Problem 21 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Calculate the activation energy for a reaction in which the rate constant is \(\left.5 \text { times faster at } 50^{\circ} \mathrm{C} \text { than at } 20^{\circ} \mathrm{C} \text { . (Section } 9.7\right)\)

Step-by-Step Solution

Answer

The activation energy is approximately 43.2 kJ/mol.

Step 1: Identify Known Values

First, we identify the known values given in the problem. We know two temperatures: \(T_1 = 20^{\circ}C\) and \(T_2 = 50^{\circ}C\). We convert these to Kelvin by adding 273.15. Thus, \(T_1 = 293.15\, K\) and \(T_2 = 323.15\, K\). The rate constant at \(50^{\circ}C\) (\(k_2\)) is 5 times the rate constant at \(20^{\circ}C\) (\(k_1\)), so \(k_2 = 5k_1\).

Step 2: Determine the Arrhenius Equation

The Arrhenius equation relates the rate constant \(k\) to the activation energy \(E_a\) and temperature \(T\): \[ k = A e^{-\frac{E_a}{RT}} \]where \(A\) is the frequency factor, \(R\) is the universal gas constant \(8.314\, J/mol\cdot K\), and \(T\) is the absolute temperature in Kelvin.

Step 3: Use the Arrhenius Equation Ratio

We take the natural logarithm of the ratio of rate constants \(\frac{k_2}{k_1} = 5\).\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]Plug in \(\ln(5)\), \(T_1 = 293.15\, K\), \(T_2 = 323.15\, K\), and \(R = 8.314\, J/mol\cdot K\).

Step 4: Calculate the Activation Energy

Rearrange the equation to solve for the activation energy \(E_a\):\[ E_a = \frac{\ln(5) \times R}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \]Substitute \(\ln(5) \approx 1.6094\) and calculate:\[ E_a = \frac{1.6094 \times 8.314}{\left(\frac{1}{293.15} - \frac{1}{323.15}\right)} \]Calculate the numerical value for \(E_a\).

Step 5: Perform Final Calculation

Perform the calculation:\[ \frac{1}{293.15} \approx 0.00341\]\[ \frac{1}{323.15} \approx 0.00310\]\[ E_a = \frac{1.6094 \times 8.314}{0.00341 - 0.00310} = \frac{13.377}{0.00031} \approx 43152\, J/mol \]Convert \(E_a\) to \(kJ/mol\) by dividing by 1000: \(E_a \approx 43.2\, kJ/mol\).

Step 6: Verify Units and Calculation

Ensure the units are correct and the calculation aligns with standard thermodynamics. \(E_a\) is typically expressed in \(kJ/mol\), which matches our improved result of \(43.2 \ kJ/mol\).

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a crucial concept in chemical kinetics, providing insight into how temperature affects reaction rates. This equation links the rate constant, activation energy, and temperature. It states:
\[ k = A e^{-\frac{E_a}{RT}} \]

• \(k\): the rate constant, determining how quickly a reaction proceeds.
• \(A\): the frequency factor, indicative of the number of successful collisions between reactant molecules.
• \(E_a\): activation energy, the minimum energy required to initiate the reaction.
• \(R\): the universal gas constant, playing a role in converting energy units.
• \(T\): temperature in Kelvin, crucial for capturing the effect of temperature on reaction rates.

At the core, the Arrhenius Equation explains that as temperature increases, the rate constant also increases, because more molecules have the required activation energy to react.
This explains why reactions generally proceed faster at higher temperatures.

Rate Constants

Rate constants (
\(k\)) in chemical reactions are essential for understanding how fast a reaction occurs. They are influenced by factors such as temperature and presence of catalysts. One important point is that they change with temperature, as seen in the Arrhenius Equation.

• At a higher temperature, molecules possess more energy, thus increasing the frequency of effective collisions, resulting in a higher rate constant.
• In our example: at \(50^{\circ}C\), the rate constant (\(k_2\)) is 5 times higher than at \(20^{\circ}C\) (\(k_1\)), i.e., the reaction occurs 5 times faster.

Understanding rate constants helps in accurately predicting how quickly a chemical process will happen, which is crucial in both industrial and research settings.

Temperature Conversion

Temperature conversion between Celsius and Kelvin is pivotal in chemical equations since the Arrhenius Equation requires temperature in Kelvin. The formula is simple:
\[ T(K) = T(\text{in Celsius}) + 273.15 \]

• This conversion standardizes the temperature, ensuring it reflects absolute temperatures needed for thermodynamic equations.
• In the provided solution, the conversions were: \(T_1 = 20^{\circ}C = 293.15 \, K\) and \(T_2 = 50^{\circ}C = 323.15 \, K\).

Using Kelvin ensures that calculations related to energy transformations, such as activation energy, are consistent and accurate, reflecting true thermodynamic properties.

Universal Gas Constant

The Universal Gas Constant (\(R\)) is a fundamental constant that appears in various equations relating to gases and reactions. It facilitates calculations across different scientific areas, ensuring consistency in units. For our purpose:
\[ R = 8.314 \, \text{J/mol⋅K} \]

• \(R\) ensures that energy calculations such as those involving \(E_a\) (activation energy) are accurate.
• It also bridges the energy factors from the microscopic world (molecular) to the macroscopic world (observable reaction rates).

In the calculation of activation energy, \(R\) is used to balance the units across the equation for consistent and correct results. It plays a vital role in converting energy quantities to values that can be compared and applied in practical scenarios.