Problem 54 - Chemistry: The Central Science

Question

(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 \(500 \mathrm{~K} ?\)

Step-by-Step Solution

Answer

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 1: (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.

Step 2: (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 \[ P_{500} = e ^ {(-160,000) /(8.314 \times 500)} \]

Step 3: (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 \[ P_{520} = e ^ {(-160,000) /(8.314 \times 520)} \]

Step 4: (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: \[ \text{Ratio} = \frac{P_{520}}{P_{500}} \]

Step 5: (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 × 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 × 10^{-14}. The ratio of the fraction at 520 K to that at 500 K is approximately 4.42.

Key Concepts

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.