Question

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of $1/T$. How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and (b) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section $14.5$

Short answer

In summary, for case (a) where the activation energy is different but the frequency factor is the same, the two graphs will have the same intercept on the vertical axis but different slopes, with the graph of the reaction with a higher activation energy having a steeper slope. In case (b), where the activation energy is the same but the frequency factors are different, the two graphs will have the same slope but different intercepts, with the graph of the reaction with a higher frequency factor having a higher intercept on the vertical axis.

Step-by-step solution

Recall the Arrhenius Equation

The Arrhenius equation is given by: $$k=A{e}^{\frac{-E_a}{RT}}$$ where $k$ is the rate constant, $A$ is the frequency factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature (in Kelvin). We are asked to graph the natural logarithm of the rate constant ($\ln k$) as a function of $1/T$. So, we must first take the natural logarithm of the Arrhenius equation.

Taking the natural logarithm of the Arrhenius Equation

Take the natural logarithm of both sides of the equation: $$\ln k=\ln (A{e}^{\frac{-E_a}{RT}})$$ Using the properties of logarithms, we can simplify the equation: $$\ln k=\ln A+\ln (e^{\frac{-E_a}{RT}})$$ Now, since $\ln (e^x)=x$, we get: $$\ln k=\ln A-\frac{E_a}{RT}$$

Addressing the Two Cases

We have two cases that we need to analyze: (a) Same frequency factor ($A$) and different activation energy ($E_a$). (b) Same activation energy ($E_a$) and different frequency factors ($A$).

Case (a): Comparing activation energies

In this case, both reactions have the same frequency factor ($A$) but different activation energies ($E_{a1}$ and $E_{a2}$) where $E_{a2}>E_{a1}$. For the first reaction: $$\ln k_1=\ln A-\frac{E_{a1}}{RT}$$ For the second reaction: $$\ln k_2=\ln A-\frac{E_{a2}}{RT}$$ Since $E_{a2}>E_{a1}$, the slope of the graph for the second reaction will be steeper than that of the first reaction. Thus, when plotted as $\ln k$ against $1/T$, the two graphs will have the same intercept (due to the same value of $\ln A$) on the vertical axis and different slopes.

Case (b): Comparing frequency factors

In this case, both reactions have the same activation energy ($E_a$) but different frequency factors ($A_1$ and $A_2$) where $A_2>A_1$. For the first reaction: $$\ln k_1=\ln A_1-\frac{E_a}{RT}$$ For the second reaction: $$\ln k_2=\ln A_2-\frac{E_a}{RT}$$ Since both reactions have the same activation energy, their slopes will be the same. However, due to the difference in the frequency factors, the second reaction will have a higher intercept on the vertical axis (as $\ln A_2>\ln A_1$). Thus, when plotted as $\ln k$ against $1/T$, the two graphs will have different intercepts and the same slope.

Key concepts

Activation Energy

The concept of activation energy ($E_a$) is critical to understanding how chemical reactions take place. Activation energy is the minimum amount of energy required to initiate a chemical reaction. Simply put, it's the 'energy barrier' that reactants must overcome to transform into products.

Let's visualize activation energy with a real-world analogy: Imagine trying to push a heavy boulder over a hill. The energy you need to exert to get the boulder to the top is similar to the activation energy required for a chemical reaction. Once at the top, the boulder can roll down the other side—a representation of the reaction proceeding to product formation.

In our exercise scenario, when comparing two reactions with different activation energies, the higher the activation energy, the steeper the slope will be on the graph of $\ln k$ versus $1/T$. This represents that more energy is required to start the reaction, meaning it will be slower to begin at the same temperature compared to a reaction with a lower activation energy.

Frequency Factor

The frequency factor (A), often called the pre-exponential factor, relates to the number of times reactant particles collide with the correct orientation per unit of time. It is a measure of the likelihood that a reaction will occur when the activation energy barrier is overcome.

This factor is important because it implies that not every collision between reactant particles leads to a reaction; they must collide in the right way and with enough energy. Think of it like this: If you and a friend are trying to hit a bell with thrown balls, the frequency factor would be how many throws you two make in a minute, and the correct orientation would be hitting the bell at its sweet spot.

In the case of our exercise, even if two reactions have the same activation energy, the reaction with the higher frequency factor will have a graph with a higher $\ln k$ intercept. This means that more of the collisions per unit time will be successful, hence that reaction will tend to occur at a faster rate.

Rate Constant

The rate constant (k) is an essential component in the study of reaction kinetics, as it quantifies the speed of a chemical reaction. For any given reaction at a certain temperature, this constant can be thought of like the pace at which the 'reaction clock' ticks; the higher the value of $k$, the faster the reaction proceeds.

To explain how it is influenced, imagine you're riding a bike: the rate constant is like the speed of cycling. If you're facing an uphill path (high activation energy) or you're just learning to ride (low frequency factor), you'll cycle more slowly. If the path is downhill (low activation energy) or you're an experienced cyclist (high frequency factor), you'll cycle faster. That's similar to how reactants react to form products—more favorable conditions lead to a higher 'cycling speed' or rate constant.

Reaction Rates Temperature Dependence

The temperature dependence of reaction rates is best represented by the Arrhenius equation, which shows how the rate constant $k$ varies with temperature ($T$) and has implications for both activation energy and frequency factor.

As the temperature increases, particles move faster and collide more often, which generally increases the reaction rate. This effect can be thought of in terms of a crowded dance floor—turning up the music (increasing temperature) gets people (molecules) to move and interact (collide) more. For reactions, a higher temperature means the reactants have greater kinetic energy, which increases the likelihood they'll have sufficient energy to overcome the activation energy barrier.

In the exercise, plotting $\ln k$ against $1/T$ actually generates a straight line, and the slope of this line is inversely proportional to the activation energy of the reaction. This relationship highlights the exquisite sensitivity of rate constants to changes in temperature, embodying the strong temperature dependence of chemical reaction rates.