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What is a zero-order reaction?

Short Answer

Expert verified
A zero-order reaction has a constant rate, independent of reactant concentration.

Step by step solution

01

Defining Reaction Order

A zero-order reaction is defined by how the rate of reaction depends on the concentration of the reactants. In a zero-order reaction, the rate is independent of the concentration of the reactants.
02

Understanding Rate Law

For a zero-order reaction, the rate of the reaction is given by the equation: \[ ext{Rate} = k \] where \(k\) is the rate constant. This means the rate remains constant throughout the reaction.
03

Implications of Zero-order

Since the rate does not change with the concentration of reactants, doubling or halving the concentration will not affect the reaction rate. The reaction proceeds at a constant rate until the reactant is depleted.
04

Graphical Representation

A plot of the concentration of reactants versus time for a zero-order reaction yields a straight line with a negative slope. The slope of this line is equal to \(-k\), showing that the concentration decreases linearly with time.

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Key Concepts

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

Reaction Order
When we talk about reaction order, we're discussing how the concentration of reactants affects the rate of a chemical reaction. The reaction order is determined experimentally and it helps us understand the relationship between concentration and the rate. A zero-order reaction is particularly interesting because the rate does not depend on the concentration of the reactants at all. This means the rate stays exactly the same regardless of how much reactant you start with. In everyday terms, you can think of reaction order as an instruction for how the reaction rate behaves as the reactants change over time.
Rate Law
The rate law defines the exact relationship between the reaction rate and the concentration of reactants. For a zero-order reaction, the rate law is beautifully simple. It is expressed as \[ \text{Rate} = k \]where \( k \) is the rate constant. Unlike higher-order reactions, you do not see any reactant concentration terms in this equation. The implication is that for zero-order reactions, the rate is constant and not affected by any change in reactant concentration. This is why zero-order reactions are predictable and often used in systems where a constant rate is desirable, such as in controlled drug release systems.
Rate Constant
The rate constant, denoted as \( k \), is a critical component in the rate law. In the context of a zero-order reaction, \( k \) holds particular significance. It represents the rate of reaction when there's no dependency on reactant concentration. Thus, in a zero-order reaction, \( k \) is numerically equivalent to the reaction rate. It's worth noting that the units of \( k \) vary depending on the order of the reaction, and for zero-order reactions, the units are concentration/time (e.g., M/s). The rate constant is unique to every reaction and depends on environmental factors like temperature.
Graphical Representation
Understanding graphical representation is essential for visualizing how a zero-order reaction proceeds. Imagine you are plotting the concentration of reactants against time. For a zero-order reaction, this graph will be a straight line descending diagonally. The line's negative slope directly corresponds to \(-k\), the negative of the rate constant. This visualization helps to easily see that as time progresses, the concentration of reactants decreases linearly. Such straightforward plotting allows you to predict how much reactant will remain at any given time, making it a powerful tool for planning and analysis in chemical reactions. Graphs for zero-order reactions are simple yet highly informative, offering clear insights into the reaction kinetics.

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Most popular questions from this chapter

A proposed mechanism for the reaction described by the equation $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ is (1) \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \frac{k_{1}}{\vec{k}_{-1}} \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) (2) \(\mathrm{NO}_{2}(g)+\mathrm{NO}_{5}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{NO}(g)\) (3) \(\mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \stackrel{k_{3}}{\longrightarrow} 3 \mathrm{NO}_{2}(g)\) Assuming a steady-state condition for \(\left[\mathrm{NO}_{3}\right]\), show that the predicted rate law for the overall reaction is rate of reaction \(=\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}=\frac{k_{1} k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{k_{2}+k_{-1}}\)

The reaction described by $$ 2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is postulated to occur via the mechanism (1) \(\mathrm{NO}_{2}(g)+\mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{NO}_{3}(g) \quad\) (slow) (2) \(\mathrm{NO}_{3}(g) \rightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad\) (fast) The rate law for the reaction is $$ \text { rate of reaction }=k\left[\mathrm{NO}_{2}\right]^{2} $$ Is this mechanism consistent with the observed rate law? Justify your answer.

What is meant by an "elementary reaction equation"? What information can we gain from an elementary reaction equation that we cannot gain from the overall reaction equation? Can an elementary reaction equation and the overall reaction equation ever be the same?

The activation energy for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) according to the equation $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ is \(102 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\). At \(45.0^{\circ} \mathrm{C}\) the rate constant is \(5.0 \times 10^{-4} \mathrm{~s}^{-1} .\) What is the rate constant at \(65.0^{\circ} \mathrm{C}\) ?

Determine the value of \(K_{\mathrm{M}}\), the MichaelisMenten constant, from the following data for the myosin-catalyzed hydrolysis of ATP. $$ \begin{array}{l|ccccc} {[\mathbf{A T P}] / \mathbf{\mu} \mathbf{m o l} \cdot \mathbf{L}^{-1}} & 7.5 & 12.5 & 20.0 & 43.5 & 62.5 \\ \hline \boldsymbol{R} / \mathbf{p m o l} \cdot \mathbf{L}^{-1} \cdot \mathbf{s}^{-1} & 67 & 95 & 119 & 155 & 166 \end{array} $$

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