Question

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

The Arrhenius parameters for the unimolecular decomposition of urea in aqueous solution are \(A ≈ 1.37 × 10^{11} s^{-1}\) and \(Ea ≈ 103383 \, J\,mol^{-1}\). Using the Eyring equation, we determined the values of \(ΔH‡ ≈ 102783 \,J\,mol^{-1}\) and \(ΔS‡ ≈ -174 \, J\,mol^{-1}\,K^{-1}\).

Step-by-step solution

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: \( k_1 = 1.20 × 10^{-7} \ s^{-1} \), so \( ln(k_1) = ln(1.20 × 10^{-7}) ≈ -16.328 \) For Trial 2: \( k_2 = 4.40 × 10^{-7} \ s^{-1}\), so \( ln(k_2) = ln(4.40 × 10^{-7}) ≈ -14.741 \)

Calculate the inverse of the temperature

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

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: \( ln(k_1) = ln(A) - \frac{Ea}{R × T_1} \) \( -16.328 = ln(A) - \frac{Ea}{8.314 × 333.15} \) For Trial 2, we get: \( ln(k_2) = ln(A) - \frac{Ea}{R × T_2} \) \( -14.741 = ln(A) - \frac{Ea}{8.314 × 344.65} \) Solve these two linear equations to find the values of \(A\) and \(Ea\). We get \(A ≈ 1.37 × 10^{11} s^{-1}\) and \(Ea ≈ 103383 J\,mol^{-1}\).

Determine ΔH‡ and ΔS‡ using the Eyring equation

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

Key concepts

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:\[k = \frac{k_B T}{h} e^{-\Delta H‡/RT} e^{\Delta S‡/R}\]where:- \(k_B\) is the Boltzmann constant.- \(h\) is the Planck constant.- \(T\) represents the temperature in Kelvin.- \(\Delta H‡\) is the change in enthalpy (or activation enthalpy) during the transition state.- \(\Delta S‡\) 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 \(\Delta H‡\) and \(\Delta S‡\) 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.