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Chapter 14: Problem 54

(a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is \(160 \mathrm{~kJ} / \mathrm{mol}\). Calculate the fraction of methyl isonitrile molecules that has an energy of \(160.0 \mathrm{~kJ}\) or greater at \(500 \mathrm{~K}\). (b) Calculate this fraction for a temperature of \(520 \mathrm{~K}\). What is the ratio of the fraction at \(520 \mathrm{~K}\) to that at \(500 \mathrm{~K} ?\)

Short Answer

Expert verified
In conclusion, at 500 K, approximately \(2.91 \times 10^{-15}\) of the methyl isonitrile molecules have an energy of 160 kJ/mol or greater. At 520 K, the fraction is approximately \(1.29 \times 10^{-14}\). The ratio of the fraction at 520 K to that at 500 K is approximately 4.42.

Step by step solution

01

(Step 1: Identify constants and formulas)

For this exercise, we'll need the following constants and formulas: - Activation energy (Ea) = 160 kJ/mol - Gas constant (R) = 8.314 J/(mol K) - Temperature 1 (T1) = 500 K - Temperature 2 (T2) = 520 K Boltzmann factor formula: \[ P = e ^ {-Ea/(RT)} \] We will first calculate the fraction of molecules with an energy of 160 kJ/mol or greater at 500 K, followed by the same calculation at 520 K, and finally find the ratio of the two fractions.
02

(Step 2: Calculate the fraction at 500 K )

Using the Boltzmann factor formula, plug in the values for Ea, R, and T1, and solve for the probability (P) at 500 K: Ea = 160,000 J/mol (convert from kJ to J) R = 8.314 J/(mol K) T1 = 500 K \[ P_{500} = e ^ {(-160,000) /(8.314 \times 500)} \]
03

(Step 3: Calculate the fraction at 520 K )

Using the Boltzmann factor formula, plug in the values for Ea, R, and T2, and solve for the probability (P) at 520 K: Ea = 160,000 J/mol R = 8.314 J/(mol K) T2 = 520 K \[ P_{520} = e ^ {(-160,000) /(8.314 \times 520)} \]
04

(Step 4: Calculate the ratio of the two fractions)

Now, we need to calculate the ratio of the fraction at 520 K to that at 500 K. To do this, divide P520 by P500: \[ \text{Ratio} = \frac{P_{520}}{P_{500}} \]
05

(Step 5: Finishing the calculations)

Now, calculate the fractions and the ratio using a calculator, and round the results: \[ P_{500} \approx 2.91 \times 10^{-15} \] \[ P_{520} \approx 1.29 \times 10^{-14} \] \[ \text{Ratio} \approx 4.42 \] In conclusion, at 500 K, approximately 2.91 脳 10^{-15} of the methyl isonitrile molecules have an energy of 160 kJ/mol or greater. At 520 K, the fraction is approximately 1.29 脳 10^{-14}. The ratio of the fraction at 520 K to that at 500 K is approximately 4.42.

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Key Concepts

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

Understanding Activation Energy
Activation energy, symbolized as Ea, is the minimum amount of energy required for a chemical reaction to proceed. In simple terms, it's the energy needed to push reactants over an energy barrier so that they can transform into products. For instance, lighting a match involves providing activation energy to initiate combustion.

In the specific case of methyl isonitrile isomerization, the activation energy is given as 160 kJ/mol. This means that each mole of methyl isonitrile molecules requires 160 kJ of energy to start isomerizing under given conditions. Understanding the activation energy is crucial for chemists, as it helps in controlling reactions by adjusting temperatures and using catalysts to lower the Ea, thus influencing the reaction speed.
Methyl Isonitrile Isomerization Chemistry
Methyl isonitrile (CH3NC) is an organic compound known for its ability to isomerize into acetonitrile (CH3CN). Isomerization is a process where a molecule is transformed into another molecule with the same molecular formula but a different structural arrangement of atoms. The isomerization of methyl isonitrile to acetonitrile is an example of a reaction where the functional group, in this case, a cyano group (-CN), rearranges itself within the molecule.

Understanding the energy landscape of such a reaction can aid in the prediction and control of the isomerization process, which is pivotal in various fields, including pharmaceuticals and organic synthesis. A higher activation energy will generally require higher temperatures to achieve a significant rate of reaction.
Temperature Effect on Reaction Rate
The reaction rate of chemical processes increases with temperature. When the temperature is raised, molecules move faster and collide more often and with greater energy, increasing the chances of overcoming the activation energy barrier. The Boltzmann factor, a term in statistical mechanics, describes the fraction of particles in a system that have an energy at or above a certain level at a given temperature. In our exercise, the Boltzmann factor equation is used to calculate the number of methyl isonitrile molecules capable of isomerizing at different temperatures.

As the temperature rises from 500 K to 520 K, the fraction of active molecules increases significantly. This exemplifies the strong dependency of reaction rates on temperature, which, according to the Arrhenius equation, approximately doubles for every 10 degrees Celsius increase. The ratio computed in the exercise, 4.42, quantifies this substantial increase in active molecules, reflecting the increased rate at which isomerization occurs at the higher temperature.

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Most popular questions from this chapter

Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide \((\mathrm{NO})\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is second order with a rate constant of \(0.0796 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(737^{\circ} \mathrm{C}\) and \(0.0815 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(947{ }^{\circ} \mathrm{C}\). Calculate the activation energy for the reaction.

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{~nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{~nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ?

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the halflife for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol}\) ? (b) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\) What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ? (c) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{~h}\) if the gas is collected in a \(10.0\)-L container. (Assume that the products do not dissolve in chloroform.)
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