Problem 57 - Chemistry The Molecular Science

Question

For the gas-phase reaction $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}(\mathrm{g}) \longrightarrow \mathrm{CH}_{2} \mathrm{CH}_{2}(\mathrm{~g})+\mathrm{HI}(\mathrm{g}) $$ the activation energy \(E_{\mathrm{a}}\) is \(221 \mathrm{~kJ} / \mathrm{mol}\) and the frequency factor \(A\) is \(1.2 \times 10^{14} \mathrm{~s}^{-1}\). If the concentration of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}\) is \(0.012 \mathrm{~mol} / \mathrm{L},\) calculate the rate of the reac- tion at (a) \(400 .{ }^{\circ} \mathrm{C}\) $$ \text { (b) } 800 .{ }^{\circ} \mathrm{C} $$

Step-by-Step Solution

Answer

At 400°C, the rate is \(5.15 \times 10^{-6} \, mol/(L\,s)\); at 800°C, it is \(0.0404 \, mol/(L\,s)\).

Step 1: Convert Temperature to Kelvin

Firstly, convert the given temperatures in Celsius to Kelvin. The formula to do this is: \[ T(K) = T(°C) + 273.15 \]For part (a), \( 400°C \) becomes: \[ T_1 = 400 + 273.15 = 673.15 \, K \]For part (b), \( 800°C \) becomes: \[ T_2 = 800 + 273.15 = 1073.15 \, K \]

Step 2: Use Arrhenius Equation

The rate constant \( k \) for a reaction is determined using the Arrhenius equation:\[ k = A e^{-E_a / (RT)} \]Where:- \( A \) is the frequency factor \( 1.2 \times 10^{14} \, s^{-1} \)- \( E_a \) is the activation energy \( 221 \, kJ/mol = 221,000 \, J/mol \)- \( R \) is the universal gas constant \( 8.314 \, J/(mol\,K) \)- \( T \) is the temperature in Kelvin

Step 3: Calculate Rate Constant for 400°C

Substitute the values for the first temperature (673.15 K) into the Arrhenius equation:\[ k_1 = 1.2 \times 10^{14} \cdot e^{-221000 / (8.314 \times 673.15)} \]Calculate the exponent:\[ -\frac{221000}{8.314 \times 673.15} = -39.545 \]Then:\[ k_1 = 1.2 \times 10^{14} \cdot e^{-39.545} \approx 4.29 \times 10^{-4} \, s^{-1} \]

Step 4: Calculate Rate Constant for 800°C

Substitute the values for the second temperature (1073.15 K) into the Arrhenius equation:\[ k_2 = 1.2 \times 10^{14} \cdot e^{-221000 / (8.314 \times 1073.15)} \]Calculate the exponent:\[ -\frac{221000}{8.314 \times 1073.15} = -24.801 \]Then:\[ k_2 = 1.2 \times 10^{14} \cdot e^{-24.801} \approx 3.37 \, s^{-1} \]

Step 5: Calculate Reaction Rate

The reaction rate is calculated using the rate constant and the concentration of the reactant:\[ \text{Rate} = k \times [\text{CH}_3\text{CH}_2\text{I}] \]For part (a):\[ \text{Rate at 400°C} = 4.29 \times 10^{-4} \, s^{-1} \times 0.012 \, mol/L \approx 5.15 \times 10^{-6} \, mol/(L\,s) \]For part (b):\[ \text{Rate at 800°C} = 3.37 \, s^{-1} \times 0.012 \, mol/L \approx 0.0404 \, mol/(L\,s) \]

Step 6: Conclusion

We have computed the reaction rates at two different temperatures. The calculations show that the rate of reaction increases significantly as the temperature increases from 400°C to 800°C, reflecting the exponential dependence of the rate on temperature.

Key Concepts

Activation Energy

Activation energy is a crucial concept in the study of chemical reactions. It represents the minimum energy required for a reaction to occur. Think of it as a barrier that reactants must overcome to transform into products. This energy barrier ensures that only molecules with sufficient kinetic energy can participate in the reaction.

• The activation energy, represented by \(E_a\), is often measured in joules per mole \( (J/mol) \) or kilojoules per mole \( (kJ/mol) \).
• A higher activation energy means a slower reaction rate at a given temperature, as fewer molecules have the necessary energy to overcome the barrier.
• Conversely, a lower activation energy implies that more molecules can overcome the barrier, leading to a faster reaction.

In our exercise, the activation energy for the decomposition of ethyl iodide, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \), is \( 221 \mathrm{~kJ} / \, \mathrm{mol} \). This information is vital for determining the reaction rate at different temperatures using the Arrhenius equation.

Reaction Rate

The reaction rate is an indication of how fast a chemical reaction occurs. It can be expressed as the change in concentration of a reactant or product per unit time. Understanding reaction rates allows us to predict how quickly products will form, which is essential in both industrial and laboratory settings.

• In our exercise, the rate of the reaction at specific temperatures is calculated using the rate constant \( k \) and the concentration of the reactant.
• The rate formula is \( \text{Rate} = k \times [\text{Reactant}] \), where \([\text{Reactant}] \) is the concentration.
• An increase in the rate constant \( k \) indicates a faster reaction. This typically happens when the temperature rises or the activation energy decreases.

Understanding how to calculate the reaction rate using given information can help predict the behavior of the reaction under different conditions, as shown in the solutions.

Temperature Conversion

Temperature conversion is a fundamental step in many chemical calculations, particularly when using the Arrhenius equation. The Arrhenius equation requires temperature to be in Kelvin for calculations. To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature.

• The formula is: \[ T(K) = T(°C) + 273.15 \]
• This straightforward conversion is crucial for accurate results in calculations involving reaction rates and temperature.

For our exercise:

• At 400°C, the temperature in Kelvin becomes 673.15 K.
• At 800°C, it becomes 1073.15 K.

Always ensure temperature units are consistent when plugging values into equations to avoid errors.

Frequency Factor

The frequency factor, denoted as \( A \), is a constant that represents the number of times that reactants collide with the correct orientation for a reaction to occur. In the context of the Arrhenius equation, the frequency factor is crucial for calculating the rate constant \( k \).

• The formula for the Arrhenius equation is: \[ k = A e^{-E_a / (RT)} \]where \( A \) is the frequency factor.
• It is expressed in units of \( s^{-1} \) or \( mol^{-1} \, L \cdot s^{-1} \), indicating how frequently reactions occur per second.
• A higher frequency factor implies more frequent collisions and thus potentially a faster reaction rate.

In the given exercise, the frequency factor \( A \) is \( 1.2 \times 10^{14} \mathrm{~s}^{-1} \), which influences the rate constant \( k \) significantly by determining how many successful reactions can occur in a given time period.