Question

The decomposition of NH3 to N2 and H2 was studied on two surfaces: Without a catalyst, the activation energy is 335 \(\mathrm{kJ} / \mathrm{mol}\) . a. Which surface is the better heterogeneous catalyst for the decomposition of \(\mathrm{NH}_{3} ?\) Why? b. How many times faster is the reaction at 298 \(\mathrm{K}\) on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor \(A\) is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form $$ |text {Rate} =k \frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{H}_{2}\right]} $$ How can you explain the inverse dependence of the rate on the \(\mathrm{H}_{2}\) concentration?

Short answer

a. We cannot determine the better heterogeneous catalyst without information about the activation energy with the catalysts on both surfaces. b. The reaction on the W surface is approximately 7.4 million times faster than the reaction with no catalyst at 298 K. c. The inverse dependence of the rate on the H2 concentration can be explained by the adsorption of H2 and NH3 onto the surface of the catalyst. As H2 concentration increases, more H2 molecules get adsorbed onto the surface, occupying more sites and blocking NH3 from adsorbing onto those sites, resulting in fewer available sites for NH3 adsorption, which slows down the reaction.

Step-by-step solution

a. Identifying the better heterogeneous catalyst

For the first part of the question, we have to identify which surface is the better heterogeneous catalyst. A good catalyst lowers the activation energy for a reaction, thus increasing the reaction rate. Since the activation energy without a catalyst is given as 335 kJ/mol, we need information about the activation energy with the catalysts on both surfaces. If not given, we cannot determine the better heterogeneous catalyst and explain why.

b. Comparing the reaction rate on the W surface with no catalyst

To find out how many times faster the reaction on the W surface is at 298 K compared to the reaction without a catalyst, we will use the Arrhenius equation to find the ratio of the rate constants. Remember that the Arrhenius equation is given by: \[k = A e^{-Ea / RT}\] Where k is the reaction rate constant, A is the frequency factor (assumed to be the same for both reactions), Ea is the activation energy, R is the ideal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. In this case, we have both activation energies, no catalyst (Ea, no catalyst = 335 kJ/mol) and W surface (Ea, W = 100 kJ/mol), and the temperature T = 298 K. Now let's find the ratio of the rate constants without catalyst and on the W surface: \[\frac{k_{W}}{k_{no\;catalyst}} = \frac{A e^{-Ea_{W} / RT}}{A e^{-Ea_{no\;catalyst} / RT}}\] By cancelling out A and simplifying, \[\frac{k_{W}}{k_{no\;catalyst}} = \frac{e^{-Ea_{W} / RT}}{e^{-Ea_{no\;catalyst} / RT}}\] Now we can plug in the values of Ea and T to calculate the ratio: \[\frac{k_{W}}{k_{no\;catalyst}} = \frac{ e^{-\frac{100 \times 10^3}{8.314 \times 298}} }{ e^{-\frac{335 \times 10^3}{8.314 \times 298}} }\] After calculating the ratio: \[\frac{k_{W}}{k_{no\;catalyst}} \approx 7.4 \times 10^6\] This means the reaction on the W surface is approximately 7.4 million times faster than the reaction with no catalyst.

c. Explaining the inverse dependence of the rate on H2 concentration

The given rate law for the decomposition reaction on both surfaces is: \[Rate = k \frac{[NH_3]}{[H_2]}\] To explain the inverse dependence of the rate on the H2 concentration, let's consider the mechanism of the reaction on these surfaces. A heterogeneous catalyst usually works by adsorbing the reactants onto its surface, weakening their bonds, and facilitating the reaction between them. In this case, NH3 and H2 are both adsorbed onto the surface of the catalyst, and the reaction between adsorbed NH3 and adsorbed H2 produces adsorbed N2 and H2. As the concentration of H2 increases, more H2 molecules get adsorbed onto the surface, occupying more sites and blocking NH3 from adsorbing onto those sites. As a result, there are fewer available sites for NH3 adsorption, which slows down the reaction between adsorbed NH3 and adsorbed H2. Hence, the rate of the reaction is inversely proportional to the concentration of H2.

Key concepts

Activation Energy

Activation energy is a crucial concept in chemistry, particularly when discussing catalysis and reaction rates. It refers to the minimum amount of energy that reactant molecules need to successfully undergo a chemical transformation. This energy acts as a barrier that molecules must overcome, analogous to having to push a boulder up a hill before it can roll down the other side.

In reactions, the role of a catalyst, such as a heterogeneous catalyst, is to lower this activation energy, thereby increasing the reaction rate. By providing an alternative reaction pathway with a lower energy requirement, catalysts enable more molecules to have the required energy for the reaction to occur, even at lower temperatures.

For instance, if the activation energy without a catalyst is 335 kJ/mol and a specific catalyst surface reduces it to 100 kJ/mol, the reaction becomes significantly faster. This is because more molecules can overcome the much-reduced energy barrier, leading to an increased frequency of successful collisions. Understanding activation energy is vital for controlling and optimizing chemical reactions, especially in industrial and laboratory settings.

Rate Law

The rate law of a reaction provides an equation that links the rate of a chemical reaction to the concentration of its reactants. It is typically expressed in terms of a rate constant, k, and the concentrations of the reactants raised to some power, which refers to their respective reaction orders.

In the case presented, the rate law is given as: \[ \text{Rate} = k \frac{[\mathrm{NH}_3]}{[\mathrm{H}_2]} \] This form of the rate law indicates that the reaction rate is directly proportional to the concentration of ammonia \([\mathrm{NH}_3]\) and inversely proportional to the concentration of hydrogen \([\mathrm{H}_2]\).

The inverse relationship here suggests that as more hydrogen is present, it potentially occupies the catalyst's active sites, hindering the adsorption of ammonia and thus slowing down the reaction. It's essential to consider how reactants interact with a catalyst, as in this scenario, where competition for the catalyst sites affects the reaction rate. This concept highlights the importance of understanding surface dynamics in catalyzed reactions, as well as the stoichiometry reflected in the rate expression.

Arrhenius Equation

The Arrhenius equation is an essential formula in physical chemistry that describes how the reaction rate constant, k, varies with temperature and activation energy. It is given by: \[ k = A e^{-Ea / RT} \] where:

• k is the reaction rate constant.
• A is the frequency factor, indicating the frequency of collisions and their correct orientation.
• \( Ea \) is the activation energy.
• R is the ideal gas constant (8.314 J/mol·K).
• T is the temperature in Kelvin.

The exponential part of the equation, \( e^{-Ea / RT} \), is particularly important as it shows the fraction of molecules with enough energy to surpass the activation energy barrier.

In practical applications, the Arrhenius equation helps chemists understand and predict how changes in temperature affect the speed of a reaction. Since the activation energy \( Ea \) is exponentiated and negatively affects the rate, even small decreases in \( Ea \) by catalysts can drastically increase the reaction rate, as more molecules will have the energy needed to react.

This equation is also vital for comparative studies, such as evaluating how a reaction on a catalyst, like a W surface with a lower \( Ea \), can be significantly faster compared to an uncatalyzed reaction, reflecting the catalyst’s effectiveness in speeding up chemical processes.