Question

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .^{\circ} \mathrm{C},\) respectively. What is the value of the activation energy?

Short answer

The activation energy (Ea) for the first-order reaction can be found using the Arrhenius equation and the given rate constants at two different temperatures. By rearranging and simplifying the equation, we can calculate Ea as: \[Ea = -R \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} ln\left(\frac{k_2}{k_1}\right)\] Plugging in the given values of rate constants, temperatures, and the gas constant R, we find that the activation energy for this reaction is approximately: \[Ea \approx 9.17 \times 10^3 \, \mathrm{J/mol}\]

Step-by-step solution

Write down the Arrhenius equation

The Arrhenius equation is given by: \[k = Ae^{-\frac{Ea}{RT}}\] where: - k is the rate constant - A is the pre-exponential factor (also known as the frequency factor) - Ea is the activation energy (in J/mol) - R is the gas constant (8.314 J/mol K) - T is the temperature in Kelvin

Arrhenius equation for both temperatures

Since we have two sets of rate constants and temperatures, we can write down the Arrhenius equation for both cases: For the first set of data (T1 = 0°C and k1 = 4.6 x 10⁻² s⁻¹): \[k_1 = Ae^{-\frac{Ea}{R(T_1)}}\] For the second set of data (T2 = 20°C and k2 = 8.1 x 10⁻² s⁻¹): \[k_2 = Ae^{-\frac{Ea}{R(T_2)}}\]

Convert temperatures to Kelvin

We need to convert the temperatures from Celsius to Kelvin because the Arrhenius equation uses the Kelvin scale. We can do this using the following formula: \[T(K) = T(°C) + 273.15\] T1 in Kelvin: \(T_1 = 0 + 273.15 = 273.15 K\) T2 in Kelvin: \(T_2 = 20 + 273.15 = 293.15 K\)

Use the Arrhenius equation, taking the ratio

Now, we will take the ratio of the Arrhenius equations for both temperatures: \[\frac{k_2}{k_1} = \frac{Ae^{-\frac{Ea}{R(T_2)}}}{Ae^{-\frac{Ea}{R(T_1)}}}\] Note that the 'A' terms in the numerator and the denominator cancel out: \[\frac{k_2}{k_1} = \frac{e^{-\frac{Ea}{R(T_2)}}}{e^{-\frac{Ea}{R(T_1)}}}\]

Simplify and solve for Ea

Now let's simplify the equation by taking the natural logarithm (ln) of both sides: \[ln\left(\frac{k_2}{k_1}\right) = ln\left(e^{-\frac{Ea}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}\right)\] Now, apply the logarithm property \(ln(e^x) = x\): \[ln\left(\frac{k_2}{k_1}\right) = -\frac{Ea}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\] To find the activation energy (Ea), we will rearrange the equation: \[Ea = -R \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} ln\left(\frac{k_2}{k_1}\right)\]

Calculate the activation energy (Ea)

Plug in the given values for the rate constants (k1 and k2) and the temperatures in Kelvin (T1 and T2) into the equation, and substitute the values of R: \[Ea = - 8.314\left(\frac{1}{293.15} - \frac{1}{273.15}\right)^{-1} ln\left(\frac{8.1 \times 10^{-2}}{4.6 \times 10^{-2}}\right)\] After performing the calculation, we get the value of Ea: \[Ea \approx 9.17 \times 10^3 \, \mathrm{J/mol}\] So, the activation energy for this first-order reaction is approximately 9.17 x 10³ J/mol.

Key concepts

Arrhenius Equation

The Arrhenius equation is a fundamental formula that relates the rate of a chemical reaction to temperature. It takes the form:
\[k = Ae^{-\frac{Ea}{RT}}\]

• k represents the rate constant of the reaction.
• A is the pre-exponential factor, also known as the frequency factor.
• Ea is the activation energy of the reaction, usually in joules per mole (J/mol).
• R is the universal gas constant, which has a value of 8.314 J/mol*K.
• T is the temperature in Kelvin (K).

In essence, this equation shows that as the temperature increases, so does the rate constant, thereby speeding up the rate of the reaction. The activation energy is a critical part of this equation as it is the minimum energy required for the reactants to form the transition state, ultimately leading to products. By manipulating the Arrhenius equation, it is possible to determine Ea through an experiment by measuring the rate constants at different temperatures, which is exactly what the original exercise involved.

Rate Constant

The rate constant, denoted by the symbol k, is a proportionality factor that relates the reaction rate to the reactant concentrations for a given reaction at a particular temperature. It's an intrinsic value that is determined by the reaction's specific characteristics, including the activation energy and frequency factor. The rate constant is also an indicator of the speed of a reaction; a higher value of k implies a faster reaction under the same conditions.

For a first-order reaction, where the reaction rate is directly proportional to the concentration of one reactant, the rate constant has units of s^-1. This aligns with our exercise problem, where the rate constants are given for two temperatures. By comparing the rate constants at these different temperatures, we can learn about how sensitive the reaction is to temperature changes and we can calculate the activation energy, which serves as an energy barrier that the reactants must overcome to react.

First-Order Reaction

A first-order reaction is one in which the rate of reaction is directly proportional to the concentration of one reactant. Mathematically, this can be represented as:
\[\text{Rate} = k[A]\]
where:

• Rate is the speed of the reaction.
• k is the rate constant.
• [A] represents the concentration of the reactant A.

First-order reactions have simple rate laws and their rate constants hold significance as they are used directly to calculate the reaction rates. In the context of the textbook problem, understanding that the reaction is first-order allows us to appropriately apply the Arrhenius equation. We take the temperature-dependent rate constants at two temperatures and use them to derive the activation energy for the reaction. This simplicity in the rate law for a first-order reaction makes the calculation straightforward once we convert temperatures to Kelvin and apply the Arrhenius equation.