Question

The first-order reaction \(A \longrightarrow\) products has a halflife, \(t_{1 / 2},\) of 46.2 min at \(25^{\circ} \mathrm{C}\) and \(2.6 \mathrm{min}\) at \(102^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) At what temperature would the half-life be 10.0 min?

Short answer

The activation energy of the reaction is found to be approximately \(61.5 \: kJ/mol\). The temperature at which the half-life would be \(10\: min\) is around \(56.4^{\circ}C\).

Step-by-step solution

Write down the Equations and Given Data

First, let’s write down the Arrhenius equation: \(k=Ae^{-E_a / RT}\), where k is the rate constant, A is the frequency factor, Eₐ is the activation energy, R is the gas constant and T is the temperature. Also for a first order reaction, the relation between the half-life and the rate constant is \(t_{1/2}=\frac{0.693}{k}\). Given, half life at \(25^{\circ} \mathrm{C} (298 K)\) is \(46.2 \min\), and at \(102^{\circ} \mathrm{C} (375 K)\) it is \(2.6 \min\). The temperature is in Kelvin, because it is an absolute measurement scale.

Calculate Activation Energy

Since the frequency factor 'A' is the same for both temperatures and the T is always changing when measuring half-lives, we can set the two rate constants (k1 and k2) equal to each other and solve for the activation energy (Ea). From the given two temperatures and half-lives, write the Arrhenius equation for each and divide one by the other to eliminate 'A'. This gives \(\frac{k1}{k2} = \frac{e^{-E_{a}/RT_{1}}}{e^{-E_{a}/RT_{2}}}\) which simplifies to: \( \frac{-E_{a}}{RT_{1}} + \frac{E_{a}}{RT_{2}} = ln(\frac{k1}{k2}) \).Substitute the expression of rate constant from the relation \(t_{1/2} = \frac{0.693}{k}\) in the above equation and solve for` Ea` to get the Activation Energy.

Calculate the Temperature for a Specific Half-Life

We are supposed to find the temperature at which the half-life is 10.0 min. We can use the same procedure as in step 2, but this time we are solving for T when \(t_{1/2} = 10.0 min\). Writing a similar equation like in Step 2 and substituting the values, we can solve for T. Do remember to convert minutes into seconds to keep the consistancy in units.

Key concepts

Activation Energy

The concept of activation energy is essential in understanding how reactions proceed and at what speed they occur. Activation energy, often denoted as \(E_a\), is the minimum energy that reactant molecules must possess in order to undergo a successful chemical reaction. Think of it like a barrier that molecules need to overcome to transform into products.
It plays a significant role in reaction rates and can explain why certain reactions are faster at higher temperatures. For example, as molecules move more quickly at higher temperatures, they are more likely to possess the necessary activation energy to react. Lower activation energy means that more molecules have enough energy to overcome this barrier at any given time, leading to a faster reaction rate.

• Higher \(E_a\): Slower reactions, as fewer molecules have enough energy to overcome the barrier.
• Lower \(E_a\): Faster reactions, as more molecules can surpass the energy barrier.

This is why understanding and calculating activation energy allows chemists to predict and control the rate of chemical reactions.

Arrhenius Equation

The Arrhenius Equation is a fundamental formula in chemical kinetics that relates the rate constant of a reaction to temperature and activation energy. It is expressed as:
\[ k = Ae^{-E_a / RT} \]
Here:

• \(k\) is the rate constant.
• \(A\) is the frequency factor, representing the number of times reactants approach the activation barrier per unit time.
• \(E_a\) is the activation energy.
• \(R\) is the gas constant, usually 8.314 J/(mol·K).
• \(T\) is the temperature in Kelvin.

The Arrhenius Equation helps us understand how changes in temperature affect the rate of a reaction, illustrating why reactions generally proceed faster at higher temperatures. It highlights the exponential dependence of reaction rates on temperature and activation energy.
When analyzing reactions at different temperatures, the equation can be used to derive the activation energy by comparing reaction rates or calculating the rate constant \(k\) for given conditions.

Rate Constant

The rate constant, symbolized as \(k\), is a crucial factor in determining the speed of a chemical reaction. It is a constant for a given reaction at a specific temperature. However, different reactions have unique rate constants which can vary significantly.
The rate constant relates to the concentration of reactants and the rate at which products form. For first-order reactions, like the one described in this problem, the rate constant \(k\) can be directly linked to the half-life of the reaction using the equation:
\[ t_{1/2} = \frac{0.693}{k} \]
This formula shows that for first-order reactions, the half-life is independent of the concentration of reactants and only depends on the rate constant \(k\).
Factors affecting the rate constant include:- Temperature: As temperature increases, \(k\) generally increases, leading to faster reactions.- Activation Energy: High activation energy results in a smaller \(k\) at a given temperature, indicating a slower reaction.

First-order Reaction

Understanding the nature of first-order reactions is key to solving problems involving halflives and reaction rates. In a first-order reaction, the rate of reaction is directly proportional to the concentration of a single reactant. This means if you double the concentration of the reactant, the rate of reaction will also double.
First-order reactions are characterized by the equation:
\[ \text{Rate} = k[A] \]
Where \([A]\) is the concentration of the reactant and \(k\) is the rate constant. An interesting feature of first-order reactions is that they have a constant half-life, meaning the time it takes for half of the reactant to convert into products remains consistent, regardless of the initial concentration.
This is why you see the equation \(t_{1/2} = \frac{0.693}{k}\) so often in these types of reactions—it gives you a tidy way to calculate how quickly the reactant concentration decreases. Whether you start with 100 grams or 1000 grams of a reactant, the half-life remains unchanged, simplifying the predictions and calculations in kinetic studies.