Problem 94 - Chemistry: An Atoms-Focused Approach

Question

The compound 1,1 -difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: $$\mathrm{CH}_{3} \mathrm{CHF}_{2}(g) \rightarrow \mathrm{CH}_{2} \mathrm{CHF}(g)+\mathrm{HF}(g)$$ At \(460^{\circ} \mathrm{C}, k=5.8 \times 10^{-6} \mathrm{s}^{-1}\) and \(E_{\mathrm{a}}=265 \mathrm{kJ} / \mathrm{mol} .\) To what temperature would you have to raise the reaction to make it go four times as fast?

Step-by-Step Solution

Answer

Answer: Approximately 593.4°C

Step 1: Find initial rate constant value

Given the initial rate constant (k) at 460°C, we have: $$k_1 = 5.8 \times 10^{-6} \, \text{s}^{-1}$$

Step 2: Calculate final rate constant value

We know that the rate of the reaction needs to be four times faster, so we need to find the new rate constant value (k2): $$k_2 = 4k_1 = 4 \times 5.8 \times 10^{-6} \, \text{s}^{-1} = 2.32 \times 10^{-5} \, \text{s}^{-1}$$

Step 3: Convert given temperatures to Kelvin

Convert the initial given temperature (460°C) to Kelvin by adding 273.15: $$T_1 = 460 + 273.15 = 733.15 \, K$$

Step 4: Use the Arrhenius equation

The Arrhenius equation is: $$k = A \, e^{\frac{-E_a}{RT}}$$ For this problem, we need to rearrange the equation to solve for the final temperature (\(T_2\)). Divide the equations for final and initial rate constant values: $$\frac{k_2}{k_1} = \frac{A \, e^{\frac{-E_a}{R T_2}}}{A \, e^{\frac{-E_a}{R T_1}}}$$ The pre-exponential factor (A) will cancel out. Simplifying and taking the natural logarithm of both sides, we get: $$\ln{\frac{k_2}{k_1}} = -\frac{E_a}{R}(\frac{1}{T_2} - \frac{1}{T_1})$$ Now we can solve for the final temperature (\(T_2\)).

Step 5: Plug in the known values and solve for the final temperature (\(T_2\))

Given that activation energy (Ea) is 265 kJ/mol, we need to convert it to J/mol (1 kJ = 1000 J): $$E_a = 265 \times 10^3 \, \text{J/mol}$$ The gas constant (R) is 8.314 J/(mol K). Now, plug in all the known values and solve for \(T_2\): $$\ln{\frac{2.32 \times 10^{-5}}{5.8 \times 10^{-6}}} = -\frac{265 \times 10^3 \, \text{J/mol}}{8.314 \, \text{J/(mol K)}}(\frac{1}{T_2} - \frac{1}{733.15 \, K})$$ Solving for \(T_2\), we get: $$T_2 \approx 866.55 \, K$$

Step 6: Convert the final temperature to Celsius

Convert the final temperature in Kelvin to Celsius by subtracting 273.15: $$T_2 \approx 866.55 - 273.15 = 593.4 \, ^\circ \text{C}$$ So, to make the reaction go four times as fast, the temperature must be raised to approximately $$593.4 \, ^\circ \text{C}$$.

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a fundamental concept in chemical kinetics, providing a quantitative basis for understanding how reaction rates change with temperature. This equation is represented as:\[ k = A e^{-\frac{E_a}{RT}} \]Where:

• \( k \) is the rate constant, which quantifies the speed of the reaction.
• \( A \) is the pre-exponential factor, also known as the frequency factor. It represents the rate at which collisions happen with the correct orientation for a reaction.
• \( e \) is the base of the natural logarithm.
• \( E_a \) is the activation energy, or the minimum energy that must be overcome for a reaction to occur.
• \( R \) is the gas constant, valued at 8.314 J/(mol K).
• \( T \) is the absolute temperature in Kelvin.

Understanding the Arrhenius Equation allows chemists to predict how changes in temperature will influence the rate constant \(k\). By increasing the temperature, \(T\), the \(\frac{E_a}{RT} \) term decreases, indicating a higher rate constant and a faster reaction. The step-by-step application in the exercise shows how this equation can determine the temperature increase needed to achieve a desired reaction speed.Calculations often involve using the natural logarithm to rearrange the equation for unknown values, especially when seeking new temperatures like in the given problem.

Rate Constant

The rate constant \(k\) plays a critical role in determining the speed at which chemical reactions proceed. It can be thought of as a proportionality factor in the rate equation that relates reactant concentrations to the rate of chemical reactions.In the context of the Arrhenius Equation, \(k\) is clearly temperature-dependent. As shown in the exercise, with the initial rate constant given as \(5.8 \times 10^{-6} \, \text{s}^{-1}\) at 460°C, we're tasked to find the new rate constant if we want the reaction to go four times as fast. This means that we need to calculate:\[ k_2 = 4k_1 = 4 \times 5.8 \times 10^{-6} \, \text{s}^{-1} = 2.32 \times 10^{-5} \, \text{s}^{-1} \]This new \(k_2\) represents the desired rate constant that the reaction must achieve under new conditions. Adjusting temperatures to reach this rate constant is a common objective in adjusting reaction conditions.The rate constant is unique to each reaction and depends not only on temperature but also the specific nature of the physiological or chemical system involved, including factors such as state, pressure, and presence of catalysts.

Activation Energy

Activation Energy, denoted as \(E_a\), is the barrier that reactants must overcome for a reaction to occur. It is essentially the "energy hurdle" that leads to the formation of an activated complex or transition state before the products are formed.In the Arrhenius Equation, \(E_a\) serves as a critical determinant of reaction rates:\[ k = A e^{-\frac{E_a}{RT}} \]A higher activation energy signifies a slower reaction rate at a given temperature, as the exponential factor \(e^{-\frac{E_a}{RT}}\) decreases significantly with increasing \(E_a\). Conversely, a lower \(E_a\) suggests that less energy is needed for the reaction to proceed, leading to higher rates.For the problem exercise, the activation energy is given as \(265 \, \text{kJ/mol}\), which when converted to \( \text{J/mol} \) equals \(265 \times 10^3 \, \text{J/mol}\). This conversion is necessary for consistency with the gas constant \(R\) which is in energy units \(\text{J/(mol K)}\).Understanding and manipulating \(E_a\) is crucial for controlling reactions. Lowering \(E_a\) through catalysts, for instance, helps speed up reactions at lower temperatures, which is vital in many industrial and biological processes.