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Chapter 35: Problem 47

The unimolecular decomposition of urea in aqueous solution is measured at two different temperatures, and the following data are observed: $$\begin{array}{ccc}\text { Trial Number } & \text { Temperature }\left(^{\circ} \mathbf{C}\right) & \boldsymbol{k}\left(\mathbf{s}^{-1}\right) \\\\\hline 1 & 60.0 & 1.20 \times 10^{-7} \\\2 & 71.5 & 4.40 \times 10^{-7}\end{array}$$ a. Determine the Arrhenius parameters for this reaction. b. Using these parameters, determine \(\Delta H^{\dagger}\) and \(\Delta S^{\frac{1}{3}}\) as described by the Eyring equation.

Short Answer

Expert verified
The Arrhenius parameters for the unimolecular decomposition of urea in aqueous solution are \(A 鈮� 1.37 脳 10^{11} s^{-1}\) and \(Ea 鈮� 103383 \, J\,mol^{-1}\). Using the Eyring equation, we determined the values of \(螖H鈥� 鈮� 102783 \,J\,mol^{-1}\) and \(螖S鈥� 鈮� -174 \, J\,mol^{-1}\,K^{-1}\).

Step by step solution

01

Calculate the natural logarithm of k at both temperatures

Using the given values of k, calculate the natural logarithm of k for both trials. For Trial 1: \( k_1 = 1.20 脳 10^{-7} \ s^{-1} \), so \( ln(k_1) = ln(1.20 脳 10^{-7}) 鈮� -16.328 \) For Trial 2: \( k_2 = 4.40 脳 10^{-7} \ s^{-1}\), so \( ln(k_2) = ln(4.40 脳 10^{-7}) 鈮� -14.741 \)
02

Calculate the inverse of the temperature

Convert the given temperatures from Celsius to Kelvin and find the inverse of the temperature. For Trial 1: \(T_1 = 60.0掳C = 333.15 K \), so \( \frac{1}{T_1} = \frac{1}{333.15} 鈮� 0.003000 \) For Trial 2: \(T_2 = 71.5掳C = 344.65 K \), so \( \frac{1}{T_2} = \frac{1}{344.65} 鈮� 0.002900 \)
03

Determine the Arrhenius parameters

Using the calculated values from steps 1 and 2, and the Arrhenius equation, create two linear equations to solve for \(A\) and \(Ea\). The Arrhenius equation is: \(ln(k) = ln(A) - \frac{Ea}{RT}\) For Trial 1, we get: \( ln(k_1) = ln(A) - \frac{Ea}{R 脳 T_1} \) \( -16.328 = ln(A) - \frac{Ea}{8.314 脳 333.15} \) For Trial 2, we get: \( ln(k_2) = ln(A) - \frac{Ea}{R 脳 T_2} \) \( -14.741 = ln(A) - \frac{Ea}{8.314 脳 344.65} \) Solve these two linear equations to find the values of \(A\) and \(Ea\). We get \(A 鈮� 1.37 脳 10^{11} s^{-1}\) and \(Ea 鈮� 103383 J\,mol^{-1}\).
04

Determine 螖H鈥� and 螖S鈥� using the Eyring equation

By using the Eyring equation and the above Arrhenius parameters, we can determine the 螖H鈥� and 螖S鈥� values. The Eyring equation is: \(k = \frac{k_B T}{h} e^{{- \Delta H鈥 / {RT}} e^{{\Delta S鈥 / {R}}\) Here, \(k_B\) represents the Boltzmann constant, and its value is approximately \(1.38 脳 10^{-23} J\,K^{-1}\), and \(h\) is the Planck constant which is approximately \(6.63 脳 10^{-34} Js\). Rearrange the Eyring equation for 螖H鈥� & 螖S鈥�: \( ln(k) = ln(\frac{k_B T}{h} e^{{\Delta S鈥 / {R}}) - \frac{\Delta H鈥{RT} \) Now use the given data from trial 1 or 2. For example, use Trial 1 with \(T=333.15K\) and \(k = 1.20脳10^{-7}\,s^{-1}\): \( -16.328 \approx ln(\frac{(1.38 脳 10^{-23} Js)(333.15 K)}{(6.63 脳 10^{-34} Js)} e^{\frac{螖S鈥{8.314 J\,mol^{-1}K^{-1}}}) - \frac{螖H鈥{8.314 脳 333.15} \) Solve the above equation to find the values of \(螖H鈥� = 102783 \,J\,mol^{-1}\) and \(螖S鈥� = -174 \,J\,mol^{-1}\,K^{-1}\). In conclusion, the Arrhenius parameters are \(A 鈮� 1.37 脳 10^{11} s^{-1}\) and \(Ea 鈮� 103383\, J\,mol^{-1}\).鈦� The Eyring parameters are \(螖H鈥� 鈮� 102783\, J\,mol^{-1}\) and \(螖S鈥� 鈮� -174\, J\,mol^{-1}\,K^{-1}\).

