaction is independent of the concentration of the reactants. This might seem counterintuitive at first because we might expect that the more reactant we have, the faster a reaction should proceed. However, in a zero-order reaction, there are other factors, such as surface area of a catalyst or radiation intensity, that control the reaction rate.

Mathematically, we describe a zero-order reaction with the rate law \( R = k \), where \( R \) represents the reaction rate and \( k \) is the reaction rate constant. Notice that the rate is constant and thus independent of the concentration of the reactant. The units of \( k \) for a zero-order reaction are typically \(M/s\) (molarity per second), which is unique to zero-order kinetics.

According to the rate law, the decrease in concentration of reactant A over time (\( -\Delta[A] \) ) is directly proportional to the elapsed time (\( \Delta t \) ). Consequently, as a result of this constant rate, the concentration of the reactant decreases linearly with time, a hallmark of zero-order reactions. The integrated rate law we derive \( [A] = [A]_{0} - kt \) illustrates this linear relationship between concentration and time, supporting the concept that regardless of how much reactant A is present, the reaction rate remains constant over time.
Chemical Kinetics
Chemical kinetics delves into the rate at which a chemical process occurs and the factors that affect this rate. It's important to emphasize that the 'rate of a reaction' refers to the speed at which reactants are converted into products. In studying chemical kinetics, we look at various types of reactions such as zero-order, first-order, and second-order, each with its own unique rate law.

In a broader context, chemical kinetics also looks at the pathway of the reaction, called the reaction mechanism, and energy changes that occur, which are represented through activation energy and the potential energy surface. The principles of kinetics are used to understand how different conditions – such as temperature, pressure, concentration, and the presence of catalysts – influence the speed of a reaction.

Additionally, kinetics theories can explain why certain reactions are instantaneous while others might take years to reach completion. By applying mathematical models and equations like the integrated rate law provided in the original exercise, chemists can predict the behavior of a chemical reaction over time and under varying conditions, which is invaluable in both research and industrial applications.
Reaction Rate Constant
The reaction rate constant (\( k \) ) is a proportionality factor that features prominently in the rate laws of chemical reactions. Its value is crucial as it provides insight into the speed of a reaction and is determined experimentally. Importantly, the reaction rate constant is not affected by the concentration of reactants, but it is highly sensitive to temperature changes and can be influenced by the presence of a catalyst.

For a zero-order reaction, as discussed in the solution steps, the reaction rate constant \( k \) has units of concentration/time, specifically \(M/s\). However, for reactions of different orders, \( k \) would have different units. For example, in a first-order reaction, the units would be \(1/s\), while for a second-order reaction, the units would be \(M^{-1}s^{-1}\).

Using the integrated rate law derived from the exercise, \( [A] = [A]_{0} - kt \), the reaction rate constant links the time variable with the change in concentration, allowing us to predict how much reactant will be left at a given point in time. It's fundamental to the study of reaction kinetics, as knowing \( k \) allows chemists to calculate the time it will take for a reaction to reach a certain completion or to determine the necessary conditions to achieve a desired reaction rate.

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Most popular questions from this chapter

At high temperatures, the decomposition of cyclobutane is a first-order reaction. Its activation energy is \(262 \mathrm{~kJ} / \mathrm{mol}\). At \(477^{\circ} \mathrm{C}\), its half-life is \(5.00 \mathrm{~min}\). What is its half-life (in seconds) at \(527^{\circ} \mathrm{C}\) ?

WEB When boron trifluoride reacts with ammonia, the following \(T\) reaction occurs: for $$\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{BF}_{3} \mathrm{NH}_{3}(g)$$ The following data are obtained at a particular temperature: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{BF}_{3}\right]} & {\left[\mathrm{NH}_{3}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\\\\hline 1 & 0.100 & 0.100 & 0.0341 \\ 2 & 0.200 & 0.233 & 0.159 \\ 3 & 0.200 & 0.0750 & 0.0512 \\ 4 & 0.300 & 0.100 & 0.102 \\\\\hline\end{array}$$

For a reaction involving the decomposition of \(\mathrm{Y}\), the following data are obtained: $$\begin{array}{lllll}\hline \text { Rate }(\mathrm{mol} / \mathrm{L} \cdot \min ) & 0.288 & 0.245 & 0.202 & 0.158 \\ {[\mathrm{Y}]} & 0.200 & 0.170 & 0.140 & 0.110 \\ \hline\end{array}$$ (a) Determine the order of the reaction. (b) Write the rate expression for the decomposition of Y. (c) Calculate \(k\) for the experiment above.

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

Two mechanisms are proposed for the reaction $$\begin{array}{cl}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \\ \text { Mechanism 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} & \text { (fast) } \\ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \\ \text { Mechanism 2: } \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \end{array}$$ Show that each of these mechanisms is consistent with the observed rate law: rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{O}_{2}\right]\).

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