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Chapter 14: Problem 42

Ammonia decomposes on the surface of a hot tungsten wire. Following are the half-lives that were obtained at \(1100^{\circ} \mathrm{C}\) for different initial concentrations of \(\mathrm{NH}_{3}:\left[\mathrm{NH}_{3}\right]_{0}=0.0031 \mathrm{M}, t_{1 / 2}=7.6 \mathrm{min} ; 0.0015 \mathrm{M}\) \(3.7 \mathrm{min} ; 0.00068 \mathrm{M}, 1.7 \mathrm{min.}\) For this decomposition reaction, what is (a) the order of the reaction; (b) the rate constant, \(k ?\)

Short Answer

Expert verified
The order of reaction is 2nd order, and the rate constant \(k\) approximates to 43.0 M-1min-1.

Step by step solution

01

Identifying the reaction order

The initial concentrations of NH3 and their corresponding half-life times are given. Observe the relationship between the concentration of NH3 and its half-life. If the half-life is constant and does not change with concentration, the reaction is first order. However, if the half-life is variable and changes with concentration, the reaction is more likely to be of second or zero order. Here, it can be seen that as the concentration decreases, the half-life also decreases, indicating that the reaction is probably not first order as its half-life is not constant and depends on concentration.
02

Confirming Reaction Order

To further confirm, you can look at the relationship between the half-life and the concentration. If it's a direct relationship (increased concentration leads to increased half-life), the reaction is second order. If it's an inverse relationship (increased concentration leads to decreased half-life), the reaction is zero order. Here, as the concentration increases, the half-life also increases, indicating that this reaction is second-order.
03

Finding the Rate Constant, \(k\)

For a second order reaction, the half-life is given by \( t_{1/2} = \frac{1}{{k[N]_0}} \). So, it can be rearranged to find k: \( k = \frac{1}{{t_{1/2} [N]_0}} \). Using the given concentration and half-life values, calculate k. For example, when [NH3]0 = 0.0031 M and t1/2 = 7.6 minutes, \( k = \frac{1}{{7.6 \times 0.0031}} \approx 43.0 \, \text{M}^{-1}\text{min}^{-1} \). The unit of k for second order reactions is M-1min-1. Repeating calculation for all given data points can cross-check the rate constant value and its consistency.

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Most popular questions from this chapter

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 \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0050 \mathrm{M} ?\)

Explain the important distinctions between each pair of terms: (a) first-order and second-order reactions; (b) rate law and integrated rate law; (c) activation energy and enthalpy of reaction; (d) elementary process and overall reaction; (e) enzyme and substrate.

For the reaction \(\mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D},\) the rate law is rate of reaction \(=k[\mathrm{A}][\mathrm{B}]\) (a) Show that the following mechanism is consistent with the stoichiometry of the overall reaction and with the rate law. $$\begin{array}{l} \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{I} \quad(\text { slow }) \\ \mathrm{I}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad(\text { fast }) \end{array}$$ (b) Show that the following mechanism is consistent with the stoichiometry of the overall reaction, but not with the rate law. $$\begin{array}{c} 2 \mathrm{B} \stackrel{k_{1}}{\mathrm{k}_{1}} \mathrm{B}_{2} \text { (fast) } \\\ \mathrm{A}+\mathrm{B}_{2} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}+\mathrm{D} \text { (slow) } \end{array}$$

One of the following statements is true and the other is false regarding the first-order reaction $2 \overrightarrow{\mathrm{A}} \longrightarrow \mathrm{B}+\mathrm{C} .$ Identify the true statement and the false one, and explain your reasoning. (a) A graph of [A] versus time is a straight line. (b) The rate of the reaction is one-half the rate of disappearance of A.

Acetoacetic acid, \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH},\) a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and \(\mathrm{CO}_{2}(\mathrm{g}).\) $$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}(\mathrm{aq}) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})$$ This first-order decomposition has a half-life of 144 min. (a) How long will it take for a sample of acetoacetic acid to be \(65 \%\) decomposed? (b) How many liters of \(\mathrm{CO}_{2}(\mathrm{g}),\) measured at \(24.5^{\circ} \mathrm{C}\) and 748 Torr, are produced as a \(10.0 \mathrm{g}\) sample of \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}\) decomposes for 575 min? [Ignore the aqueous solubility of \(\mathrm{CO}_{2}(\mathrm{g}) \cdot \mathrm{l}.\)
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