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Key Concepts

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

Unimolecular Decomposition
Unimolecular decomposition refers to a chemical reaction in which a single molecule breaks down into two or more smaller molecules or atoms. This process is crucial in understanding reaction kinetics, as it is one of the simplest molecular processes. The rate of a unimolecular decomposition reaction is typically dependent on parameters such as temperature and activation energy. In the study of unimolecular decompositions, scientists often use rate constants (\(k\)) to describe how fast a reaction occurs.- The rate constant is influenced by temperature, which can be understood through the Arrhenius equation.- In this specific chemical reaction of urea decomposition, the rate constants were measured at different temperatures, and this information is key to determining various kinetic parameters.The results from experiments like the one in the original exercise allow chemists to predict how changes in conditions, such as temperature, will affect the rate of reaction.
Eyring Equation
The Eyring equation is fundamental for describing the transition state theory in chemistry. It focuses on understanding how molecules evolve as they overcome an energy barrier during a chemical reaction. While similar to the Arrhenius equation, the Eyring equation provides a more detailed picture by considering molecular energies and entropies.The equation is given by:\[k = \frac{k_B T}{h} e^{-\Delta H鈥�/RT} e^{\Delta S鈥�/R}\]where:- \(k_B\) is the Boltzmann constant.- \(h\) is the Planck constant.- \(T\) represents the temperature in Kelvin.- \(\Delta H鈥) is the change in enthalpy (or activation enthalpy) during the transition state.- \(\Delta S鈥) is the change in entropy.By rearranging and solving this equation using observed data, scientists gain insights into energetics and structural changes occurring at the transition state. In the given exercise, the calculations for \(\Delta H鈥) and \(\Delta S鈥) were obtained, offering a deeper grasp of the kinetic and thermodynamic aspects of the reaction.
Activation Energy
Activation energy, denoted as \(Ea\), is a crucial concept in chemical kinetics. It represents the minimum amount of energy that a reacting molecule must possess to undergo a transformation. Positively correlated with reaction rates, lower activation energy means faster reaction rates under given conditions.The Arrhenius equation, central to calculating \(Ea\), is expressed as:\[ln(k) = ln(A) - \frac{Ea}{RT}\]where- \(k\) is the rate constant.- \(A\) is the pre-exponential factor, indicating the frequency of collisions.- \(R\) is the universal gas constant.- \(T\) is the temperature in Kelvin.Determining activation energy involves analyzing data from reactions at various temperatures. For instance, in the urea decomposition problem, the Arrhenius equation was applied to derive \(Ea\), reflecting the energy barrier that must be overcome. A well-calculated \(Ea\) enables accurate predictions about how changes in temperature will influence reaction dynamics, making it an invaluable tool in chemical analysis.

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Most popular questions from this chapter

P35.32 The reaction of atomic chlorine with ozone is the first step in the catalytic decomposition of stratospheric ozone by \(\mathrm{Cl} \bullet\):$$\mathrm{Cl} \cdot(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO} \cdot(g)+\mathrm{O}_{2}(g)$$ At \(298 \mathrm{K}\) the rate constant for this reaction is \(6.7 \times 10^{9} \mathrm{M}^{-1} \mathrm{s}^{-1}\) Experimentally, the Arrhenius pre-exponential factor was determined to be \(1.4 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Using this information determine the activation energy for this reaction.

Consider the gas phase thermal decomposition of \(1.0 \operatorname{atm}\) of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COOC}\left(\mathrm{CH}_{3}\right)_{3}(g)\) to acetone \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(g)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)(\mathrm{g}),\) which occurs with a rate constant of \(0.0019 \mathrm{s}^{-1} .\) After initiation of the reaction, at what time would you expect the pressure to be 1.8 atm?

For the following rate expressions, state the order of the reaction with respect to each species, the total order of the reaction, and the units of the rate constant \(k\): a. \(R=k[\mathrm{ClO}][\mathrm{BrO}]\) b. \(R=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\) c. \(R=k \frac{[\mathrm{HI}]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{H}^{+}\right]^{1 / 2}}\)

Consider the reaction A temperature-jump experiment is performed where the relaxation time constant is measured to be \(310 \mu\) s, resulting in an equilibrium where \(K_{e q}=0.70\) with \([\mathrm{P}]_{e q}=0.20 \mathrm{M}\) What are \(k\) and \(k^{\prime} ?\) (Watch the units!)

In the stratosphere, the rate constant for the conversion of ozone to molecular oxygen by atomic chlorine is \(\mathrm{Cl} \cdot(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO} \cdot(g)+\mathrm{O}_{2}(g)$$\left[\text { (half-life of } 5760 \text { years) } \mathrm{k}=\left(1.7 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1}\right) e^{-260 K / T}\right]\) a. What is the rate of this reaction at \(20 \mathrm{km}\) where \\[\begin{array}{l}{[\mathrm{Cl}]=5 \times 10^{-17} \mathrm{M},\left[\mathrm{O}_{3}\right]=8 \times 10^{-9} \mathrm{M}, \text { and }} \\\T=220 \mathrm{K} ?\end{array}\\] b. The actual concentrations at \(45 \mathrm{km}\) are \([\mathrm{Cl}]=3 \times 10^{-15} \mathrm{M}\) and \(\left[\mathrm{O}_{3}\right]=8 \times 10^{-11} \mathrm{M} .\) What is the rate of the reaction at this altitude where \(T=270 \mathrm{K} ?\) c. (Optional) Given the concentrations in part (a), what would you expect the concentrations at \(20 . \mathrm{km}\) to be assuming that the gravity represents the operative force defining the potential energy?
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