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(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

Short Answer

Expert verified
(a) For a first-order reaction, plotting \(\ln[A]\) versus time (t) will yield a straight line. The equation for this linear relationship is \(\ln[A] = -kt + \ln[A]_0\). (b) To calculate the rate constant (k) for a first-order reaction from the graph, use the slope of the straight line representing the relationship between ln[A] and time. The rate constant k is the negative of the slope: k = -slope.

Step by step solution

01

Part (a): Identify the linear form for the first-order reaction

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant A. The rate equation can be written as: \(r = k [A]\) where r is the rate of reaction, k is the rate constant, and [A] is the concentration of A. To find a linear relationship for this reaction, we will use the integrated rate law for a first-order reaction: \(\ln[A] = -kt + \ln[A]_0\) where [A] is the concentration of A at time t, and [A]₀ is the initial concentration of A. This equation resembles the formula for a straight line (y = mx + b), where: - y = \(\ln[A]\) - m = -k - x = t - b = \(\ln[A]_0\) Thus, plotting ln[A] versus time (t) will yield a straight line for a first-order reaction.
02

Part (b): Calculate the rate constant from the graph

Now that we've determined that the linear relationship for a first-order reaction involves plotting ln[A] versus time, we can use the slope of the straight line from the graph to calculate the rate constant (k). As previously stated, the equation of the straight line for the first-order reaction is: \(\ln[A] = -kt + \ln[A]_0\) From this equation, we can see that the slope of the straight line (-k) is the rate constant multiplied by -1. To calculate the rate constant k, we take the negative of the slope of the straight line from the graph, which is the relationship between ln[A] and time: k = -slope So, by plotting ln[A] versus time for a first-order reaction, we can obtain the rate constant k by taking the negative of the slope of the resulting straight line.

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Key Concepts

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

Rate Constant
In chemistry, the rate constant (often denoted as \(k\)) is a crucial factor in determining how fast a reaction proceeds. In a first-order reaction, this constant has units of time\(^{-1}\) (such as \( ext{s}^{-1}\)), which indicates that it describes how the concentration of a reactant changes over time solely dependent on the concentration itself. The rate constant gives us an idea about:
  • **The reaction speed:** A larger rate constant suggests a faster reaction.
  • The temperature dependency:** Typically, as the temperature increases, \(k\) also increases.
For a given reaction, if we know the rate law, we can find the rate constant by experimental methods. For a first-order reaction, once we plot \(\ln[A]\) (natural log of concentration) versus time, the slope of this line (-slope = -k) directly gives us the rate constant. So, the rate constant is determined as the negative of the slope of this graph. If the reaction is **exponentially decaying**, it aligns with a first-order behavior, which is crucial for accurately determining \(k\) in the lab.
Integrated Rate Law
The integrated rate law is a way to relate the concentration of a reactant to time. For a first-order reaction, the integrated rate law is derived from the basic rate equation:\[\text{rate} = -\frac{d[A]}{dt} = k[A]\]By integrating this equation, we arrive at:\[\ln[A] = -kt + \ln[A]_0\]This form resembles the equation of a straight line \((y = mx + b)\):
  • \(y = \ln[A]\)
  • \(m = -k\)
  • \(x = t\)
  • \(b = \ln[A]_0\)
One of the key points of this law is that it confirms how the concentration of \([A]\) decreases exponentially over time in a first-order reaction. By plotting \(\ln[A]\) against time, we can directly observe this exponential decay **graphically** and use the straight line to calculate the rate constant \(k\). The simplicity of this approach is one reason why the integrated rate law is so powerful in chemical kinetics, making it invaluable for both theoretical and experimental chemists alike.
Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical reactions occur and the factors that affect these rates. This branch of physical chemistry not only helps determine how quickly a reaction will proceed, but also the mechanism by which it happens. **Key Considerations in Reaction Kinetics:**
  • **Concentration of reactants:** Higher concentrations usually lead to a higher reaction rate.
  • **Temperature:** Increasing temperature generally speeds up a reaction by providing more energy to reactant molecules.
  • **Presence of a catalyst:** Catalysts can significantly enhance the rate of a reaction without being consumed in the process.
  • **Reaction order:** Determines how the rate is affected by the concentration of reactants. First-order reactions depend on the concentration of one reactant.
In the context of a first-order reaction, the rate is directly proportional to the concentration of a single reactant, making it very straightforward to examine: just measure the concentration over time. The insights gained from reaction kinetics are crucial for processes ranging from industrial synthesis to biological enzyme function, helping predict how systems behave under different conditions.

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Most popular questions from this chapter

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm} .\) (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min} .\) Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .(\mathbf{a})\) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?\) (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol}\) ? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}\), \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c),\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\). and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

The reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\) \(\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) was studied with the following results: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{CIO}_{2}\right](M)} & {\left[\mathrm{OH}^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.060 & 0.030 & 0.0248 \\ 2 & 0.020 & 0.030 & 0.00276 \\ 3 & 0.020 & 0.090 & 0.00828 \\ \hline \end{array} $$ (a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{ClO}_{2}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)

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