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Chapter 14: Problem 46

From the following data for the second-order gas-phase decomposition of HI at \(430^{\circ} \mathrm{C}\), calculate the second-order rate constant and half- life for the reaction: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { [HI]/mol } \mathrm{dm}^{-3} \\ \hline 0 & 1 \\ 100 & 0.89 \\ \hline 200 & 0.8 \\ \hline 300 & 0.72 \\ \hline 400 & 0.66 \\ \hline \end{array} $$

Short Answer

Expert verified
The second-order rate constant for the gas-phase decomposition of HI is approximately \(1.25 \times 10^{-3} \ \text{mol}^{-1}\text{dm}^3\text{s}^{-1}\), and the half-life of the reaction is 800 s.

Step by step solution

01

Write the second-order rate equation

For a second-order reaction, the rate equation can be written as: $$ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt $$ Where: - \([A]_t\) is the concentration of the reactant at time \(t\) - \([A]_0\) is the initial concentration of the reactant - \(k\) is the second-order rate constant - \(t\) is the time
02

Calculate the rate constant (k) using the given data

We can choose any of the given data points (other than the initial point) to calculate the rate constant (k). Let's choose the data point at t = 200 s, where [HI] = 0.8 mol/dm³. Using the second-order rate equation: $$ \frac{1}{0.8} - \frac{1}{1} = k(200) $$ Solving for k, we get: $$ k = \frac{\frac{1}{0.8} - \frac{1}{1}}{200} = \frac{(1 - 0.8) / 0.8}{200} = 1.25 \times 10^{-3} \ \text{mol}^{-1} \text{dm}^3 \text{s}^{-1} $$ So, the second-order rate constant (k) is approximately \(1.25 \times 10^{-3}\) mol⁻¹ dm³ s⁻¹.
03

Calculate the half-life (t1/2) using the rate constant (k) and initial concentration [HI]₀

The half-life of a second-order reaction can be calculated as: $$ t_{1/2} = \frac{1}{k[A]_0} $$ where: - \(t_{1/2}\) is the half-life - \(k\) is the second-order rate constant - \([A]_0\) is the initial concentration of the reactant Using the calculated rate constant (k) and the initial concentration [HI] = 1 mol/dm³: $$ t_{1/2} = \frac{1}{(1.25 \times 10^{-3})(1)} = \frac{1}{1.25 \times 10^{-3}} = 800 \ \text{s} $$ So, the half-life (t1/2) of the reaction is 800 s. In conclusion, the second-order rate constant for the gas-phase decomposition of HI is approximately \(1.25 \times 10^{-3} \ \text{mol}^{-1}\text{dm}^3\text{s}^{-1}\), and the half-life of the reaction is 800 s.

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Most popular questions from this chapter

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\) are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2}\), and \(\mathrm{N}_{2} .(\mathbf{b})\) Using a resource such as Table 8.3 , look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? \((\mathbf{c})\) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

(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)?

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{array}{l} \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{array} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3} .\) (b) Do we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? (d) Is this an example of homogeneous catalysis or heterogeneous catalysis?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.

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.
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