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The following data were obtained for the dimerization of 1,3 -butadiene, \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{g}),\) at 600 K: \(t=0 \min ,\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0169 \mathrm{M} ; 12.18 \mathrm{min}, 0.0144 \mathrm{M} ; 24.55 \mathrm{min}, 0.0124 \mathrm{M} ; 42.50 \mathrm{min}, 0.0103 \mathrm{M}, 68.05 \min , 0.00845 \mathrm{M}.\) (a) What is the order of this reaction? (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.00423 \mathrm{M} ?\) (d) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0050 \mathrm{M} ?\)

Short Answer

Expert verified
The order of the reaction is first order, with a rate constant \(k = 0.0924 \min^{-1}\). The concentrations of 1,3-Butadiene will reach 0.00423 M at approximately 102.54 minutes, and 0.0050 M at approximately 85.33 minutes.

Step by step solution

01

Determine the Order of the Reaction

In this step, use the concentration and time data provided to ascertain whether the chemical reaction is zeroth, first or second order. Calculate the change in concentration per unit time for any two consecutive data points. If this value remains the same, its order of reaction is zero; if this value variation is linear, then it is first order; if this variation is inversely proportional to the concentration, order of reaction is second. Here, comparing concentration changes over time it could be seen that the rate of reaction decreases linearly over time, therefore, this reaction is first order.
02

Calculate the Rate Constant (k)

We know that the integrated first order rate law is \(\ln [A] = -kt + \ln [A_0]\), where \([A]\) is the concentration of the reactant, \(k\) is the rate constant, \(t\) is the time, and \([A_0]\) is the initial concentration. By rearranging, we have \(k = (\ln[A_0] - \ln[A])/t)\). From the data, choose two points and substitute the values into this formula to solve for \(k\). Since the reaction is first order and constant irrespective of which points are chosen, picking any two points should yield the same value for \(k\). Let's choose the first and last data points: \(t = 0 \min, [C_4H_6] = 0.0169 M\) and \(t = 68.05 \min, [C_4H_6] = 0.00845 M\). Therefore, calculating \(k = (\ln[0.0169] - \ln[0.00845])/68.05\) gives us \(k = 0.0924 \min^{-1}\).
03

Calculate the Time for a Specific Concentration

Use the first-order integrated rate equation with \(A = 0.00423 M\) to calculate the time at which this concentration is achieved. From the formula, \(t = (\ln[A_0] - \ln[A]) / k\). Inserting the values, we get \(t = (\ln[0.0169] - \ln[0.00423])/0.0924\) which results in \(t \approx 102.54\) minutes.
04

Calculate the Time for another Specific Concentration

Similarly, substitute \(A = 0.005 M\) into the rate equation and solve for \(t\). On substituting the values, we get \(t = (\ln[0.0169] - \ln[0.005])/0.0924\) which gives us \(t \approx 85.33\) minutes.

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Key Concepts

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

First-Order Reactions
First-order reactions are a type of chemical reaction where the rate depends linearly on one reactant's concentration. In simple terms, if you double the concentration of this reactant, the reaction rate also doubles.

First-order reactions are often seen in processes like radioactive decay and some chemical reactions, such as the dimerization of 1,3-butadiene in our example problem. This behavior can be identified by plotting the natural logarithm of the reactant's concentration against time. A straight line indicates a first-order reaction, as the rate of change of the concentration should remain constant when plotted this way.

Understanding the order of reactions helps predict how changes in concentration affect reaction rates, which is crucial for controlling industrial chemical processes and experimental laboratory reactions. This knowledge allows chemists to optimize conditions to make reactions more efficient.
Rate Constant Calculation
The rate constant, denoted as \( k \), is a crucial parameter in the rate equation of chemical reactions. It gives us insights into the speed of a reaction under specific conditions.

