Question

Chemists commonly use a rule of thumb that an increase of 10 \(\mathrm{K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C} ?\)

Short answer

The activation energy for this reaction, considering that the rate of reaction doubles every 10 K increase in temperature, is approximately \(52.6\thinspace kJ\thinspace mol^{-1}\).

Step-by-step solution

Write down the Arrhenius equation

The Arrhenius equation is given by: \(k = A \cdot e^{-\frac{E_a}{RT}}\) where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

Set up equations for doubling the rate constant

We know that the rate constant doubles every 10 K increase in temperature, so we can set up two equations with \(T_1 = 298K\) and \(T_2 = 308K\) (temperature conversions from Celsius to Kelvin). Our equations will be: \(k_1 = A \cdot e^{-\frac{E_a}{R \cdot T_1}}\) \(k_2 = 2 \cdot k_1 = A \cdot e^{-\frac{E_a}{R \cdot T_2}}\) This step is crucial to show the relationship between temperatures and rate constants according to the problem statement.

Divide the second equation by the first equation

Dividing the second equation by the first equation will allow us to eliminate the pre-exponential factor \(A\) and solve for the activation energy \(E_a\). The equations will look like this: \(\frac{k_2}{k_1} = \frac{A \cdot e^{-\frac{E_a}{R \cdot T_2}}}{A \cdot e^{-\frac{E_a}{R \cdot T_1}}}\)

Simplify the equation and isolate E_a

Now we can simplify the equation as follows: \(2 = \frac{e^{-\frac{E_a}{R \cdot T_2}}}{e^{-\frac{E_a}{R \cdot T_1}}}\) Taking natural logarithm on both sides, we get: \(ln\thinspace 2 = ln\thinspace(e^{\frac{E_a}{R \cdot T_1} -\frac{E_a}{R \cdot T_2}}) \Longrightarrow ln\thinspace 2 = \frac{E_a}{R \cdot T_1} -\frac{E_a}{R \cdot T_2}\) Now isolating E_a we get: \(E_a = R \cdot \frac{T_1\thinspace T_2}{T_2 - T_1} \cdot ln\thinspace 2\)

Substitute the values and solve for E_a

Finally, we can substitute the values of \(T_1\), \(T_2\), and \(R\) into the formula to calculate the activation energy: \(E_a = (8.314\thinspace J\thinspace mol^{-1}K^{-1}) \cdot \frac{(298K)(308K)}{308K - 298K} \cdot ln\thinspace 2\) By calculating the above expression, we obtain: \(E_a \approx 52.6\thinspace kJ\thinspace mol^{-1}\) Thus, the activation energy for this reaction is approximately \(52.6\thinspace kJ\thinspace mol^{-1}\).

Key concepts

Arrhenius Equation

In the field of chemistry, the Arrhenius Equation plays a critical role when determining the speed at which reactions occur. This equation, formulated as \( k = A \cdot e^{-\frac{E_a}{RT}} \), links the reaction rate constant \( k \) with the temperature \( T \) and activation energy \( E_a \). The symbol \( A \) represents the pre-exponential factor, a constant indicating the frequency of collisions and the orientation of reacting particles.

One of the remarkable features of the Arrhenius Equation is its ability to describe how likely a reaction is to proceed. An increase in temperature typically increases the reaction rate since more molecules have the kinetic energy required to overcome the activation energy barrier. This relationship highlights the importance of the activation energy \( E_a \), which essentially serves as a threshold energy that reactants must overcome to transform into products.

Understanding the Arrhenius Equation not only helps chemists predict reaction rates but also allows them to manipulate reaction conditions effectively to achieve desired outcomes.

Reaction Rate

The reaction rate is a fundamental concept in chemistry that defines how quickly a reaction proceeds. It can be influenced by several factors including the concentration of reactants, temperature, presence of a catalyst, and the activation energy of the reaction. An increase in temperature, for instance, generally results in a higher reaction rate. This is because higher temperatures provide particles with more energy, thus increasing the frequency of effective collisions.

Often, chemists are interested in knowing how quickly a reaction will complete. This requires understanding the rate constant \( k \) within the Arrhenius Equation. This constant is not a fixed value; it changes with temperature and provides insight into how readily reactants are converted into products under specific conditions.

When solving problems involving reaction rates, it is essential to consider all influencing factors to predict how the reaction will behave in a real-world setting. This helps in enhancing the efficiency and yield of chemical processes.

Temperature Dependence

Temperature dependence is a critical concept in chemistry that helps explain how the rate of a chemical reaction is affected by changes in temperature. An increase in temperature generally leads to an increase in the reaction rate. For many reactions, a rise of 10 K might double the rate, a rule of thumb often used by chemists.

This concept is intricately linked to the Arrhenius Equation, as temperature directly influences the exponential term \( e^{-\frac{E_a}{RT}} \). When the temperature \( T \) increases, the factor \( \frac{E_a}{RT} \) decreases, resulting in a higher value for the rate constant \( k \). This means that reactants have more energy to surpass the activation energy barrier, enhancing the likelihood of reaction occurrence.

By understanding how reaction rates change with temperature, chemists can predict and control the speed of reactions, which is especially important in industrial applications where precise control over reaction rates is necessary.

Chemistry Problem Solving

Solving chemistry problems, especially those involving reaction kinetics, requires a clear understanding of the principles involved, such as the Arrhenius Equation, reaction rates, and temperature dependence. Successful problem solving in chemistry involves several steps:

• Clearly identifying the problem and the relationships between variables.
• Using the correct equations to describe the system. For reaction kinetics, the Arrhenius Equation is often crucial.
• Paying attention to units and making sure conversions, especially for temperature from Celsius to Kelvin, are handled correctly.
• Breaking down complex equations by isolating variables to simplify calculations.

Each of these steps relies heavily on understanding the concepts and applying them correctly. For instance, when dealing with a problem requiring the calculation of activation energy, it is important to use known data effectively and ensure accurate substitution into the simplified Arrhenius-derived equations. This structured approach improves accuracy and reliability when tackling diverse chemistry problems.