Question

Would the slope of a \(\ln (k)\) versus \(1 / T(\mathrm{~K})\) plot for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus \(1 / T(\mathrm{~K})\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

Short answer

The slope of a \(ln(k)\) versus \(1/T(\mathrm{~K})\) plot for a catalyzed reaction would be less negative than the slope of the \(ln(k)\) versus \(1/T(\mathrm{~K})\) plot for the uncatalyzed reaction. This is because a catalyst lowers the activation energy of the reaction, resulting in a less negative slope due to the relationship given by the Arrhenius equation: \(\ln(k) = \ln(A) - \frac{Ea}{R} * \frac{1}{T}\).

Step-by-step solution

Understand the Arrhenius equation

To compare the slopes of the two plots, we first need to understand the Arrhenius equation, which relates the rate constant (k) of a reaction to the temperature (T) and activation energy (Ea) via the Arrhenius factor (A). The Arrhenius equation is given by: \[k = Ae^{-\frac{Ea}{RT}}\] Where: - k is the rate constant - A is the Arrhenius factor (also called the pre-exponential factor) - Ea is the activation energy of the reaction - R is the gas constant (8.314 J/(mol K)) - T is the temperature in kelvins

Understand the plots of ln(k) vs 1/T

To compare the slopes of the plots, we first need to get the equation in the form of ln(k) vs 1/T. To do that, we can take the natural logarithm of the Arrhenius equation on both sides. \[\ln(k) = \ln(Ae^{-\frac{Ea}{RT}})\] Using the property \(\ln(ab) = \ln(a) + \ln(b)\), we can rewrite the equation as: \[\ln(k) = \ln(A) - \frac{Ea}{R} * \frac{1}{T}\] Now, the equation is in the form of y = mx + b, where y = ln(k), x = 1/T, and the slope (m) = -Ea/R.

Compare the slopes for the catalyzed and uncatalyzed reactions

A catalyst reduces the activation energy of a reaction by providing an alternate pathway with a lower activation energy barrier. Thus, the activation energy for the catalyzed reaction (Ea_c) is less than the activation energy for the uncatalyzed reaction (Ea_u). Since the slope of the plots is equal to -Ea/R, the slope of the catalyzed reaction is: Slope_c = -Ea_c/R And the slope of the uncatalyzed reaction is: Slope_u = -Ea_u/R

Determine whether the slope of the catalyzed reaction plot is more or less negative

Since Ea_c < Ea_u, we can conclude that the slope of the catalyzed reaction plot (Slope_c) is less negative than the slope of the uncatalyzed reaction plot (Slope_u). To summarize, the slope of a ln(k) versus 1/T plot for a catalyzed reaction would be less negative than the slope of the ln(k) versus 1/T plot for the uncatalyzed reaction. This is because a catalyst lowers the activation energy of the reaction.

Key concepts

Activation Energy

Activation energy, denoted as \(E_a\), is a crucial concept in understanding chemical kinetics. It's the energy barrier that reactant molecules must overcome in order to transform into products during a chemical reaction. Think of it as the 'gatekeeper' of a reaction; higher activation energy means the reactants need more energy to start converting into products, thus slowing down the rate of the reaction.

When you look at a reaction's energy profile, activation energy is represented by the peak of the curve separating reactants and products. Only a fraction of molecules with sufficient thermal energy can exceed this peak at a given temperature.

As stated in the Arrhenius equation \(k = Ae^{-\frac{Ea}{RT}}\), the relationship between activation energy and rate constant is exponential. This means even a small decrease in activation energy can lead to a significant increase in the reaction rate constant, and vice versa. Hence, a plot of \(\ln(k)\) versus \(1/T\) includes activation energy in its slope, which equates to \(-Ea/R\). This negative slope indicates that as temperature increases, the natural logarithm of the rate constant also increases, albeit inversely with respect to temperature.

Rate Constant

The rate constant, symbolized by \(k\), is a pivotal parameter in the field of reaction kinetics. It quantifies the speed of a chemical reaction; a larger rate constant signifies a faster reaction rate. This constant is influenced by various factors, including temperature, which is showcased in the Arrhenius equation.

One can derive the rate constant's relationship with temperature by taking the natural logarithm of the Arrhenius equation, which results in \(\ln(k) = \ln(A) - \frac{Ea}{R} * \frac{1}{T}\), casting the equation into a linear form. The negative correlation indicates that as the temperature increases, the rate constant also increases.

When examining a plot of \(\ln(k)\) versus \(1/T\), the negative slope is directly proportional to the activation energy while being inversely proportional to the temperature, emphasizing the sensitive nature of the rate constant to these variables. Understanding this relationship is crucial for chemists who aim to control reaction rates, whether to hasten them in industrial processes or slow them down for better precision.

Catalyzed vs Uncatalyzed Reactions

Catalysts are substances that accelerate chemical reactions without undergoing permanent chemical change themselves. They operate by offering a different route for the reaction with a lower activation energy. This alternate pathway increases the proportion of reactant molecules that can surpass the energy barrier and convert to products, effectively boosting the reaction rate.

When comparing catalyzed versus uncatalyzed reactions, the key difference lies in their activation energies. A catalyzed reaction will have a lower activation energy (\(Ea_c\)) compared to an uncatalyzed one (\(Ea_u\)). Consequently, the slope of a ln(k) vs 1/T plot for a catalyzed reaction (denoted as Slope_c) is mathematically less negative than that of an uncatalyzed reaction (Slope_u), as seen in the exercise solution where \(Slope_c = -Ea_c/R\) and \(Slope_u = -Ea_u/R\).

Hence, in a graph comparing the two, the line representing the catalyzed reaction would appear flatter compared to the steeper slope of the uncatalyzed reaction, illustrating the dramatic impact a catalyst can have on a chemical process. This visualization helps students grasp the concept of catalysis and its profound effect on reaction rates.