For first-order reactions, the integrated rate law is expressed as: \[\ln [A] = -kt + \ln [A_0]\]where \([A]\) is the concentration at time \( t \), \([A_0]\) is the initial concentration, and \( k \) is the rate constant.

To calculate \( k \) for a first-order reaction, select two data points from your concentration vs. time data. The equation rearranges to: \[k = \frac{\ln [A_0] - \ln [A]} {t}\]

By using the provided data for 1,3-butadiene dimerization, you can determine \( k \) accurately regardless of which pairs of data points are chosen. In the given example, using points from \( t = 0 \min\) to \( t = 68.05 \min\), the rate constant was calculated as \( k = 0.0924 \min^{-1} \). This consistency confirms the first-order nature of the reaction, because \( k \) remains constant over various intervals when the reaction is first-order.



Knowing the rate constant allows scientists to predict the time required for a reactant to reach a certain concentration and to understand better the reaction kinetics.
Integrated Rate Law
The integrated rate law for first-order reactions is a mathematical expression that describes the relationship between the concentration of a reactant and time. It's particularly useful when you need to calculate how long it takes for a reactant to drop to a particular concentration.

The first-order integrated rate law formula is:

  • \( \ln [A] = -kt + \ln [A_0] \)
This equation can be rearranged to solve for \( t \), the time, as follows:
  • \( t = \frac{\ln [A_0] - \ln [A]} {k} \)
With this arrangement, you can insert the initial concentration \([A_0]\), the desired concentration \([A]\), and the rate constant \( k \) to find \( t \).

In our example, to find out when the concentration of 1,3-butadiene reaches 0.00423 M, we substituted these values into the formula. These calculations show that the concentration reaches this value at approximately 102.54 minutes. Similarly, solving for 0.005 M resulted in about 85.33 minutes.

This ability to predict concentration changes over time makes the integrated rate law a powerful tool in chemical kinetics, allowing chemists to plan and control reactions more effectively.

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Most popular questions from this chapter

One of the following statements is true and the other is false regarding the first-order reaction 2A \(\longrightarrow \mathrm{B}+\mathrm{C}\). Identify the true statement and the false one, and explain your reasoning. (a) The rate of the reaction decreases as more and more of \(\mathrm{B}\) and \(\mathrm{C}\) form. (b) The time required for one-half of substance \(A\) to react is directly proportional to the quantity of A present initially.

The following rates of reaction were obtained in three experiments with the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}).\) $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial Rate of } \\ \text { Expt } & \text { [NO], M } & \text { [Cl }_{2} \text { ], M } & \text { Reaction, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 1 & 0.0125 & 0.0255 & 2.27 \times 10^{-5} \\ 2 & 0.0125 & 0.0510 & 4.55 \times 10^{-5} \\ 3 & 0.0250 & 0.0255 & 9.08 \times 10^{-5} \\ \hline \end{array}$$ What is the rate law for this reaction?

The decomposition of \(\mathrm{HI}(\mathrm{g})\) at \(700 \mathrm{K}\) is followed for \(400 \mathrm{s},\) yielding the following data: at \(t=0,[\mathrm{HI}]=\) \(1.00 \mathrm{M} ;\) at \(t=100 \mathrm{s},[\mathrm{HI}]=0.90 \mathrm{M} ;\) at \(t=200 \mathrm{s}, [\mathrm{HI}]=0.81 \mathrm{M} ; t=300 \mathrm{s},[\mathrm{HI}]=0.74 \mathrm{M} ;\) at \(t=400 \mathrm{s}, [\mathrm{HI}]=0.68 \mathrm{M} .\) What are the reaction order and the rate constant for the reaction: $$\mathrm{HI}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g}) ?$$ Write the rate law for the reaction at 700 K.

The following statements about catalysis are not stated as carefully as they might be. What slight modifications would you make in them? (a) A catalyst is a substance that speeds up a chemical reaction but does not take part in the reaction. (b) The function of a catalyst is to lower the activation energy for a chemical reaction.

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

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